Instructor’s Manual

to accompany

MODERN ELECTRONIC COMMUNICATION

Ninth Edi t ion

Jef f rey S. Beasley Gary M. Mil ler

Upper Saddle River, New Jersey Columbus, Ohio

__________________________________________________________________________________ Copyr ight © 2008 by Pearson Educat ion, Inc., Upper Saddle River, New Jersey 07458. Pearson Prentice Hall. All rights reserved. Printed in the United States of America. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department. Pearson Prent ice Hal l™ is a trademark of Pearson Education, Inc. Pearson® is a registered trademark of Pearson plc Prent ice Hal l® is a registered trademark of Pearson Education, Inc. Instructors of classes using Beasley/Miller, Modern Electronic Communication, Ninth Edition, may reproduce material from the solutions manual for classroom use.

10 9 8 7 6 5 4 3 2 1

ISBN-13: 978-0-13-225080-1 ISBN-10: 0-13-225080-2

Instructor’s Manual to Accompany

MODERN ELECTRONIC COMMUNICATION / 9e

Chapter Overviews Test Item File

Answers to Chapter Problems Troubleshooting with Electronics Workbench - Solutions

Laboratory Manual Experiment Results Electronic Workbench Multisim - Solutions

Jeffrey S. Beasley and Gary M. Miller

i

CONTENTS Part I: Chapter Overviews Chapter 1: Introductory Topics 1 Chapter 2: Amplitude Modulation: Transmission 4 Chapter 3: Amplitude Modulation: Reception 7 Chapter 4: Single-Sideband Communication 10 Chapter 5: Frequency Modulation: Transmission 12 Chapter 6: Frequency Modulation: Reception 15 Chapter 7: Communications Techniques 17 Chapter 8: Digital Communication: Coding Techniques 19 Chapter 9: Wired Digital Communication 21 Chapter 10: Wireless Digital Communications 23 Chapter 11: Network Communications 25 Chapter 12: Transmission Lines 28 Chapter 13: Wave Propagation 30 Chapter 14: Antennas 32 Chapter 15: Waveguides and Radar 34 Chapter 16: Microwaves and Lasers 37 Chapter 17: Television 39 Chapter 18: Fiber Optics 42 Part II: Test Item File Chapter 1 44 Chapter 2 53 Chapter 3 64 Chapter 4 78 Chapter 5 91 Chapter 6 106 Chapter 7 112 Chapter 8 117 Chapter 9 122 Chapter 10 126 Chapter 11 134 Chapter 12 143 Chapter 13 151 Chapter 14 157 Chapter 15 162 Chapter 16 168 Chapter 17 172 Chapter 18 179

ii

Part III: Answers to Chapter Problems Chapter 1 188 Chapter 2 199 Chapter 3 208 Chapter 4 216 Chapter 5 222 Chapter 6 229 Chapter 7 233 Chapter 8 238 Chapter 9 243 Chapter 10 247 Chapter 11 251 Chapter 12 259 Chapter 13 283 Chapter 14 289 Chapter 15 297 Chapter 16 305 Chapter 17 311 Chapter 18 319 PART IV: Troubleshooting with EWB Multisim - Solutions Chapter 1: Understanding the Frequency Spectra 324 Chapter 2: AM Measurements 324 Chapter 3: AM Demodulation 325 Chapter 4: Single-Sideband Generation 325 Chapter 5: Generating and Analyzing the FM Signal 326 Chapter 6: FM Receiver Blocks 326 Chapter 7: Mixer and Squelch Circuits 327 Chapter 8: Sampling the Audio Signal 327 Chapter 9: Sequence Detector 328 Chapter 10: BPSK Transmit-Receive Circuit 328 Chapter 11: Audio Signal Measurements 329 Chapter 12: Network Analyzer 329 Chapter 13: Crystals and Crystal Oscillators 330 Chapter 14: Dipole Antenna Simulation and Measurements 330 Chapter 15: Lossy Transmission Lines and Low-Loss Waveguide 331 Chapter 16: Characteristics of High-Frequency Devices 331 Chapter 17: The Television RF Spectrum 332 Chapter 18: Light Budget Simulation 332

iii

Part V: Laboratory Manual Experiment Results 1 Active Filter Networks 333 2 Frequency Spectra of Popular Waveforms 341 3 Tuned Amplifiers and Frequency Multiplication 346 4 Low-Pass Impedance Transformation Networks 350 5 Phase-Shift Oscillator 354 6 LC Feedback Oscillator 358 7 Colpitts RF Oscillator Design 361 8 Hartley RF Oscillator Design 365 9 Swept-Frequency Measurements 368 10 Nonlinear Mixing Principles 374 11 AM Modulation Using an Operational Transconductance Amplifier 378 12 RF Mixers and Superhetereodyne Receivers 386 13 Cascode Amplifiers 394 14 Sideband Modulation and Detection 403 15 Frequency Modulation: Spectral Analysis 412 16 Phase-Locked Loops: Static and Dynamic Behavior 418 17 FM Detection and Frequency Synthesis Using PLLs 424 18 Pulse Amplitude Modulation and Time Division Multiplexing 429 19 Pulse Width Modulation and Detection 436 21 Digital Communication Link Using Delta Modulation Codecs 440 22 Electronics Workbench Multisim - dB Measurements in Communications 446 23 Electronics Workbench Multisim – Smith Chart Measurements 449 Using the EWB Network Analyzer 24 Tone Decoder 452 32 Using a Spectrum Analyzer 454 33 Using Capacitors for Impedance Matching 455 34 Electronics Workbench Multisim – Impedance Matching 457 35 AM Generation Using an Electronic Attenuator 459 36 Generating FM from a VCO 460 37 Upconversion and Downconversion 461

iv

Part VI Electronics Workbench Multisim

EWB Complementary Exercises

Experiment 1 – EWB 462 Simulation of an ACTIVE FILTER NETWORKS

Experiment 2 – EWB 464 USING THE SPECTRUM ANALYZER and the SIMULATION

and ANALYSIS of COMPLEX WAVEFORMS Experiment 3 – EWB 467 Simulation of Class C Amplifiers and Frequency Multipliers Experiment 5 – EWB 469 Simulation of a Phase-Shift Oscillator Experiment 6 – EWB 471 Simulation of an LC Feedback Oscillator Experiment 11 – EWB 473 Percentage of Modulation Measurement of an Amplitude Modulated Waveform Notes to the Instructor 474

1

OVERVIEW – CHAPTER 1 INTRODUCTORY TOPICS

1-1 INTRODUCTION Following a brief introduction to the field of electronic communications, the concept of modulation is introduced. At this early stage very basic words such as a carrier “carrying” the information are used. Equation 1-1 shows the three characteristics of a carrier that could be modified to carry the information include the amplitude, frequency, and phase. These concepts form the basis for Chapters 2-6. Table 1-1 describes the sub-divisions within the radio-frequency spectrum and Fig. 1-1 presents a simple communication system in block diagram form. A discussion of it should get your students thinking and whet their appetite for the chapters that follow. 1-2 The dB in COMMUNICATIONS The dB (decibel) is an extremely important measure in communications. Decibels are used to specify measured and calculated values in communications system measurements. The equations for calculating dB using power and voltage ratios are provided in equations 1-2 and 1-3. Examples 1-1 to 1-3 demonstrate the method for calculating and converting dB values. Another technique for converting many common dB values is provided in Table 1-2. Tables and computer programs are often used on the job for performing most dB calculations or conversions. A list of common decibel terms is provided in Table 1-3. 1-3 NOISE A fundamental limitation in communication systems is noise. This, of course, is due to the fact that the signal picked up by the receiver is very small. Separating it from all the various sources of noise is a critical task. The various types of external noise (man-made, atmospheric, and space) and internal noise (thermal, transistor, and flicker) are described. The calculations associated with thermal and noise voltage are facilitated with Examples 1-4 and 1-5. 1-4 NOISE DESIGNATION AND CALCULATION The important concept of signal-to-noise ratio is introduced as a simple ratio (Eq. 1-12) and in decibel form (Eq. 1-13). This is followed by defining the noise figure as the ratio of S/N at the input over the S/N at the output. Some practice on the related calculations is provided in Example 1-6. The concepts of reactance noise effects, noise due to amplifiers in cascade, equivalent noise temperature, and equivalent noise resistance close out this section. The importance of related calculations is indicated by the many examples provided to help your students master these topics. 1-5 NOISE MEASUREMENT The use of a diode noise generator to make some basic noise measurements is introduced. A simple yet effective measurement technique using the diode noise generator is illustrated in Example 1-10. A quick and useful measurement technique using a basic dual-trace oscilloscope is detailed in Fig. 1-7. It is termed the tangential noise measurement technique.

2

1-6 INFORMATION AND BANDWIDTH There are two fundamental limitations on the performance of a communication system.

Besides the noise effects just introduced, the bandwidth allocated for transmission is the other basic limitation. Hartley’s law states that the amount of information that can be transmitted is proportional to the bandwidth times the time of transmission. To help the student understand the bandwidth that various signals occupy, an introduction to understanding the frequency spectra is provided. This basically non-mathematical approach promotes understanding of a signal’s sinusoidal harmonics and how they combine to form complex signals. The square wave waveform analysis in Fig. 1-9 and 1-10 provides graphic illustration of this process. Visual examples of the frequency components making up complex waveforms are provided in Figures 1-11 and 1-12. These are the FFT representations for a sine wave and a square wave. Table 1-4 gives Fourier expressions for some common periodic waveforms. Figure 1-13 demonstrates the effect a bandwidth-limited signal has on a square wave. 1-7 LC CIRCUITS Sections 1-7 and 1-8 cover some basic characteristics of LC circuits and oscillators. If your students have a good background to this from previous studies, you may wish to omit them. The characteristics of inductors and capacitors are introduced including the concepts of quality (Q) and dissipation (D) factors. This is followed by the concept of resonance and bandpass filters. Examples 1-13 and 1-14 provide practice calculating bandwidth, Q, required component values, and resonant frequencies. 1-8 OSCILLATORS Oscillators are key elements in communication systems. The concept of creating a sine wave via the “flywheel” effect is introduced with the help of Fig. 1-21. An analysis of some common LC oscillators follows including the Hartley, Colpitts, and Clapp oscillators. The very important crystal oscillator is then detailed. Table 1-5 provides stability and cost information for four different crystal oscillator configurations. A useful crystal test circuit is shown in Fig. 1-29. 1-9 TROUBLESHOOTING All chapters of this text are concluded with a troubleshooting section. The importance of developing good troubleshooting skills cannot be over-emphasized. Employers and accrediting agencies are in strong agreement on this matter. Each one of these sections provides troubleshooting skills related to the chapter’s topics. Often some general troubleshooting techniques are also included as is the case with this section. This section opens with a comprehensive overview of troubleshooting. This is followed by detail on the four types of circuit failures. Detail on the four basic troubleshooting techniques (symptoms, signal tracing and injection, voltage and resistance measurements, and substitution) concludes the general troubleshooting material. Testing a crystal with the aid of the block diagram in Fig. 1-32 is discussed. This section is concluded with information on testing the inductors and capacitors in a Clapp oscillator. A section on understanding digital sampling oscilloscope waveforms is also included in this section. This section discusses the importance of selecting the sample frequency and the effect an improperly selected sample frequency has on the displayed waveform.

3

1-10 TROUBLESHOOTING WITH ELECTRONICS WORKBENCH MULTISIM Representative computer simulations using Electronics Workbench (EWB) Multisim are provided in each chapter in this text. The computer files are provided in the CD-ROM which comes with the text. The use of virtual instruments is incorporated into each chapter’s presentation on using EWB Multisim. Detailed steps are used in the text to lead the student through each of the virtual experiments.

In Chapter 1, the oscilloscope and the spectrum analyzer virtual instruments are used to examine the properties of a square wave. The performance and operation of these virtual instruments closely resemble real test equipment and the user has the ability to make connections and adjustments comparable to that made on equipment when working on a bench. Three EWB exercises are included to further develop the student’s understanding of the simulation tool and the virtual instruments. The files provided in the text’s CDROM support both the newer Multisim 9 (.ms9), Multisim 7 (.ms7) and Multisim 6 (.msm) formats. The files are located in the ms9, ms7 or msm folders in each respective chapter.

4

OVERVIEW – CHAPTER 2

AMPLITUDE MODULATION: TRANSMISSION 2-1 INTRODUCTION This is a critical point in your students' study. While modulation has been introduced in Chapter 1, it is now time for the student to really come to an understanding of what it is all about. It may be a good idea to give a quiz after covering Sections 1 through 4 just to make sure that a reasonable level of comprehension has been attained. 2-2 AMPLITUDE MODULATION FUNDAMENTALS A good way to introduce the basic AM process is to compare the linear combination of two signals in Fig. 2-1 with the nonlinear combination in Fig. 2-2. Emphasize that only the non-linear combination produces an AM signal. You also might want to explain why the transmission of the linear combination would leave just the carrier at the receiver while the AM signal should be received basically as transmitted. The equation defining the AM waveform is provided in equation 2-1. This is also a good opportunity to review or introduce the importance trigonometric relationship (sin x)(sin y), equation 2-2. You will find that a thorough discussion of the transmission of a range of modulation frequencies will now be possible. A detailed study of Example 2-1 should be most helpful here. The phasor analysis is also important, as for many students this is when something finally falls together, the light bulb goes on, and now they have seen the light. 2-3 PERCENTAGE MODULATION The concept of percentage modulation is usually mastered with ease. In fact, your students will probably enjoy making some quantitative calculations such as illustrated in Example 2-2. This is also a good time to introduce the concept of overmodulation and talk about the problems that it causes. 2-4 AM ANALYSIS Your students may be somewhat resistant to the brief mathematical analysis at the beginning of this section but it is important to their overall understanding. This is their first exposure to a form of modulation and they need to realize that this is not some form of magic—it does withstand analysis with some basic mathematical tools. The importance of transmitting a high-percentage of modulation is now understandable—just make sure they remember that overmodulation is taboo. It is now time to indicate that the carrier in AM systems effectively wastes a lot of transmitter power. Examples 2-3 through 2-8 illustrate a number of useful calculations regarding percentage modulation, total power, carrier power and sideband power.

5

2-5 CIRCUITS FOR AM GENERATION It is now time to introduce some circuits used to create AM. The whole key here is that it takes a nonlinear combination of carrier and intelligence to generate AM. The difference between high-level and low-level modulation is discussed. Be sure to stress that this has nothing to do with high-percentage modulation. Use Fig. 2-12 to help explain the difference between high and low-level modulation. It is certainly true that IC modulators are used in the majority of new designs. Introduction of several discrete device designs is still important to overall understanding and in working with older equipment. However, there is certainly nothing wrong with emphasizing the linear integrated circuit designs at this time. 2-6 AM TRANSMITTER SYSTEMS At this point it is appropriate to talk about a complete transmitter system as opposed to just the AM modulator. The citizens band transmitter described here is simple enough so that the student can comprehend the various system aspects without getting bogged down with too many details. The concept of coupling transmitter power to an antenna is introduced as is some detail on the fabrication and tuning of this compact transmitter. 2-7 TRANSMITTER MEASUREMENTS At this point your students may be anxious to learn some laboratory measurements useful in AM analysis. The trapezoid pattern technique is very good for measuring percentage modulation and for pinpointing some specific problems with the modulator. It is also important to realize that some meaningful measurements can be made with a dc ammeter. The spectrum analyzer is also introduced at this point. It is one of the most important instruments available for communication’s equipment and it may be new to some of your students. Its use in making harmonic distortion measurements is provided and Example 2-9 provides a sample computation. This section is concluded with some precautions to take when making measurements on RF circuits. It is often troublesome for the beginner to understand that the measurement tool can be changing the measurement. It is important to also understand why this is happening. 2-8 TROUBLESHOOTING The first discussion has to do with the importance of initially inspecting a piece of equipment when repair is necessary. The novice is surprised at how much time is saved by this process. If inspection by itself has not cleared up the problem then a strategy for repair should be developed. This includes verification that a problem exists, isolation of the defective stage, isolation of the defective component, and replacement of the defective component. Troubleshooting a simple self-biased RF amplifier is then provided. This includes looking at the effects of various components being opened or shorted. It is very important for the student to start thinking about shorts and opens as this is such a prevalent type of failure. The process of checking an entire transmitter is the next topic. Be sure to emphasize the material on safety provided when working on high voltage systems. Troubleshooting topics covered include improper frequency of operation, measurement of output power, and how to remedy these parameters when they are not in specification.

6

2.9 TROUBLESHOOTING WITH ELECTRONICS WORKBENCH MULTISIM The EWB Multisim tools are used to simulate an AM modulator circuit. The student will gain additional experience measuring the modulation index of an AM signal. This exercise also enhances the students understanding of the carrier and modulating signal components of the AM envelope. The oscilloscope virtual instrument is used extensively in this exercise for making measurements on the waveform. The EWB exercises provide the opportunity for the student to test their ability to determine the modulation index and the carrier frequency of an AM signal.

7

OVERVIEW – CHAPTER 3 AMPLITUDE MODULATION: RECEPTION

3-1 RECEIVER CHARACTERISTICS You may find it helpful to get your students thinking about a receiver by asking them to think about a desirable design for its block diagram. A likely outcome is the TRF concept introduced in this section. This can then lead to a discussion of receiver sensitivity and selectivity, important concepts to carefully consider at this point. The basic problem of variable selectivity in a TRF receiver is the main reason that it is not a viable design today. Variable selectivity can easily be shown by working through Example 3-1. 3-2 AM DETECTION A key to the detection process is to stress that it takes a nonlinear device just as a non-linear device was necessary to create the AM signal. The basic diode detector is described and Fig. 3-3 can be used to explain the waveforms in a typical circuit. The potential problem of diagonal clipping is introduced and the synchronous detector is also provided. An Electronics Workbench Multisim implementation of a synchronous AM detector is provided in Figure 3-5(a). This is a good place to reinforce the concept of a mixer circuit and the function of the synchronous detection circuit. 3-3 SUPERHETERODYNE RECEIVERS The variable selectivity problem with TRF receivers explained in Section 3-1 should be mentioned again at this point because now the solution is at hand. The superheterodyne (superhet) receiver, first used in the l930’s, has proven to be the dominant receiver format to this day. It should therefore be carefully introduced, as your relatively unsophisticated communication students will be a bit confused by some of the details of this concept. The frequency conversion process explanation will be enhanced by using the block diagrams of Figs. 3-7 and 3-8. 3-4 SUPERHETERODYNE TUNING The bulk of today’s receivers use frequency synthesis tuning. This is mentioned here but its detailed discussion is in Chapter 7. It is good to stick to the basics for now as the student’s mastery skills are developing. The variable ganged capacitor tuning is used for the introduction in Section 3-3 and a varicap diode electronic tuning circuit is provided in this section. These tuning methods are still prevalent in low cost receivers. The tracking problems of these tuning systems is explained and the tracking adjustment process described.

8

3-5 SUPERHETERODYNE ANALYSIS The problem of image frequency is a bit confusing but by the time the student attains comprehension, the whole superheterodyne concept will usually click into place. If you discuss Figs. 3-13 and 14 an understanding of when images are and are not going to be a problem should be accomplished. The use of double conversion to circumvent the image problem is mentioned but left for a full discussion in Chapter 7. The remainder of this section introduces various circuits commonly used for RF amplifiers, IF amplifiers and the mixer/local oscillator. You may want to use transparencies of these circuits to enhance your explanation and discussions. 3-6 AUTOMATIC GAIN CONTROL To initiate AGC discussion, ask your students if all received signals are at the same level. The obvious answer is no, but then ask them why the speaker output is basically constant for all those different signals. This should establish the need for AGC and is a natural lead-in to how it is accomplished. The first step is to obtain the AGC level—some dc level proportional to signal strength. The next step is to control the gain of an amplifier to allow maintaining a relatively constant output level. A discrete AGC circuit illustration is provided in Fig. 3-19 while Fig. 3-20 provides an IC AGC system. 3-7 AM RECEIVER SYSTEMS We’re now in a position to “put it all together” by looking at a complete AM receiver system. A discussion of the discrete system shown in Fig. 3-21 can be kind of exciting to the student when realization that comprehension of such a complicated schematic is possible. Once again, an overhead transparency display to accompany the discussion (and Fig. 3-22 for an IC receiver) is very helpful. The AM stereo receiver shown in Fig. 3-25 is also appropriate at this time. This section concludes providing a receiver analysis with respect to the power gain or attenuation of all the various stages. A discussion of Example 3-3 should facilitate an understanding of the power levels throughout a receiver. It also serves to illustrate the use of dbm and dbW in this type of analysis. 3-8 TROUBLESHOOTING Some detailed troubleshooting is provided regarding the self-excited mixer shown in Fig. 3-27. Additionally, Table 3-1 provides experience using a troubleshooting chart. The use of these types of troubleshooting aids is common in the industry. A basic series-pass electronically regulated power supply is provided in Fig. 3-28 and detailed techniques for its repair are included. This section is concluded with some tips for troubleshooting audio and RF amplifiers.

9

3-9 TROUBLESHOOTING WITH ELECTRONICS WORKBENCH MULTISIM The diode detector circuit is simulated using EWB Multisim in this exercise. A virtual AM source, provided by the EWB Multisim tool, is used to generate the AM signal for inputting to the diode detector. The instructor and the student will like the ease that the AM source can be adjusted. The diode detector circuit also contains a “virtual” variable capacitor. The student will be able to increase or decrease the value of the capacitor by pressing c or C to determine if varying the capacitor value effects the recovered signal. The chapter 3 EWB Multisim exercise introduces a virtual exercise in troubleshooting a communications circuit. EWB Multisim provides a feature which allows for the addition of a component fault in the circuit. The student will be able to test their troubleshooting knowledge and their understanding of the circuit operation to determine which component(s) have failed. The student will have many virtual instruments to use in the troubleshooting process including the oscilloscope and the multimeter. Component repair is easily accomplished by double-clicking on the component, and correct the setting under the Fault tab. The EWB exercises include two virtual circuits which contain faults. These exercises can be used to test the student’s troubleshooting ability and circuit knowledge. A third exercise requires the student to adjust virtual capacitors to minimize the RF noise and to record their settings.

10

OVERVIEW – CHAPTER 4 SINGLE-SIDEBAND COMMUNICATION

4-1 SINGLE-SIDEBAND CHARACTERISTICS This section begins with a bit of history on the development of SSB. As was mentioned in the AM chapters, the carrier contains no information and yet contains most of the power in the AM transmission. The sidebands contain the information but with (at most) 1/3 the power. A discussion of peak envelope power is also provided. A number of different types of SSB are used. The remainder of this section introduces them and concludes with a discussion of the advantages of SSB over standard AM. 4-2 SIDEBAND GENERATION: THE BALANCED MODULATOR The balanced modulator is a widely used circuit in electronic communications. Since this is the first encounter a detailed look at it is given here. The balanced ring modulator and a linear integrated circuit balanced modulator are described. The output of these modulators is a double-sideband signal but the carrier has been suppressed. The next two sections will address changing from this double-sideband signal to a single-sideband signal. 4-3 SSB FILTERS The elimination of one of the sidebands in a DSB signal can be accomplished with a high-Q filter. Example 4-1 provides clear illustration of this fact. Crystal filters are used in the most demanding of these requirements. Since they were covered in Chapter 1, a relatively brief review is provided at this point. Ceramic filters are widely used in today’s communication circuits. They offer Q’s that are less than crystal filters but much better than standard LC filters. Surprisingly, mechanical filters are still used in some SSB applications. This section concludes with a discussion of their characteristics. 4-4 SSB TRANSMITTERS There are two basic systems used to create SSB, the filter method and the phase method. The block diagram in Fig. 4-8 is useful as a transparency to help the student understand the filter method. It is pretty straightforward, but a discussion of the block diagram should make it crystal clear. The phase method is a bit more difficult for the student but the block diagram of Fig. 4-10 along with a mathematical discussion using Eqs. 4-2, 3 and 4 should take care of it. Also introduced in this section is the concept of amplitude compandoring. This allows the lower level signals to be transmitted with greater power while staying within the PEP ratings of the transmitter. This section concludes with a discussion of linear power amplifiers. You may wish to use Fig. 4-13 as a transparency for a classroom analysis. 4-5 SSB DEMODULATION Up to this point the many advantages of SSB have been extolled. The demodulation of SSB exposes a fundamental drawback. The carrier must somehow be recombined with a sideband through a non-linear device in order to recover the intelligence. So the carrier that we removed at the transmitter must somehow be recreated at the receiver. Some techniques for doing this are the subject of this section.

11

4-6 SSB RECEIVERS The block diagram of a typical SSB receiver is given in Fig. 4-17. It basically looks like a superhet AM receiver except for second mixer (detector) and the block that reinserts the carrier-often referred to as the beat frequency oscillator (BFO). The complete receiver schematic in Fig. 4-18 makes a good discussion circuit when displayed as a transparency. 4-7 TROUBLESHOOTING The first topic of this section involves a basic balanced modulator circuit. Testing for carrier leakthrough is described using an oscilloscope. Checking carrier suppression using a spectrum analyzer is also introduced. A technique for checking the filter used to suppress one of the sidebands is the next troubleshooting topic. This is followed by an explanation of the two-tone test for testing the linearity of a linear amplifier. The final topic in this chapter’s troubleshooting section is to discuss a technique for testing an entire receiver by using signal injection. Table 4-1 summarizes the procedure in conjunction with the block diagram shown in Fig. 4-26. 4-8 TROUBLESHOOTING WITH ELECTRONIC WORKBENCH MULTISIM

This exercise demonstrates an important fundamental concept of communications which is the production of the sum and difference frequencies. A multiplier, provided in the Multisim tools, is used in this example to produce the sum and difference frequencies however a balanced modulator and “mixer” circuits produce the same result. The complex waveform output, generated by the multiplier, is analyzed using the virtual spectrum analyzer. The exercise includes a demonstration of removing the lower sideband using a five-element Chebyshev filter to produce a SSB signal. The spectrum analyzer is used to verify that the lower sideband has been significant attenuated. The exercise includes a troubleshooting example in which a fault has been introduced in the filter. The students are guided through the troubleshooting exercise and are reminded to perform a visual check of the circuit. The EWB questions include an exercise where the student must tune a high-pass filter circuit and troubleshoot two circuits which contain faults

12

OVERVIEW – CHAPTER 5 FREQUENCY MODULATION: TRANSMISSION

5-1 ANGLE MODULATION Angle modulation includes phase modulation and frequency modulation. Phase modulation is not widely used but has importance as a means of generating frequency modulation. The study of FM is crucial to the communication student. A relatively slow start in this study may be beneficial as the initial concepts are a bit more difficult than for amplitude modulation. 5-2 A SIMPLE FM GENERATOR The capacitor microphone FM generator in Fig. 5-1 provides an excellent starting point for an understanding of FM basics. This circuit is not normally used in practical FM systems but is an excellent learning device. With it, the student can grasp the concept that intelligence amplitude determines the amount of frequency change and intelligence frequency determines the rate of frequency change. Equation 5-1 describes the relationship for the FM signal generate by the capacitor microphone. A discussion of Ex. 5-1 should help to reinforce these concepts as the students work at reorienting themselves from AM systems. We have found it beneficial to remind them as we move on through Chapters 5 and 6 to think back on this simple capacitor mike circuit whenever confusion on the FM basics sets in. 5-3 FM ANALYSIS This section opens with some basic mathematical analysis of FM and PM. The use of Bessel functions to determine amplitude of the side-frequencies is introduced. Examples 5-3 through 5-5 provide illustration of how to apply this analysis in determination of bandwidth, modulation index and the power in the various side-frequencies. A bandwidth approximation, known as Carson’s rule, is presented in equation 5-7. This important relationship should be included along with the discussion on the Bessel functions. It is interesting to discuss the fact that at various levels of modulation index the carrier power goes to zero and all the power is in the side-frequencies. The zero carrier amplitude discussion is followed by introduction of broadcast and narrowband FM systems. Example 5-6 does some calculations relative to the broadcast and narrowband systems. 5-4 NOISE SUPPRESSION One of the main reasons that FM has become so popular is its ability to suppress noise that can be troublesome to AM systems. The concept of a noise spike not passing through a limiter compared to AM systems (where limiter use is not possible since it would destroy the intelligence) is easily demonstrated with the help of Fig. 5-6. Unfortunately, even though the spike is eliminated, not all of its effect is gone since it has also created a phase shift which means an undesired frequency shift. The analysis of how that effect is minimized is explained, and working with Exs. 5-8 and 9 will greatly enhance understanding this process. Once the noise reduction process is understood it becomes easy to introduce the concept of the capture effect. It also eases one into the need for preemphasis in FM systems and the usefulness of dynamic preemphasis which is what the Dolby system is all about.

13

5-5 DIRECT FM GENERATION The two basic categories of FM generation are direct and indirect. In this section some common means of direct generation are introduced. The use of a varactor diode in direct generation is explained with the use of Fig. 5-12. The reactance modulator is also introduced. The easily used linear integrated circuit, voltage-controlled oscillator (LIC-VCO) is then provided and the spec sheet for the 566 VCO is provided in Fig. 5-14. The Crosby direct FM transmitter is detailed next. A discussion using the block diagram in Fig. 5-15 is very helpful in allowing the student to get the big picture for this type of transmitter. This section concludes with a discussion of basic frequency doubler operation. 5-6 INDIRECT FM GENERATION Indirect FM generation is accomplished by first generating phase modulation and then making a conversion to FM. This allows you to apply the modulating signal to a stable crystal oscillator. This type of system is called an Armstrong modulator. The conversion from PM to FM and obtaining enough deviation makes this a fairly complex system. A complete system block diagram with frequencies for a 90.2 MHz carrier is provided in Fig. 5-18. 5-7 PHASE-LOCKED LOOP FM TRANSMITTER The use of a PLL to generate FM is the topic of this section. The system illustrated in Fig. 5-20 starts with a varactor-controlled crystal oscillator/modulator. Its output is an input of the phase detector in the PLL. A complete PLL FM transmitter schematic is provided in Fig. 5-21. A description of circuit operation is given and detail on alignment and operation on other bands is also discussed. A single-chip FM transmitter is show in Fig. 5-22 using the Maxim 2606 IC. This IC is designed for home audio use and an FM system is very simple for the student to construct. 5-8 STEREO FM Students are usually quite interested in how stereo FM is implemented. The discussion of how two separate signals can be transmitted in the same bandwidth and still be compatible with a mono broadcast is a good way to introduce the student to a frequency division multiplexing system. You will find it helpful to have the information contained in Figs. 5-23 and 24 on transparency to help student comprehension. A discussion on why the stereo signal is more prone to noise than the mono broadcast is sometimes helpful to overall student understanding of this system. 5-9 FM TRANSMISSIONS This section introduces the five major categories of FM transmissions. The ability for FM to use highly efficient Class C power amplifiers as compared to the need for less efficient linear amplifiers for AM and SSB is also discussed. 5-10 TROUBLESHOOTING The first troubleshooting concept is dealing with FM transmitter systems. Techniques for dealing with no RF output, incorrect frequency and incorrect deviation are provided. A reactance modulator is then analyzed in detail and Table 5-3 provides a summary of problems and probable causes. The next topic deals with troubleshooting the stereo/SCA FM signal generator. This is a good exercise to help the student fully understand its operation as well as troubleshooting techniques.

14

5-11 TROUBLESHOOTING WITH ELECTRONICS WORKBENCH MULTISIM A simulated voltage controlled oscillator is used to generate an FM signal in this EWB exercise. The student will use the VCO to investigate the effects changes on the input have on the FM signal. The virtual oscilloscope is used to examine the generated FM signal. An exercise is also included which uses the virtual spectrum analyzer to measure the bandwidth of the FM signal. Next, the measured result is compared to the result obtained using Carson’s rule. The exercises reinforce the use the spectrum analyzer to analyze the FM signal.

15

OVERVIEW – CHAPTER 6 FREQUENCY MODULATION: RECEPTION

6-1 BLOCK DIAGRAM This section provides an introduction to a basic FM receiver block diagram. The similarities to AM receivers is noted and the differences are also mentioned. These include the fact that we generally use the term discriminator rather than detector for an FM receiver. Additionally, FM receivers often do not include the AGC function so necessary for AM systems. This is because the limiter can take care of amplitude changes and in FM the intelligence is not contained in the amplitude. The intelligence is carried by the amount of frequency deviation and the rate of that frequency deviation. 6-2 RF AMPLIFIERS AM receivers often operate without an RF amplifier, but that is seldom the case with FM receivers. For that reason a discussion of these amplifiers is provided here. The advantage of FET vs. BJT designs is provided and relates to the question of intermodulation distortion. The square-law input/output relationship of FETs compared to the exponential relationship for BJTs is the key. The distortion components are much easier to filter out for the FET as a result. A typical design using a MOSFET is illustrated in Fig. 6-2. The accompanying table provides the component value to change between 100 MHz and 400 MHz operation. 6-3 LIMITERS The use of a limiter in FM receivers has already been mentioned. A transistor limiter is shown in Fig. 6-3. A key to its operation is the collector resistor, R

c , which is shown in blue. It

serves to limit the collector voltage which limits the range of its ac output. The result is the clipped (limited) output shown in Fig. 6-4. As indicated in the figure, passing this clipped signal through an LC tank tuned to its center frequency restores the limited signal to its sinusoidal form. This section concludes with a discussion of the relationship between limiting and the sensitivity of an FM receiver. This important concept is reinforced by thinking through Ex. 6-1. 6-4 DISCRIMINATORS Extracting the intelligence from the FM signal is sometimes a difficult concept for the student. You will find that reminding them of the capacitor mike generator from Chapter 5 will often help clear the air when they get bogged down in their analysis. An easy way to break into the discriminator discussion is to first introduce the slope detector provided in Fig. 6-6. This is followed by details on the Foster-Seely and ratio detector circuits. Since they are not used often in new designs, you may elect to not cover them. Input from reviewers is mixed on whether they should still be included. A discussion of the quadrature detector concludes this section. A transparency of Fig. 6-10 is helpful in explaining its operation. A block diagram of the Philips TEA 5767 single-chip FM stereo radio is provided in Fig. 6-20.

16

6-5 PHASE-LOCKED LOOP The PLL was briefly introduced in Chapter 5 showing its use as an FM generator. We now discuss its use as a demodulator—a much more important application. Because of the importance, a more detailed look at PLL operation is now provided. The three states of operation are discussed—free-running, capture and locked (tracking). A description of how a PLL demodulates an FM signal is provided followed by a complete specification sheet for the LM565 PLL. Analysis for calculations of its VCO center frequency (free-running frequency), loop gain and hold-in range are provided. 6-6 STEREO DEMODULATION Figure 6-15 graphically shows the difference between mono and stereo receivers with block diagrams. Students are usually quite interested in the stereo concept so this is a good opportunity for student interaction. The FCC authorized subsidiary communication authorization (SCA) is then introduced and the use of a 565 PLL as an SCA decoder is illustrated in Fig. 6-18. This section is concluded with an analysis of a rather complex LIC, the CA 3090 which is used as a stereo decoder. 6-7 FM RECEIVERS The single-chip FM stereo radio receiver shown in Fig. 6-20 makes a very good discussion circuit at this point. The architecture of this IC has dramatically reduced the number of external components. Have the students compare this circuit to the older style FM receiver shown in Fig. 6-21. 6-8 TROUBLESHOOTING The FM receiver block diagram in Fig. 6-21 is used as the basis for a discussion on locating the defective stage. This step-by-step process involves the signal injection process. Specific detail is given for troubleshooting quadrature detectors and the stereo demodulator. This section is concluded with a detailed description of the process for testing diodes and transistors using a digital multimeter (DMM). 6-9 TROUBLESHOOTING WITH ELECTRONICS WORKBENCH MULTISIM

An implementation of an FM receiver using Multisim is provided in this section. This example includes a block level implementation so that the student can gain additional understanding of an FM receiver. The students should be able to follow the signal from the RF input through to the output amplifier. Each stage is clearly identified and the block level implementation facilitates easy examination using the oscilloscope. The exercises include a problem in troubleshooting an FM receiver, a limiter, and an exercise which reexamines the relationship sin A x sin B.

17

OVERVIEW – CHAPTER 7 COMMUNICATIONS TECHNIQUES

7-2 FREQUENCY CONVERSION The problem of image frequencies is often a problem with communication transceivers as was discussed in Chapter 3. A solution is the use of double conversion which is illustrated with the block diagram in Fig. 7-1. A second technique is the use of up-conversion systems. A method of calculating the amount of image frequency rejection is the use of Equation 7-1, and Example 7-6 shows how to apply that equation. 7-3 SPECIAL TECHNIQUES Since AGC systems somewhat reduce the gain of even weak stations, the use of delayed AGC is utilized in some receivers. Figure 7-5 provides a graphic representation of no AGC, simple AGC and delayed AGC. Auxiliary AGC is also explained and it provides a reduced gain for very strong signals to prevent overloading the receiver. The Analog Devices AD8369 variable gain amplifier is shown in Fig. 7-7(b) and (c). The next topics covered in this section are variable sensitivity and variable selectivity. This is followed by the use of noise limiters. The schematic of an automatic noise limiter is provided in Fig. 7-9. This section is concluded with introductions to receiver metering and squelch circuitry. A discussion of squelch techniques has been provided at the end of this section. This discussion examines the different methods used to implement the squelch feature in communication receivers. 7-4 RECEIVER NOISE, SENSITIVITY, AND DYNAMIC RANGE RELATIONSHIPS The effects of noise are a dominating factor in the design of high quality receivers. Equation 7-2 relates the sensitivity with a receivers noise factor and the desired signal to noise ratio. The important concepts of dynamic range and third order intercept are introduced at this point. This is followed by four examples to help tie this all together. The concept of intermodulation distortion follows this discussion to conclude the section. You will find Figs. 7-12 and 13 helpful in explaining IMD to your students. 7-5 FREQUENCY SYNTHESIS The use of frequency synthesis to replace the local oscillator is now occurring in even modest cost/performance receivers. The basics of frequency synthesis are provided here with the block diagram of Fig. 7-14 serving as a useful starting point. The concept of programmable division is provided and this is followed by a discussion of two-modulus dividers. It is helpful to discuss the three synthesizer configurations shown in Fig. 7-16 with your students. This will give them a better overall understanding of the frequency synthesis process. A schematic for a CB transceiver synthesizer is shown in Fig. 7-19 and a detailed discussion of its operation is provided. The section is concluded with a discussion of a UHF receiver incorporating a PLL frequency synthesizer circuit. A block diagram of the circuit is provided in Fig. 7-21 and a schematic of the receiver is provided in Fig. 7-22.

18

7-6 DIRECT DIGITAL SYNTHESIS The block diagram for a direct digital synthesizer is shown in Fig. 7-23. These synthesizers have become popular in recent years as the availability and pricing of the complex ICs has become more favorable. The DDS synthesizers offer extremely small frequency increments and have the ability to shift frequencies very quickly. They are limited by operation at the higher frequencies and exhibit more phase noise problems than their analog counterparts. 7-7 HIGH FREQUENCY COMMUNICATION MODULES This section addresses how the performance of electronic communication circuits changes extensively with changes in frequency especially when high frequencies are being used. This includes a discussion on assembling high frequency circuits and printed circuit boards. The important issue of how miniaturization of circuitry has changed electronic communication systems is addressed. The section concludes with a discussion on using Mini-Circuits® modules and includes three examples of using modular electronic systems (Mini-Circuits® modules) to implement electronic communication circuitry (AM modulator, FM modulator, and a mixer circuit.) 7-8 TROUBLESHOOTING The block diagram for a typical FM transmitter is shown in Fig. 7-32. Troubleshooting detail is provided regarding the microphone/audio, modulator, TR switch, power amp, and oscillator. The next topic is a discussion of some basic logic problems you might encounter in a typical communications transceiver. The troubleshooting section concludes with some tips on handling the frequency synthesizer shown in Fig. 7-14. 7-9 TROUBLESHOOTING WITH ELECTRONICS WORKBENCH MULTISIM The concept of a mixer circuit is explored in this Multisim exercise. It is very important for the students to understand this relationship. The students will also gain experience examining the spectral output of a mixer circuit using the EWB spectrum analyzer. An exercise working with an implementation of a squelch circuit is provided. The circuit is easy for the student to follow and includes a virtual squelch control. The students enjoy adjusting the control and analyzing the effect its changes have on the circuit. The exercises include troubleshooting the squelch circuitry to determine why the circuit is not functioning. One exercise simulates what might happen if someone dropped a screwdriver into the receiver cabinet with power on. The student is asked to troubleshoot the circuit and correct the problems. Just a note: I never dropped a screwdriver into a receiver chassis …. Well okay, maybe I did once.

19

OVERVIEW – CHAPTER 8 DIGITAL COMMUNICATION: CODING TECHNIQUES

8-1 INTRODUCTION The three basic forms of digital communications are shown in Fig. 8-1. They include an analog signal converted to digital for transmission, a digital signal transmitted as a baseband (unmodulated), and a digital signal converted to analog for transmission. The ability of regeneration for a digital signal is discussed with respect to its ability to deal with noise problems. The use of digital signal processing to perform various operations on a digital communications signal is also described. 8-2 ALPHANUMERIC CODES The ASCII code is first described here. The concept of parity is introduced. This is followed by introduction of the EBCDIC code and the Baudot code. The Baudot code is presented so that the student will understanding that there are many ways to present alphanumeric text in a digital format. The various characteristics of these alpha-numeric codes is provided as well as some specific applications. This section concludes with detail on the grey code commonly used in telemetry systems. 8-3 PULSE-CODE MODULATION

The most common technique for representing an analog signal in digital format is PCM. This section presents an overview of the sample and hold circuit and the generation of the pulse amplitude modulated waveform. The sample frequency, as defined by the Nyquist rate is provided in equation 8-1. Figure 8-11 provides a visual reference of selecting the sample frequency. Terms such as aliasing, fold-over distortion, and anti-aliasing filters should be introduced to the student. Be sure and convey their importance. The concept of quantization and dynamic range are presented for a PCM system. Equation 8-4 is a useful relationship for the student to know. The section concludes with a discussion on companding and D/A and A/D convertors. These topics should be used as needed to supplement the students understanding of the PCM system.

8-4 DIGITAL SIGNAL ENCODING FORMATS The codes used for PCM systems are covered in this section. First the non-return to zero codes are introduced—NRZ-L, NRZ-M, and NRZ-S. The RZ codes are then presented— RZ(unipolar), RZ(bipolar), and RZ-AMI. The phase-encoded and delay-modulated codes (bi-phase and Miller) codes are provided. They are commonly used in optical systems, satellite telemetry links, and magnetic recording systems. A description of the multilevel binary codes (Dicode NRZ and Dicode RZ) concludes this section. 8-5 CODING PRINCIPLES This section introduces the concept of increasing the Hamming distance in digital data. This technique of adding data bits to a digital word facilitates error detection and correction. Figure 8-27 and 8-28 provide a visual representation of why bits are added to data.

20

8-6 CODE ERROR DETECTION AND CORRECTION Error detection and correction is very important when dealing with code transmission. Parity is explained as the first issue here. Concepts such as even and odd parity, symbol substitution, and protocols are introduced. The block check character process is explained and followed by detail on cyclic redundancy check. The final topic in this section has to do with forward error-correcting using the Hamming code and Reed-Solomon codes. 8-8 DIGITAL SIGNAL PROCESSING (DSP) The basics of digital signal processing are presented in this section. The section begins with a basic review of both passive and active analog filters. A block diagram of the DSP process in presented in Fig. 8-38. A discussion of how the sampled signal is processed using a difference equation is presented. This includes the definition of a recursive and non-recursive filters. The section concludes with an example of using a spreadsheet program to demonstrate how a second-order, recursive, low-pass Butterworth filter is processed. 8-9 TROUBLESHOOTING The digital waveform is initially discussed here. The effect of noise on the pulse is presented along with an analysis of impedance mismatch effects. The effect of low and high-frequency attenuation on a pulse is explained. It is necessary, when troubleshooting digital communication systems, to be able to recognize pulse distortion and what causes it. 8-10 TROUBLESHOOTING WITH ELECTRONICS WORKBENCH MULTISIM This exercise demonstrates the use of EWB Multisim to simulate the sampling process. The students will have the opportunity to experiment with the selection of the sample frequency and examine the frequency spectra of a sampled signal that contains aliased frequencies.

21

OVERVIEW – CHAPTER 9 WIRED DIGITAL COMMUNICATION

9-1 INTRODUCTION Six possible methods for transmitting analog and digital information are provided in

Fig. 9-1. This figure helps the student to understand the variety of methods possible for transmitting information. 9-2 BACKGROUND MATERIAL FOR DIGITAL COMMUNICATION This section introduces some of the basic concepts needed by the student to understand and appreciate fully the material in chapter 9. This section introduces the basic concepts of binary coding, code noise immunity, the concept of bit error rate (BER), the probability of a bit error, and the definitions of the fundamental concepts of simplex, half duplex, full duplex, synchronous and asynchronous data communications. 9-3 BANDWIDTH CONSIDERATIONS Begin your discussion on bandwidth with an introduction to the Shannon-Hartley theorem and the calculation for channel capacity. Example 9-4 demonstrates the calculation of the capacity of a telephone channel given a S/N of 60 dB. This examples shows that there is a limitation to the amount of data that can be transmitted down a telephone line. Equation 9-6 defines how to calculate the minimum bandwidth of a digital communications link. 9-4 DATA TRANSMISSION The basic concepts of high serial data transmission are presented in this section. The objective is to make sure the student understands how data communications is accomplished over the Telco connection. Concepts presented include data rates and line coding formats. The student will need to understand the concept of the point of presence and the requirements of using a CSU/DSU in an interface to a communications carrier. The concept of packet switching and frame relay are presented and should be briefly reviewed for the student. The basic operating concept of an ATM system is also introduced. 9-5 TIME DIVISION MULTIPLE ACCESS (TDMA) The TDMA concept is extremely important and the student should understand how many digital communication sources can share a single channel. The data process is demonstrated in Figs. 9-9 and 9-10. The concepts of slot time, guard times, and inter-symbol interference are important.

22

9-6 DELTA AND PULSE MODULATION The concepts of delta and pulse modulation are introduced. These systems are relatively simple and easy to implement. Limitations of such a system, such as slope overload are defined. Pulse modulation techniques including pulse-amplitude modulation, pulse-width modulation, and pulse-position modulation are presented. 9-7 COMPUTER COMMUNICATION This section presents a look at computer based serial communications. Topics include USB and Firewire. Figure 9-26 shows an example of using the MAX3451 transceiver for establishing a USB connection. RS-232 is still extremely important and plan to devote some time to this topic in class. The signaling concepts are important for the student to understand. Demonstrate an RS-232 link from one computer to another if possible. Most computers have the hyper-terminal software available for making the link. This will make for an exciting demonstration of the concept. A brief coverage of facsimile machines is good at this point. 9-8 TROUBLESHOOTING Techniques for troubleshooting telemetry systems are presented. This section presents a good overview of basic troubleshooting techniques including possible problems at the transmitter and the encoder/decoder. 9-9 TROUBLESHOOTING WITH ELECTRONICS WORKBENCH MULTISIM This section examines the use of Electronics Workbench Multisim to simulate a sequence detector. These type of circuits are often used in communication circuits to detect the beginning of a serial data stream. The sections guide the student through the use of the Multisim Word Generator and Logic Analyzer. The students are given the opportunity to troubleshoot circuit problems using the included Multisim files on the text CDROM.

23

OVERVIEW – CHAPTER 10 WIRELESS DIGITAL COMMUNICATIONS

10-1 INTRODUCTION This section provides an introduction into wireless technologies. An overview of fixed wireless, mobile wireless, and IR wireless is presented. It is important that the student understands that although wireless communications is not new, the term “wireless” is most often used today in conjunction with modern digital communication systems. 10-2 DIGITAL MODULATION TECHNIQUES Frequency shift keying is first explained in this section. This is an easy concept for the students to understand since they have already been introduced to analog FM transmission and reception. Phase shift keying is next introduced. Fig. 10-3 illustrates a BPSK constellation. Techniques for generating and receiving a BPSK signal are presented in Figs. 10-5 and 10-6. The students should understand the concept of a coherent carrier recovery circuit. The QPSK phase constellation is next presented. A block diagram of a QPSK demodulating circuit is provided in Fig. 10-9. A QPSK constellation with noise is presented in Fig. 10-11. This figure illustrates decision boundaries and data recovery. QAM techniques are also presented and illustrated in Figs. 10-14 and 10-15. The eye pattern is helpful in diagnosing the performance of a digital modulation system. Figure 10-16 shows what the eye pattern looks like for various digital communication conditions. 10-3 SPREAD SPECTRUM TECHNIQUES This section emphasizes the techniques used in modern digital communications using spread spectrum techniques. Each block of this very important technology is presented. The first block is the generation of Pseudonoise (PN) codes. The mechanics and terminology of PN code generation are presented. An EWB Multisim implementation of a seven-bit PN sequence generator is demonstrated in Fig. 10-18. Frequency Hopping Spread Spectrum (FHSS) is next presented. A simplified RF spectrum for FHSS is provided in Fig. 10-22. Direct Sequence Spread Spectrum (DSSS) is next examined. By this point in the chapter the student should have a good understanding of PN codes and spread spectrum. This section includes a thorough discussion of DSSS accompanied by an EWB Multisim implementation of how DSSS signals are created. Fig.s 10-24 to 10-26 can be used to guide the student through the process of generating a DSSS signal. The spreading of a BPSK signal is demonstrated in Figures 10-28 to 10-30. Example 10-3 demonstrates how to calculate the spreading of the DSSS signal.

24

10-4 Orthogonal Frequency Division Multiplexing (OFDM) Many modern wireless digital systems are using this technique for transporting data. This technique is used in the 802.11a wireless LAN technologies, DSL, and cable modems. The basic concept of what it means for signals to be orthogonal should be discussed. An example of an OFDM transmission are presented in Fig. 10-31 and 10-32. OFDM is not a spread spectrum technology but a version called FLASH OFDM is considered to be a spread spectrum technology. Another technology that uses OFDM for transporting the digital data is HD radio. HD radio is a digital radio technology that operates in the same frequency bands as broadcast AM (530 – 1705 kHZ) and FM (88 – 108 MHz). The RF spectrum for a hybrid (analog and digital) AM and FM transmission is presented in Figure 10-33 and 10-34. A block diagram of an HD radio is provided in Figure 10-35. 10-5 TELEMETRY The block diagram for a telemetry systems is provided in Fig. 10-36. Telemetry systems can use FDM and TDM techniques as shown in Fig. 10-37. 10-6 TROUBLESHOOTING This can be a fun section for the students. The topic focuses on the steps used by technicians to troubleshoot wireless telephone problems. This section explains the concepts of the wireless phone’s electronic serial number, the mobile identification number, and the master subsidy lock. The section also explains how water damage is detected by the technician. Of course, none of us have ever dropped our phone in water or worse, the toilet……(sigh …). 10-7 TROUBLESHOOTING WITH ELECTRONICS WORKBENCH MULTISIM An EWB Multisim simulation of a BPSK transmit-receive circuit is demonstrated. The simulation is based on the BPSK block diagrams presented in Figs. 10-4 to 10-6. The student should be able to follow the data through the entire path and develop a good understanding of generating the digital data and data recovery. The student will have the opportunity to troubleshoot the BPSK circuit in the exercises.

25

OVERVIEW – CHAPTER 11 NETWORK COMMUNICATIONS

11-1 INTRODUCTION A network is an interconnection of users that permits communication. The world-wide telephone grid fits into that description and will be introduced in this chapter. Computer networks have rapidly become a major interconnected system and that subject is the topic of the remainder of this chapter. The Internet is the largest of computer networks, and that is a subject of a textbook on its own. It is introduced in Section 11-8. 11-2 BASIC TELEPHONE OPERATION A short introduction to telephone basics starts this section. The representation of the telephone and the so-called BORSCHT function are illustrated in Fig. 11-1. A complete telephone system block diagram is provided in Fig. 11-3, which makes a good transparency for classroom discussion. Line quality and attenuation distortion are discussed as well as the concept of delay distortion. The problem of dealing with telephone traffic is broadly introduced by comparing it to vehicular traffic. Telephone traffic is either expressed in erlangs or in hundred-call-seconds. A general discussion of telephone congestion is followed by some detail on traffic observation and measurement. 11-3 TELEPHONE SIGNALING SYSTEMS This section addresses the role that SS7 plays in the process of establishing and “tearing down” telephone calls over the PSTN (Public Switched Telephone Network). A description of each SS7 layer as well as the service it provides is describe. A section on troubleshooting SS7 networks is also included which describes how a “Protocol Analyzer” is used to sort through many data messages to identify a problem with the telephone network. 11-4 MOBILE TELEPHONE SYSTEMS The Mobile Telephone System has radically changed in the last few years. This section examines the fundamental concepts of a cellular phone system layout (Fig. 11-11) and how the MTSO (mobile telephone switching office) is used to link two mobile users (Fig. 11-13). The section includes material on the latest in mobile communications, GSM and CDMA. This includes discussions on the control signals used in mobile telephone communications. An example of troubleshooting CDMA systems is discussed and includes a discussion and picture of Walsh codes (Fig. 11-18). The use of a “real-time spectrum analyzer” is presented which shows how the instrument can be used to isolate interference problems (Figs. 11-19 and 11-20). The section concludes with a look at the path to 3G wireless in the United States. 11-5 LOCAL AREA NETWORKS

This section provides an introduction to computer local area networks. The students should understand the topologies, particularly the star. Figs. 11-21 to 11-23 graphically show examples of the token-ring, the bus, and the star topology. It is very important that the student understand the CSMA/CD ethernet protocol and as a

26

minimum, how to construct an ethernet LAN. Make sure the student understands the concept of a MAC address. 11-6 ASSEMBLING A LAN This sections describes the techniques for assembling an office and a building LAN. Wireless LANs, based on the IEEE 802.11b. 802.11a, and 802.11g standards, are discussed. The section also includes discussions on the broadband wireless system WiMAX and Bluetooth. 11-7 LAN INTERCONNECTION The connection of local area networks is described with the help of the open systems interconnection (OSI) reference model. The three basic interconnection methods are described and they include bridge, router, and gateway methodology. Figures 11-32, 11-33, and 11-34 illustrate these methods. 11-8 INTERNET

The basic concepts of the INTERNET and IP addressing are introduced. Table 11-9 defines the three classes of IP networks. 11-9 IP TELEPHONY

The technique for using computer networks to carry telephone traffic is introduced. A 3COM NBX100 system is introduced and the features of the NBX system are compared to a traditional PBX. The concept of QoS (Quality of Service) is defined and an example IP telephone network is provide in Fig. 11-35. 11-10 INTERFACING THE NETWORKS This section provides and overview of the technologies and protocols being used to interface the networks. Advancements in modem technologies are reviewed including cable modems. The ISDN digital data link is examined. The DSL services are defined in Table 11-9. The frequency spectrum of the ADSL transmission is provided in Fig.11-39. The protocol developed to bridge mobile communications and the Internet is the wireless application protocol (WAP). The operating principles of this protocol are presented, including the wireless markup language (WML). 11-11 WIRELESS SECURITY The availability of the internet and the mobility provided by the cell phone, wireless laptop computers, WiMAX, Bluetooth devices, and many other related wireless technologies provide the user with an easier way to share information (e.g. data, images, etc.). But with this flexibility, the threat of the eavesdropping, jamming or even theft of information over the communication channel has increased. This section provides an overview of the basic communication security concepts a person responsible for maintaining a communications link should know. The five aspects of wireless security (confidentiality, integrity, authentication, non-repudiation , and availability of the network) are first defined followed by a discussion on how these

27

apply to wireless security. The section discusses data encryption and an example is presented demonstrating the need for properly securing data transmitted over a communication link by examining how easy it is to decode simple encrypted messages (cipher-text). 11-12 TROUBLESHOOTING Some typical LAN problems are described along with some techniques for troubleshooting and solving LAN problems are presented. The procedure for using the ping and trace route commands are covered. The section includes troubleshooting unshielded twisted-pair networks. This section concludes with some general cabling tips. 11-13 TROUBLESHOOTING WITH ELECTRONICS WORKBENCH MULTISIM

Audio dB level and total harmonic distortion (THD) measurements are presented using EWB Multisim. The student will have the opportunity make dBm level measurements and measure the THD of various input signals.

28

OVERVIEW – CHAPTER 12 TRANSMISSION LINES

12-2 TYPES OF TRANSMISSION LINES A transmission line can be defined as the conductive elements that conduct signal power between system elements. At low frequencies that is very straightforward, but at higher frequencies the make-up of the connection starts having appreciable effect on circuit action. That is the subject of this chapter. The physical characteristics of the basic types of transmission lines are described in this section. These include the two-wire open line, twisted pair, shielded pair, and coaxial lines. The concluding topic discusses the concept of balanced and unbalanced lines. The unshielded twisted pair cable used in CAT6 network installations is examined, including the T568A/T568B wiring standards. The categories for twisted pair cable are provided in Table 12-2. 12-3 ELECTRICAL CHARACTERISTICS OF TRANSMISSION LINES The electrical equivalent circuit for a two-wire transmission line is described with the help of Fig. 12-9. This discussion leads to looking at what happens when this line becomes infinitely long. The resulting circuit analysis leads to development of the equation for the line’s characteristic impedance (Equation 12-8). Example 12-1 provides a simple illustration of calculating characteristic impedance for a common cable and vividly demonstrates that it is independent of the line’s length. Calculating characteristic impedance using the physical (as opposed to electrical) characteristics is then presented, followed by an introduction to transmission line losses. 12-4 PROPAGATION OF DC VOLTAGE DOWN A LINE The introduction to propagating an ac voltage is facilitated by the dc discussion first. The velocity of propagation is mathematically derived leading to Equation 12-20. The concept of using a transmission line as a delay line naturally follows as shown in Example 12-3. The concept of velocity factor is also demonstrated here. The concept of wavelength and its formula is provided. The fact that wavelength in the line is less than free-space wavelength (due to reduced velocity in the line) is also introduced. 12-5 NONRESONANT LINE A nonresonant line is of infinite length or terminated with its characteristic impedance. A discussion of traveling dc waves is followed by an ac wave traveling down the line. Since none of this incident energy is reflected by the perfectly matched load, the waveform at any point on this line will be a replica of the generator waveform. There will be a finite delay as discussed in Section 12-4. 12-6 RESONANT TRANSMISSION LINE The student should now be ready for the imperfect case where the load is not perfectly matched to the line. This is termed a resonant line and now the resulting reflections must be dealt with. An incident dc level on an open and shorted line are explainable with the aid of Figs. 12-17 and 12-18. This eases them in to the more conceptually difficult process of the ac wave analysis provided in Fig. 12-19. Walking your students through Example 12-6 in conjunction with Fig. 12-22 should help in their understanding.

29

12-7 STANDING WAVE RATIO The reflection coefficient (Γ) is defined in terms of incident/reflected voltages and load/characteristic impedances. The voltage standing wave ratio is then defined and the equation for its calculation using reflection coefficient is given (Equation 12-27). This is followed by a discussion of the ill effects of mismatches and the resulting reflections. Example 12-7 provides a good reinforcement of these concepts. An introduction to the quarter-wavelength transformer is then provided. This section closes with an explanation of when transmission line effects must be taken into account. In general these effects become important when a line is greater than 1/16 of a wavelength. 12-8 THE SMITH CHART Transmission line calculations are rather cumbersome and even with the availability of special software packages, the use of Smith charts is still common. The chart is introduced and the process of normalizing impedances and admittances is explained. The actual use of the chart is easily accomplished by using transparencies and working through a number of examples. The examples in the text deal with finding a line’s input impedance, matching a line to its load using a quarter-wave-length transformer, and the use of stub tuners for matching. 12-9 TRANSMISSION LINE APPLICATIONS The use of transmission line sections to simulate discrete circuit elements opens this section. The transmission line balun is explained followed by a description of how lines are used as filters. The slotted line is introduced and the section is concluded with an explanation of time-domain reflectometry. Your TDR description can be enhanced by using Fig. 12-33 as an overhead. 12-10 TROUBLESHOOTING The troubleshooting begins with a discussion of lines used for interconnection of local area networks. The various types of losses are described and crosstalk is defined. Minimizing crosstalk in twisted pair lines is explained. The use of a basic ohmmeter for some limited testing is provided. This section is concluded with some hints on dealing with a troublesome transmission line between a TV and its antenna. 12-11 TROUBLESHOOTING WITH ELECTRONICS WORKBENCH MULTISIM

Impedance measurements, using the EWB Multisim network analyzer are presented. This easy to follow procedure for using a network analyzer is accompanied with many figures and examples for resistive, RC, and RL networks. The student will find that this virtual instrument makes making impedance measurements relatively painless.

30

OVERVIEW – CHAPTER 13 WAVE PROPAGATION

13-1 ELECTRICAL TO ELECTROMAGNETIC CONVERSION The concept of electromagnetic waves is often a tough concept to for the student to grasp. You might want to start by explaining how an electrical signal creates light when driven into a tungsten filament. Light is an electromagnetic wave at a much higher frequency than the electromagnetic waves that function as radio waves. As will be discussed in Chapter 14, we create radio electromagnetic waves by driving an electrical signal into an antenna. 13-2 ELECTROMAGNETIC WAVES The electric and magnetic fields are first described. The direction of the electric field is defined as giving the polarization of the electromagnetic field. A wavefront is then defined as the plane joining all points of equal phase of an electromagnetic field. The characteristic impedance of free space is then derived as being equal to 377 ohms. It is also shown to equal the square root of the ratio of permeability to permittivity. 13-3 WAVES NOT IN FREE SPACE The process of reflection of an electromagnetic wave is described. The phase reversal that occurs at reflection is also noted. This is followed by a discussion of the refraction process. Snell’s law is provided and students may be familiar with it from a study of optics. Since light waves and radio waves are both electromagnetic, it is logical that it applies to both. The process of diffraction is then discussed. The creation of a shadow zone by diffraction is illustrated with the help of Fig. 13-5. 13-4 GROUND- AND SPACE-WAVE PROPAGATION The ground wave is descriptively also called a surface wave. It travels along the earth’s surface. They are possible only at low frequencies—typically less than 2-MHz. The space wave is either direct or reflected as illustrated in Fig. 13-6. Equation 13-7 provides an approximation of the maximum direct space wave based on curvature of the earth limitations. Example 13-1 examines ghosting in TV caused by reception of a direct and reflected space wave. 13-5 SKY-WAVE PROPAGATION Long distance propagation is possible by refraction of waves by the ionosphere. The various layers of the ionosphere are described as is their variation between day and evening. The concepts of critical frequency, critical angle, maximum usable frequency and skip zone are then discussed. The difference between skip zone and skip distance can easily be shown with the help of Fig. 13-13. Two causes of fading with sky-wave propagation is illustrated with Fig. 13-14. The section concludes with a discussion of tropospheric scatter. Because of fading problems, the various types of diversity reception are introduced. They are space diversity, frequency diversity and angle diversity.

31

13-6 SATELLITE COMMUNICATIONS This section has been completely revised to address the latest in satellite communications. A detailed view of a Boeing 601HP satellite is presented in Figure 13-16. Orbital patterns of satellites are examined as well as low-earth orbit satellites. The steps for making azimuth and elevation angles are presented followed by a discussion of multiplexing techniques. Figure 13-21 and 22 provide an illustration of how TDMA is used to allow a single satellite to service multiple earth stations simultaneously and steps for calculating the distance to/from the satellite are next presented. This is followed by an introduction to VSAT and MSAT systems. Satellite radio service is a common buzz word in electronic communication today and this topic is addressed at the end of the section. 13-7 FIGURE OF MERIT AND LINK BUDGET ANALYSIS This section presents the equations for calculating the figure of merit (G/T) and the satellite link budget. Note: An online satellite system calculator has been developed specifically for this textbook based on the satellite calculations presented in the text. The URL for the online calculator is:

http://web.nmsu.edu/~jbeasley/Satellite/

The figure of merit is a way to compare different earth station receivers. The figure of merit takes into consideration the technical quality of each piece of the satellite earth station equipment and enables the end user to obtain some measure of performance for the entire system. The final figure of merit can then be used as a comparison to other earth stations. In regards to the satellite link budget, satellite receivers will have a required carrier-to-noise (C/N) at the input. The satellite link budget is used to verify the required C/N and signal level to the satellite receiver will be met to ensure the satellite receiver outputs a signal that meets specifications. 13-8 TROUBLESHOOTING Radio interference is the main topic of this troubleshooting section. The various types of interference are discussed and three methods for reducing interference are provided. A discussion of EMI and RFI is presented. 13-9 TROUBLESHOOTING WITH ELECTRONICS WORKBENCH MULTISIM

This exercise examines the behavior of the crystals used to generate high frequency signals. The student will examine the RLC equivalent circuit and the series resonant point using the EWB Bode plotter. The exercises include an example of troubleshooting a Pierce crystal oscillator.

32

OVERVIEW – CHAPTER 14 ANTENNAS

14-1 BASIC ANTENNA THEORY Antennas serve to generate or collect electromagnetic energy. An antenna receives energy with the same efficiency and characteristics as it would transmit. This is termed reciprocity. The received signal strength is usually described in terms of the electric field strength. If an antenna receives a signal that generates 1-mV in a 1-meter antenna, the field strength is 1-mV/m. 14-2 HALF-WAVE DIPOLE ANTENNA The development of a 1/2 wavelength dipole antenna from an open-ended 1/4 wavelength transmission line opens this section. The impedance starts at 73 ohms at the center and increases to 2500 ohms at the ends. A discussion of its radiation pattern follows with Fig. 14-4 showing the two-dimensional pattern and Fig. 14-5 the three-dimensional pattern. A description of its gain is provided and the formula for power received by an antenna (Equations 14-1a-c) are introduced. Example 14-1 and 14-2 provide practice in applying this formula. 14-3 RADIATION RESISTANCE The portion of an antenna’s input impedance that is the result of power radiated into space is called the radiation resistance. The antenna’s efficiency is also introduced at this point. The effects of antenna length and ground effects are then discussed with respect to radiation resistance. This section concludes with a discussion of electrical versus physical length and the effects of nonideal antenna length. 14-4 ANTENNA FEED LINES Resonant antenna feed lines are not commonly used but they do find some convenience in certain high-frequency applications. The length of this line is critical to efficient operation. On the other hand, the nonresonant feed line functions properly with any length—the only need is to match the antenna’s impedance to the line’s characteristic impedance. Four examples of feeding antennas with non-resonant lines are provided in Fig. 14-10. The delta match and quarter-wave matching are included here. The quarter-wave technique is most useful for narrowband operation while the delta section is more useful for broadband situations. 14-5 Monopole (vertical) ANTENNA The vertical antenna is a 1/4 wavelength and used mainly for frequencies below 2 MHz. The transmitter is connected between the antenna and ground. The ground forms the image antenna making half-wave operation possible. The impedance is 36.6 ohms, however, instead of the 73 ohms for the Hertz antenna. The use of a counterpoise to provide good earth conductivity is explained. The use of loaded antennas to allow reasonable efficient operation with less than 1/4 wavelength is also explained.

33

14-6 ANTENNA ARRAYS The half-wave dipole antenna with a parasitic array is described. The change in radiation pattern with respect to the basic antenna is shown in Fig. 14-16. This leads to the Yagi-Uda antenna that includes a reflector and a director for its parasitic elements. Its physical dimensions and radiation pattern is shown in Fig. 14-17. The driven collinear and broadside arrays are subsequently explained. The vertical array that is commonly used by standard AM broadcast stations is described. Examples of the countless possible radiation patterns are shown in Fig. 14-20. 14-7 SPECIAL-PURPOSE ANTENNAS The log-periodic antenna is shown to be a versatile wideband device. The design ratio is defined and generally equals about 0.7. The loop antenna is then described. Its simplicity and directional characteristics makes it ideal for direction finding (DF) applications. The addition of a ferrite core to the loop makes it ideal for small AM receivers. The folded dipole is shown to be similar to the half-wave dipole antenna but offers an impedance of 300 ohms instead of 73 ohms. The description of slot antenna arrays concludes this section. Their usefulness with airplane radars is explained. 14-8 TROUBLESHOOTING The proper antenna installation procedures are introduced. Special emphasis on safety considerations and proper grounding techniques is provided. Typical problems encountered with installation are described. The grid-dip meter is introduced and its ability to make measurements on unpowered circuits is explained. The SWR meter is also described and information on troubleshooting with it is provided. The use of an anechoic chamber in making antenna measurements is also included. 14-9 TROUBLESHOOTING WITH ELECTRONICS WORKBENCH

MULTISIM The first Multisim example demonstrates the modeling of a dipole antenna. The

antenna model is examined with the EWB network analyzer. The second examples demonstrates how a single stub tuner can be used to provide a better impedance match of the antenna to the load. The exercises include designing a 92 MHz dipole, a 450 MHz dipole, and providing an impedance match for a 100 MHz dipole antenna.

34

OVERVIEW – CHAPTER 15 WAVEGUIDES AND RADAR

15-1 COMPARISON OF TRANSMISSION SYSTEMS We’ve now studied the use of antennas and transmission lines for transmitting electrical information. We’ll look at waveguides in this chapter and fiber optics in Chapter 18. At this point it will give a bit of perspective to your students to look at the comparison of transmission systems highlighted by Fig. 15-1. As discussed in the accompanying text, various transmission schemes make sense depending on the circumstances. 15-2 TYPES OF WAVEGUIDES A rigorous mathematical analysis is beyond our intentions here, but a discussion of the modes of operation and subsequent E and H field configuration is introduced. The concept of transverse electric (TE) and transverse magnetic (TM) modes of operation is introduced. The formula for the dominant mode transverse electric waveguide is provided. The cut-off frequency for a common waveguide, RG-52U, is determined using this formula. 15-3 PHYSICAL PICTURE OF WAVEGUIDE PROPAGATION The reflection of energy off the side walls of rectangular waveguide as it travels down the guide is described. Since the resulting wave velocity is less than free space velocity, the guide wavelength is different than free space wavelength for the frequency being transmitted. Equations are presented that relate the free space and guide wavelength. The concept of phase velocity is introduced and its relationship to guide velocity and the velocity of light is given with Equation 15-7. 15-4 OTHER TYPES OF WAVEGUIDES Rectangular waveguide is by far the most used type of waveguide. Other types are used in special applications as described in this section. Circular waveguide is commonly found in situations where a rotating section is required such as in radar systems. Ridged waveguide is useful when space is at a premium since it allows operation at lower frequencies for the volume that it occupies compared to rectangular or circular waveguide. Flexible waveguide is used when constant flexing of the guide is occurring such as in laboratory applications. 15-5 OTHER WAVEGUIDE CONSIDERATIONS Waveguide is able to propagate huge amounts of power. Standard X-band waveguide can handle 1 million watts. The standard H and E bends are then described as well as the twist section. The shunt, series and hybrid tees are described. Your classroom discussion here can be enhanced with transparencies of Figs. 15-12 and 13. The slide screw tuner and double-slug tuner are then described. Analogies to their transmission line counterparts are provided. 15-6 TERMINATION AND ATTENUATION Equation 15-8 shows that the characteristic impedance of a waveguide depends on its size and the frequency (or wavelength) of the signal involved. Recall that the characteristic impedance of a transmission line does not depend on frequency. Various ways to terminate a waveguide in a specific impedance are illustrated in Fig. 15-15. The two common methods of

35

providing variable attenuation are shown in Fig. 15-16. They are the flap attenuator and the vane attenuator. 15-7 DIRECTIONAL COUPLER The two-hole directional coupler is illustrated in Fig. 15-17. The text explains how it transfers energy from a primary to an adjacent waveguide for energy traveling in one direction only. It can therefore distinguish between waves traveling in opposite directions. It finds common application in measurement of power and for determining the SWR. 15-8 COUPLING WAVEGUIDE ENERGY AND CAVITY RESONATORS Coupling energy into or out of a waveguide is obviously an important topic. Basically it can be accomplished in one of three ways. The first is probe (capacitive) coupling. The use of loops (inductive) is also commonly used. The third category is called aperture or slot coupling. These three techniques are illustrated in Figs. 15-18 through 20. This section concludes with an introduction to cavity resonators and methods for tuning them. 15-9 RADAR Radar was the first major application of waveguides. After a general introduction, the typical radar waveform and range determination is provided. The important concepts of maximum unambiguous range, minimum range and duty cycle are also developed. Figure 15-26 provides a block diagram of a typical radar system that can be useful in classroom discussions. The section is concluded with an introduction to the Doppler effect. Equation 15-15 can be used to determine the velocity of an object when using Doppler radar. 15-10 RFID – RADIO FREQUENCY IDENTIFICATION This section examines RFID – Radio Frequency Identification which is a technique that uses radio waves to track and identify people, animal, objects, and shipments. This is done by the principle of modulated backscatter. The term “backscatter” is referring to the reflection of the radio waves striking the RFID tag and reflecting back to the transmitter source with its stored unique identification information. There are three parameters that define an RFID system. These include the means of powering the tag, frequency of operation, and the communications protocol (also called the air interface protocol). The basic block diagram of an RFID system is provided in Fig. 15-27 and pictures of RFID tags are provided in Fig. 15-28. 15-11 MICROINTEGRATED CIRCUIT WAVEGUIDING The use of stripline and microstrip waveguiding has become common with modern mass production circuit fabrication techniques. The physical geometry of stripline and the two formats for microstrip are shown in Fig. 15-31. The formulas for their characteristic impedances depend on some specific dimensions and spacings and are given in Fig. 15-32. The use of microstrip to simulate discrete circuit elements is described and the section is concluded with an introduction to dielectric waveguide.

36

15-11 TROUBLESHOOTING Some problems caused by joints and flanges in waveguide systems are presented. The possibility of arcs and how they might occur is also introduced. Some techniques for troubleshooting rotary joints in radar systems are detailed. The use of directional couplers for measuring power and the VSWR is presented with the assistance of Figs. 15-35 and 15-36. Equation 15-7 is a convenient formula to calculate the VSWR based on the incident and reflected power measurements. 15-12 TROUBLESHOOTING WITH ELECTRONICS WORKBENCH

MULTISIM The students will be able to examine the characteristics of a lossy transmission line in this EWB Multisim section. The EWB network analyzer will be used to examine the characteristics of transmission lines and waveguides. This section will help the student to better understand the Smith chart impedance information being displayed by the network analyzer. The exercises provide opportunities for the students to test their ability to use the network analyzer to obtain impedance characteristics.

37

OVERVIEW – CHAPTER 16 MICROWAVES AND LASERS

16-1 MICROWAVE ANTENNAS Microwave antennas typically have a different look than the ones studied in Chapter 14. The higher frequencies involved allow for highly directive designs with reasonable sizes. The horn antennas are usually used with waveguides and basically involve a flaring out of the guide to allow the launch of the wave. The parabolic antenna has become commonplace in today’s world of communications. Figure 16-3 makes a good transparency to illustrate the three methods of feed. Example 16-2 shows how to calculate power gain and beamwidth for the parabolic antenna. The lens antenna is a peculiar design for a microwave antenna based on “zoning” a dielectric material. This section concludes with discussions of patch antennas and phased arrays for SATCOM. 16-2 MICROWAVE TUBES Students are often surprised to learn that tubes are still an important part of microwave electronics. The magnetron is used as a source of microwaves when fairly high powers are required. The microwave oven is a major application of this device. The traveling wave tube is the workhorse of SATCOM amplifiers. Detailed operation of the TWT is provided. This section concludes with a short introduction to the klystron. 16-3 SOLID-STATE MICROWAVE DEVICES The Gunn oscillator is a solid state bulk-effect source of microwave energy. The operation and applications for this device are described. This is followed by an introduction to the IMPATT diode. If you plan to cover this device in detail, Figs. 16-13, 14, and 15 would be helpful as transparencies. The p-i-n diode is then introduced. They are widely used as microwave switches. A general discussion of microwave transistors and integrated circuits concludes this section. 16-4 FERRITES At microwave frequencies, ferrites exhibit high resistance and resonance takes place within the iron atoms. These effects make them useful as microwave devices. A basic explanation of the theory is followed by discussion of the various applications. Their use as attenuators, isolators, in Faraday rotation and circulators is provided. Their use as cores for windings is discussed and their use as ferrite beads is explained. 16-5 LOW-NOISE AMPLIFICATION Microwave amplifiers are often dealing with a very low-level input signal. When the ultimate in low-noise amplification is required, one of the two amplifiers detailed in this section can be utilized. The parametric amplifier provides amplification via the variation of a reactance. A simplified 10-GHz parametric amplifier is detailed in Fig. 16-28. The maser is an even lower noise amplifier. A ruby maser amplifier is provided in Fig. 16-29. This section concludes with an introduction to a rubidium atomic clock.

38

16-6 LASERS The laser has become a very important device in the field of electronic communication. Its use in fiber optic communication is covered in chapter 18. A brief explanation of a laser’s operation is provided here. The direct use of lasers (without fiber optics) is illustrated with the help of Fig. 16-32. The section concludes with a discussion of an optical switch called a transphasor. This may lead to a computer that is optically based rather than electronic and would offer much higher operating speeds. 16-7 TROUBLESHOOTING The two basic types of regulated power supplies are the switching and the linear. Since virtually all electronic equipment (except battery powered) contains one or the other, an introduction to both types is included here. Some typical schematics/block diagrams are shown in Figs. 16-34 and 35 and pertinent troubleshooting material is provided. This section concludes with information on troubleshooting a traveling wave tube amplifier. Included are some initial considerations, typical dc problems, and typical RF troubles. 16-8 TROUBLESHOOTING WITH ELECTRONICS WORKBENCH MULTISIM

This section uses EWB Multisim to examine the RF capacitor, the RF inductor, and the RF transistor. The students should understand the behavior of these devices at high frequencies. An equivalent model of the RF capacitor is provided in Fig. 16-38. This is how the computer simulation is modeling the high frequency behavior of the RF capacitor. The Bode plot analysis helps the student to understand each devices high frequency behavior. The exercises provided the student with an opportunity to explore the characteristics of the RF inductor and troubleshooting an RF amplifier.

39

OVERVIEW – CHAPTER 17 TELEVISION

17-2 DIGITAL TELEVISION An exciting development in television is the deployment of digital television in the United States. This section includes an overview of the basics of digital television including the changes in the aspect ratio of the video image, the digital video signal, the AC-3 digital audio signal, and transmission of the digital signals. Figs. 17-2 and 17-4 show the structure of the 8VSB segment and frame sync. A chip set for the front-end of a hybrid TV receiver system using the Philips TDA8980 and 8961 ICs is provided in Fig. 17-7. 17-3 MONITORING THE DIGITAL TELEVISION SIGNAL This section will provide a look at the steps and measurements typically taken when monitoring the ATSC digital transport stream and the transmitted 8VSB signal. The steps and measurements for monitoring both the ATSC digital transmission system and the 8VSB transmitted signal are examined. The digital signal path is followed through the multiplexing part through real-time measurement and analysis of the digital signal. The section concludes with a look at the DTV RF spectrum. 17-4 NTSC TRANSMITTER PRINCIPLES The simplified block diagram in Fig. 17-16 is a good place to start your discussion of television. It clearly illustrates the concept of the separate transmitter sections for the aural (sound) and video sections. This discussion is followed by an introduction to TV cameras and the scanning process. Relating the scanning of the scene in Fig. 17-18a to the resulting electrical signal in Fig. 17-18c is often a very good learning experience for the student. 17-5 NTSC TRANSMITTER/RECEIVER SYNCHRONIZATION Synchronization will be a new concept for many of your students. It obviously is a key to the video portion of TV operation and merits close attention. This section starts off with a discussion of horizontal and vertical retrace. This leads to the use of interlaced scanning and the concepts of frame frequency, flicker, and the field. The sync pulses for horizontal retrace are provided in Fig. 17-20 while those for the vertical retrace are in Fig. 17-21. It is good to show a transparency of both during a classroom discussion of the sync signals. 17-6 RESOLUTION Resolution is the ability to resolve detailed picture elements. The concept of how to determine vertical and horizontal resolution is provided and the relationship to required bandwidth is introduced. Examples 17-1 and 17-2 provide good exercises to help in understanding these concepts. They are also helpful in proving Hartley’s Law that was introduced in Chapter 1.

40

17-7 THE TELEVISION SIGNAL The television signal uses a 6 MHz bandwidth and is detailed in Fig. 17-22. The frequency allocations for all standard broadcast channels is given in Table 17-3. It is interesting to note the small bandwidth of the audio signal compared to the video. The concept of vestigial-sideband is introduced in this section. While using somewhat more bandwidth than SSB, it does not require the relatively expensive process of carrier reinsertion at the receiver. 17-8 TELEVISION RECEIVERS The TV receiver block diagram of Fig. 17-23 makes a very good discussion base for classroom discussion. This relatively complicated piece of electronic equipment incorporates much of the technology covered up to this point in the book and therefore can serve to tie together a lot of loose ends for the student. 17-9 THE FRONT END AND IF AMPLIFIERS The VHF/UHF tuner block diagram in Fig. 17-24 shows the TV front end up to the IF amps. The portion in dashed lines represents the portion used for UHF operation exclusively. The four functions performed by the tuner other than the obvious one of selecting the desired frequency range (channel) are detailed in this section. The IF amplifiers are then discussed and the concept of stagger tuning is introduced. The use of surface acoustic wave (SAW) filters is provided and some detail on their operation is provided. This section concludes with discussions of the ideal IF amplifier response and wavetraps. 17-10 THE VIDEO SECTION A good starting point for a discussion of the video section is the block diagram in Fig. 17-29. This is the time to talk about the contrast control that varies the amplitude of the signal applied to the CRT. It is analogous to the volume control for the audio. The takeoff point for the AGC signal occurs in the video section. The brightness control is also shown in Fig. 17-29. It varies the overall brightness of the picture without changing the difference between the white and black signal as does the contrast control. 17-11 SYNC AND DEFLECTION The sync separator process is shown in Fig. 17-30. From it are generated the signals to keep the horizontal and vertical oscillators exactly synchronized with the transmitter. Also covered in this section is a block diagram showing a typical horizontal deflection system as driven by the horizontal sync pulses. The generation of the anode and focus high-voltages from the horizontal deflection system is also covered. A complete schematic for a horizontal deflection system is given in Fig. 17-33. 17-12 PRINCIPLES OF NTSC COLOR TELEVISION The use of the interleaving process to add the necessary color information is the starting point for this section. The system block diagram in Fig. 17-36 should be carefully studied at this point. The processing of the I, Q, and Y signals is very important. The construction and operation of the color CRT is discussed to close out this section. 17-13 SOUND AND PICTURE IMPROVEMENTS The audio and video improvements of recent years is the subject of this section. The technique used to create a stereo channel for television is discussed. The spectrum for stereo TV transmission is provided in Fig. 17-42.

41

17-14 TROUBLESHOOTING This chapter’s troubleshooting section deals with the identification of defective stages in a TV receiver based upon the observable symptoms. Table 17-7 provides detail on symptoms, causes for those symptoms, and the probable section of the receiver with the failure. This table makes a good transparency for stimulating classroom discussion. 17-15 TROULESHOOTING WITH ELECTRONICS WORKBENCH MULTISIM In this simulation exercise, the student will be able to use the spectrum analyzer to examine the RF spectrum of a UHF television signal. The students will also be given the opportunity experiment with and troubleshoot a wavetrap circuit.

42

OVERVIEW – CHAPTER 18 FIBER OPTICS

18-1 INTRODUCTION Figure 18-1 gets the ball rolling here, showing a basic fiber-optic communication system. The text provides a step-by-step description of each element in the system. This introductory section concludes with a detailed listing of the advantages of fiber optics with respect to waveguides or copper conductors for transmitting electrical signals. The extremely wide bandwidths afforded by fiber optics is listed first for good reason. 18-2 THE NATURE OF LIGHT The basics of reflection and refraction are the first topics of this section. Snell’s law is reintroduced—it was first discussed in Chapter 13 when the refraction of electromagnetic waves by the ionosphere was discussed. Example 18-1 illustrates the calculation of the wavelength of light and Fig 18-4 provides a chart of the electromagnetic spectrum. This is an important concept for the student to understand. Make sure the student knows where the fiber optic wavelengths of (850, 1310, 1550 and 1625) nm are found related to visible light. 18-3 OPTICAL FIBERS This is an important section in that it explains the common types of fiber used in electronic communication. The different types used greatly affect the characteristics of the system. The three basic categories are discussed, and a pictorial summary of them is given in Fig. 18-9. It makes a good transparency to facilitate classroom discussion. A comparison of single-mode and multimode fiber is provided in Table 18-4. 18-4 FIBER ATTENUATION AND DISPERSION This section starts with detail on the two distance-limiting parameters in fiber-optic transmission: attenuation and dispersion. Fig. 18-10 provides a chart of absorption loss due to scattering of the optical signal. The loss is shown for the three major wavelengths used in optical communications. Fig. 18-11 will make a good overhead to help explain the concept of pulse broadening to the student. Figs. 18-12 and 13 are also helpful when explaining dispersion. The section concludes with a discussion on dispersion compensation including a new device called fiber bragg grating. The student should become familiar with this concept. 18-5 OPTICAL COMPONENTS The two choices for fiber optic communications light sources are detailed in this section. They are the diode laser and the high-radiance light-emitting diode. The one chosen depends on system requirements including power levels, temperature sensitivities, response time, lifetime to failure, and characteristics of failure. Figure 18-14 provides the physical geometry of a semiconductor laser. The intermediate components used in the optical span are identified in pages 875-876.

18-6 FIBER CONNECTIONS AND SPLICES Due to the high purity of the glass in fiber optics, the connections (glass to glass and light source/detector to glass) become critical to the overall attenuation in the system. The sources of excessive connector loss include axial misalignment, angular misalignment, an air gap, or rough surfaces. Connector techniques and splice techniques are provided here with the assistance of

43

Fig. 18-20. Fiber connectorization is discussed including the SC, ST, and small form factors MT-RJ and VF-45 18-7 SYSTEM DESIGN AND OPERATIONAL ISSUES

The primary issues for optical links are bite error rate for a digital system and the carrier-to-noise ratio for an analog system. The issues for long-haul and short-haul systems are discussed. This sections a discussion on system design as depicted in Figs. 18-22, 23 and 24. The student must understanding the concept of developing the light budget, defining the receiver dynamic range, and meeting the specifications defined by the RSL (received signal level) 18-8 CABLING AND CONSTRUCTION

This section provides a brief outline of the issues associated with exterior and interior installation of fiber cable. The section includes a discussion on understanding the traces generated by an OTDR (optical time-domain reflectometer). A set of OTDR traces is provided in Fig. 18-25. 18-9 OPTICAL NETWORKS The use of fiber optics in local area networks has increased dramatically in recent years. The many advantages they have over copper connections are introduced and are followed by a discussion of network topology. The basic concepts of fiber to the home and fiber to the curb are discussed. An example of connections for an optical LAN are provided in Fig. 18-27 Pictures of the North American fiber-optic long haul routes are provided on pages 898 and 899. 18-10 SAFETY The discussion on fiber is not complete without at least a brief discussion on safety issues with fiber. A summary of safety issues, related to fiber, is provided on page 900. 18-11 TROUBLESHOOTING The problems involved with losses and power requirement calculations are described. The use of an optical power meter is introduced. Problems with connectors, cable, LEDS, and DLs are explained. A simple, self-constructed, light probe (Fig. 18-29) is described as is its application to troubleshooting situations. 18-11 TROUBLESHOOTING WITH ELECTRONICS WORKBENCH

MULTISIM This section includes a simulation of a fiber system design. The students will be able to experiment with received signal levels and see how they impact the optical communication system. The exercises include troubleshooting the system to determine why the system is not functional.

44

1. A technique for resolving complex repetitive waveforms into sine or cosine waves and a dc component is known as:

a. Harmonic Analysis b. Armstrong's Principle c. Fourier Analysis d. Frequency Analysis

Answer: c 2. Shot noise is directly related to:

a. dc current. b. ac voltage. c. resistor value. d. boltzman constant.

Answer: a 3. The statement "information is proportional to bandwidth" is known as:

a. Simpson Rule. b. Shannon's Law. c. Fourier's Law. d. Hartley's Law.

Answer: d 4. Given that a filter has a Q of 50 and a BW of 500 Hz, determine its resonant or center frequency.

a. 100 kHz b. 25,000 kHz c. 2500 kHz d. 250 kHz

Answer: b 5. Determine the frequency of a Clapp oscillator if L4 = 35 mH and C3 = 20pf.

a. 200 kHz b. 20 kHz c. 190 kHz d. 1.9 MHz

Answer: c 6. The device which superimposes information onto a high-frequency signal for transmission is called:

a. a demodulator. b. the carrier. c. a modulator. d. the intelligence.

Answer: c

TEST ITEM FILE - CHAPTER 1INTRODUCTORY TOPICS

45

TEST ITEM FILE - Chapter 1 7. A device that converts energy from one form to another is called a:

a. transducer. b. transmitter. c. modulator. d. detector.

Answer: a 8. The device which extracts the information signal from the high frequency carrier in a receiver is called a:

a. transducer. b. transmitter. c. modulator. d. demodulator.

Answer: d 9. Frequencies between 3 and 30 Mhz are referred to as:

a. Ultra-High Frequencies (UHF'). b. High Frequencies (HF). c. Very-High Frequencies (VHF'). d. Medium Frequencies (MF).

Answer: b 10. Ultra-High Frequency signals are between:

a. 30 Mhz and 300 MHz. b. 3 Mhz and 30 GHz. c. 3 Mhz and 30 MHz. d. 300 Mhz and 3 GHz.

Answer: d

11. The type of noise that is often produced by spark-producing equipment is known as:

a. man-made noise. b. atmospheric noise. c. thermal noise. d. transistor noise.

Answer: a

12. Which is NOT an example of internal noise?

a. Noise produced by resistors b. Noise produced by transistors and integrated circuits c. Excess noise d. Space noise

Answer: d

46

TEST ITEM FILE - Chapter 1 13. Which is NOT a term for noise produced by resistors?

a. White noise b. Shot noise c. Johnson noise d. Thermal noise

Answer: b 14. If the noise produced by a resistor in a radio receiver is 2.4 uV and the resistance is then doubled, the resulting noise produced by the new resistance is approximately:

a. 3.4 uV. b. 2.4 uV. c. 1.2 uV. d. 4.8 uV.

Answer. a 15. Two types of noise that occur at extremely low and high frequencies in amplifiers containing transistors are known as:

a. excess noise and transit-time noise. b. flicker noise and shot noise. c. 1/f noise and thermal noise. d. Johnson noise and pink noise.

Answer: a 16. A figure of merit that best describes how much noise a device creates is known as:

a. signal-to-noise ratio. b. Noise figure. c. Friis noise. d. effective noise bandwidth.

Answer: b 17. A transistor amplifier has a measured S/N power of 25 at its input and 8 at its output. Its noise figure in decibels is:

a. 118. b. 17. c. 9.10. d. 4.95.

Answer: d 18. A popular, convenient way of representing noise calculations, mainly involved with microwave receivers, is:

a. SIN ratio. b. Noise figure. c. Equivalent noise temperature. d. Equivalent noise resistance.

Answer: c

47

TEST ITEM FILE - Chapter 1 19. In a three-stage amplifier.

a. the noise ratio of all three stages is mainly due to the noise figure of the first stage. b. Friss’s formula can be used to determine the overall noise effect of all three stages. c. the noise ratio of the last stage is usually insignificant. d. all of the above.

Answer. d 20. When using a noise diode generator in making noise measurements, the noise diode's current in ma. is numerically equal to the:

a. noise ratio of the DUT no matter what its impedance is. b. noise figure in db of the DUT no matter what its impedance is. c. noise ratio of the DUT only if its impedance is 50 ohms. d. noise figure in decibels of the DUT only if its impedance is 50 ohms.

Answer: c 21. Hartley's law states that:

a. bandwidth is directly proportional to both the amount of information and time of transmission. b. the amount of information is directly proportional to both the system bandwidth and time of transmission. c. the time of transmission is directly proportional to both the system bandwidth and the amount of information. d. the bandwidth is directly proportional to both the amount of information and the amount of noise present.

Answer: b 22. A ramp waveform of the type shown in Figure 1-la has a peak-to-peak amplitude of 2V and a frequency of 50 Hz.

The term representing the fifth harmonic has a coefficient of a. 0.2 b. 2.5

c. 0.127 d. 0.254

Answer: c

48

TEST ITEM FILE - Chapter 1

23. A square wave of the type shown in Figure 1-lc has a peak-to-peak amplitude of 10V and a frequency of 200

Hz. The term representing the fifth harmonic is approximately:

a. 1.27 sin 6283t b. 0.2 sin 6283t c. 4 sin 3142t d. 6.35 sin 6283t

Answer: a 24. A 4.7 mH inductor has a Q of 3500 at a frequency of 50 MHz. Its internal resistance is approximately:

a. 422 ohms. b. 67.14 ohms. c. 14.9 milliohms. d. 2.37 milliohms.

Answer: a

49

TEST ITEM FILE - Chapter 1

25. The resonant frequency of the resonant circuit given in Figure 1-2 is:

a. 129 kHz. b. 41.1 kHz. c. 20.6 kHz. d. 2.65GHz.

Answer: c 26. The impedance of the resonant circuit given in Figure 1-2 is:

a. 1553 ohms. b. 2053 ohms. c. 500 ohms. d. 1632 ohms.

Answer: c 27. The bandwidth of the resonant circuit given in Figure 1-2 is:

a. 1.59 x 10'° Hz. b. 2.4 X 106 Hz. c. 41.6 X 103 Hz. d. 6635 Hz.

Answer: d 28. The parallel resonant circuit of Figure 1-3 has a resonant frequency of approximately:

a. 259 kHz b. 10.7 GHz c. 41.3 kHz d. 82.6 kHz

Answer: c

50

TEST ITEM FILE - Chapter 1 29. The parallel resonant circuit of Figure 1-3 has Q and BW equal to:

a. 233 and 176. Hz, respectively b. 233 and 241 Hz, respectively c. 1111 and 37.2 Hz, respectively d. 171.2 and 241 Hz, respectively

Answer: a 30. The maximum impedance at the resonant frequency for the circuit of Figure 1-3 is approximately:

a. 10.8 kilo-ohms b. 1.165 kilo-ohms c. 5.825 kilo-ohms d. 271.4 kilo-ohms Answer: d

31. The oscillator design that is characterized by having a "tapped" inductor is the:

a. Hartley design. b. Clapp design. c. Colpitts design. d. Crystal design.

Answer: a 32. The oscillator design that uses two capacitors in the tank circuit is the:

a. Hartley design. b. Clapp design. c. Colpitts design. d. Crystal design.

Answer: c 33. The oscillator design that has the highest frequency stability is the:

a. Hartley design. b. Clapp design. c. Colpitts design. d. Crystal design.

Answer: d 34. The main requirements that must be met for an oscillator to successfully oscillate are known as:

a. the flywheel effect. b. Barkhausen criteria. c. the piezoelectric effect. d. frequency synthesis.

Answer: b

51

TEST ITEM FILE - Chapter 1 35. The repetitive exchange of energy from the magnetic field of an inductor to an electric charge on a capacitor in a

resonant circuit is known as:

a. the flywheel effect. b. Barkhausen criteria. c. the piezoelectric effect. d. frequency synthesis.

Answer: a

36. A measured value of 10mW will result in what dBm power level?

a. 0dBm b. 3dBm c. 10dBm d. –3dBm Answer: c

37. Determine the voltage level required to produce a +10dBm level. Assume a 600Ω system.

a. 1.947V b. 2.45V c. .775V d. none of the above

Answer: b 38. Convert 300W to dBW.

a. 54.77 dBW b. –24.77 dBW c. –10.0 dBW d. 24.77 dBW Answer: d

39. A laser diode outputs +8 dBm. Convert this value to Watts.

a. 0.0063W b. 0.063W c. 0.63W d. 0.00063W Answer: a

40. A square wave is made up of a summation of:

a. ramps b. sinusoids c. rectangle waves d. pulses Answer: b

52

TEST ITEM FILE - Chapter 1

41. The time series and FFT waveforms shown in Figure 1-4 show what a square wave can look like when passed through:

a. rectifier circuit b. highpass filter c. bandstop filter d. a bandwidth limited channel Answer: d

42. Figure 1-5 was obtained from a DSO. What are the frequencies of the third and fifth harmonics?

a. 12.5 and 25kHz b. 37.5 and 62.5 kHz c. 10 and 20 kHz d. 30 and 50 kHz Answer: b

Figure 1-4

Figure 1-5

53

TEST ITEM FILE – CHAPTER 2 AMPLITUDE MODULATION: TRANSMISSION

1. Which of the following is NOT produced when two sine waves are combined

through a nonlinear device?

a. Components of each of the two original frequencies b. Components at the sum and difference frequencies c. Harmonics of the two original frequencies d. ac level

Answer: d

2. Which of the following is considered a nonlinear device?

a. Resistor b. Capacitor c. Potentiometer d. Transistor

Answer: d

3. Which of the following modulation techniques is the most economical?

a. High-level b. Low-level c. Medium-level d. Ultra low-level

Answer: a

4. If the % modulation of an AM transmitter is 60% and the unmodulated antenna current

is 10 A, what is the modulated current?

a. 11 A b. 14 A c. 5 A d. 13 A

Answer: a

5. If the carrier transmits 12 kW, what is the modulated power from Problem 4?

a. 12 kW b. 10 kW c. 14 kW d. 16 kW Answer: c

54

TEST ITEM FILE – Chapter 2 6. The reason modulation is used in electronic communication is:

a. Since all intelligence signals occur at approximately the same frequency, there would be catastrophic interference problems if these frequencies were used. b. Audio frequency radio waves do not propagate long distances very reliably. c. Efficient transmission and reception of radio waves are not possible unless extremely large antennas are used. d. All of the above.

Answer: d

7. Which of the following is not created by nonlinear mixing?

a. The original two frequencies b. Harmonics of the sum and difference frequencies c. The sum and difference of the two original frequencies d. dc (0 Hz) Answer: b

8. A 2 kHz sinewave is mixed with a 1.5 MHz carrier sinewave through a nonlinear device. Which frequency is not present in the output signal?

a. 3 MHz b. 1.502 MHz c. 3.004 MHz d. 1.498 MHz

Answer: c

9. A 2.5 MHz carrier is modulated by a music signal that has frequency components ranging

from 100 Hz to 5 kHz. What is the range of frequencies generated for the upper sideband? a. 2.5 MHz to 2.505 MHz b. 2.495 MHz to 2.505 MHz c. 2.5001 MHz to 2.505 MHz d. 2.495 MHz to 2.499 MHz

Answer: c

10. Overmodulation: a. results when the modulation index exceeds unity. b. is undesirable because it produces sideband splatter. c. causes the AM signal to become distorted so that the receiver cannot produce a clean replica of the original intelligence signal. d. all of the above.

Answer: d

55

TEST ITEM FILE – Chapter 2

Figure 2-1

11. In Figure 2-1, A is 220 mVp-p and B is 350 mVp-p. The percent modulation of the AM signal is: a. 62.9% b. 1.59% c. 22.8% d. 4.38%

Answer: c

12. In Figure 2-1, Ei is 530 mV peak and Ec is 780 mV peak. The percent modulation of the

AM signal is: a. 67.9% b. 32.1% c. 19.1% d. 14.7%

Answer: a

13. In Figure 2-1, A is 1.5Vp-p and B is 3.3Vp-p. If the carrier signal is set at 1.2V peak, the

intelligence signal is: a. 3.2V peak. b. 0.55V peak. c. 0.9V peak. d. 0.45V peak. Answer: d

56

TEST ITEM FILE – Chapter 2

14. The total output power of an AM transmitter that is being operated at 50% modulation is

measured to be 1800 watts. What is the carrier power? a. 1440 watts b. 1600 watts c. 900 watts d. 2025 watts Answer: b 15. A 250W carrier is to be modulated at an 85% modulation level. What is the total

transmitted power? a. 340.3 watts b. 183.7 watts c. 430.6 watts d. 356.3 watts Answer: a

16. An AM broadcast station operates at its maximum allowed output power of 80W at a

percent modulation of 60%. What is the upper sideband power? a. 6.1 watts b. 18.47 watts c. 9.23 watts d. 12.2 watts Answer: a 17. The antenna current of an AM transmitter is 5A when it is not modulated. It increases

to 6A when it is modulated. Its modulation index expressed as a percentage is: a. 83.3% b. 63.2% c. 69.4% d. 93.8% Answer: d 18. An intelligence signal is amplified by a 65% efficient amplifier before being combined

with a 250W carrier to generate an AM signal. If it is desired to operate at 100% modulation, what must be the dc input power to the final intelligence signal amplifier?

a. 384.6W b. 192.3W c. 162.5W d. 83.3W Answer: b

57

TEST ITEM FILE – Chapter 2 19. High-level modulation is used: a. when the intelligence signal is added to the carrier at the last possible point before the transmitting antenna. b. in high-power applications such as standard radio broadcasting. c. when the transmitter must be made as power efficient as possible. d. all of the above. Answer: d 20. The process of neutralization is: a. placing a negative feedback capacitor in an RF amplifier to reduce the tendency for self-oscillation. b. a technique for filtering out all of the undesired frequencies produced by mixing action in a

nonlinear amplifier except for the carrier, sum, and difference frequencies. c. the process of adjusting the tank circuit so that the transmitter produces the proper output frequency. d. the process of adjusting the percent modulation to its desired level in a modulator stage. Answer: a

Figure 2-2 21. The transmitter scheme in Figure 2-2 is set up for: a. high-level modulation. b. low-level modulation. c. medium-level modulation. Answer: a 22. In Figure 2-2, the audio amplifier and modulator stages: a. are biased Class A or B for low distortion. b. are biased Class A or B for high power efficiency. c. are biased Class C for good mixing action. d. are biased Class C for high power efficiency. Answer: a

58

TEST ITEM FILE – Chapter 2 23. In Figure 2-2, which stages contain "tuned" amplifiers? a. stages b and c b. stages e and f c. stages c and f d. stages b and f Answer: a 24. In Figure 2-2, the AM waveform is created in: a. stage b b. stage c c. stage e d. stage f Answer: b

Figure 2-3

25. The transmitter scheme in Figure 2-3 is set up for: a. high-level modulation. b. low-level modulation. d. medium-level modulation. Answer: b 26. In Figure 2-3, which stages contain tuned amplifiers? a. stages b, c, and d b. stages b and f c. stages c and f d. stages e and f Answer: a

59

TEST ITEM FILE – Chapter 2

27. In Figure 2-3, the AM waveform is created in: a. stage b. b. stage c. c. stage d. d. stage f. Answer: a 28. In Figure 2-3, which stages use linear amplification? a. stages b, c, and d b. stages c, d, and f c. stages b and f d. stages b and c Answer: b 29. The O.T.A. is: a. a special type of op amp used to create an AM signal. b. an operational transconductance amplifier. c. a linear integrated circuit that creates AM with an absolute minimum of design

considerations. d. all of the above. Answer: d 30. The last stage of intelligence amplification before mixing with the carrier occurs in: a. the modulator. b. the modulated amplifier. c, the buffer. d. the RF linear amplifier. Answer: a

60

TEST ITEM FILE – Chapter 2

Figure 2-4a

31. In Figure 2-4a, the trapezoidal display indicates: a. improper bias or low carrier signal power. b. proper in-phase trapezoidal pattern for typical AM signal. c. poor linearity of the modulator. d. lack of an intelligence signal. Answer: b

Figure 2-4b

32. In Figure 2-4b, the trapezoidal display indicates: a. improper bias or low carrier signal power. b. proper in-phase trapezoidal pattern for typical AM signal. c. poor linearity of the modulator. d. lack of an intelligence signal. Answer: d

61

TEST ITEM FILE – Chapter 2

Figure 2-4c

33. In Figure 2-4c, the trapezoidal display indicates: a. improper bias or low carrier signal power. b. proper in-phase trapezoidal pattern for typical AM signal. c. poor linearity of the modulator. d. lack of an intelligence signal. Answer: c

Figure 2-4d

34. In Figure 2-4d, the trapezoidal display indicates: a. improper bias or low carrier signal power. b. proper in-phase trapezoidal pattern for typical AM signal. c. poor linearity of the modulator d. lack of an intelligence signal. Answer: a

62

TEST ITEM FILE – Chapter 2

35. In Figure 2-5, the carrier frequency is shown as being 50.003 MHz. The frequencies of the spurs on either side of the carrier are: Figure 2-5 a. approximately 24 kHz away from the carrier. b. 48.0034 MHz and 52.0034 MHz. c. approximately 15 dB below the carrier level. d. approximately 14 dB above the noise floor. e. none of the above. Answer: c 36. The equation defining the AM envelope is

a. e =(Ec +Ei sin ωit) sin ωit b. e = Ec sin ωct c. e = Ei sin ωit + sin ωct d. e = Ec sin ωct × e = Ei sin ωit e. none of the above Answer: a

37. The result of the trigonometric identity (sin x)( sin y) is

a. cos(x-y) – cos(x+y) b. –0.5 cos(x-y) + cos(x+y) c. –0.5 cos(x-y) + 0.5cos(x+y) d. 0.5 cos(x-y) – 0.5 cos(x+y) e. none of the above Answer: d

38. Determine the side frequency voltage if the modulation index is 70%and the carrier amplitude is 50V. a. 25 b. 17.5 c. 35 d. 50 e. none of the above Answer: b

39. Determine the frequency of the AM carrier shown in Figure 2-6. a. 1 MHz b. 10 kHz c. 10 MHz d. none of the above

Answer: d

Figure 2-6

63

TEST ITEM FILE – Chapter 2 40. The typical output impedance for an RF transmitter is

a. 75 Ω b. 50 Ω c. 8 Ω d. 16 Ω e. none of the above Answer: b

64

TEST ITEM FILE - CHAPTER 3

AMPLITUDE MODULATION: RECEPTION

1. Which of the following is not an advantage of a synchronous detection? a. Low distortion b. Eliminate diagonal clipping c. Greater ability to follow fast-modulated signals d. Ability to produce gain Answer: b 2. The mixer is often referred to as: a. RF amplifier. b. oscillator generator. c. second detector. d. first detector. Answer: d 3. Varactor diodes are used for tuning by: a. capacitance adjustment through a reverse bias diode. b. capacitance adjustment through forward bias. c. temperature compensation of diodes. d. all of the above. Answer: a 4. In a varactor diode, as voltage increases, capacitance: a. increases. b. stays the same. c. decreases. d. none of the above. Answer: c 5. The only roadblock to having complete receivers on a chip aside from station selection and volume controls is: a. limiting factors of tuned circuits. b. local oscillator. c. mixer circuits. d. IF amplifier. Answer: a

65

TEST ITEM FILE - Chapter 3

6. The radio receiver that simply consists of an RF amplifier, detector, and audio amplifier is known as: a. a superheterodyne receiver. b. a TRF receiver. c. a selective receiver. d. a sensitive receiver. Answer: b 7. A receiver’s sensitivity is: a. the extent to which a receiver is capable of differentiating between the desired signal and other signals. b. its ability to drive the output speaker to an acceptable level. c. the ability of the receiver to demodulate a modulated signal. d. the ability of a receiver to attenuate noise signals. Answer: b 8. A receiver’s selectivity is: a. the extent to which a receiver is capable of differentiating between the desired signal and other signals. b. its ability to drive the output speaker to an acceptable level. c. the ability of the receiver to demodulate a modulated signal. d. the ability of a receiver to attenuate noise signals. Answer: a 9. If a receiver is overly selective: a. too much noise is picked up and amplified by the receiver. b. only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results. c. the tank circuits within the tuned amplifiers have insufficient Q. d. when the volume control is turned up to maximum, the desired station is very weak. Answer: b 10. If a receiver is underselective: a. only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results. b. the tank circuits within the tuned amplifiers have too high a Q. c. when the volume control is turned up to maximum, the desired station is very weak. d. more than one radio station on different frequencies may be picked up by the receiver at the same time. Answer: d

66

TEST ITEM FILE - Chapter 3

11. A TRF receiver is to be designed with a single tuned circuit using an 8.2 uH inductor. If the frequency is to be tuned from 55 kHz to 1600 kHz, find the BW that results at 550 kHz if there is exactly 10 kHz BW at a frequency of 1050 kHz. a. 105 kHz b. 15.24 kHz c. 5.24 kHz d. 10 kHz Answer: c 12. The diode detector: a. is one of the simplest and most effective AM detectors. b. consists of a nonlinear diode and low-pass filter. c. is sometimes referred to as an envelope detector. d. all of the above. Answer: d 13. Which is not an advantage of diode detectors? a. Power absorbed from the tuned circuit by the diode detector reduces the Q of the tuned circuit. b. They develop a readily usable dc voltage for automatic gain control circuits. c. They are highly efficient. d. Distortion decreases as the amplitude of the AM signal increases. Answer: a 14. Diagonal clipping: a. occurs if the time constant of the low-pass filter is too large compared to the period of the RF waveform. b. is a type of distortion that occurs with diode detectors. c. is characterized by having the capacitor voltage not follow the full changes of the envelope of the AM waveform. d. all of the above. Answer: d 15. Synchronous detectors: a. are often called product detectors. b. offer low distortion compared to diode detectors. c. have the ability to provide gain. d. all of the above. Answer: d

67

TEST ITEM FILE - Chapter 3

16. The superheterodyne receiver design is superior to the TRF design: a. since it allows for a constant selectivity over the entire tuning range of the receiver. b. since it always uses synchronous detectors instead of diode detectors. c. since it uses many RF amplifier stages before the RF signal is mixed with the local oscillator signal. d. all of the above. Answer: a 17. An AM signal having a carrier frequency of 560 kHz is to be mixed with a local oscillator signal at a frequency of 1035 kHz. What does the output of the IF amplifier consist of? a. a 455 kHz carrier b. a 475 kHz sinewave c. a 475 kHz AM signal d. the original intelligence signal Answer: c 19. In Figure 3-1, the output signal of stage (e) is: a. an AM signal with a carrier frequency of 490 kHz. b. an AM signal with a carrier frequency of 1850 kHz. c. a 490 kHz sinewave. d. an 1850 kHz sinewave. Answer: d 20. In Figure 3-1, the output signal of stage (d) is: a. an AM signal with a carrier frequency of 490 kHz. b. an AM signal with a carrier frequency of 1360 kHz. c. a 490 kHz sinewave. d. a 1 kHz sinewave. Answer: d

68

TEST ITEM FILE - Chapter 3 21. In Figure 3-1, the output of stage (a) is: a. an AM signal with a carrier frequency of 1360 kHz. b. an AM signal with a carrier frequency of 1850 kHz. c. a 490 kHz sinewave. d. an 1850 kHz sinewave. Answer: a 22. In Figure 3-1, the output signal of stage (c) is: a. an AM signal with a carrier signal of 490 kHz. b. an AM signal with a carrier frequency of 1360 kHz. c. a 490 kHz sinewave. d. a 1 kHz sinewave. Answer: a 23. In Figure 3-1, the receiver design is known as: a. regenerative. b. superheterodyne. c. TRF. d. synchronous. Answer: b 24. In Figure 3-1, the stage sometimes referred to as the first detector is: a. stage a. b. stage b. c. stage c. d. stage d. Answer: b 25. In Figure 3-1, the stages that contain tuned circuits are: a. stages a, b and d. b. stages a, b and c. c. stages a, d and e. d. stages a, c and d. Answer: b

26. In Figure 3-1, the stages that must contain nonlinear devices are: a. stages a, b and c. b. stages a and e. c. stages b and d. d. stages b and c. Answer: c

69

TEST ITEM FILE - Chapter 3 27. A padder capacitor: a. is placed in series with the tank inductor to provide tracking at the low end of a large frequency band. b. is placed in parallel with each section of the ganged capacitors of the tank to provide tracking at the high end of a large frequency band. c. is placed in an RF amplifier to provide for proper neutralization. d. is placed in a tank circuit to provide for electronic tuning. Answer: a 28. A trimmer capacitor: a. is placed in series with the tank inductor to provide tracking at the low end of a large frequency band. b. is placed in parallel with each section of the ganged capacitor of the tank to provide tracking at the high end of a large frequency band. c. is placed in an RF amplifier to provide for proper neutralization. d. is placed in a tank circuit to provide for electronic tuning. Answer: b 29. A varicap: a. is placed in series with the tank inductor to provide tracking at the low end of a large frequency band. b. is placed in parallel with each section of the ganged capacitors of the tank to provide tracking at the high end of a large frequency band. c. is placed in an RF amplifier to provide for proper neutralization. d. is placed in a tank circuit to provide for electronic tuning. Answer: d 30. In Figure 3-1, the image frequency would be: a. 980 kHz. b. 2340 kHz. c. 1850 kHz. d. 870 kHz. Answer: b

31. Image frequency rejection on a standard AM broadcast band receiver is not a major problem since: a. the image frequency is not close to the IF frequency. b. the image frequency is not close to the LO frequency. c. the image frequency is not produced by mixing action. d. the image frequency is so far away from the RF amplifier stage’s tuned frequency. Answer: d

70

TEST ITEM FILE - Chapter 3

32. Which of the following is not a major benefit of using RF amplifier stages in superheterodyne receiver design? a. improved image frequency rejection b. larger frequency tuning range c. more gain resulting in improved sensitivity d. improved noise characteristics Answer: b 33. Which of the following is not an advantage of FETs over BJTs in RF amplifier usage? a. Their input impedance does not load down the Q of the circuit preceding the FET stage. b. The availability of dual gate FETs provides an isolated injection point for the AGC. c. Their input/output square-law relationship allows for lower distortion levels. d. They have improved image frequency rejection. Answer: d 34. An autodyne mixer is: a. a stage that provides the mixing and generates the LO at the same time. b. a mixer that uses a dual-gate FET. c. a mixer that automatically provides for AGC action. d. a stage that mixes the LO with the AM signal without the use of a transistor. Answer: a 35. In a superheterodyne receiver the bulk of the receiver’s sensitivity and selectivity is due to the: a. RF amplifier stages. b. converter stages. c. IF amplifier stages. d. local oscillator. Answer: c 36. Double conversion is: a. a receiver design that uses two superheterodyne receivers to receive a weak signal. b. a technique used to reduce image frequency problems in a superheterodyne receiver. c. a technique used to solve the TRF tuning problems. d. a method that ensures that a superheterodyne receiver does not break into oscillations due to stray positive feedback. Answer: b

71

TEST ITEM FILE - Chapter 3

37. The circuit of Figure 3-2 is an example of: a. an RF mixer, local oscillator, and IF filter. b. an autodyne mixer. c. a receive converter. d. all of the above. Answer: d 38. In Figure 3-2, the tank circuit made up of L1 and C1 is tuned at: a. the IF frequency. b. the LO frequency. c. the RF carrier frequency. d. the image frequency. Answer: c 39. In Figure 3-2, the tank circuit made up of L4 and C4 is tuned at: a. the IF frequency. b. the LO frequency. c. the RF carrier frequency. d. the image frequency. Answer: b 40. In Figure 3-2, the tank circuit made up of L5 and C5 is tuned at: a. the IF frequency. b. the LO frequency. c. the RF carrier frequency. d. the image frequency. Answer: a

72

TEST ITEM FILE - Chapter 3

41. In Figure 3-2, the purpose of C3 is: a. to determine the frequency of oscillation of the LO. b. to couple the local oscillator frequency from the tank circuit to be amplified by Q1. c. to act as a bypass capacitor for R3. d. to neutralize the RF amplifier stage.

Answer: b

42. The AGC control voltage: a. is actually the dc voltage component produced by the mixing action in the AM demodulator stage. b. varies as the signal strength of the received signal varies. c. is a negative feedback voltage. d. is produced by an RC circuit having a much larger time constant than that of the detector. e. all of the above. Answer: e Q1 T1 Q2 T2 Q3 T3 Q4 L1

L4

43. In Figure 3-3, the tank circuit made up of L1, A, and B is tuned to: a. the LO frequency. b. the RF carrier frequency. c. the IF frequency. d. the image frequency. Answer: b

73

TEST ITEM FILE - Chapter 3

44. In Figure 3-3, the tank circuit made up of L4, C, and D is tuned to: a. the LO frequency. b. the RF carrier frequency. c the IF frequency. d. the image frequency. Answer: a 45. In Figure 3-3, the tank circuit inside of T1 is tuned to: a. the LO frequency. b. the RF carrier frequency. c the IF frequency. d. the image frequency. Answer: c 46. In Figure 3-3, the transistor Q1 is used as: a. the nonlinear device in an RF mixer stage. b. the active part of an RF amplifier. c. the active part of an LO stage. d. all of the above. Answer: d 47. In Figure 3-3, the transistor Q2 is used as: a. an RF mixer stage transistor. b. an IF amplifier stage transistor. c. a detector transistor. d. an audio amplifier stage transistor. Answer: b 48. In Figure 3-3, the transistor Q3 is used as: a. an RF mixer stage transistor. b. an IF amplifier stage transistor. c. a detector transistor. d. an audio amplifier stage transistor. Answer: b 49. In Figure 3-3, the transistor Q4 is used as: a. an RF mixer stage transistor. b. an IF amplifier stage transistor. c. a detector transistor. d. an audio amplifier stage transistor. Answer: d

74

TEST ITEM FILE - Chapter 3

50. In Figure 3-3, the AM demodulation is accomplished by: a. transistor - Q3. b. diode - E1. c. diode - E2. d. transistor - Q4. Answer: c

51. In Figure 3-3, the filter that produces the AGC voltage consists of: a. R11 and C11. b. R5 and C4. c. R11 and C12. d. R10 and C10. Answer: b 52. In Figure 3-3, the inductors L1 and L2 function as: a. an IF transformer. b. a loopstick antenna. c. part of the local oscillator. d. a nonlinear mixer. Answer: b 53. In Figure 3-3, the transformer, T3, is tuned to: a. the intelligence frequencies. b. the RF carrier frequency of the received station. c. the IF frequency. d. the local oscillator frequency. Answer: c 54. In Figure 3-3, the selectivity is accomplished by: a. T1, T2, and T3. b. L1 and L2. c. L3 and L4. d. R11 and C11. Answer: a 55. In Figure 3-3, E1 functions as: a. an auxillary AGC diode. b. a mixer diode. c. an AM detector diode. d. an IF amplifier diode. Answer: a

75

TEST ITEM FILE - Chapter 3

56. In Figure 3-3, the volume is controlled by adjusting: a. T3. b. R12. c. capacitors B and D. d. R17. Answer: b

57. The reference level for the unit, dBm, is: a. the milliwatt. b. the milliampere. c. the watt. d. the millivolt. Answer: a

58. In Figure 3-4, the power driven into the audio amplifier stage is: a. 2.51W. b. 2.51 mW. c. 0.398 mW. d. 398 mW. Answer: c 59. The conversion gain of the mixer in Figure 3-4 is: a. -81 dBm. b. 3 dBm. c. -3 dBm. d. -78 dBm. Answer: b

76

TEST ITEM FILE - Chapter 3

60. The total gain of the entire receiver in Figure 3-4 from the antenna input to the audio amplifier output is: a. 89 dB. b. 59 dB. c. 119 dB. d. 30 dBm. Answer: c 61. The power gain of the audio amplifier in Figure 3-4 expressed as a ratio quantity is approximately: a. 1000. b. 2512. c. 34 dB. d. 30 dBm. Answer: b 62. A receiver has a dynamic range of 65 dB. It has a sensitivity of 0.88 nW. The maximum allowable input signal is approximately: a. 1.56 uW. b. 278 mW. c. 2.78 mW. d. 156 uW. Answer: c 63. A receiver has a maximum input signal of 75 mW before distortion occurs. Its sensitivity is measured to be 1.5 nW. Its dynamic range is approximately: a. 47 dB. b. 77 dB. c. 154 dB. d. 87 dB. Answer: b 64. Good troubleshooting practice says:

a. perform a visual check and check the power supply voltages. b. prepare a trouble report c. log the serial and model number d. check the power supply voltages e. none of the above Answer: a

77

TEST ITEM FILE - Chapter 3

65. Electronics Workbench Multisim provides a feature that allows for the addition of a component fault in a circuit. This is accomplished by: a. replacing the part with an F-prefix part b. replacing the part with a non model part c. double-clicking on the component, select fault and specify the type of failure d. all of the above e. none of the above Answer: c

78

TEST ITEM FILE - CHAPTER 4 SINGLE-SIDEBAND COMMUNICATIONS

1. SSB transmitters are usually rated in terms of: a. power gain. b. voltage gain. c. modulation index. d. peak envelope power. Answer: d 2. Which of the following is required for SSB transmission? a. carrier elimination. b. elimination of one sideband. c. a and b d. none of the above Answer: c 3. Which of the following is NOT an advantage that the phase method of generating has

over the filter method? a. requires high Q filters. b. lower intelligence frequencies can be used. c. greater ease in switching from one sideband to the other. d. SSB generated directly at desired transmitting frequency. Answer: a 4. The major disadvantage of the phase method of SSB generation is: a. requires two balance modulators b. does not use high Q filters c. 90° phase-shifting network for intelligence signal d. all of the above Answer: c 5. Which of the following can be used as an SSB demodulator? a. RF amplifier b. audio amplifier c. AGC circuitry d. mixer Answer: d

79

TEST ITEM FILE – Chapter 4

6. At 100% modulation, an AM signal has a total power of 1200 watts. If it was converted to SSB and the sideband component had the same amplitude as before, the total power would be:

a. 100 watts. b. 200 watts. c. 400 watts. d 800 watts. Answer: b 7. The PEP rating of an SSB transmitter is often misleading due to: a. the fact that with normal voice signals, an SSB transmitter develops an average of

only one-third to one-fourth of its PEP rating. b. the SSB transmitter's power supplies and components are not always designed for

continuous duty PEP power levels. c. the PEP rating is the same rating no matter what type of waveform the transmitter is

providing. d. all of the above. Answer: d 8. If the carrier and one of the sidebands are completely eliminated, the resulting signal is

referred to as: a. SSB. b. SSBSC. c. ACSSB. d. vestigial sideband. Answer: a 9. If the carrier is reduced in amplitude and one of the sidebands is completely eliminated, the

resulting signal is referred to as: a. independent sideband. b. SSBSC c. ACSSB. d. vestigial sideband. Answer: b 10. The type of sideband that involves having the upper sideband contain different information

than the lower sideband is referred to as: a. independent sideband. b. SSBSC c. ACSSB. d. vestigial sideband. Answer: a

80

TEST ITEM FILE – Chapter 4

11. The type of sideband that is used for television video transmission is called: a. independent sideband. b. SSBSC c. ACSSB. d. vestigial sideband. Answer: d 12. A more recently developed system of sideband that involves compressing the amplitude of

the intelligence signal before modulation occurs is called: a. independent sideband. b. SSBSC c. ACSSB. d. vestigial sideband. Answer: c 13. The most important advantage of SSB over AM is: a. that it is a more effective utilization of the available frequency spectrum. b. that it is less subject to the effects of selective fading. c. the resulting savings in power. d. simplicity in design. Answer: a 14. Which is not an advantage of SSB over AM? a. SSB provides a more effective utilization of the available frequency spectrum. b. SSB is less subject to the effects of selective fading. c. SSB has a resulting savings in power. d. simplicity in design. Answer: d 15. The output of a balanced modulator in an SSB transmitter being modulated with a voice

intelligence signal is: a. conventional AM. b. SSB. c. DSB-SC. d. a sinewave at the carrier frequency. Answer: c

16. Which of the following will not work as a balanced modulator? a. a dual-gate FET and a band-pass filter b. a push-pull modulator c. a ring-diode modulator

d. an LM1496 integrated circuit stage

Answer: a

81

TEST ITEM FILE – Chapter 4

17. In a push-pull modulator, the carrier suppression is accomplished by: a. a dual-gate FET having symmetry. b. center-tapped transformers causing canceling magnetic fields. c. the nonlinearity of the diodes that are used. d. symmetrical differential amplifier stages. Answer: b 18. In a linear integrated circuit balanced modulator such as the LM1496, carrier suppression

is accomplished by: a. center-tapped transformers causing canceling magnetic fields. b. a dual-gate FET having symmetry. c. the nonlinearity of the diodes that are used. d. symmetrical differential amplifier stages. Answer: d 19. Voice transmission requires an audio frequency range of: a. 20 Hz to 20 kHz. b. 30 Hz to 10 kHz. c. 100 Hz to 3 kHz. d 500 Hz to 1 kHz. Answer: c 20. In an SSB transmitter having voice modulation using audio frequencies ranging from

100 Hz to 3 kHz, the upper and lower sidebands generated by the balanced modulator are separated by:

a. 500 Hz. b. 200 Hz. c. 6 kHz. d. 40 Hz. Answer: b

Figure 4-1

82

TEST ITEM FILE – Chapter 4 21. Calculate the required Q for the bandpass filter depicted in Figure 4-1 for 80 dB sideband suppression

if the carrier frequency component of the IF is 455 kHz. a. 1137.5 b. 56,875 c. 227,500 d. 113,750 Answer: b 22. Crystal filters: a. have Q values as high as 50,000. b. employ phasing capacitors to produce a rejection notch of undesired resonant frequencies. c. are quite delicate in construction. d. all of the above. Answer: c 23. Which is not an advantage of ceramic filters over crystal filters? a. more rugged in construction. b. smaller in size. c. higher Q values. d. less expensive. Answer: c 24. The ratio of the 60 dB and 6 dB bandwidths for a ceramic filter is known as: a. the shape factor. b. the ripple amplitude. c. the quality factor. d. the roll-off rate of the skirt. Answer: a 25. The variation in attenuation within the bandpass frequencies of a ceramic filter is called: a. the shape factor. b. the ripple amplitude. c. the quality factor. d. the roll-off rate of the skirt. Answer: b 26. The first designed sharp filter used in SSB transmitters to produce SSB is the: a. crystal filter. b. ceramic filter. c. mechanical filter. d. tank circuit. Answer: c

83

TEST ITEM FILE – Chapter 4

27. The filter consisting of resonant disks, driving rods, bias magnets and transducer coils is the: a. crystal filter. b. ceramic filter. c. mechanical filter. d. tank circuit. Answer: c

Figure 4-2

28. In Figure 4-2, the filter Q required in the linear amplifier is approximately: a. 20. b. 10. c. 15. d. 100. Answer: b 29. In Figure 4-2, the output signal of stage (b) is: a. DSB-SC at 500 kHz. b. DSB-SC at 10 MHz. c. SSB at 500 kHz. d. SSB at 10 MHz. Answer: a 30. In Figure 4-2, the ouput signal of stage (d) is: a. DSB-SC at 500 kHz. b. DSB-SC at 10 MHz. c. SSB at 500 kHz. d. SSB at 10 MHz. Answer: b

84

TEST ITEM FILE – Chapter 4

31. In Figure 4-2, the first mixer and filter are necessary to: a. produce SSB at the final transmitter output frequency. b. produce SSB using filters having adequately high Q values. c. produce DSB-SC at the final transmitter output frequency. d. produce DSB-SC using filters having adequately high Q values. Answer: b 32. In Figure 4-2, the second mixer and filter are necessary to: a. produce SSB at the final transmitter output frequency. b. produce SSB using filters having adequately high Q values. c. produce DSB-SC at the transmitter output frequency. d. produce DSB-SC using filters having adequately high Q values. Answer: a

Figure 4-3

85

TEST ITEM FILE – Chapter 4

33. In Figure 4-3, the purpose of Q1 and Q2 are to: a. mix the carrier with the intelligence signal to produce DSB-SC. b. amplify the carrier signal. c. filter out one of the sidebands. d. amplify the audio intelligence signals. Answer: a 34. In Figure 4-3, the required push-pull action of Q1 and Q2 is furnished by: a. the balance potentiometer, R2. b. the transformer, T1. c. the fact that one of them is a common gate amplifier and the other is a common source amplifier. d. the 0.1 µf capacitor that connects their sources together. Answer: b 35. In Figure 4-3, Q4 is needed: a. to provide gain to the carrier output signal. b. to act as an oscillator to produce a carrier signal. c. to upset the balance of the balanced modulator in order to produce a carrier output signal. d. to help filter out the undesired sideband. Answer: c 36. In Figure 4-3, the 9 MHz crystal filter can produce either USB or LSB due to: a. its bandwidth being wide enough to filter out both sidebands. b. the 30 pf capacitors across its input and output terminals. c. the IF amplifier transistor, Q3, having sufficiently high bandwith. d. the carrier frequency being slightly altered to create either sideband at 9 MHz. Answer: d

37. Which is not a general method used to generate SSB in transmitter design? a. filter method b. lattice method c. amplitude-compandoring method d. phase method Answer: b 38. Which is the most popular method used to generate SSB in transmitter design? a. filter method b. lattice method c. amplitude-compandoring method d. phase method Answer: a

86

TEST ITEM FILE – Chapter 4 39. The main disadvantage of the phase method in generating SSB is: a. the complex method of mixing the carrier with the intelligence signal. b the complex method used to amplify the resulting SSB signal. c. the complex design of the 90 degree phase shifting network for the intelligence signal. d. the complex design of the 90 degree phase shifting network for the carrier signal. Answer: c 40. ACSSB is: a. a method of compressing the audio before modulation and to expand it following

demodulation. b. an acronym for amplitude compandored single-sideband. c. a new method of allowing narrow band voice communications with the performance

of FM systems for the land-mobile communication industry. d. all of the above. Answer: d 41. The signal processor that increases all negative dBm power levels and at the same time

decreases all positive dBm power levels is: a. a signal expander. b. a linear amplifier. c. a signal compressor. d. a phase-locked loop. Answer: c

42. The signal processor that decreases all negative dBm power levels and at the same time

increases all positive dBm power levels is: a. a signal expander. b. a linear amplifier. c. a signal compressor. d. a phase-locked loop. Answer: a 43. Why is the carrier signal needed in an SSB receiver? a. It is needed for the receiver to maintain its sensitivity and selectivity. b. It is needed for the demodulator to be able to recreate the intelligence. c. The carrier actually contains the intelligence signal. d. Without a carrier, the signal strength of the received station is zero. Answer: b 44. The standard SSB demodulator consists of: a. an RF amplifier, nonlinear mixer, and IF amplifier stage. b. a nonlinear diode and a low-pass filter. c. a sharp bandpass filter and an audio amplifier stage. d. a beat frequency oscillator, nonlinear mixer, and low-pass filter. Answer: d

87

TEST ITEM FILE – Chapter 4 45. In an SSB receiver, if the BFO drifts slightly off of the carrier frequency: a. the output intelligence is badly distorted. b. the output intelligence amplitude is reduced. c. the noise level increases. d. harmonics are produced by mixing action of the nonlinear device. Answer: a

Figure 4-4

46. In Figure 4-4, the purpose of stage (b) is: a. to demodulate the 20 MHz USB signal. b. to create a 1 MHz SSB IF signal. c. to recreate the 100 Hz to 3 kHz intelligence signal. d. to act as a product detector. Answer: b 47. In Figure 4-4, what stages have tracking capacitors? a. stages d and f b. stages d and e c. stages b and e d. stages a and c Answer: d 48 In Figure 4-4, what frequency must the BFO be set at for optimum results? a. 455 kHz b. 20 MHz c. 1 MHz d. 21 MHz Answer: c 49. In Figure 4-4, most of the selectivity and sensitivity are furnished by: a. stage a. b. stage b. c. stage e. d. stage d. Answer: d

88

TEST ITEM FILE – Chapter 4 50. In Figure 4-4, the SSB demodulation is accomplished by: a. stages b and c. b. stages d and f. c. stages d and e. d. stages e and f. Answer: d

51. In which stages of Figure 4-4 should an AGC control signal be applied? a. stages a and d b. stages b and e c. stages c and f d. stages e and g Answer: a

52. In Figure 4-5, transistor Q1 functions: a. to recreate the original intelligence signal from the original SSB signal. b. as a product detector. c. to mix the VFO output signal with the received RF signal in order to produce a 453.55 kHz IF signal. d. as an RF amplifier stage of the received SSB signal. Answer: c

Figure 4-5

89

TEST ITEM FILE – Chapter 4 53. In Figure 4-5, transistor Q2 acts as: a. an IF amplifier stage. b. a beat frequency oscillator. c. a variable frequency oscillator. d. a mixer stage. Answer: a 54. In Figure 4-5, transistor Q3 acts as: a. an IF amplifier stage. b. a beat frequency oscillator. c. a variable frequency oscillator. d. a mixer stage. Answer: c 55. In Figure 4-5, transistor Q4 functions as: a. a product detector. b. an RF mixer to create the IF frequency. c. a variable frequency oscillator. d. a beat frequency oscillator. Answer: a 56. In Figure 4-5, transistor Q5 functions as: a. a product detector. b. an RF mixer to create the IF frequency. c. a variable frequency oscillator. d. a beat frequency oscillator. Answer: d 57. In Figure 4-5, the purpose of switches S1A and S1B is to: a. ensure that the USB or LSB mixer output signal remains in the middle of the IF bandwidth filter. b. alter the BFO frequency slightly by selecting separate crystals. c. alter the VFO frequency slightly by adding bias to the switching diode, D2. d. all of the above. Answer: d

90

TEST ITEM FILE – Chapter 4

58. The image shown in figure 4-6 is an example of: a. double sideband output spectrum b. DSB-SC c. Upper sideband with a suppressed lower sideband d. Lower sideband with a suppressed upper sideband e. SSB-FC

Answer: c

Figure 4-6

91

TEST ITEM FILE – CHAPTER 5 FREQUENCY MODULATION: TRANSMISSION

1. The modulation index is: a. top envelope divided by center frequency. b. deviation divided by intelligence frequency. c. VCO voltage divided by center frequency. d. all of the above. Answer: b 2. Noise is usually clipped by: a. amplifiers. b. phase detectors. c. limiter circuits d. ARC circuits. Answer: c 3. If the S/N of the input signal is 4 and the intelligence signal is 10 kHz, determine the deviation. a. 145 kHz b. 160 kHz c. 200 kHz d. 100 kHz Answer: a 4. The standard time constant used for pre-emphasis in a non-Dolby system is: a. 75 µs. b. 25 µs. c. 50 µs . d. 175 µs. Answer: a 5. Which of the following is indirect FM generation? a. varactor diode b. reactance modulator c. Crosby modulator d. Armstrong modulator Answer: d 6. The two types of angle modulation are: a. amplitude and frequency modulation. b. phase and frequency modulation. c. pulse and frequency modulation. d. phase and amplitude modulation. Answer: b

92

TEST ITEM FILE – Chapter 5 7. Phase modulation is an indirect form of: a. amplitude modulation. b. pulse modulation. c. frequency modulation. d. angle modulation. Answer: c 8. The first working FM system was developed by: a. Marconi. b. Armstrong. c. Sarnoff. d. Hartley. Answer: b 9. In an FM modulator, the amplitude of the intelligence signal determines the: a. rate of the carrier frequency deviation. b. power level of the FM signal. c. phase angle of the carrier frequency component. d. amount of carrier frequency deviation. Answer: d 10. In an FM modulator, the intelligence frequency determines the: a. rate of carrier frequency deviation. b. power level of the FM signal. c. phase angle of the carrier frequency component. d. amount of carrier frequency deviation. Answer: a 11. An FM signal has a center frequency of 154.5 MHz, but is swinging between 154.45 MHz

and 154.55 MHz at a rate of 500 times per second. Its input intelligence frequency is: a. 100 kHz. b. 50 kHz. c. 154.5 MHz. d. 500 Hz. Answer: d 12. An FM signal has a center frequency of 154.5 MHz but is swinging between 154.45 MHz

and 154.55 MHz at a frequency of 500 times per second. Its input carrier frequency is: a. 100 kHz. b. 50 kHz. c. 154.5 MHz. d. 500 Hz. Answer: c

93

TEST ITEM FILE – Chapter 5

13. An FM signal has a center frequency of 154.5 MHz but is swinging between 154.45

MHz and 154.55 MHz at a rate of 500 times per second. Its index of modulation is: a. 50,000. b. 100. c. 500 d. 100,000. Answer: b 14. The amount of frequency deviation is proportional to the amplitude of the intelligence

signal in: a. an FM signal. b. a PM signal. c. both FM and PM signals. d. neither FM nor PM signals. Answer: c 15. To solve for the frequency components of an FM signal, what high-level mathematical

tool is used? a. Laplace transforms b. Calculus c. Bessel functions d. Fourier transforms Answer: c 16. An FM signal has an intelligence frequency of 5 kHz and a maximum deviation of 30 kHz.

Its index of modulation is: a. 60. b. 35. c. 150. d. 6. Answer: d

94

TEST ITEM FILE – Chapter 5

Figure 5-1

17. An FM signal has an intelligence frequency of 5 kHz and a maximum deviation of 30 kHz.

Its bandwidth, using the Bessel chart of Figure 5-1, is: a. 6 kHz. b. 60 kHz. c. 90 kHz. d. 45 kHz. Answer: c 18. An FM signal has an intelligence frequency of 5 kHz and a maximum deviation of 30 kHz. How

many frequency components are there in the output spectra? (Use the Bessel Chart of Figure 5-1.) a. 9 b. 18

c. 19 d. 6 Answer: c 19. An FM signal has an intelligence frequency of 2 kHz and a maximum deviation of 10 kHz.

If its carrier frequency is set at 162.4 MHz, what is the frequency of its highest frequency component within its bandwidth? (Use the Bessel Chart of Figure 5-1.)

a. 162.401 MHz b. 162.400 MHz c. 162.408 MHz d. 162.404 MHz e. 162.416 MHz Answer: c

95

TEST ITEM FILE – Chapter 5 20. An FM signal has an intelligence frequency of 2 kHz and a maximum deviation of 10 kHz.

If its carrier frequency is set at 162.4 MHz, what is its bandwidth? (Use the Bessel Chart of Figure 5-1.)

a. 32 kHz. b. 20 kHz. c. 16 kHz. d. 10 kHz. Answer: a 21. Carson's rule is: a. a quick approximation method to find the bandwidth of an FM signal.

b. a quick approximation method to find the index of modulation of an FM signal. c. a quick approximation method to find the bandwidth of a PM signal. d. a quick approximation method to find the index of modulation of a PM signal. Answer: a

Figure 5-2

22. The FM signal given in Figure 5-2 is applied to a 50 ohm load. Its carrier frequency is: a. 15 kHz. b. 300 MHz. c. 150 kHz. d. 150 MHz. Answer: d 23. The FM signal given in Figure 5-2 is applied to a 50 ohm load. Its output power level is

approximately: a. 450 watts. b. 2025 watts. c. 2863 watts. d. 4050 watts. Answer: b

96

TEST ITEM FILE – Chapter 5 24. The FM signal given in Figure 5-2 is applied to a 50 ohm load. Its index of modulation is: a. 4 b. 450 c. 9.42. d. 112.5. Answer: a 25. The FM signal given in Figure 5-2 is applied to a 50 ohm load. It has an intelligence frequency of: a. 30 kHz. b. 150 MHz. c. 15 kHz. d. 10 kHz. Answer: c 26. The FM signal given in Figure 5-2 is applied to a 50 ohm load. It has a frequency deviation of: a. 60 kHz. b. 15 kHz. c. 30 kHz. d. 3.75 kHz. Answer: a 27. The FM signal given in Figure 5-2 is applied to a 50 ohm load. Using Carson's

rule, it has a bandwidth of: a. 120 kHz. b. 150 kHz. c. 60 kHz. d. 75 kHz. Answer: b 28. Viewing an FM signal on a spectrum analyzer, if the carrier frequency component is zero and

there are four or five sidebands on either side of the carrier frequency, the index of modulation is: a. 0. b. 2.2. c. 5.5 d. 8.65. Answer: b 29. Standard FM broadcast stations use a maximum intelligence frequency of: a. 5 kHz. b. 15 kHz. c. 75 kHz. d. 150 kHz. Answer: b

97

TEST ITEM FILE – Chapter 5 30. Standard FM broadcast stations use a maximum frequency deviation of: a. 5 kHz. b. 200 kHz. c. 75 kHz. d. 150 kHz. Answer: c 31. Narrowband FM stations: a. use a maximum deviation of 10 kHz. b. use intelligence frequencies ranging from 100 Hz to 3 kHz. c. are found in police, aircraft, taxicabs, weather service, and industrial applications. d. all of the above. Answer: d 32. The modulation index that occurs when the deviation and intelligence frequencies are

maximum is called: a. the maximum bandwidth. b. the maximum modulation index. c. the deviation ratio. d. the maximum side frequency component. Answer: c 33. The most important advantage of FM over AM is: a. its limited bandwidth. b. its conservation of energy. c. its superior noise characteristics. d. its frequency stability. Answer: c 34. The noise characteristics of an FM communication system are mainly due to: a. its modulator stage. b. its narrow bandwidth characteristics. c. its low level of modulation index. d. its limiter and detector stages. Answer: d 35. The worst case signal-to-noise ratio at the output of an FM detector occurs when: a. the desired signal is 90 degrees out of phase with the noise signal. b. the desired signal is 90 degrees out of phase with the resultant signal of

adding the signal to the noise. c. the noise signal is 90 degrees out of phase with the resultant signal of adding the signal to

the noise. d. the desired signal is 90 degrees out of phase with the intelligence signal. Answer: c

98

TEST ITEM FILE – Chapter 5 36. Determine the worst case output signal-to-noise ratio for a broadcast FM receiver with a

maximum deviation of 75 kHz and a maximum intelligence frequency of 15 kHz if the input signal-to-noise ratio is 2:1.

a. 5:1 b. 10:1 c. 14.7:1 d. 3:1 Answer: b 37. Determine the worst case output signal-to-noise ratio for a broadcast FM receiver with a

maximum deviation of 75 kHz and a maximum intelligence frequency of 10 kHz if the input signal-to-noise ratio is 2:1.

a. 10:1 b. 15:1 c. 30:1 d. 2:1 Answer: b 38. Determine the worst case output signal-to-noise ratio for a broadcast FM receiver with a

maximum deviation of 75 kHz and a maximum intelligence frequency of 15 kHz if the input signal-to-noise ratio is 3:1.

a. 5:1 b. 10:1 c. 14.7:1 d. 3:1 Answer: c 39. Pre-emphasis is:

a. increasing the relative strength of low-frequency components before being fed into the modulator of an FM transmitter.

b. decreasing the relative strength of low-frequency components of the output signal of an FM detector in an FM receiver.

c. decreasing the relative strength of high-frequency components at the output signal of an FM detector in an FM receiver.

d. increasing the relative strength of high-frequency components before being fed into the modulator of an FM transmitter.

Answer: d

99

TEST ITEM FILE – Chapter 5 40. De-emphasis is: a. increasing the relative strength of low-frequency components before being fed into the

modulator of an FM transmitter. b. decreasing the relative strength of low-frequency components of the output signal of an FM

detector of an FM receiver. c. decreasing the relative strength of high-frequency components of the output signal of an FM

detector in an FM receiver. d. increasing the relative strength of high frequency components before being fed into the

modulator of an FM transmitter.

Answer: c 41. A de-emphasis network has: a. an upper cutoff frequency of 2.120 kHz. b. a time constant of 75 microseconds. c. a high-frequency roll-off rate of –20 db per decade. d. all of the above. Answer: d 42. The main purpose of pre-emphasis and de-emphasis networks in FM communication systems is to: a. provide a near constant noise reduction capability between low and high frequency intelligence signals. b. allow for a reduction in bandwidth of the FM communication channel. c. allow for stereo broadcasts to be received by a monaural receiver. d. filter out noise produced by the FM transmitter's modulator stage. Answer: a 43. Which of the following is not an example of a direct FM modulator? a. varactor diode modulator. b reactance modulator. c. 566 VCO modulator. d. Armstrong modulator. Answer: d 44. Which best describes how a varactor diode modulator creates FM? a. The intelligence signal creates mixing action in the nonlinear varactor diode to

create an FM signal. b. The intelligence signal alters the amount of forward bias of the varactor diode to create an FM signal. c. The intelligence signal alters the capacitance of the diode to shift the resonant frequency of a tank circuit.

d. The intelligence signal causes the diode to create phase shift which indirectly creates FM. Answer: c

100

TEST ITEM FILE – Chapter 5 45. In this modulator, FM is created by having the intelligence signal cause a change in the

transconductance of a JFET, which causes a change in the JFET amplifier's input capacitance, which shifts the resonant frequency of a tank circuit.

a. varactor diode modulator b. reactance modulator c. VCO modulator d. PLL modulator Answer: b 46. The main disadvantage of direct FM modulators is: a. they have very limited frequency stability. b. they have insufficient frequency deviation. c. they can only work at low radio frequencies. d. they work reliably only with low-level intelligence signals. Answer: a 47. FM systems utilizing a direct FM modulator with an automatic frequency control (AFC)

are known as: a. Hartley systems. b. Armstrong systems. c. Crosby systems. d. PLL systems. Answer: c 48. Frequency multipliers: a. are used to multiply the frequency of the carrier signal of an FM signal. b. are typical of a Class C amplifier followed by a tank circuit which filters out a single

harmonic. c. are used to multiply the frequency deviation of an FM signal. d. all of the above. Answer: d 49. A stage that produces a specific dc output voltage level based on the exact frequency of the

input signal is known as: a. a discriminator. b. a direct FM modulator. c. an indirect FM modulator. d. a frequency multiplier. Answer: a

101

TEST ITEM FILE – Chapter 5 50. FM systems that utilize an indirect FM modulator consisting of a phase modulator stage

are known as: a. Hartley systems. b. Armstrong systems. c. Crosby systems. d. PLL systems. Answer: b 51. The advantage of indirect FM modulation over direct FM modulation is: a. in their improved frequency stability. b. in their increased frequency deviation. c. in their ability to work with small signal amplitudes. d. in their ability to work at higher intelligence frequencies. Answer: a 52. The disadvantage of indirect FM modulators is: a. they have very limited frequency stability. b. they have insufficient frequency deviation. c. they can only work at low intelligence frequencies. d. they only work with large signal amplitudes. Answer: b 53. A phase modulator can be made to create FM indirectly by: a. integrating the intelligence signal prior to phase modulation. b. feeding the intelligence signal through a frequency correcting network. c. ensuring that the audio signal has an amplitude that varies inversely proportional

to its frequency. d. all of the above. Answer: d 54. An indirect FM wideband transmitter can have its output signal be at the proper carrier

frequency and at the same time have the required deviation by: a. using a balanced modulator and a 90 degree phase shifter. b. using combinations of mixers and multiplier stages. c. using a reactance modulator. d. using a frequency correcting network. Answer: b

102

TEST ITEM FILE – Chapter 5

55. In Figure 5-3, the diode CR3 functions as: a. the varactor diode used in the FM modulator stage. b. the varactor diode used in the PM modulator stage. c. the varactor diode used in the VCO. d. the varactor diode used in the frequency synthesizer. Answer: a 56. In Figure 5-3, the diode CR2 functions as: a. the varactor diode used in the FM modulator stage. b. the varactor diode used in the PM modulator stage. c. the varactor diode used in the VCO. d. the varactor diode used in the frequency synthesizer. Answer: c

103

TEST ITEM FILE – Chapter 5 57. In Figure 5-3, the 4044 integrated circuit functions as: a. the VCO portion of the phase-locked loop used as the frequency synthesizer. b. a high speed digital divider. c. the phase detector portion of the phase-locked loop used as a frequency synthesizer. d. an audio amplifier stage for the intelligence signal. Answer: c 58. In Figure 5-3, the 82S90 integrated circuit functions as: a. the VCO portion of the phase-locked loop used as a frequency synthesizer. b. a high speed digital divider. c. the phase detector portion of the phase-locked loop used as a frequency synthesizer. d. an audio amplifier stage for the intelligence signal.

Answer: b 59. In Figure 5-3, the CA3130 integrated circuit functions as: a. the VCO portion of the phase-locked loop used as a frequency synthesizer. b. a high speed digital divider. c. the phase detector portion of the phase-locked loop used as a frequency synthesizer. d. an audio amplifier stage for the intelligence signal. Answer: d 60. Multiplex transmission is: a. using two carriers to transmit two separate channels of information. b. using two transmitters to transmit two separate channels of information. c. the technique of separating the L and R signals into L+R and L-R in a stereo

FM transmitter. d. the simultaneous transmission of two or more signals on one carrier. Answer: d 61. In FM stereo broadcasts, why are the left and right audio signals first converted to

L+R and L-R signals before being frequency modulated? a. This is necessary to be able to be multiplexed properly. b. This permits higher intelligence frequencies to be used. c. This permits compatibility between monaural and stereo systems. d. This produces higher signal-to-noise ratios. Answer: c 62. In an FM stereo transmitter, the L+R audio signal applied to the FM modulator stage: a. extends from 0 to 15 kHz. b. extends only from 23 kHz to 38 kHz..

c. extends only from 38 kHz to 53 kHz. d. extends from 23 kHz to 53 kHz.

Answer: a

104

TEST ITEM FILE – Chapter 5 63. In an FM stereo transmitter, the L-R audio signal applied to the FM modulator stage: a. extends from 0 to 15 kHz. b. extends from 23 kHz. to 38 kHz. c. extends from 38 kHz to 53 kHz d. extends from 23 kHz to 53 kHz. Answer: d 64. An FM stereo broadcast is an example of: a. time-division multiplexing. b. frequency-division multiplexing. c. amplitude-division multiplexing. d. phase-division multiplexing. Answer: b 65. In a stereo FM broadcast transmitter, when the 38 kHz pilot carrier is mixed with the

L-R signal, what is produced? a. an FM signal containing L+R and L-R sidebands b. a 38 kHz AM signal c. an L-R double sideband suppressed carrier signal d. a 19 kHz pilot carrier Answer: c 66. Why are stereo FM broadcasts more prone to noise than are monophonic FM broadcasts? a. The L-R signal is weaker than the L+R signal. b. The L-R signal is at higher modulating frequencies than the L+R signal. c. Stereo has two separate channels that can become covered by noise rather than the single channel of the monophonic broadcast. d. All of the above. Answer: d

67. Why can low-level modulated FM signals be amplified by class C power amplifier stages

without distorting unlike AM or SSB signals? a. These FM signals do not vary in amplitude. b. These FM signals have only one upper and lower sideband frequency components. c. The FM signals do not have a carrier frequency component. d. These FM signals have larger bandwidths than do AM or SSB signals. Answer: a 68. The deviation return for NTSC broadcast television is

a. 1.67 b. 5 c. .5 d. 2.5 e. 6.71 Answer: a

105

TEST ITEM FILE – Chapter 5 69. The deviation ratio for broadcast FM radio is:

a. 1.67 b. 5 c. .5 d. 2.5 e. 6.71

Answer: b

70. The spectrum analyzer view of an FM signal is shown in figure 5-4. Estimate the 3dB bandwidth.

a. 240 kHz b. 120 MHz c. 24 kHz d. 120 kHz e. none of the above Answer: d

Figure 5-4

106

TEST ITEM FILE - Chapter 6 FREQUENCY MODULATION: RECEPTION

1. In FM reception, the circuitry used to extract the intelligence from the carrier is: a. VCO. b. Crosby modulator. c. limiter. d. discriminator. Answer: d 2. A form of receiver noise called cross modulation is caused by: a. noise caused by using non-square law device. b. noise caused by using JFETs. c. noise caused by the mixer. d. noise caused by high impedance device. Answer: a 3. The minimum required voltage for limiting is called: a. quieting. b. threshold. c. limiting knee. d. all of the above. Answer: d 4. Which of the following does provide amplitude limiting? a. ratio detector b. Foster-Seely discriminator c. PLL d. all of the above Answer: a 5. Input into the PLL is at the: a. VCO. b. low-pass filter. c. phase detector. d. comparator circuit. Answer: c 6. The basic difference between the block diagram of an AM receiver and FM receiver is that: a. an FM receiver uses a limiter and discriminator instead of a diode detector. b. an FM receiver does not use the superheterodyne design. c. an FM receiver cannot ever use automatic gain control. d. an FM receiver must use automatic frequency control. Answer: a

107

TEST ITEM FILE – Chapter 6 7. A standard IF frequency in FM receivers is: a. 455 kHz. b. 4.5 MHz. c. 9.0 MHz. d. 10.7 MHz. Answer: d 8. Why do most quality FM receivers employ FETs in their RF amplifiers? a. FETs have a very high input impedance. b. FETs have a very low input capacitance. c. FETs have a square-law input/output relationship. d. FETs have a superior frequency response. Answer: c 9. The square-law FET characteristics: a. allow for greater sensitivity in an FM receiver. b. minimize the cross-modulation distortion in an FM receiver. c. allow for narrower bandwidth in the tuned amplifier that it is used in. d. reduces the shot noise produced by the FET. Answer: b 10. A limiter stage in an FM receiver functions to: a. limit the amount of frequency deviation in the received FM signal. b. remove any residual amplitude modulation in the RF signal before being fed to

the detector stage. c. filter out any cross modulation frequency components produced by the mixer stage. d. reduce noise produced by transistors in the RF and IF amplifier stages. Answer: b 11. Which of the following does not occur in a limiter stage found in an FM receiver? a. amplifying a signal with an easily overdriven amplifier b. clipping action of the FM signal c. recreating the intelligence signal d. recreating the FM sinewave due to the flywheel effect Answer: c 12. The measure of how much signal is required in an FM receiver to produce a specific level of quieting is referred to as: a. the receiver's stereo separation. b. the receiver's noise reduction. c. the receiver's sensitivity. d. the receiver's selectivity. Answer: c

108

TEST ITEM FILE – Chapter 6 13. Which of the following FM discriminators produces excessive distortion due to its nonlinear roll-off characteristics? a. slope detector b. Foster-Seely detector c. ratio detector d. quadrature detector e. PLL detector Answer: a 14. Which of the following FM detectors is a classical design dating back to vacuum tube technology? a. quadrature detector b. Foster-Seely detector c. PLL detector d. diode detector Answer: b 15. The ratio detector design is superior to the Foster-Seely design in that: a. amplitude changes in the input FM signal have no effect on the output. b. it offers superior linear response to wideband FM deviation. c. it can be built using digital integrated circuits. d. it uses less diodes than does the Foster-Seely design. Answer: a 16. The FM detector that can be implemented using digital exclusive-or gates is the: a. slope detector. b. Foster-Seely design. c. ratio detector. d. quadrature detector. e. PLL FM detector.

Answer: d 17. The term "quadrature" in a quadrature detector refers to: a. the fact that the two input signals to the exclusive-or gate are 90 degrees out of phase

with one another. b. the fact that a four input gate is part of the detector stage. c. their use in producing four-channel stereo detection. d. the fact that it was the fourth FM detector design to gain popularity. Answer: a 18. Which of the following stages is not a building block of a phase-lock loop? a. phase comparator b. frequency multiplier c. low-pass filter d. voltage-controlled oscillator Answer: b

109

TEST ITEM FILE – Chapter 6 19. A PLL is set up so that its VCO free-runs at 8.9 MHz. The VCO does not change frequency

unless its input is within plus or minus 75 kHz of 8.9 MHz. After it does lock, the input frequency can be adjusted within plus or minus 120 kHz of 8.9 MHz without having the PLL start to free-run again. The tracking range of the PLL is:

a. 75 kHz. b. 120 kHz. c. 150 kHz. d. 240 kHz. Answer: d 20. When a PLL is set up as an FM detector, the input FM signal is applied to the input of: a. the phase detector of the PLL. b. the low-pass filter of the PLL. c. the frequency multiplier of the PLL. d. the VCO of the PLL. Answer: a 21. When a PLL is set up as an FM detector, the output intelligence signal appears at the output of: a. the phase detector of the PLL. b. the low pass filter of the PLL. c. the frequency multiplier of the PLL. d. the VCO of the PLL. Answer: b 22. The FM detector design that does not require intricate coil adjustments to be tuned is the: a. slope detector. b. Foster-Seely detector. c. ratio detector. d. quadrature detector. e. PLL detector. Answer: e 23. In an FM stereo receiver, what is the purpose of the matrix network? a. to filter out the SCA signal at the output of the discriminator b. to filter out the L-R signal at the output of the discriminator c. to filter out the L+R signal at the output of the discriminator d. to produce separate L and R signals from the L+R and L-R signals Answer: d 24. In an FM stereo receiver, why is there a frequency doubler stage in the stereo demodulation section? a. It doubles the frequency of the L+R audio signals. b. It doubles the 19 kHz pilot carrier so that the AM detector will work properly. c. It is required in the de-emphasis network. d. It is part of the FM discriminator stage. Answer: b

110

TEST ITEM FILE – Chapter 6

25. In an FM stereo receiver, what is the purpose of the AM detector stage? a. It converts the L-R DSB signal back into an L-R signal at audio frequencies. b. It mixes the L+R and L-R signals to produce separate L and R audio signals. c. It converts the L+R DSB signal back into an L+R signal at audio frequencies. d. It filters out the 19 kHz pilot carrier from the complex discriminator output signal. Answer: a

26. Subscription services that provide industry, department stores, and restaurants with commercial free background music use what type of transmission? a. usually a 67 kHz carrier with 7.5 kHz deviation b. a third channel frequency multiplexed on the FM modulated signal of a standard broadcast FM station c. subsidiary communication authorization d. all of the above Answer: d 27. An example of an FM stereo demodulator integrated circuit is the: a. 3090. b. 565. c. 3089. d. 560. Answer: a 28. The Bode plot shown in Figure 6-1 is the plot of a:

a. band –stop filter b. low pass filter c. band-pass filter d. high-pass filter e. none of the above

Answer: c

Figure 6-1

111

TEST ITEM FILE – Chapter 6

29. The expected output frequency of a mixer circuit with an input frequency of 92.1 MHz and a local oscillator frequency of 102.8MHz is:

a. 194.9 MHz b. 10.1 MHz c. 11.7 MHz d. 9.7 MHz e. none of the above Answer: e

112

TEST ITEM FILE - CHAPTER 7 COMMUNICATIONS TECHNIQUES

1. The process of stepping down the RF signal to a first IF frequency and then mixing down again to a second IF signal is known as: a. double conversion. b. down conversion. c. mixing. d. none of the above. Answer: a 2. Which of the following is not considered man-made noise? a. ignition systems b. switching of high current loads c. resistor noise d. motor systems Answer: c 3. The input power range over which the receiver or amplifier provides a useful output is known as: a. noise floor. b. S/N ratio. c. power range. d. dynamic range. Answer: d 4. The S meter is often found in which of the following circuits? a. IF stage b. AGC circuitry c. audio output d. RF input amplifiers Answer: b 5. Circuitry used to minimize noise output that occurs when tuning between stations is known as: a. noise limiters. b. noise filters. c. squelch circuitry. d. all of the above. Answer: c

113

TEST ITEM FILE – Chapter 7

6. Determine the S/N for a receiver with a 10 MHz bandwidth, a 20 dB noise figure, and a sensitivity of –60 dBm. a. 24 dB b. 20 dB c. 10 dB d. 14 dB Answer: a 7. Determine the sensitivity of a receiver with a 5 MHz bandwidth, 25 dB noise figure, and S/N of 40 dB. a. -70 dBm b. -84 dBm c. -20 dBm d. -42 dBm Answer: d 8. If the above system had a +5dBm third order intercept point, determine its dynamic range. a. 66 dB b. 25 dB c. 31 dB d. 58 dB Answer: c 9. A transceiver is: a. a transmitter that can be tuned to several bands of frequencies. b. a transmitter that transmits digital data signals. c. a receiver that receives digital data signals. d. a transmitter and receiver in a single package. Answer: d 10. Up-conversion is: a. a technique used to reduce the image frequency problem from occurring in superheterodyne receivers. b. the process of having the IF frequency in a superheterodyne receiver be higher than the range of RF frequencies being received by the receiver. c. a superheterodyne receiver design that is feasible due to the availability of VHF crystal filters (30 – 120 MHz) for the IF circuitry. d. all of the above. Answer: d

114

TEST ITEM FILE – Chapter 7

11. An AGC that does not provide any gain reduction until some arbitrary signal level is attained and therefore has no gain reduction for weak signals is known as: a. arbitrary AGC. b. auxiliary AGC. c. delayed AGC. d. variable AGC. Answer: c 12. The ability of an FM receiver to totally mute the receiver audio output except for when a carrier is received is known as: a. automatic noise limiting (ANL). b. automatic gain control (AGC). c. squelch. d. variable sensitivity. Answer: c 13. A receiver has a 30 dB noise figure, a 1.5 MHz bandwidth, a 6 dBm third intercept point, and a 3 dB signal-to-noise ratio. Its dynamic range is: a. 56.8 dB. b. 48.8 dB. c. 66.7 dB. d. 58.7 dB. Answer: a 14. The tuned circuit prier to the mixer is often referred to as the :

a. pre-mixer b. pre-selector c. post-processor d. age circuit

Answer: b

15. An AGC circuit is shown in Figure 7-1. If diode D1 is open, the squelch circuit will:

a. operate normally b. cause Q1 to short c. only operate over a limited frequency range d. not work

Answer: d

115

TEST ITEM FILE – Chapter 7

16. An AGC circuit is shown in Figure 7-1. If diode C1 is shorted, the squelch circuit will:

a. operate normally b. cause Q1 to not function properly c. work normally but at a reduced amplitude d. work normally but at an increased level

Answer: b

17. A parasitic is a. helps to provide circuit stability b. an unwanted component of an electronic circuit that is a byproduct of fabrication, component assembly, or both c. is used to increase high frequency circuit performance d. all of the above Answer: b

Figure 7-1

116

TEST ITEM FILE – Chapter 7

18. The spacing of parallel lines on a 0.10” thick printed circuit board should be a. 0.40” b. 0.20” c. 0.80” d. 0.56” Answer: a 19. The five basic issues to consider when assembling a printed circuit board for use at high frequency include which of the following?

a. Wires become transmission lines. b. The length and geometry of the wires becomes an issue. c. Bends in the circuit board layout are an issue. \ d. At high frequencies you have to transition from some coaxial structure into a circuit board to properly

launch the wave. e. Impedance match all elements in your circuit to minimizing ghosting within the circuit board.Make

sure your components are impedance matched to minimize ghosting with. f. all of the above g. only a,c,d h. only a,b,e

Answer: f 20. RF choke are used to? a. isolate the AC signal b. block DC voltages c. isolates the RF signal from the antenna d. isolate the DC circuit from the RF circuitry. Answer: d

117

TEST ITEM FILE – CHAPTER 8 DIGITAL COMMUNICATION: CODING TECHNIQUES

1. Find the generating polynomial for a generating function of 1100111 for an 8-bit message. a. X7 + X5 + X3 + 1 b. X6 + X5 + X2 + X + 1 c. X6 + X4 + X3 + X + 1 d. X7 + X4 + X2 + 1 Answer: b 2. Find the dynamic range of a PCM system of 16 bits. a. 96.32 dB b. 48.16 dB c. 24.08 dB d. 12.04 dB Answer: a 3. To achieve a dynamic range of 78 dB, how many bits will be used? a. 16 b. 6 c. 10 d. 13 Answer: d 4. Which of the following is a baseband type of signal? a. PCM b. PAM c. RZ d. PWM Answer: c 5. Which coding is commonly used as Ethernet for local area networks? a. Manchester coding b. RZ c. NRZ d. Multilevel binary codes Answer: a

118

TEST ITEM FILE – Chapter 8

6. The number of bits required for a binary code to represent 305 different possibilities is: a. 8 bits. b. 9 bits. c. 2 bits. d. 3 bits. Answer: b 7. An early alphanumeric 5-bit code that was developed in the days of teletype machines that

has limited power is known as: a. ASCII. b. EBCDIC. c. the Baudot code. d. the Gray code. Answer: c

8. An alias signal is: a. a signal that forms in a PCM system that has failed to meet the Nyquist sampling rate. b. a signal that has a frequency equal to the difference between the applied analog

frequency and the sampling frequency. c. usually filtered out by anti-aliasing filters in PCM systems. d. all of the above. Answer: d 9. A single LSI integrated circuit that contains the A/D and D/A circuitry in a PCM

system is known as: a. a repeater. b. a codec. c. a quantizer. d. a time-division multiplexer. Answer: b 10. The varying amplitude values which are used in order to code an analog intelligence

signal in a PCM system are called: a. compression levels. b. quantizing levels. c. companding levels. d. sample levels.

Answer: b

11. The type of PCM coding scheme that is popular in optical systems, satellite telemetry links, and magnetic recording systems is:

a. NRZ – non-return to zero. b. RZ – return to zero. c. phase-encoded and delay modulation. d. multilevel binary. Answer: c

119

TEST ITEM FILE – Chapter 8

12. Which of the following error detection schemes does not require retransmission of the data for correction to occur?

a. Hamming code b. block check character c. parity d. cyclic redundancy check Answer: a 13. Which of the following is an application of PCM? a. digital audio recording b. telephone systems

c. CD laser disks d. all of the above Answer: d 14. A popular serial data communication standard used on computers to connect to a mouse.

a. IEEE-488 b. RS-435 c. RS-232 d. UART 16550A Answer: c 15. In which of the following is the sample signal voltage held constant between samples? a. flat-top sampling b. natural sampling c. sample and hold d. none of the above Answer: a 16. The rule that a signal sampled at a minimum of twice the rate of its highest significant frequency component can be fully reconstructed at the receiver with a high degree of accuracy is known as: a. the error algorithm. b. Barkenhausen criteria. c. Nyquist criteria. d. the Shannon-Hartley theorem. Answer: c

17. The amount of time it takes the hold capacitor in A/D system to reach its final value.

a. hold time b. sample time c. aperture time d. acquisition time Answer: d

120

TEST ITEM FILE – Chapter 8

18. The time that the sample and hold circuit must hold the sampled voltage. a. hold time b. sample time c. aperture time d. acquisition time Answer: c

19. The phenomenon associated with the generation of an erroneously created signal in the sampling

process is called: a. aliasing b. fold over distortion c. both a and b d. none of the above Answer: c

20. The logical distance between defined states is called

a. Viterbi distance b. hamming distance c. binary distance d. none of the above Answer: b

21. The distance of a received code (00111) from a valid code of (00101);

a. 1 b. 2 c. 3 d. 4 Answer: a

22. Given a (7,4) cyclic code, the length of the message is: a. 1 b. 2 c. 3 d. 4 Answer: d

23. Given a (7,4) cyclic code, the length of the BCC is:

a. 1 b. 2 c. 3 d. 4 Answer: c

24. Reed-Solomon codes are:

a. forward error-correcting b. seldom used c. also called CRC codes d. none of the above Answer: a

121

TEST ITEM FILE – Chapter 8

25. The oscilloscope trace shown in Fig 8-1 was obtained from a sample-and-hold circuit. This is an

example of: a. natural sampling b. flat-top sampling c. over-sampling d. under sampling Answer: b

Figure 8-1

26. A non-recursive filter a. algorithm that includes the previous output values to generate the current output b. algorithm that include the previous and present output values to generate the current output c. algorithm that does not include previous output values to generate the current output d.. none of the above Answer: c 27. What is the order of a difference equation that requires three past input and output values? a. second b. third order c. sixth order d. first order

Answer: b

28. Given the following difference equations, state the order of the filter and identify the values of the coefficients. 3213200 0359.22021.35495.202008.004016.002008.0 yyyxxxy +−++−= a. third order b. sixth order c. second order d. none of the above Answer: a

122

TEST ITEM FILE – CHAPTER 9

WIRED DIGITAL COMMUNICATIONS

1. Which of the following can be used to generate PWM? a. PCM b. PPM c. PAM d. PTM Answer: b 2. In what type of transmission is bit rate and baud rate the same? a. analog b. pulse amplitude modulation c. binary transmission d. hexadecimal transmission Answer: c 3 Baseband analog data channels are characterized by: a. the fact that a modem has converted the intelligence to an analog signal for use on an analog channel. b. the fact that a modulator has converted the analog intelligence into a modulated analog signal on another frequency range. c. the fact that a coder has converted the digital intelligence into a specific coding format for subsequent transmission. d. the fact that the intelligence is being transmitted at its base frequencies; no modulation to another frequency range has occurred. Answer: d 4 An advantage of PPM over PWM is: a. its simplicity in design. b. its frequency response of the intelligence signal. c. its greater resistance to any error produced by noise. d. its limited bandwidth. Answer: c 5 The most desired pulse modulation scheme is: a. pulse amplitude modulation (PAM). b. pulse width modulation (PWM). c. pulse frequency modulation (PFM). d. pulse position modulation (PPM). Answer: d

123

TEST ITEM FILE – CHAPTER 9

6 A two-tone amplitude modulation system uses 400 Hz for a "one" and 550 Hz for a "zero" and

a carrier frequency of 9.35 MHz. The required bandwidth for this communication channel is: a. 400 Hz. b. 550 Hz. c. 800 Hz. d. 950 Hz. e. 1100 Hz. Answer: e 7 In a QAM transmitter, binary data is converted to two signals: a. called the I and Q channels.

b. that are 90 degrees out of phase with one another. c. that are half the original bit rate. d. all of the above. Answer: d 8 A certain language has 18 letters in its alphabet. How many bits are required to represent

the 18 letters? a. 6.1 b. 4.2 c. 3.8 d. 5 Answer: b

9 What is the coding efficiency of problem 11? a. 83% b. 87% c. 95% d. none of the above Answer: a

10. Another name for delta modulation is:

a. slope modulation b. pulse modulation c. codec d. all of the above

Answer: a

11. Which is not an advantage of digital data communication systems over analog

communications? a They have narrower bandwidths. b. They have superior noise performance. c. They eliminate errors in the received signal. d. They have the ability to process the signal at the transmitter and receiver using

digital signal processing techniques. Answer: a

124

TEST ITEM FILE – CHAPTER 9

12. A digital transmission has an error probability of 1 x 10-5 and is 1 x 108 bits long. Its BER is: a. 1 x 10-3

b. 1 x 10-5

c. 1 x 10-4 d. 1 x 10-8

Answer: b 13. The major advantage of a delta modulator over PCM circuitry is its:

a. simplicity in design b. frequency response of the intelligence signal c. bandwidth characteristics d. greater resistance to any error produced by noise Answer: a

14. A technique used to transport data from multiple users over the same data channel is:

a. CDMA b. FDMA c. Fractional T1 d. TDMA Answer: d

15. Inter-symbol interference is a serial digital communication link can be minimized by:

a. adding data b. increasing the data rate c. adding guard times d. increasing the BER Answer: c

16. The T1data rate is:

a. 56 kbps b. 512 kbps c. 1.544 Mbps d. 44.736 Mbps Answer: c

17. A term used to indicate that only a portion of the data bandwidth of a T1 line is being used is:

a. fractional T1 b. Multiplexed T1 c. Companded T1 d. T1/2 Answer: a

125

TEST ITEM FILE – CHAPTER 9

18. The point where the communications carrier brings in service to a facility is called the: a. CSU/DSU b. termination line c. patch line d. point of presence Answer: d

19. An example of a packet switching network designed to carry data traffic over public data network is:

a. IEEE 802.3 b. frame relay c. IEEE 802.5 d. none of the above Answer: b

20. A cell relay technique designed for voice, data and video traffic and considered to be an evolution of

packet switching is: a. ACR-3 b. MPEG-2 c. ATM d. POP Answer: c

21 Which of the following is not a major protocol function? a. framing b. line control c. flow control d. character insertion Answer: d

22. Identify the type of USB connector shown. In Fig. 9-1.

a. male b. female c. Type A d. Type B

Answer: c Fig. 9-1 23. Identify the type of USB connector shown in Fig. 9-2.

a. male b. female c. Type A d. Type B

Answer: d Fig. 9-2

126

TEST ITEM FILE – CHAPTER 10 WIRELESS DIGITAL COMMUNICATIONS

1. The three groups of wireless technologies are: a. wired wireless, mobile wireless, IR wireless b. wired wireless, mobile wireless, wired IR c. fixed wireless, mobile wireless, IR wireless d. fixed wireless, mobile wireless, wired wireless Answer: c 2. Determine the channel bandwidth required for a wideband FSK system. a. 5 to 10 kHz b. 5 to 15 kHz c. 20 to 25 kHz d. not enough information to answer Answer: d Use the following equation to answer questions 3 – 5

Mitvtv co

)1(2sin[)( −+=

πω

3. Find vo(t) for a phase shift keying system given i = 1, M = 2 a. Vsinωct b. -Vsinωct + c. 2Vsinωct d. not enough information to solve. Answer: a 4. Find vo(t) for a phase shift keying system given i = 3, M = 8 a. Vsinωct b. -Vsinωct c. Vsinωct + π/2 d. not enough information to solve. Answer: c 5. Find vo(t) for a phase shift keying system given i = 2, M = 4 a. Vsinωct b. -Vsinωct c. Vsinωct + π/2 d. not enough information to solve. Answer: c 6. The output signal of a M-ary system is called a a. array b. constellation c. star d. none of the above Answer: b

127

TEST ITEM FILE – Chapter 10 7. The output data of a digital communication link is shown in Figure 10-1. The modulation technique

being used is:

a. BPSK b. QPSK c. 8PSK d. 8QAM Answer: a

8. The vector representation of the constellation shown in Figure 10-2 is an example of :

a. BPSK b. QPSK c. 8PSK d. 16QAM Answer: b

9. The constellation shown in figure 10-3 is an example of:

a. BPSK b. QPSK c. 8PSK d. 16QAM Answer: d

Figure 10-1

Figure 10-2

Figure 10-3

128

TEST ITEM FILE – Chapter 10 10. Which of the eye patterns shown in Figure 10-4 shown a normal system.

a. a b. b c. c d. d Answer: a

11. Which eye pattern shown in Figure 10-4 shows timing jitter?

a. a b. b c. c d. d

Answer c

12. The terms sin2A can be written as a. ½ [1 + cos 2A] b. ½ [1 x cos 2A] c. ½ [1 – cos 2A] d. ½ [1 – cos A] Answer: c 13. The output of a QPSK phase detector is 0.5A cos (0 + Φd) – (0.5A cos 2ωct). Identify the high frequency component for this output. a. 0.5 is the high frequency component b. Φd is the high frequency componrnt c. 2ωct is the high frequency component d. none of the above Answer: c

Figure 10-4

129

TEST ITEM FILE – Chapter 10 14. How is the high frequency component removed for the output of a QPSK phase detector with a value of 0.5A cos (0 + Φd) – (0.5A cos 2ωct). a. low-pass filter b. band-stop filter c. high-pass filter d. notch filter Answer: a 15. Determine the voltage vpd for the data vector point defined by (0,0) for a QPSK system with a unit value of 1. a. 0.707V, 0.707V b. 0.707V, 0.35V c. 0.35V, 0.707V d. 0.35V, 0.35V Answer: d 16. What is the purpose of a loop back when used in digital modulation systems. a. the loop back is compared with the original data to indicate system performance b. the loop back is compared with the original data to indicate system data rate c. the loop back is never compared with the original data d. the loop back is only compared with the original data to indicate a system noise figure Answer: a 17. This type of code has an output that appears to be noise-like. a. pseudo noise code b. pseudo code c. pseudo spectrum d. spread code Answer: a 18. A PN code of length 2n – 1 is said to be of a. repetitive b. minimal length c. maximal length d. lengthy Answer: b 19. Determine the sequence length of a properly connected PN sequence generator containing 5 shift registers. a. 32 b. 69 c. 36 d. 31 Answer: d

130

TEST ITEM FILE – Chapter 10 20. Determine the sequence length of a properly connected PN sequence generator containing 31 shift registers. a. 2,147,483,648 b 2,147,483,649 c. 2,147,483,647 d. 2,147,483,646 Answer: c 21. The RF spectrum shown in Figure 10-5 is an example of a. direct sequence spread spectrum b. orthogonal frequency division multiplexing c. frequency division spread spectrum d. frequency hopping spread spectrum Answer: d

Figure 10-5

22. DSSS uses a pseudorandom sequence of pulses shorter than the message bit. These are called a. fractals b. parce bits c. fractures d. chips Answer: d 23. The term for the situation when two spread spectrum transmitters momentarily transmit at the same

time is called a. lost data b. design error c. a hit d. a data burst Answer: c

131

TEST ITEM FILE – Chapter 10 24. The signature sequence in DSSS is a. eliminates the need for QPSK transmission b. the pseudo random digital sequence used to spread the signal c. aids with the reception of weak signals d. limits the data rate Answer: b 25. A general rule is that the chip rate is what relative to the data modulation rate? a. approximately 5 to 7 times the modulation rate b. must be less that 12% of the modulation rate c. is independent of the modulation rate d. must be greater than 70.7% of the modulation rate Answer: a 26. Determine the spreading of a DSSS signal given the following parameters Modulation bit rate: 56 kbps Chip rate: 280 kbps a. spreading = 0.2 b. spreading = 1.2 c. spreading = 1.0 d. spreading = 5.0 Answer: d 27. Compare the two RF spectrums shown. What has changed for from Figure 10-6 A to B?

Figure 10-6

a. the QPSK signal has been spread b. the BPSK signal has been spread c. the 16QAM signal has been spread d. the 8PSK signal has been spread Answer: b

132

TEST ITEM FILE – Chapter 10 28. Another name for OFDM is a. multitone modulation b. discrete modulation c. DSSS d. QPSK Answer: a 29. If two signals are sent over the same media without interference the two signals are said to be a. symmetrical b. vertical c. orthogonal d. parallel Answer: c 30. A cyclic prefix in OFDM means a. the very beginning of a symbol is copied thus leaving no gap b. the very end of a symbol is copied to the beginning of the data stream thus leaving a gap c. the very end of a symbol is copied to the beginning of the data stream thus leaving a defined gap d. the very end of a symbol is copied to the beginning of the data stream thus leaving no gap Answer: d 31. FSK is a form of: a. AM b. FM c. PM d. none of the above Answer: b 32. The LM1871 and LM1872 integrated circuits: a. can be used to create a radio-telemetry system that operates at a frequency of approximately 49 MHz. b. use PWM as a pulse modulation technique. c. can handle more than one channel of digital data. d. all of the above. Answer: d 33. Which of the following is an advantage of transmitting in a digital format? a. less sensitive to noise b. less crosstalk c. lower distortion levels

d. all of the above Answer: d

133

TEST ITEM FILE – Chapter 10

34. The term for an analog digital FM signal that shares the same bandwidth is a. spread spectrum FM b. SSB FM c. hybrid FM d. PCM AM Answer: c 35. The digital data rate for AM digital is a. 36 kbps b. 56 kbps c. 1.544 Mbps d. 96 kbps Answer: a 36. The digital data rate for FM digital is a. 36 kbps b. 96 kbps c. 1.544 Mbps d. 56 kbps Answer: b

134

TEST ITEM FILE – CHAPTER 11 NETWORK COMMUNICATIONS

1. Which transformer allows full duplex operation by providing isolation between the transmit

and receive legs of the system? a. primary b. step-up c. line hybrid d. secondary Answer: c 2. The process of using the same carrier frequency in different cells that are geographically

separated is known as: a. frequency reuse. b. narrowband FM bandwidth that is used. c. Rayleigh fading d. cell splitting. Answer: a 3. Which of the following is considered a wide area network? a. two or more LANs connected over a wide geographical area b. metropolitan area network c. open systems interconnection d. none of the above Answer: a 4. Repeaters are used in T1 lines to carry voice channels when the lines stretch more than: a. 6,000 feet. b. 6,000 miles. c. 50 miles. d. 500 feet. Answer: a 5. The ringing signal voltage of the telephone service is set at: a. 2V rms. b. –48VDC. c. 90V rms. d. 2 mV rms. Answer: c

135

TEST ITEM FILE – CHAPTER 11 6. The voltage difference between the tip and ring lines of the phone line when it is on-hook is: a. 0V. b. 48 VDC. c. 90V rms. d. 2 mV rms. e. 2V rms. Answer: b 7. An integrated circuit that contains both encoding PCM circuitry and decoding PCM circuitry is

known as: a. a modem. b. an analog to digital converter. c. a codec. d. a digital to analog converter. Answer: c

8. The FM capture effect in cellular telephone systems is hindered from minimizing co-channel

interface effects due to: a. the use of frequency reuse. b. the narrowband FM bandwidth that is used. c. Rayleigh fading. d. the cell splitting process. Answer: c 9. Which is not a basic mode of switching digital data? a. circuit switching b. network switching c. message switching d. packet switching Answer: b 10. Which is not a popular topology for a local area network? a. star b. ring c. bus d. broadband Answer: d 11. Which OSI level is associated with handling error recovery, flow control, and sequencing? a. physical b. data link c. network d. transport e. session Answer: b

136

TEST ITEM FILE – CHAPTER 11

12. Which OSI level accepts outgoing messages and combines messages into packets, adding a header?

a. physical b. data link c. network d. transport e. session Answer: c 13. Which OSI level is concerned with message integrity between the source and destination? a. data link b. network c. transport d. session e. presentation Answer: c 14. A PBX is a:

a. Public Branch Exchange b. Private Branch Exchange c. Private Bit Exchange d. Public Bit Exchange Answer: b

15. A circuit connecting one central office to another is

a. line b. link c. trunk d. interface Answer: c

16. A cable with added inductance every 6000, 4500 or 3000 feet is called a:

a. loaded cable b. “L” cable c. minus c cable d. balanced cable Answer: a

17. A situation in a telephone switching office when calls are unable to reach their destination is referred

to as:

a. throughput problems b. lag c. delay d. congestion Answer: d

137

TEST ITEM FILE – CHAPTER 11

18. The process of changing channels for a new cell site is called a:

a. tradeoff b. handoff c. token passing d. frequency transfer Answer: b

19. The most common reason for sending a REL message is a. the destination receives a broadcast b. the source receives a broadcast c. one party hangs up d. the ringer voltage has not been set Answer: c 21. CSMA/CD stands for:

a. Carrier Sense Many Applications Common Data b. Carrier Sense Multiple Access with Collision Detection c. Carrier Sense Multiple Access with Common DATA d. None of the above

Answer: b

22. The length of a MAC address is :

a. 56 bytes b. 4 bytes c. 6 bytes d. 3 bytes Answer: c

23. Setting the destination MAC address to all 1’s is called a:

a. broadcast address b. multiple delivery address c. common address d. router address Answer: a

24. The 8-pin modular connector used to connectorize CAT5/5e/6 cable is:

a. RJ-11 b. RJ-48 c. RJ-45 d. RJ-40 Answer: c

138

TEST ITEM FILE – CHAPTER 11 25. The term 100 baseT means:

a. 100 meters, baseband twisted pair b. 100 Mbps, baseband, twisted pair c. 100 meter, broadband, terminated d. none of the above Answer: b

26. The CSMA/CA protocol avoids collisions by

a. limiting transmission b. using separate cables c. waiting for acknowledgement that a packet has arrived d. none of the above

Answer: c

27. A wireless Ethernet standard is the:

a. IEEE 802.11b b. IEEE 802.3 c. IEEE 802.5b d. IEEE 802.6b

Answer: a

28. 128.123.1.10 is an example of a:

a. class A network b. class B network c. class C network d. none of the above Answer: b

29. The V.92/90 standard for high speed modems is

a. combination of digital and analog b. data rate to the service provider is typically at V.34 speeds c. data rate from the service provider is 56 kbps d. all of the above Answer: d

30. A technique used by cable modems to determine the time it takes for data to travel to the cable head-

end is called.

a. pinging b. ranging c. ringing d. roaming Answer: b

139

TEST ITEM FILE – CHAPTER 11 31. The data rate for a (2B+D) ISDN system is:

a. 144 kbps b. 192 kbps c. 64 kbps d. 128 kbps Answer: a

32. A DSL service that provides 1.544 Mbps form the user to the service provider and up to 8 Mbps back to the user from the service provider is:

a. IDSL b. HDSL c. ADSL d. VDSL Answer: c

33. For ADSL, a multi-carrier technique called is used to carry the data over the copper

lines.

a. DPSK b. QPSK c. Spectrum modulation d. Discrete multi-tone Answer: d

34. The protocol that has been developed to bridge the gap between mobile communications, the internet,

and corporate intranets is the :

a. wireless application protocol b. wireless markup language c. wireless telephony applications d. wireless transmission protocol Answer: a

35. The rms voltage of a 0dBm sine wave is:

a. 2.188V b. 1.094V c. 0.774V d. 1.414V Answer: c

36. When making a total harmonic distortion measurement of a 1kHz sinewave, the fundamental

frequency of the test instrument must be set to:

a. 1 kHz b. 3 kHz c. 500 kHz d. 2 kHz Answer: a

140

TEST ITEM FILE – CHAPTER 11

37. The 3G wireless system supports data rates up to

a. 2 Mbps b. 1 Mbps c. 1.544 Mbps d. none of the above

Answer: a 38. CDMA 2000 a. is a 2G wireless development in the United States b. is a 2.5G wireless development in the United States c. is a 21G wireless development in the United States d. is a 3G wireless development in the United States Answer: d 39. A wireless Ethernet standard is the:

e. IEEE 802.11g f. IEEE 802.3g g. IEEE 802.5b h. IEEE 802.6b

Answer: a

40. A wireless Ethernet standard is the:

a. IEEE 802.11b b. IEEE 802.11g c. IEEE 802.11a d. all of the above e. none of the above Answer: d

41. A number that identifies the switch nearest to the caller is called which of the following. a. Destination Point Code b. Circuit Identification Code c. Source Identification Code d. Origination Point Code Answer: d 42. A number that identifies the switch nearest to the called party is called which of the following. a. Destination Point Code b. Circuit Identification Code c. Source Identification Code d. Origination Point Code Answer: a

141

TEST ITEM FILE – CHAPTER 11

43. In CDMA, which Walsh code is the synchronizing signal. a. 0 b. 16 c. 32 d. 5 Answer: c 44. In CDMA, the frequency reference is provided by? a. GPS satellites b. UPS clocking c. UTC time code d. SSBSC Answer: a 45. How does WiMAX differ from WiFi? (select all that apply) a. WiMAX is a wired technology b. Frequency assignments differ c. data rates differ d. WiMax unit only has to compete once to gain entry into the network. Answers: b,c,d 46. In what frequency band does Bluetooth operate? a. the 5.2 GHz ISM band. b. the 2.2 GHz ISM band. c. the 2.4 GHz ISM band d. the 5.4 GHz ISM band. Answer: c 47. This is used by Bluetooth to discover other Bluetooth devices or to allow itself to be discovered. a. inquiry procedure b. paging procedure c. pinging procedure d. all of the above Answer: a 48. What are the five aspects to consider when securing a communication link? (select all that apply) a. authentication b. confidentiality c. integrity d. non-repudiation e. availability of the network f. all of the above g. a,b,c,e Answers: f

142

TEST ITEM FILE – CHAPTER 11 49. In a passive active a. the bad guy is disrupting the communications link b. the bad guy is outside the system c. the bad guy is inside the system d. the bad guy is just listening Answer: d 50. This is a secret code that is used in the encryption algorithm to both create and decode the message a. pn code b. cipher-key c. pin key d. CRC code Answer: b

143

TEST ITEM FILE – CHAPTER 12 TRANSMISSION LINES

1. Which of the following is a definition of characteristic impedance? a. coaxial line impedance b. impedance at any point c. impedance of infinite line d. impedance at front end Answer: c 2. When moving from the source to load current and voltage traveling waves move: a. 30° phase shift b. 60° phase shift c. 90° phase shift d. no phase shift Answer: d 3. One use of a quarter-wavelength transformer is: a. step down current. b. match line resistance to load resistance. c. shift voltage. d. all of the above. Answer: b 4. Which of the following is not an application of a transmission line? a. impedance match b. filters c. balanced lines d. discrete circuit simulation Answer: d 5. Which of the following can be determined using a slotted line method? a. VSWR b. generator frequency c. unknown load impedance d. all of the above Answer: d 6. What makes conductors behave differently at high frequencies than they do at low frequencies? a. At high frequencies, there are many sources of line loss. b. At high frequencies, energy sent down a wire may be reflected back. c. At high frequencies, there are inductive, capacitive, and resistive effects to transmission lines. d. all of the above. Answer: d

144

TEST ITEM FILE – Chapter 12

7. What is an advantage of open-wire line over coaxial cable? a. low radiation loss b. low noise pickup c. simple construction d. low inductive loss Answer: c 8. A commonly used coaxial cable, RG-58, has a characteristic impedance of 50 ohms.

If its length is doubled, its characteristic impedance will: a. increase. b. decrease. c. remain the same. Answer: c 9. A commonly used coaxial cable, RG-58, has a characteristic impedance of 50 ohms. If

the spacing between the conductors is doubled, its characteristic impedance will: a. increase. b. decrease. c. remain the same. Answer: a 10. In a conductor at high frequencies, most of the current flow is at the outer surface of the

conductor. This phenomena is known as: a. radiation loss. b. the skin effect. c. induction loss. d. resistive attenuation. e. surface loss. Answer: b 11. Which will reduce the radiation loss of transmission line? a. use coaxial transmission line b. use better shielding c. terminate the line with a resistive load equal to the transmission line's characteristic impedance d. all of the above Answer: d 12. Determine the amount of delay in a 200 ft. section of RG-8A/U coaxial cable if its

inductance is 73.75 nH per ft. and its capacitance is 29.5 pf per ft. a. 217.5 ns b. 295 ns c. 49.9 ns d. 9.983 us Answer: b

145

TEST ITEM TILE – Chapter 12 13. If the length of a section of RG-59 coaxial cable is doubled, its velocity factor will: a. increase. b. decrease. c. remain the same. Answer: c 14. A transmission line that is terminated with a resistive load that is equal to the characteristic

impedance of the line is called: a. reactive. b. resonant. c. lossy. d. nonresonant. Answer: d 15. A transmission line that is terminated with an impedance that is not equal to its characteristic

impedance is called: a. reactive b. resonant. c. lossy. d. nonresonant. Answer: b

16. A transmission line having a delay time of 45 microseconds and terminated with an open is

applied with a 4 VDC incident voltage. At the load end of the line: a. the voltage changes from 0V to 8V at t = 45 microseconds. b. the voltage changes from 0V to 4V at t = 45 microseconds. c. the voltage changes from 0V to –4V at t = 45 microseconds. d. the voltage changes from 0V to 8V to t = 90 microseconds. e. the voltage changes from 0V to 4V at t = 45 microseconds. Answer: a 17. A non-lossy transmission line terminated with an open circuit will have: a. an in-phase reflected voltage that is equal in magnitude to the incident voltage. b. an opposite-phase reflected voltage that is equal in magnitude to the incident voltage. c. an in-phase reflected voltage that is smaller in magnitude than the incident voltage. d. an opposite-phase reflected voltage that is smaller in magnitude than the incident voltage. Answer: a 18. A non-lossy transmission line terminated with an open circuit will have: a. an in-phase reflected current that is equal in magnitude to the incident current. b. an opposite-phase reflected current that is equal in magnitude to the incident current. c. an in-phase reflected current that is smaller in magnitude than the incident current. d. an opposite-phase reflected current that is smaller in magnitude than the incident current. Answer: b

146

TEST ITEM FILE – Chapter 12 19. A non-lossy transmission line terminated with a short circuit will have: a. an in-phase reflected current that is equal in magnitude to the incident current. b an opposite-phase reflected current that is equal in magnitude to the incident current. c. an in-phase reflected current that is smaller in magnitude than the incident current. d. an opposite-phase reflected current that is smaller in magnitude than the incident current. Answer: a 20. A transmission line terminated with an open has a VSWR of: a. –1. b. 0. c. +1. d. infinity. Answer: d 21. A transmission line terminated with a short has a VSWR of: a. –1. b. 0. c. +1. d. infinity. Answer: d 22. A transmission line terminated with a load resistance that is equal to the transmission line's

characteristic impedance has a VSWR of: a. –1. b. 0. c. +1 d. infinity. Answer: c 23. A transmission line terminated with an open has a reflection coefficient at the load of: a. –1. b. 0. c. +1. d. infinity. Answer: c 24. A transmission line terminated with a short has a reflection coefficient at the load of: a. –1. b. 0. c. +1 d. infinity. Answer: a

147

TEST ITEM FILE – Chapter 12

25. A transmission line terminated with a load resistance that is equal to the transmission line's characteristic impedance has a reflection coefficient at the load of: a. –1. b. 0. c. +1. d. infinity. Answer: b 26. A 50 ohm transmission line that has a load impedance of 300 ohms has a reflection

coefficient of: a. 6. b. 0.166. c. –0.714. d. 0.714. e. 0.666 Answer: d 27. The disadvantage of not having a perfectly matched transmission line in a communication

system is: a. the fully generated power does not reach the load. b. the cable dielectric may break down. c. the existence of reflections increases the power loss. d. noise problems are increased. e. none of the above. Answer: e 28. The input impedance of a quarter-wavelength section of 50 ohm transmission line that is

terminated with an open is: a. 50 ohms. b. 0 ohms (a short). c. infinite (open). d. 100 ohms. Answer: b 29. The input impedance of a half-wavelength section of 50 ohm transmission line that is

terminated with a short is: a. 50 ohms. b. 0 ohms. c. infinite (open). d. 100 ohms. Answer: b

148

TEST ITEM FILE – Chapter 12

30. The input impedance of a half-wavelength section of 50 ohm transmission line that is terminated with an open is:

a. 50 ohms. b. 0 ohms (a short). c. infinite (open). d. 100 ohms. Answer: c 31. Moving a full revolution around an SWR circle on the Smith chart represents: a. moving a quarter-wavelength down the transmission line. b. moving a half-wavelength down the transmission line. c. moving a full wavelength down the transmission line. d. moving two wavelengths down the transmission line. Answer: b 32. The center of the Smith chart represents: a. an open circuit load impedance. b. a short circuit load impedance. c. a matched load impedance. d. a 1 ohm load impedance. Answer: c 33. The perimeter of the Smith chart graph represents: a. the SWR of a transmission line terminated with an open circuit. b. the SWR of a transmission line terminated with a short circuit. c. an infinite SWR circle. d. all of the above. Answer: d 34. A quarter-wavelength transmission line is terminated with a 300 ohm load. Its input

impedance is 75 ohms. Its characteristic impedance is: a. 50 ohms. b. 75 ohms. c. 150 ohms. d. 187.5 ohms. e. 300 ohms. Answer: c 35. A pulse-technique used in determining the location of a fault in a transmission line is called: a. slotted-line measurements. b. SWR measurements. c. time-domain reflectometry. d. stub tuning. Answer: c

149

TEST ITEM FILE – Chapter 12 36. Near-end crosstalk is best defined as

a. a measure of the level of crosstalk within a cable b. a measure of the signal coupling outside the cable c. a measue of the level in side the cable d. a desired perormance objective e. none of the above Answer: a

37. CAT 5e cable can handle a 100MHz bandwidth up to a length of:

a. 22 meters b. 110 meters c. 330 meters d. 90 meters e. 100 meters Answer: e

38. Wiring of the RJ-45 connector CAT6 cable is defined by the telecommunications industry standard:

a. 802.3 b. 802.11 c. ACR d. TIA 568B e. None of the above Answer: d

39. ACR is a combined measurement of

a. T568A and T568B b. attenuation and crosstalk. c. crosstalk and PSNEXT d. CAT3 and CAT5 e. none of the above

Answer: b

40. The CAT1 specification for twisted pair:

a. same as CAT3 b. same as TIA 568B c. never existed d. CAT 4 e. None of the above Answer: c

150

TEST ITEM FILE – Chapter 12

41. An impedance value of Z11=1+j0 indicates it is:

a. capacitive b. inductive c. 50Ω d. purely resistive e. none of the above Answer: d

42. The characteristic impedance of a network analyzer is set to 50Ω. A Z11=0.5+j0.5 translates to:

a. 25+j0 b. 50+j25 c. 25+j25 d. 25+j50 e. 50+j50 Answer: c

43. At resonance, a series RLC circuit will be:

a. resistive only b. capacitive only c. inductive only d. reactive e. none of the above Answer: a

151

TEST ITEM FILE - CHAPTER 13

WAVE PROPAGATION 1. The polarization of an electromagnetic wave is determined by the direction of: a. B field. b. H field. c. E field. d. all of the above. Answer: c 2. The diffraction process allows reception beyond a mountain in all but a small area known as: a. dead zone. b. shadow zone. c. masking. d. open space. Answer: b 3. A term used to describe variations in signal strength that occur at a receiver during the time a signal is being received is known as: a. skipping. b. refraction. c. bouncing. d. fading. Answer: d 4. An electronic system performing the reception, frequency translation, and retransmission is called a: a. TRF. b. superheterodyne. c. satellite. d. transponder. Answer: d 5. Another term for groundwave is: a. space wave. b. sky wave. c. surface wave. d. direct wave. Answer: c

152

TEST ITEM FILE - Chapter 13

6. A device that converts electrical energy into electromagnetic energy is known as: a. a light bulb. b. an antenna. c. a transducer. d. all of the above. Answer: d 7. The characteristic impedance of free space is: a. 50 ohms. b. 75 ohms. c. 300 ohms. d. 377 ohms. Answer: d 8. The bending of electromagnetic waves due to its passing through regions of different densities is known as: a. radiation. b. reflection. c. refraction. d. diffraction. Answer: c 9. The type of wave that is limited to line-of-sight transmission distances is: a. ground wave. b. space wave. c. sky wave. Answer: b 10. RFI is:

a. polarization b. reactive c. radiofrequency Interference d. all of the above e. none of the above Answer: c

11. A point in space that radiates equally in all directions is:

a. parabolic reflector b. isotropic Point source c. electromagnetic wave d. an antenna e. none of the above Answer: b

153

TEST ITEM FILE - Chapter 13 12. The term for the case when the same signal arrives at the TV receiver at the two different times is:

a. ghosting b. phasing c. tagging d. pulsing e. reflecting Answer: a

13. The constant temperature stratosphere is also called the:

a. ionosphere b. perisphere c. critical frequency d. sunspot region e. none of the above Answer: e

14. The highest frequency transmitted vertically under given ionosphere conditions that will be returned to

the earth is called the:

a. reflected frequency b. damped frequency c. critical angle d. critical frequency e. none of the above Answer: d

15. Between the point where the ground wave is completely dissipated and the point where the fist sky

wave returns is called:

a. quiet zone b. skip zone c. fading d. a and b e. a and c Answer: d

16. What is used to accommodate severe fading problems?

a. space diversity b. frequency diversity c. angle diversity d. diversity reception e. all of the above

Answer: d

154

TEST ITEM FILE - Chapter 13

17. When a satellite is parked in orbit approximately 22,000 miles of the earth’s surface, it is said to be in:

a. synchronous orbit b. geo-synchronous orbit c. lunar orbit d. a and c e. a and b Answer: e

18. The C-band frequency range is approximately.

a. 1-2 GHz b. 2-4 GHz c. 4-8 GHz d. 8-12 GHz e. 12-18 GHz Answer: c

19. The Ku band frequency range is approximately

a. 1-2 GHz b. 2-4 GHz c. 4-8 GHz d. 8-12 GHz e. 12-18 GHz Answer: d

20. GPS currently uses a constellation of

a. 28 satellites b. 22 satellites c. 18 satellites d. 33 satellites e. 40 satellites Answer: a

21. The term for the situation when two or more broadcasting channels bleed into each other is called.

a. multi-path fading b. diversity reception c. multi-frequency fading d. co-channel interference e. non of the above Answer: d

155

TEST ITEM FILE - Chapter 13

22. At resonance, the impedance of a crystal should be; a. maximum b. minimum c. variable d. a and c e. none of the above Answer: b

23. The typical satellite orbits being used for the satellite radio service are

a. polar b. geostationary c. inclined d. a and b e. b and c f. all of the above

Answer: e 24. The apogee of a satellite orbit is a. is the average distance to earth b. farthest point from earth c. closest point to earth d. none of the above Answer: b 25. The perigee of a satellite orbit is a. is the average distance to earth b. farthest point from earth c. closest point to earth d. none of the above Answer: a 26. An earth station is located at 101oW longitude and 28oN latitude. Determine the azimuth and elevation angles for the earth station if the antenna is to be pointed at a satellite parked at 99oW longitude. a. Azimuth = 175.7o Elevation = 45.8o b. Azimuth = 175.7o Elevation = 57.24o c. Azimuth = 153.5o Elevation = 45.82o d. Azimuth = 155.3o Elevation = 45.81o Answer: b

156

TEST ITEM FILE - Chapter 13 27. What is the distance to a satellite parked at 87o W longitude from an earth station located at 28o N latitude and 111o W longitude. Calculate the round-trip time delay for a signal traveling from the earth station to the satellite and back. a. 41,651.4 m b. 4,1651.4 m c. 41,651.4 km d. 4,1651 km Answer: c 28. Determine the figure of merit (in dB) for a satellite earth station with the following specifications: Antenna Gain – 35 dBi Reflector noise temperature – 22 K LNA noise temp – 42 K Noise Temp (various components) – 2 K a. 16.8 dB b. 15.8 dB c. 10.7 dB d. 18.3 dB Answer: a

29. Define a satellite footprint is (select one) a. a map of the satellite’s coverage area for the transmission back to the satellite b. a map of the earth station’s coverage area for the transmission back to earth

c. a map of the satellite’s coverage area for the transmission back to earth d. refers to the placement of the antennae on the satellite Answer: c

157

TEST ITEM FILE - CHAPTER 14

ANTENNAS 1. For frequencies below 2 MHz, which antenna is used? a. Hertz b. vertical c. dipole d. Helmhotz Answer: b 2. A λ/4 dipole at 500 MHz is used to receive a broadcast. Determine the length of the dipole. a. 1.5 M b. 3 M c. 0.9 M d. 0.6 M Answer: d 3. Which of the following is used to determine antenna bandwidth? a. wavelength b. power c. center frequency d. all of the above Answer: c 4. Which of the following antennas matches well with the 300-Ω input impedance terminals? a. folded dipole antenna b. slot antenna c. ferrite loop antenna d. loop antenna Answer: a

158

TEST ITEM FILE - Chapter 14

5. Which of the following is a directional antenna? a. Yagi-Uda b. dipole c. vertical d. all of the above Answer: d

6. An antenna created by taking a quarter-wavelength section of open wire line and bending

each conductor outward 90 degrees is called: a. a Marconi antenna. b. a dipole antenna. c. a vertical antenna. d. all of the above. Answer: b 7. It is desired to build a dipole to receive a 49 MHz broadcast. The optimum length for the dipole antenna is: a. 5.82 meters. b. 2.91 meters. c. 6.12 meters. d. 3.06 meters. Answer: a 8. Any non-driven element of an antenna array such as a Yagi is called a: a. reflector. b. director. c. lobe. d. parasitic element. Answer: d 9. Feeling the half-wave dipole (Hertz) antenna at the center results in an input impedance that

is purely resistive and equal to:

a. 50Ω b. 73Ω c. 300Ω d. 58Ω Answer: b

159

TEST ITEM FILE - Chapter 14

10. The half-wave dipole antenna has gain relative to an isotropic radiator of

a. 1.67dB b. 3 dB c. 2.15 dB d. 5 dB Answer: c

11. Antenna gain provided with respect to an isotropic radiator is expressed as:

a. dBm b. dBV c. dBW d. dBi Answer: d

12. The antenna gain, given in decibel with respect to a dipole is expressed as:

a. dBd b. dBV c. dBW d. dBi Answer: a

13. An impedance matching device that spreads the transmission line as it approaches

the antenna is called a:

a. wge match b. delta match c. series match d. flared match Answer: b

14. A series inductance used to tone out the capacitive appearance of an antenna is called

a. wading coil b. RF choke c. Condenser d. pole

Answer: a

160

TEST ITEM FILE - Chapter 14

15. A group of antennas or antenna elements arranged to provide the desired directional characteristics is called:

a. alumped element b. a vertical antenna c. a marconi antenna d. an antenna array Answer: d

16. A small-loop antenna has a turn of wire whose dimensions are normally much smaller than:

a. λ/2 b. 2λ c. 0.1λ d. λ Answer: c

17. A direction in space with minimal signal level is called a:

a. space b. mark c. slip d. null Answer: d

18. The characteristic impedance of twin lead is:

a. 600Ω b. 150Ω c. 300Ω d. 75Ω Answer: c

161

TEST ITEM FILE - Chapter 14

19. The model of the dipole antenna used in the Electronics Workbench Multisim simulation used an inductance of 1µH and series capacitor of 2.5pF. The resonant frequency of this antenna model is approximately:

a. 100 MHz b. 90 MHz c. 110 MHz d. 103.9 MHz Answer: a

20. The adjustment of the single stub tuner in the Electronics Workbench Multisim

simulation was accomplish by changed the effective length of the :

a. top leg b. right leg c. long leg d. ground leg Answer: d

162

TEST ITEM FILE - CHAPTER 15

WAVEGUIDES AND RADAR 1. Which of the following factors is not used to determine a mode of energy transmission? a. power level and efficiency b. reliability c. cost and maintenance d. environmental conditions Answer: d 2. The type of wave propagated by a waveguide is: a. quantum. b. electromagnetic. c. acoustic. d. mechanical Answer: b 3. A radar wave has a duty cycle of 0.7%. The pulse width is 4 µs. Determine the pulse repetition time. a. 400 µs b. 150µs c. 286 µs d. 3 ms Answer: c 4. When the target is too close to the transmitter the following is usually detected: a. peak power b. double range echoes c. maximum range echoes d. ambiguous signal Answer: b 5. The range (miles) to a target is 6 miles. What is the time interval between transmission and reception? a. 74 µs b. 40 ns c. 120 µs d. none of the above Answer: a

163

TEST ITEM FILE - Chapter 15 6. The side arm of the shunt tee is used to shunt the E field in which mode? a. TM b. HM c. TE d. TH Answer: c

7. Which of the following is used to change the plane of polarization of the wave? a. H bend b. twist c. E bend d. tees Answer: b 8. The amount of coupling (dB) is determined by: a. guide wavelength, λg. b. type of waveguide. c. type of transmission mode. d. ratio of output power and incident power. Answer: d 9. At a frequency of 1 GHz and a transmitter-receiver distance of 1500 ft., which is the most efficient device for energy transfer? a. transmission lines b. waveguides c. antennas. Answer: b 10. Which is the dominant and most widely used mode of waveguide operation? a. TE-10 b. TE-20 c. TM-11 d. TM-21 e. TE-11 Answer: a 11. One wavelength of waveguide, when compared to one wavelength of free space is: a. larger. b. smaller. c. the same length. Answer: a

164

TEST ITEM FILE - Chapter 15 12. Circular waveguide is advantageous to rectangular waveguide in: a. its cost. b. continuously rotating radar applications. c. its cross-sectional area. d. its size. Answer: b 13. In order to change the direction of propagation in waveguide, what is used? a. an H-bend b. an E-bend c. a twist d. all of the above Answer: d 14. The primary use of slide screw and double-slug tuners is in: a. minimizing VSWR in a waveguide system. b. increasing the signal level of the signal being propagated through the waveguide. c. changing the direction of propagation of the signal in the waveguide. d. minimizing the resistive losses of the waveguide. Answer: a 15. A waveguide that can distinguish between the waves traveling in opposite directions is called: a. a slider screw tuner. b. a flap attenuator. c. a directional coupler. d. a hybrid tee. Answer: c 16. Which is not a method for coupling energy into or out of a waveguide? a. probe b. tee c. loop d. aperture Answer: b 17. The process of having the frequency of a reflected signal shift due to relative motion between the source and reflecting object is known as: a. the Doppler effect. b. radar. c. directional coupling. d. cavity tuning. Answer: a

165

TEST ITEM FILE - Chapter 15 18. The result of a simulation of a wave-guide as viewed with the EWB Multisim Network Analyzer is shown in Figure 15-1. This is an example of:

a. a low-loss waveguide b. a lossy wavguide c. an open circuit d. none of the above Answer: a

Figure 15-1

166

TEST ITEM FILE - Chapter 15 19. The simulation result for a waveguide as viewed with the EWB Multisim Network Analyzer is shown in Figure 15-2. This is an example of:

a. low loss waveguide b. open circuit c. very lossy wave guide d. none of the above Answer: c

20. This term is used to describe the reflection of the radio waves striking the RFID tag and reflecting back to

the transmitter source. a. reflection coefficient b. scattering c. forward reflection d. backscatter Answer: d

Figure 15-2

167

TEST ITEM FILE - Chapter 15

21. The basic reason for including a dual-dipole antenna in an RFID tag is which of the following?

a. the two dipoles of the dual dipole antenna are oriented at 90o angles to each other which means that tag orientation is not critical.

b. the two dipoles of the dual dipole antenna are oriented at 180o angles to each other which means that tag orientation is not critical.

c. the dual-dipole enables decoding of RFID transponder codes. d. RFID tag placement requires multiple antennae. Answer: a

22. The three basic frequency bands used in RFID tags are (select three)

a. 125/134 kHz b. 88-108 MHz c. 13.56 MHz d. 860-960 MHZ e. 4.2 GHz

Answers: a,c,d

23. What are the three classifications for powering RFIP tags? (select three)

a. backscatter b. active c. passive d. semi-passive e. semi-active Answers: b, c, e

24. The air interface adopted for RFID tags is

a. Ethernet b. WiMAX c. Bluetooth d. Slotted Aloha e. 802.11g

Answer: d

168

TEST ITEM FILE - CHAPTER 16

MICROWAVES AND LASERS 1. The purpose of the low-loss dielectric material (radome) used to cover the dish is: a. for protection. b. for maintenance of internal pressure. c. to improve performance. d. a and b Answer: d 2. An apparatus whose theory of operation is based upon motion of electrons under the influence of combined E field and H field is known as: a. Gunn oscillator. b. Magnetron oscillator. c. TWT oscillator. d. Klystron oscillator. Answer: b 3. Which of the following is also known as a backward-wave oscillator (BWO)? a. Gunn oscillator. b. Magnetron oscillator. c. TWT oscillator. d. Klystron oscillator. Answer: c 4. Which of the following is similar to an IMPATT diode? a. tunnel diode b. TRAPATT diode c. Baritt diode d. Gunn diode Answer: b 5. Above 5 GHz, which device offers greater performance? a. GaAs FETs b. bipolars c. silicon FETs d. none of the above Answer: a

169

TEST ITEM FILE - Chapter 16

6. Which of the following is not an application of a ferrite? a. Faraday rotation b. isolator c. attenuator d. oscillator Answer: d 7. Degenerate mode parametric amplifiers operate with pump frequencies: a. equal to zero. b. other than two times signal frequency. c. two times the signal frequency. d. same as signal frequency. Answer: c 8. A transphasor is: a. an optical switch. b. a fiber optic amplifier. c. a type of maser. d. all of the above. Answer: b

9. Why are microwave antennas characterized by high gain and high directivity? a. because of the short wavelengths involved, the physical sizes required are small enough for designs not practical at lower frequencies. b. because there is no need for omnidirectional antennas since all microwave links are point-to-point. c. since device noise is greater at these frequencies, receivers require larger input signals. d. microwave transmitters are very limited in their output power. e. all of the above. Answer: e 10. Calculate the power gain of a microwave dish antenna with a 6 meter mouth diameter when used at 5 GHz. a. 27.8 dB b. 47.8 dB c. 95.6 dB d. 55.6 dB Answer: b

170

TEST ITEM FILE - Chapter 16 11. Which of the following is not a vacuum tube? a. traveling wave tube (TWT) oscillator b. Gunn oscillator c. Klystron oscillator d. Magnetron oscillator e. all of the above Answer: b 12. Which microwave oscillator is widely used in microwave ovens? a. a traveling wave tube oscillator (TWT) b. Gunn oscillator c. Klystron oscillator d. Magnetron oscillator Answer: d 13. Which microwave oscillator is becoming obsolete due to its large size and the complex, costly sources of dc required for its operation? a. a traveling wave tube oscillator (TWT) b. Gunn oscillator c. Klystron oscillator d. Magnetron oscillator Answer: c 14. A microwave semiconductor that is typically used as T/R switches in RF transceivers operating above 100 MHz is the: a. GaAs FET transistor. b. P-I-N diode. c. IMPATT diode. d. Gunn diode. Answer: b 15. Special ferrites that are typically used as RF chokes are called: a. yigs. b. beads. c. isolators. d. circulators. Answer: b

171

TEST ITEM FILE - Chapter 16

16. Parametric amplifiers that have pump frequencies that are double the signal frequency are operating in the: a. degenerate mode. b. nondegenerate mode. c. superconductive mode. d. quantum mode Answer: a 17. What material is most useful in the design of masers for low-noise microwave amplification? a. Gallium arsendide b. ruby c. ammonia d. manganese Answer: b 18. The model of a RF capacitor shows that the capacitor is:

a. resistive b. capacitive c. inductive d. all of the above Answer: d

19. Determine the resonant frequency of the dipole antenna model shown in figure 16-1.

a. 100 MHz b. 2.14 GHz c. 1.071 GHz d. 2.25 GHz Answer: c

Figure 16-1

172

TEST ITEM FILE – Chapter 17

TELEVISION 1. Charge coupled devices (CCD's) are typically found in which of the following camera

applications? a. camcorder cameras b. security monitors c. video phones d. all of the above Answer: d 2. The interval of time it takes the moving electron beam to move from the end of one line back to the start of the next lower line is known as: a. scanning interval. b. retrace interval. c. raster interval. d. vector interval. Answer: b 3. Which of the following frame frequencies is adequate to prevent flicker? a. 60 per second b. 30 per second c. 20 per second d. 15 per second Answer: a 4. Which of the following is a function of the sync separator? a. integrates the signal b. shapes the signal c. clips the signal d. filters the harmonics Answer: c 5. Which of the following is found in digital TV? a. D/A converter b. adaptive scanning conversion c. A/D converter d. all of the above Answer: d

173

TEST ITEM FILE – Chapter 17

6. Television had its beginnings with: a. development in the 1930's and broadcasting beginning in the 1940's. b. development in the 1920's and broadcasting beginning in the 1950's. c. development in the 1920's and broadcasting beginning in the 1940's. d. development in the 1930's and broadcasting beginning in the 1950's. Answer: c 7. The device that feeds both visual and aural TV transmitter signals to the antenna while not

allowing either to be fed back into the other transmitter is called: a. a synchronizer. b. a vidicon. c. a diplexer. d. an image orthicon. Answer: c 8. The most widely used TV camera imaging device today is: a. the CCD. b. the vidicon. c. the diplexer. d. the iconoscope. Answer: a 9. The aspect ratio for NTSC broadcast television is: a. a 4:3 ratio. b. a 7:9 ratio c. a 6:9 ration d. all of the above. Answer: a 10. For U.S. NTSC TV broadcasts, the total number of visible scanning lines is set at: a. 525 lines. b. 339 lines. c. 485 lines. d. 40 lines. Answer: c 11. The field frequency for U.S. NTSC television broadcasts is: a. 30 fields per second. b. 40 fields per second. c. 60 fields per second. d. 100 fields per second. Answer: c

174

TEST ITEM FILE – Chapter 17

12. The horizontal oscillator frequency for U.S. NTSC television broadcasts is set at approximately: a. 3.58 MHz. b. 30 Hz. c. 60 Hz. d. 15.75 kHz. Answer: d 13. The color burst frequency for U.S. NTSC television broadcasts is approximately: a. 3.58 MHz. b. 30 Hz. c. 60 Hz. d. 15.75 kHz. Answer: a 14. The type of synchronizing pulse that has a 10 microsecond pulse width with an eight cycle

sinewave burst at 3.58 MHz on its "back porch" is called: a. the color synchronizing pulse. b. the horizontal retrace pulse. c. the vertical retrace pulse. d. the equalizing pulse. Answer: b 15. Vertical resolution for U.S. NTSC television is approximately: a. 428 horizontal lines. b. 428 vertical lines. c. 339 horizontal lines. d. 339 vertical lines. Answer: c 16. Horizontal resolution for U.S. television is approximately: a. 428 horizontal lines. b. 428 vertical lines. c. 339 horizontal lines. d. 339 vertical lines. Answer: b 17. Channel 9 on U.S. television extends from 186 to 192 MHz. The channel 9 sound carrier

is approximately: a. 186.5 MHz. b. 187.25 MHz. c. 191.25 MHz. d. 191.75 MHz. Answer: d

175

TEST ITEM FILE – Chapter 17 18. The high voltage output is derived from the low voltage power supply using: a. the low voltage power supply transformer. b. the yoke coil. c. the flyback transformer. d. the vertical oscillator. Answer: c 19. Most television IF amplifiers are stagger tuned in order to: a. get a very narrow bandwidth with high Q. b. get a wide enough bandwidth for the TV channel but still have steep roll-offs at the

passband edges. c. receive both VHF and UHF TV channels. d. receive both high-band and low-band VHF TV channels. Answer: b 20. To obtain the steep attenuation curve for the sound carrier portion of the IF signal passed

by the TV receiver IF amplifiers, what is used? a. SAW filters b. ceramic filters c. mechanical filter d. wave traps Answer: d 21. In a television receiver, in order to produce sufficient contrast in the display of the picture

tube, the video amplifier stages must have a minimum frequency response of: a. 15.75 kHz. b. 4 MHz. c. 6 MHz. d. 45 MHz. Answer: b 22. By generating the color information at exactly the right color subcarrier frequency

(approximately 3.58 MHz) it is possible to produce color frequency components sandwiched in between the frequency components of the black and white video signal. This technique is known as:

a. interlaced scanning. b. interleaving. c. convergence. d. shadow masking. Answer: b

176

TEST ITEM FILE – Chapter 17

23. Which is not a color of a beam on a color picture tube? a. yellow b. blue c. green d. red Answer: a 24. Which of the following is not an improvement made in television since 1980? a. an increase in the number of UHF TV channels b. Zenith/dbx enhanced audio c. separate audio program (SAP) d. high definition TV enhanced video Answer: a 25. A faulty TV receiver having symptoms of no picture, no sound, and no raster must have

a problem in the: a. horizontal or vertical oscillator or high voltage power supply. b. main power supply. c. video amplifiers following the sound takeoff. d. RF, IF, or video amps prior to sound takeoff. Answer: b

26. A faulty TV receiver having symptoms of normal sound but no raster must have a

problem in the: a. horizontal or vertical oscillator or high voltage power supply. b. main power supply. c. video amplifiers following the sound take-off. d. RF, IF, or video amplifiers prior to sound take-off. Answer: a 27. A faulty TV receiver having symptoms of normal raster, but distorted sound and picture

must have a problem in the: a. horizontal or vertical oscillator or high voltage power supply. b. main power supply. c. video amplifiers following the sound take-off. d. RF, IF, or video amplifiers prior to sound take-off. Answer: d

177

TEST ITEM FILE – Chapter 17

28. The DTV standard based on the recommendations of the ATSC provides for the transmission of television programs in the:

a. HDTV screen format 16 × 9 b. SDTV screen format c. Multiple SDTV programs d. All of the above Answer: d

29. The video compression technique selected for DTV transmission is:

a. JPEG b. MPEG2 c. GIF d. BMP Answer: b

30. The digital audio compression technique specified for digital television is:

a. AC-3 b. MPEG2 c. NTSC d. 8VSB Answer: a

31. The RF modulation technique for ATSC DTV transmission is:

a. 8QAM b. 8VSB c. 8QPSK d. 8FSK Answer: b

32. The commercial name for the AC-3 audio standard is:

a. 5.1 channel input b. 4:4:2 c. ATSC d. none of the above Answer: a

33. The ATSC pilot carrier provides:

a. stereo indicator b. second channel programming c. frame sync d. a clock reference Answer: d

34. The 8VSB exciter inserts what to aid the 8VSB receiver in accurately recovering the transmitted data?

a. pilot carrier b. frame sync c. segment sync d. all of the above

Answer: d

178

TEST ITEM FILE – Chapter 17

35. The term for when a digital picture freezes even when there is motion is: a. freeze frame b. pixelate c. stop motion d. 8VSB Answer: b

36. The ATSC Data Transport rate is which of the following? a. 19.39kbps b. 19.39Mbps c. 3.579545 MHz d. 4.2 Mbps Answer: b 37. The purpose of the PSIP is to (select one). a. it is used to select mono and stereo audio b. it is used for closed captioning c. it is used to identify NTSC feeds d. it is used to identify the channel

Answer: d

38. This is used by the digital receiver to lock onto the digital data stream so the multiplexed data can be recovered. a. the DTV pilot signal b. the GPS data stream c. the SONET data stream d. none of the above Answer: a 39. The point where there is a lot of multiple echo and you have reached the threshold of visibility in a DTV signal is called a. cliff effect b. the receiver limit function c. the receiver gain function d. figure of merit Answer: a

40. What three things make up the ATSC transport stream?

a. Video data stream (MPE62 encoding) b. The 3.579545 MHz color subcarrier c. Audio data stream (AC-3 encoding) d. PSIP, program, Information, control, ect

Answer: a,c,d

179

TEST ITEM FILE - CHAPTER 18

FIBER OPTICS

1. The trapping of light in a fiber results from: a. total internal reflection. b. numerical aperture. c. dispersion. d. step index. Answer: a 2. The amount of bending of the light into a material depends on the __________ of that material: a. critical angle b. reflective index c. refractive index d. wavelength of light Answer: c 3. An optical fiber and its cladding have refractive indexes of 1.535 and 1.386 respectively. Calculate NA. a. 0.463 b. 0.897 c. 0.66 d. 0.369 Answer: c 4. Which of the following dispersions is not exhibited in single-mode fiber? a. modal b. material c. waveguide d. zero Answer: a

5. Which of the following is an important characteristic of light detectors? a. responsivity b. dark current c. response speed d. all of the above Answer: d

180

TEST ITEM FILE - Chapter 18

6. Determine the maximum fiber length if the maximum bit rate is 56 Kbps and the dispersion is 6 ns/km. a. 200 Km b. 10 Km c. 595 Km d. 0.000595 Km Answer: c 7. The wavelength of red light is: a. 680 nm. b. 520 nm. c. 430 nm. d. 1600 nm. Answer: a

8. The numerical aperture of a fiber: a. defines the half-angle of the cone of acceptance for propagated light in the fiber. b. is a basic specification which indicates its ability to accept light and shows how much light can be off axis and still be propagated. c. the acceptance angle where the output light is no greater than 10 db down from the peak value. d. all of the above. Answer: d 9. Which is not a major criteria in selecting a specific type of fiber optic cable? a. signal losses b. refractive index c. ease of light coupling and interconnection d. bandwidth Answer: b 10. Pulse dispersion is mainly caused by: a. variable path lengths due to the light signal traveling down the fiber in different modes. b. noise picked up by the light signal. c. excessive use of repeaters through the fiber optic cable length. d. use of too high power, highly directional laser sources to generate the light signal. Answer: a 11. The diode laser is preferred to the light-emitting diode in systems that have: a. long distance paths. b. low error rate requirements. c. moderate band to wideband requirements. d. predictable failure rates. Answer: c

181

TEST ITEM FILE - Chapter 18

12. An estimation of how long a fiber optic cable can be for a given bit rate and dispersion for successful data communications to occur is referred to as: a. a power budget. b. an error rate analysis. c. a frequency response. d. a detector system check. Answer: a 13. The material surrounding the core of the optical waveguide is called:

a. cladding b. padding c. strapping d. glasing Answer: a

14. A term used to describe the transmission of data over hundreds or thousands of miles.

a. short haul b. medium haul c. long haul d. haul Answer: c

15. Fiber cables with core diameters of about 7 to 10 µm.

a. mode field b. single mode c. multi mode d. step-index Answer: b

16. The wave length at which the material and waveguide dispersion cancel one another is called:

a. zero-dispersion wavelength b. mode-field diameter c. critical wave length d. none of the above Answer: a

17. The core size of multimode fiber is approximately:

a. 7 to 10 µm b. 25 to 35 nm c. 50 to 100 nm d. 50 to 100 µm Answer: d

182

TEST ITEM FILE - Chapter 18

18. Which of the following are typical wavelengths used in fiber optics?

a. 850 nm b. 1310 nm c. 1550 nm d. a and c e. all of the above Answer: e

19. Loss due to the light breaking up and escaping through the cladding.

a. micro bending b. absorption c. macrobending d. dispersion Answer: c

20. Loss caused by very small mechanical deflections and stress on the fiber.

a. micro bending b. absorption c. macro bending d. dispersion Answer: a

21. The attenuation of an optical signal is due to:

a. scattering b. absorption c. macro bending d. all of the above e. none of the above Answer: d

22. The broadening of an optical pulse due to different path lengths taken through the fiber by different

modes is called:

a. chromatic dispersion b. polarization mode dispersion c. modal dispersion d. all of the above Answer: c

183

TEST ITEM FILE - Chapter 18

23. The broadening of an optical pulse due to different propagation velocities of the spectral components

of the light pulse is called:

a. chromatic dispersion b. polarization mode dispersion c. modal dispersion d. all of the above Answer: a

24. A dispersion compensating fiber

a. acts like an equalizer b. cancels dispersion effects c. yields close to zero dispersion in the 1550 nm region d. all of the above Answer : d

25. A coherent light source is one that is:

a. easy to understand b. spectrally pure c. easy to decode d. due to a defective laser Answer: b

26. The term for incorporating the propagation of several wavelengths in the 1550nm range of a single fiber is called:

a. multiplexed fiber b. fiber bragg gating c. distributed feedback d. dense wave length division multiplex Answer: d

27. Common synonymous terms for a fiber optic strand are: (circle all that are correct)

a. fiber b. light pipe c. glass d. conduit Answer: a, b, c

184

TEST ITEM FILE - Chapter 18

28. RSL in regards to an optical receiver is:

a. radio signal level b. radiometry signal level c. received signal level d. none of the above Answer: c

29. A long-term method where two fibers are fused or welded together is called:

a. fusion splicing b. mechanical splicing c. core alignment d. none of the above Answer: a

30. A jelly like substance that has an index of refraction much closer to the glass than air is called:

a. framing gel b. alignment gel c. index matching gel d. splice gel Answer: c

Use Figure 18-1 to answer questions 31-33

Figure 18-1

185

TEST ITEM FILE - Chapter 18

31. The transmit power is:

a. –15 dBm b. –42 dBm c. –27 dBm d. –31.6 dBm Answer: a

32. The receiver dynamic range is:

a. –31.6 to –37.6 dBm b. –37.6 to –42 dBm c. –15 to –31.6 dBm d. –27 to –42 dBm Answer: d

33. This example is showing a length of approximately_____ for the fiber optic run:

a. 30 km b. 2.8 km c. 30 m d. 28 m Answer: a

34. A piece of test equipment that sends a light pulse down the fiber and measures the reflected light is called:

a. an OTDR b. reflecometer c. light pulser d. TDR Answer: a

Use Figure 18-2 to answer questions 35 - 39.

a. Point A is called the

a. Splice b. dead zone c. noise d. jumpers and patch panel connections Answer: b

36. Point B is

a. splice b. dead-zone c. noise d. jumper or patch panel connection Answer: a

186

TEST ITEM FILE - Chapter 18

37. Point D is a

a. splice b. dead-zone c. noise d. jumper or patch panel connection

Answer: a 38. Points F and G are:

a. splice b. end of the fiber c. noise d. jumpers and patch panel connections Answer: d

39. Point H is:

a. splice b. end of the fiber c. noise d. a jumper or patch panel connection Answer: b

Figure 18-2

187

TEST ITEM FILE - Chapter 18

40. The protocol standard for optical transmission in long-haul communication is:

a. FDDI b. STS c. SONET d. OC Answer: c

41. 1000Base LX is:

a. FDDI b. OC-3 c. FTTH d. Gigabit Ethernet Answer: d

Answers to Problems - Chapter 1

188

1. Process of putting information onto a high-frequency carrier for transmission.

2. The frequency that is used to “carry” the intelligence.

3. to provide separate carriers for different intelligencethe difficulty of transmitting signal at very low frequencies

4. amplitude, frequency, and phase

5. MF - 300kHz to 3MHzHF - 3 to 3- MHzVHF - 30 to 300MHzUHF - 300 MHz to 3GHzSHF - 3 to 30 GHz

6.

7.

8.

9.

10.

Answers to Problems - Chapter 1

189

11.

12.

13.

14.

15. any undesired voltages or currents that end up appearing in a circuit

16. external noise - noise in a received radio signal that has been introduced by thetransmitting mediuminternal noise - noise in a radio signal introduced by the receiver

17. man-made noise, atmospheric noise, space noise

18.

19.

Answers to Problems - Chapter 1

190

20. low-noise resistor - a resistor that exhibits low levels of thermal noise

21. 128.7 uV/1 Mohm equals approximately 129 pico ampsThe noise current increases with the increase in temperature

22.

23.

24.

25.

Answers to Problems - Chapter 1

191

26.

27.

28.

29.

30. The ratio of the signal + noise + distortion power out to the output signal + noise power,

Answers to Problems - Chapter 1

192

31.

32.

33.

34. A diode is used to generate a reference noise. The signal level of the device under test isthen matched to the diode's level. Equation 1-19 is then used to obtain the noise ratio.

35. DUT - Device Under Test

36. the noise signal is connected to both channels of a dual trace oscilloscope. The verticalposition is adjusted until the dark band between the traces disappears. The resultingseparation in the signals (according to the scope channel settings) is twice the rms noise.

37. information theory - concerned with the optimization of transmitted information

38. Hartley's Law - information that can be transmitted is proportional to the product of the bandwidth times the time of transmission, This means that the greater the bandwidth the more information that can be transmitted.

39. harmonic - it is a multiple of the fundamental frequency

40.

41. the square wave contains many harmonic frequencies, the sine wave justcontains the fundamental frequency.

42. use Figure 1-12 as a reference. The period and the harmonics will change because of the 2kHz sine wave signal.

Answers to Problems - Chapter 1

193

43. Fourier Analysis - method of representing complex repetitive waveforms by

sinusoidal components

44. Refer to Figure 1-13

45.

.

46.

47.

48. Inductors - store energy in the surrounding magnetic field capacitor - stores energy between the plates(Q) quality factor - ratio of the energy stored to the energy lost

49. resonance - when XC = XL

Answers to Problems - Chapter 1

194

50.

51.

52.

53. At 4KHz , e out ~ 0.96V at 6KHz e out ~ 0.8V

Answers to Problems - Chapter 1

195

54. see Fig. 1-17

55. as Q increases the filter becomes more selectivethe limiting factor is the resistance factor (see equation 1-21)

56.

57.

58.

59.

Answers to Problems - Chapter 1

196

60. constant-k filters - capacitance and inductive reactive resistance made equal to some constant k.

m-derived filters - use a tuned circuit in the filter to provide nearlyinfinite attenuation at a specific frequency

61. above 100 MHz - use LCbelow 100 MHz - use RC

62. at high frequencies the leads exhibit stray inductance and capacitance

63. number of RC or LC sections in a filterNote: There is also a mathematical relationship for filters that uses the

poles term in filter circuits.

64. they use a constant-k value for the inductive and capacitive reactance values

65. see Figures 1-23 and 1-24.

66. The capacitor in series with the inductor swamps out the transistor's internalcapacitances thereby negating transistor variation and increasing stability.

67. A Pierce crystal oscillator is provided in Fig. 1-27. The crystal has a fixed resonant frequency.

68.

69. Symptoms, Signal Tracing and Injection, Voltage and Resistance MeasurementsSubstitution or check the AC connection, check fuses, check input/output connections, ask yourself if you forgot something also, check visual inspection and all of your senses,check power supplies, verify input and output signals

70. Good components can get damaged in the substitution process.

Answers to Problems - Chapter 1

197

71. Voltages will cause incorrect resistance measurements. The power must be turned off and the component isolated.

72. complete failures, intermittent faults, poor system performance, induced failures

73. When tracing through a multiple stage circuit for a fault.

74. a distorted sine wave or a flat line

75. at series resonance the crystal should have a low resistanceat parallel resonance the crystal will have a high resistance

76. the answer should include at least two of the following issues- limitation of the electronics (eg. noise)- the bandwidth limitation of the communications channel

77.

Answers to Problems - Chapter 1

198

78. Noise temperature measurements are more common and these are based more on the system rather than the device. 79. This is verified when an oscillator won’t maintain oscillations. The level of positive

feedback might need to be adjusted for proper operation.

Answers to Problems - Chapter 2

199

1. dc. 2kHz, 1498kHz, 1500 kHz, 1502kHz, and harmonics of the last three frequencies

2. 1498, 1500, and 1502kHz

3. The non-linear mixing of the carrier and intelligence frequencies.

4. The combination of the upper and lower sideband frequencies.

5. sideband - collections of all frequencies in that bandside frequency - this just refers to one frequency

6. the modulation is at 100%, usb = lsb

7. the carrier phasor represents the peak value of the sine waveone full revolution represents 360 degrees. (see section 2-2 for more details.)

8. refer to figure 2-7

9. refer to Figure 2-8

10. sideband splatter

11.

12.

13.

14. The is shown in the first part of section 2-4

15.

Answers to Problems - Chapter 2

200

I Im

t c= +⎛⎝⎜

⎞⎠⎟1

2

212

6 7 6 2 12

212

. .= +⎛

⎝⎜

⎞

⎠⎟

m

P / P 1 12

1.5t c

2= +⎛

⎝⎜

⎞

⎠⎟ =

16.

17.

P Pm

t c= +⎛⎝⎜

⎞⎠⎟1

2

2

500 10 75

2

2

= +⎛⎝⎜

⎞⎠⎟Pc

.

18.

19. A high percentage ofmodulation isimportant so that thepower to thesidebands

is sufficient for a qualitysignal.

20.P P

mt c= +

⎛

⎝⎜⎜

⎞

⎠⎟⎟1

2

2

Check :

Answers to Problems - Chapter 2

201

P P kWt c= = +⎛⎝⎜

⎞⎠⎟ = +

⎛⎝⎜

⎞⎠⎟ =1

082

1 108

2

2 2. .1.32kW

21.

22. From Table 1-2c, the peak amplitude of each harmonic is

23. base modulation (Fig. 2-10)collector modulator (Fig.2-13)

Answers to Problems - Chapter 2

202

24. low-level modulation - in an AM transmitter, the intelligence is superimposedon the carrier and the modulated waveform is amplified before reaching the antenna.

25. high-level modulation - in an AM transmitter, the intelligence is superimposedon the carrier at the last point before the antenna.

26. high-level modulation is preferred for high-power transmissionlow-level modulation is less expensive and used for low-power applications

27. neutralizing capacitor - a capacitor that cancels feedback signals to suppress self-oscillation

28. at tuned frequencies parasitic oscillations prevent amplificationself-oscillation is a problem for all RF amplifiers; it can at the tuned frequencyor at a higher frequency.

29. parasitic oscillations are undesirable high-frequency oscillations

30. the feedback signal is in phase with the signal being amplified

31. see Fig. 2-13

32. the advantage of class C modulation is efficiency

33. PIN diodes are used at very high frequencies. Above 100 MHz and when forward biased the diodes act as variable resistors (refer to Figure 2-15). The increasein the modulating signal level increases the level of forward biasing, thusreducing the diode resistance and increasing the output carrier amplitude.

34. the crystal maintains the high accuracy of the carrier frequency requiredby the FCC.

35. see Figure 2-18

36. the buffer amplifier provides a high impedance load for the oscillator to minimize drift. It also provides enough gain to sufficiently drive the modulation amplifier.

37. The details on the amplifier shown in Figure 2-19 are discussed in the Citizen'sband Transmitter section in 2-6 AM Transmitter Systems.

38. An example of antenna coupling is provided in Fig. 2-19 using the coupling networks L3, L4, and the 250- and 150-pF capacitors.

39. The antenna coupler is necessary so that the transmitter output impedanceis matched to the antenna.

Answers to Problems - Chapter 2

203

40.

41. The tune-up procedure is used to place the transmitter on the air. The tuningcoils are typically adjusted to about 1/2 of their values. The coils arethen adjusted so that reliable oscillation is achieved and maximum power is obtained.

42. see Fig. 2-23(c)

43. The trapezoidal measurement allows easy observation of over-modulation, under-modulation,and improper phase relationships.

44. This is showing the RF spectrum of an AM signal. The carrier frequencyis 960 kHz and the modulating frequency is 2kHz. The percentage of modulation is 100% (refer to equation 2-6). The carrier voltage is 1 Volt.

45. spurs are undesired frequency components of a signal usually having to do withthe harmonics.

46.

Answers to Problems - Chapter 2

204

The first two spurs are ± 2.3 divisions from the carrier. Thus, 2.3 divisions × 5 kHz=11.5 kHzˆ50.0034 MHz ± 11.5 kHz = 50.0149 MHz and 49.9919 MHz

The second two are ±4.6 divisions from the carrier.Thus, ± 23 kHz is 50.0264 MHz and 49.9804 MHz

Power of the ± 11.5 kHz spurs is down 2.6 divisions form the carrier.

2.6 div. × 10 dB/div. = 26 dB

Similarly P of ± 23 kHz spurs •1 mW

THD without V6 is 7.348% or 0.07348with V6 is 7.416% or 0.07416

47. The carrier is 5.4 divisions above the -20 dBm noise floor. At 10 dB per division, carrier power is 54 dB above the noise floor or +34 dBm

48. The dummy antenna is used for testing the transmitter off-air. The dummy antenna isbasically a resistive load that absorbs the radiating signal.

49.

50.

Answers to Problems - Chapter 2

205

51. The purpose of this is to determine if there are any obvious reasons whythe transmitter is experiencing problems. Examples of things to look for includebroken wires, loose connections, burned up components or exploded capacitors.

52. A "hot check" is accomplished by turning on the transmitter and letting it operate a few hours off-line. This warm-up time is used to detect any additionalproblems.

53. The capacitor C1 bypasses the resistor at RF frequencies. It also serves to filterthe RF pulses and develop a DC voltage across R1.

54. The danger with high voltages is they can kill you!!! Do not enter the transmitterwithout removing power, make sure all components are discharged, use a ground-rodto verify that all connections do not have a voltage on them.

55. The dummy load replaces the antenna during testing. The dummy load impedancemust match the transmitter for accurate testing.

56. Verify the input drive signal to the transistor Q1. Verify the DC power supplyvoltage. Look for evidence of overheating of the transistor or any discolorizationof any attached components. The transistor Q1 may have to be removed to verifyit is good.

57. There will be no output since C1 will charge to the negative peak of the RF drivebecause of the rectifier action of the base-emitter junction.

58. The spectrum analyzer enables you to see the entire frequency spectrum of thetransmitted signal. This includes making sure the transmitted signal is on theproper frequency, within the proper frequency spectrum, harmonics are minimized,and the signal falls within FCC specifications.

59. L2 and Ct are used to tune the output of the RF amplifier to the antenna. The lossof L2 (short) will result in a mis-tuned output stage. Adjusting Ct will not help.The same signal will appear on both sides of the inductor.

60. In high power transmitters, burned-up or exploded components are most easilyidentified. There may also be heavily discolored components. Sometimes loose connectionsor broken wires are observed. The cooling elements (eg. fans) can also fail andthe obvious change is the sound and probably the presence of a strong burningsmell.

61. The linear combination of the two signals will not produce the desired carrierand side-frequencies.

62. The upper and lower side-frequencies should be mirror images of each other.

63. a.

Answers to Problems - Chapter 2

206

b.

c. at m = 0.2

P Pm

t c= = +⎛⎝⎜

⎞⎠⎟ = +

⎛⎝⎜

⎞⎠⎟ =1

210 1

0 22

2 2.10.2 W

at m = 0.9

P W Wt = +⎛⎝⎜

⎞⎠⎟ =10 1

0 92

14 052.

.

Answers to Problems - Chapter 2

207

d.

64. An oscilloscope is primarily used to view the time domain components; the spectrum analyzer allows thefrequency spectrum to be observed.

Answers to Problems - Chapter 3

208

At 550 kHz

At 1550 kHz

At 550kHz

1. see Figure 13-1

2. sensitivity - the minimum input RF signal to a receiver required to produce a specified audio signal at the output.

selectivity - the extent to which a receiver can differentiate between the desired signal and other signals.

These are important because they determine the requirements for the input level. They also determine how much the input signal is above the noise floorand how well the desired signal is selected.

typical units for sensitivity range in the millivolts to nanovolts for moresophisticated receivers

3. a receiver that is overly selective can result in a lack of fidelity becausepart of the intelligence is not included (filtered out)

4.

5. see Figure 3-3 and the text discussion for 3-3 (a) - (g)

Answers to Problems - Chapter 3

209

dc1.1 MHz1.1 MHz ±2kHz1.555 MHz2.655 MHz2.655 MHz ±2kHz455 kHz455 kHz ±2kHz

A accepted by IF ampand harmonics of theabove frequencies

6. Advantages- handle high power- acceptable distortion levels- efficient- provide a usable dc voltage for the AGC circuits

Disadvantages- reduced Q and selectivity- no amplification

7. Diagonal clipping - the output waveform does not follow the input (see Figure 3-4)

8. Diagonal clipping may exist if the modulated carrier exhibits a large modulationindex (change from 0 to 100%)

9. Diagonal clipping can happen if the RC time constant is too long or too short.

10. Synchronous detection- low distortion- ability to follow fast modulation waveforms- ability to provide gain

The received modulated signal is mixed with a local oscillator. The resultis the extracted intelligence and higher frequencies that are removed with filters.

11. See Figure 3-6

13.

14. first detector - the mixer stage in a superheterodyne receiver that mixes

Answers to Problems - Chapter 3

210

the RF signal with a local oscillator signal to form the IF signal.

15. The key is to make the LO frequency track with the circuits that are tuningin the incoming radio signal so that the difference is a constant frequency,the IF.

16. refer to the discussion at the beginning of section 3-4

17. An example is provided in Figure 3-12.

18.

Answers to Problems - Chapter 3

211

The intermediate frequency is 300 kHz; ˆ a 1600 kHz signal mixed with the 1300 kHz (localoscillator) would also yield an output equal to the IF (1600 kHz -1300 kHz = 300 kHz). Therefore, the image frequency is 1600 kHz.

When tuned to 20 MHz:LO = 10.7 MHz +20 MHz = 30.7 MHzImage frequency = 30.7 MHz + 10.7 MHz = 41.4 MHz

When tuned to 30 MHz:LO = 10.7 MHz + 30 MHz = 40.7 MHzImage frequency = 40.7 MHz +10.7 MHz = 51.4 MHz

Therefore, LO range = 30.7 to 40.7 MHzImage frequency range = 41.4 MHz to 51.4 MHz

19. An example is provided in Figure 3-11. The resonant frequency is approximately629 kHz. A plot of the resonant frequency for the given L and C are provided.

20.

21.

22. See Figure 3-14

23. The tuned circuit attenuates the possible image frequencies.

Answers to Problems - Chapter 3

212

The output is a constant 1.4 V rms

24. This depends on the receiver but in the case of a multistage RF section the defectivesection can be bypassed keeping in mind that the receiver selectivity can beeffected.

25. self oscillation is minimized and neutralization is not required.

26. the mixers are referred to as converters and first detectors generating the sumand difference frequencies generating the output IF frequency.

27. autodyne (self-excited mixer) - a single stage creates the LO signal and mixes itwith the applied RF signal to form the IF signal.

28. The IF amplifiers operate within a fixed bandwidth thereby providing betterselectivity and gain control.

29. The audio levels vary widely due to variations in received RF signal level.

30. An example is provided in Figure 3-19.

31. The voltage gain of a transistor amplifier is nearly directly proportionalto the dc bias current.

32.

Answers to Problems - Chapter 3

213

Image frequency is 1 MHz + (2 × 455 kHz) = 1.91 MHz

33 it provides additional gain control for strong signals

34. Not all tuning circuits can be implemented on an IC.

35.

36.

37. Dynamic range - the dB difference between the largest tolerable receiver inputlevel and its sensitivity level.

38. see Fig. 3-25. A standard for AM stereo is not established. Also, AM is still susceptible to environmental noise so it will never compete with FM for accepted quality.

39. Refer to Figure 3-24. The carrier is phase-shifted to carry the left and right channelsin AM stereo.

40. Possible problem - open winding in coil L1. A multi-meter can be used as shownin Figure 3-27.

41. The oscillator is not working probably due to a failure with Q1, L4, or C4.

42. No voltage will be detected at the emitter of Q1 with a DMM or an oscilloscope.

Answers to Problems - Chapter 3

214

When f received is minimum (4 MHz), the tuning capacitors are maximum. Thus, assume C=325 pF for LOand RF stages when receiving a 4 MHz signal.

For the RF stage,

For local oscillator, f=5.8 MHz (4 MHz +1.8 MHz)

43. Possible problem with L5 or C5.

44. Verify the output voltage, make sure that the AC voltage is connected, visually inspectfor burned or damaged components, verify the signals (voltages) at points A, B, C, andD.

45. V = [12x470]/[330+470] = 7.05

46. There will be a significant amount of ripple voltage on the output.

47. Verify powers supplies, visually inspect, verify the input to the detector, and checkfor an output from the detector stage.

48. A quick test is to touch a screwdriver or a piece of wire to the center terminalof the volume control. If the amplifier is working you should hear a loud 60Hz hum.

49. The receiver must provide for a minimum received signal level to guarantee thatthe output will meet the desired specifications.

50. Assuming that the AM signal is modulated with some low frequency intelligence then the intelligence (in the IF) can be recovered but if the AM signal contains only a high frequency carrier then only the unmodulated IF signal will be recovered.

51. Superheterodyne receivers provide constant selectivity over a wide range of receivedfrequencies. The bulk of the amplification takes place in the IF stages at a fixedfrequency enabling the use of frequency selective circuits.

52.

Answers to Problems - Chapter 3

215

At 10 MHz, the inductors stay the same, but for the RF

For the local oscillator,

Answer to Problems - Chapter 4

216

1.

2.

3. PEP - peak envelope power. This rating only occurs occasionally intime with voice transmission while a sinusoid transmission is constant.

4. ACSSB - Amplitude Compandored Single Sideband, includes a pilot carrier, speech signal is compressed at the transmitter and expanded at the receiver

SSB - Single Sideband, only one sideband is transmitted

SSBSC - the carrier and one sideband are suppressed

ISB - Independent Sideband, another name for twin-sideband suppressed carriertransmission. This involves the transmission of two independent sidebands.

5. Advantages of SSB over conventional AM- spectrum- less affected by fading- power savings- noise advantage

Disadvantage of SSB over conventional AM- SSB reception can be a problem, Donald Duck voice.

6. The main advantage of DSBSC is since there is a suppressed carrier, there is a significant power savings, so more power can be placed in the sideband that is carrying the intelligence.

7. differential inputs (+ / -) and a double side-band output

8. see Fig. 14-1 and the related discussion

Answer to Problems - Chapter 4

217

9. The LIC provides superior performance because of the matched components.

10. 100 dB, The difference is with the 10k feedback resistor (pin 14) and the compensatingcapacitor (pin 12), see Fig. 4-2 [sub Fig 9a and b].

11. The SSB output requires the use of a filter to remove one of the sidebands.

12.

13. see Fig. 14-4

14. The crystal holder capacitance Cp shunts the crystal and offers a path toother frequencies. This problem is minimized by placing an external variablecapacitance in the circuit. [see Fig. 14-4(b)]

15. they are made from quartz crystal

16. This is an example of a bandpass lattice filter. Crystals X1 and X2 are series resonant while X3 and X4 are parallel resonant. The center frequencies are at the required sideband.

17. The big advantage isstability.

18. They use a piezoelectric effect just as crystals do. See problem 18 forthe discussion of the shape factor.

Answer to Problems - Chapter 4

218

Note that the filter’s input voltage is indicated as 0.8 V p-p on the schematic

The dc emitter voltage is shown as 0.6 V

19. The ratio of the 60db to 6db bandwidth is called the shape factor.

20. The 3dB ripple amplitude is describing the peak-to valley variations.

21. The electrical energy is converted to mechanical variations and then back toelectrical variations. The mechanical filter has a resonant frequency. They provide excellent rejection characteristics, are rugged, small, and have a high Q.

22. SAW filters are typically used in high frequency applications that are not commonly used in SSB transmission.

23.

24.

25. see Figure 4-9

26.

27.

Answer to Problems - Chapter 4

219

28. Phase advantages- greater ease of switching from one sideband to the other- SSB can be generated directly- lower intelligence frequencies can be economically used since high Q filtering is not necessary

The filter method is already firmly entrenched in many systems.

29. The carrier is a single frequency where the audio signal covers a wide range of frequencies.

30. The relay K1 and associated control circuitry, Q1 and Q2, serve to engage thepower amplifier only when an RF input is present.

31. -28 dBm >>> -14 dbm +34 dBm >>> +17 dbm range (-14 to +17) dbm

with no modulation, 1/10(140W) = 14W

32. The limited dynamic range allows lower-level signals to be transmitted withgreater power.

33. 1 MHz +/- 1 kHz, 1 MHzfor USB only 1MHz + 1 kHz

34. minor drifts in the BFO frequency can cause serious problems with SSB reception.Small shifts can make the voice sound like Donald Duck.

35. The BFO should be set to 400 kHz

36. The product detector is a balanced modulator that is used to recover theintelligence in an SSB signal. The low pass filter allows only the audio(low frequencies) components to pass.

37. BFO = 1 MHz

Answer to Problems - Chapter 4

220

38.

39. There would be no modulating signal, therefore no output.

40. Make sure the ohmmeter has sufficient drive to forward bias the diode. Some metershave a diode button. When making the diode measurements, make sure each diode is isolated from the rest of the circuit. This may require un-soldering one of theleads. Each diode should display the same measurement.An unmatched diode will result in carrier leak-through.

41. carrier leak-through is caused by either an unbalanced circuit or a defectivecomponent.

42. A single-tone only verifies the amplifiers linearity within a narrow range.The two-tone test checks the linearity over a wider spectral range.

43. Point D is the input to a mixer. The expected input is an RF signal and theoutput of the mixer should be the IF. Applying the IF frequency to Dwill result in an output of zero or 2xIF frequency.

44. The problem could be with either the mixer or the local oscillator.

45. refer to Table 4-1

46. You are making the assumption that the problem is outside beforeverifying that the SSB receiver is working properly.

47. AM receivers use the carrier to tune in (lock) to the RF signal. A loss of carrier will make the receive circuits to not work. The loss of one of thesidebands should not cause a problem.

48. One sideband is eliminated, the carrier is suppressed but can be used as a pilot carrier for the receiver to lock to.

Answer to Problems - Chapter 4

221

(1)Take the same carrier and intelligence and shift 90o before feeding into balanced modulator to get

(2)

(3)

(4)

Add equation 3 to equation 4 to get

which is SSB (the lower sideband in this case) QED

49.

50. The first sections of the AM and SSB receivers are the same (see Figs. 3-6 and 4-17).The difference is the SSB receiver uses a 2nd IF and a BFO to tune into the sideband.

This would be the hardest part, modifying the AM receiver to tune into thesideband.

Answers to Problem - Chaper 5

222

1. Angle modulation is the process of superimposing the intelligence signal on a high-frequency carrier so that its phase angle or frequency is altered as a function of the intelligence amplitude. The two sub-categories are:Phase modulation (PM) and Frequency Modulation (PM)

2. PM - the amount of phase change is proportional to the intelligence amplitudeFM - the amount of frequency change is proportional to the intelligence amplitude

3. PM is often used to generate FM

4. Amplitude modulation, single-sideband, no carrier

5. see Figure 5-1 for an example

6. Deviation constant defines how much the carrier frequency will deviate for a givenmodulating input voltage level. The units for deviation are expressed as [kHz/V].

7.

8. The amplitude of the intelligence is responsible for deviating the carrier frequency. The units for deviation are expressed as [kHz/V].The rate at which the carrier is deviated is controlled by the frequency ofthe intelligence.

9.

10. The rate at which the carrier is deviated is controlled by the frequency ofthe intelligence.

11. f = fc +/- Vi * k (kHz/V)where

fc = FRM carrier frequencyvi = amplitude of the intelligence signalk = deviation constant in units of kHz/V

Answers to Problem - Chaper 5

223

J8 is last significant sideband (form Bessel Function Table)

60% of full permissible amount for broadcast FM.60% of ± 75 kHz = ±45 kHz

Double its previous swing.

Unchanged.

Same current as before since transmitter power does not change in FM.

12. modulation index equals the maximum carrier frequency shift divided by the intelligence frequency (see equation 5-4)

13. the amplitude and the frequency

14. refer to Figure 5-4, the frequency spectrum changes

15.

16. The guard bands help minimize interference with adjacent channels.

17. The maximum possible frequency deviation.

18. The center frequency is the frequency of the carrier when it is at rest.At rest implies that there is no modulating signal.

19. This refers to how much the FM channel deviates. The amount of deviation dictatesthe bandwidth allocation.

20.

21.

22.

23.

Answers to Problem - Chaper 5

224

S/N = 4Thus, worst case occurs when S and N are 90° out of phase.

noise=1noise=1

4=signalnoise=1N

R

24.

25.

26.

27. FM receivers are not sensitive to static.

28. The limiter stage removes any amplitude variations of the received FM signalbefore it reaches the discriminator.

29. Removal of some spikes cause a phase shift and thus frequency shift of theFM signal; this frequency shift can not be removed.

30.

31.

Answers to Problem - Chaper 5

225

32. The narrowband transmission has limited bandwidth and as a result onlythe fist harmonic contains significant power. All other side frequencieshave a reduced amplitude and are more susceptible to noise. The deviationof the carrier is small and there is limited range (dynamic range) fromthe intelligence to the small deviations resulting from noise.

33. The capture effect is an FM receiver phenomenon that involves locking onto thestronger of two received signals of the same frequency and suppressingthe weaker signal.

34. The voice communication channels of radio communications typically requirea limited bandwidth therefore the reduced carrier deviation has minimalimpact on the overall noise.

35. See Figure 5-10 for a pre-emphasis circuit. Pre-emphasis circuits amplify thehigher frequencies more than lower frequencies. This improves the noisereduction capability. The de-emphasis circuit returns the received signal backto the original values.

36. (a-h) refer to section 5-3 FM Analysis for the answers to these questions(i) pre-emphasis is incorporated to provide additional amplification ofthe higher frequencies that are more susceptible to noise.

37. Refer to Figure 5-12.

38. Refer to Figure 5-13. A reactance modulator is designed so that its inputimpedance has a reactance that varies as a function of the amplitude of theapplied input voltage.

39. See Figure 5-14 (sub Figure 1)

40. Crosby systems use direct FM generation with an AFC.

41. The system typically requires some AFC (Automatic Frequency Control) circuit.

42.

43. see Figure 5-15

Answers to Problem - Chaper 5

226

44. A discriminator is the opposite of a VCO in that it provides a dc level output based on the frequency input.

45. see Figure 5-18

46. The mixer output changes the center frequency without changing the deviation.The multiplier increases to deviation.

47. The circuitry is typically called a pump chain.

48. The amplified audio signal frequency modulates a crystal oscillator. The variable capacitance of the varactor diodes provides approximately +/- 200 kHz deviation(see section 5-7)

49. referring to Figure 5-24pre-emphasis - amplifies the high frequenciesmatrix network - generates the L+R and L-R audio signalsdelay network - this is added to the L+R so that the L+R and L-R signals are in phasemaster oscillator - generates the 19 kHz pilot tonefrequency doubler - doubles the frequency of the 19 kHz signal to 38 kHzbalanced modulator - provides a double sideband output and shifts the L-Rsignal up to (23-38) kHz and (38-53)kHz.

50. This is explained in section 5-8 Stereo FM. The stereo FM signal is more proneto noise. The net result is stereo FM has about 20dB less S/N than monophonic FM.

51. Frequency division multiplexing is the simultaneous transmission of two or more signalson one carrier, each having its own separate frequency range.

52. It is still FM (Frequency Modulation). The signal has simply been shifted in frequency.

53. The matrix network is used to generate the L+R and L-R audio signals. Adder circuits(summing amplifiers) and an inverter are used to create the L+R and L-R signals.

54. FM stereo has about 20dB less S/N compared to monophonic FM. Stereo receptioncan be changed to monophonic to improve reception.

55. FM has superior noise performanceFM can use low-level modulationClass C amplifiers can be used and hence more efficiency

Answers to Problem - Chaper 5

227

56. FM suffers from undesired phase-shifting when used at frequencies below30MHz.

57. improved noise performance ... and FM can use low-level modulation.

58. The multiplier stage is used to increase the frequency to the desired outputfrequency range. A spectrum analyzer can be used to verify that the stage isproducing the desired frequencies.

59. Applying a modulating signal to the base of Q1 causes a change in the reactanceof the transistor. An increase in the voltage causes the reactance to go down.If the modulating signal goes down the reactance goes up. Changes in the reactancewill causes changes in the master oscillator frequency Q2.

60. A change with R5 will cause a change with Q2's output probably producing alow modulator output. First check the power supply and visually inspect, nextyou can check the bias voltages to Q2 and you may have to de-solder the resistorsto provide an accurate check.

61. SCA is the subsidiary Communications Authorization signal extending from 60 to 70kHz above the carrier. To troubleshoot the SCA requires the use of a spectrum analyzer.Verify the presence of the SCA signal.

62. The carrier frequency can be verified with a calibrated spectrum analyzer toa frequency counter.

63. The base-emitter junction would be heavily forward biased most-likely causingtransistor Q2 to saturate and thus producing no output.

64. R1 is used to isolate the base of Q1 from the modulating signal. A shorted R1might not cause any noticeable problems. This depends on the previous stageproviding the modulating signal.

65. The L-R signal will not be shifted and in fact, there might be no output of thebalanced modulator.

66. A failure of the balanced modulator will result in no L-R output. The failedbalanced modulator could also result in problems with the summing stage.

67. The intelligence amplitude causes the frequency deviation, the intelligencefrequency affects the rate the carrier is deviating.

Answers to Problem - Chaper 5

228

When mf = 2

68. In PM, the phase of the carrier varies with the modulating signal amplitude (mp).In FM, the carrier phase is determined by the ratio of the intelligence signal amplitude to the intelligence frequency. This is the method used in the Armstrong indirect FM.

69. The deviations determine the modulation index, which in turn determines the significantsideband pairs. The bandwidth is computed by the sideband pairs and not deviationfrequency. The deviation is not the bandwidth but it does have an effect on thebandwidth.

70.

71. It can be shown, using Carson’s rule, that deviation exceeding +/- 75 kHz will result in aFM bandwidth exceeding the 200 kHz channel spacing.

Solutions to Problem - Chapter 6

229

1. It is used to recover the intelligence by converting the frequency changesto voltage amplitude variations.

2. New FM receivers have sufficient frequency stability.

3. see Fig. 6-1

4. LO = 96.5 MHz to 10.7 MHz = 107.2 MHzImage Frequency = 107.2 MHz +10 MHz = 117.9 MHz

5. In FM, it is necessary to amplify the small input voltages to get the signalup to a sufficient level for mixing. At 1 GHz and beyond, transistor noise is increasing while the gain is decreasing. It is more advantageous to feed the incoming FM signal directly to a diode mixer to step it down to a lower frequency.

6. local oscillator re-radiation - undesired radiation of the local oscillatorthrough a receiver’s antenna. The RF amplifier helps isolate the receivingantenna from the local oscillator radiation.

7. A square-law device has an output signal at the input frequency and a smallerdistortion component at 2 times the frequency. Other devices have many more distortion components.

8. The major advantage of FETs is that they have an input/output square lawrelationship.

9. The dual-gate MOSFETs allows for an isolated input for the AGC level control and they also offer increased dynamic range over JFETs.

10. The RFC is used to keep the signal frequency from appearing on the power supply.

11. In FM receivers, the limiter removes unwanted amplitide variations.

12. see Figure 6-3

13. The resistor Rc limits the dc collector supply voltage. As soon as the inputis large enough, the output will be clipped (the desired effect).

14. Sensitivity - minimum input RF signal to a receiver required to produce a specified

Solutions to Problem - Chapter 6

230

output. Quieting Voltage - the minimum FM receiver input voltage that begins thelimiting process. Limiting Knee Voltage - same as quieting voltage

15.

16. See Figure 6-6 and the accompanying discussion. Slope detection is not widely used inFM receivers because the slope characteristic of a tank circuit is not linear, especially inwideband FM.

17. Refer to Figures 6-7 and 6-8 and the related discussion.

18. The Foster-Seely discriminator is shown in Figure 6-7 and the ratio detector isshown in Figure 6-9. The Quadrature Detector is shown in Figure 6-11.

19. Refer to Figure 6-9 and the related discussion.

20. Foster-Seely offers excellent linear response to wideband FM signals but respondsto undesired amplitude variations. The ratio detector does not respond to amplitudevariations and thereby minimizes the required limiting befeore detection.

21. a) and b) see Figure 6-26c) see Figure 6-3d) see Figure 6-7

22. The full discussion is provided in section 6-4 in the Quadrature Detector section.

23. See Figure 6-12. The difference between the input frequency and the PLL VCOfrequency generates an error voltage. The error voltage output level and the rate ofchange are representative of the intelligence riding on the input FM signal.

24. The PLL's input phase comparator compares the input signal and the outputof the PLL's VCO. This difference is used to develop an error signal proportional to thedifference between the two.

Solutions to Problem - Chapter 6

231

25. Free-running - this is the nominal frequency of the PLLCapture - the point where the VCO begins to change frequencyLocked or tracking - the point where the VCO changes frequency to match the input.

26. Capture range •2 × 20 kHz = 40 kHzlock range • 2 × 150 kHz = 300 kHz

27. Refer to Figures 6-15 and 6-16.L+R is added to L-R to yield 2LL+R is added to -(L-R) to yield 2RThe levels of the 2L and 2R signals are then dropped to L and R

28. SCA - Subsidiary Communication Authorization, this is often used to transmit musicprogramming for subscription service.

29.

30.This is explained in the LIC stereo decoder discussion in section 6-6.

31.

32.

33. This can be used to verify that the limiter, discriminator, and audio amplifier circuitsare operating properly.

Solutions to Problem - Chapter 6

232

34. This is either a ceramic resonator or an LC tuned circuit.

35. Test the diode using the diode test mode. A good diode reads about 0.6 to 0.7Vin the forward direction and over range in the reverse direction. A transistor can be tested by treating it as if it is two diodes connected back-to-back. The two diodejunctions are base-emitter and base-collector.

36. The signal at point B will not be delayed 1/4 of a sinewave. This phaseproblem will cause the quadrature detector to not function.

37. Check the capacitors first, leaky caps are a common problem, and then check the diodesD1 and D2.

38. The L2L3C2 tank circuit will not operate at the desired resonant frequency.A loss in the output voltage will result.

39. The pulses at Q3's collector won't be summed and the intelligence will not be recovered.

40. Q4 will now be a significant load on Q3 and will dramatically affect the R1C3 timeconstant.

41. Yes, noise reduction is important. The noise reduction should aid withFM quieting.

42. The limiter provides an amplitude output. The constant amplitude eliminatesits need for AGC.

43. refer to Figure 6-13, use the equation fo ~ 0.13 / (RoCo) to obtain a free-runningfrequency of 455 kHz.

44. Refer to Fig. 6-19 and 6-16.

Answers to Problems - Chapter 7

233

lowest to highest

≅ −⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

20 logff

ff

Qi

s

s

i

image rejection (dB)

1. A receiver's function is to receive only. A communication transceiver transmits andreceives.

2. The advantage is the image frequency is almost totally suppressed. Image frequency forthe 1st IF = (37.7 +10>7) MHz = 48.4. MHz Image frequency for single conversion is 28MHz.

3. VHF crystal filters are now available for IF circuitry and are economicallyattractive. Additional advantages include better image suppression and less tuningrange requirement. Figure 7-3 and 7-4 provide example block diagrams.

4.

5.

6. Delayed AGC does not provide any gain reduction until some arbitrary signallevel is attained and therefore has no gain reduction for weak signals.

7. Auxiliary AGC causes a step reduction in receiver gain at some arbitrary high value ofreceived signal thus preventing very strong signals from overloading a receiver. Anexample circuit is provided in Figure 7-7(a).

8. This is called variable bandwidth tuning. A block diagram is provided in Figure 7-8. The need for this is because the bandwidth varies for different types of transmissions.

9. The output of Mixer 1 is (2050 - 2250) Hz. (2050-2100) Hz is passed to Mixer 2. Theoutput of Mixer 2 is (550-600) Hz. The system bandwidth drops to 50Hz.

Answers to Problems - Chapter 7

234

10. The noise limiter is employed to silence the receiver for the duration of a noise pulse. For the circuit in Figure 7-9, impulse noise will cause the AGC voltage toinstantaneously increase. This increases the anode voltage on D2, thus turning it off andblocking the audio from entering the audio amplifier.

11. Metering provides a visual indication of received signal strength. It can also aid withtroubleshooting.

12. The squelch circuit quiets the receiver in the absence of a carrier.

13. Quieting and Muting, Figure 7-10 shows a squelch circuit. When no signal is present, theAGC voltage is high, turning on Q1 which removes the audio signal from Q2. This turnsoff the audio out from the collector of Q2. The five different methods for providingsquelch are: 1) Fixed RF level threshold, 2) Variable level control, 3)Pilot tone control,4) Digital code control signal and, 5) Microprocessor controlled algorithm.

14. EMI creates undesired amplitude variations that can affect FM reception and destroy AMreception.

15. An ANL is employed to silence the receiver for the duration of a noise pulse.

16.

17. The 1dB compression point is where the input/output relationship has just reached a level1 dB down from the ideal linear response. From Figure 7-11, the 1 dB compression pointis approximately -10 dBm.

18. The third order intercept point is slightly greater than +20 dBm.

19.

Answers to Problems - Chapter 7

235

20.

21.

22. The output frequency of the synthesizer in Figure 7-14 is limited to the maximumfrequency of the programmable divider. The synthesizer in Figure 7-16(a) allowsnarrower channel spacing and faster lock-time. Figures 7-16(b) and (c) are subject tonoise and have wide channel spacing.

23. The circuit in Figure 7-17 overcomes the problem of high-speed programmable division. The operation of this circuit is discussed in the Two-Modulus Dividers section.

24.

25. Input code = 100011 from Table 7-1 100011 = Channel 25 = 27.245 MHz

26. The received RF signal is fed through an RF filter and amplifier. The amplifiedRF signal is then mixed with the frequency, generated by the Philips SA7025 IC 1 GHzlow-voltage fractional-N synthesizer to down-convert the receive signal to the IFfrequency. A Philips TDA7021T FM radio circuit provides the circuitry for FMreception.

27. The phase accumulator generates a phase increment of the output waveform based uponits input. The input phase determines the frequency of the output waveform. Translatingphase information is accomplished by means of a ROM look-up table. The digital outputof the NCO is converted to an analog signal via the D/A.

Answers to Problems - Chapter 7

236

28. Maximum output frequency is about 40% of 60 MHz 0.40 × 60 MHz = 24 MHz

29. A parasitic is unwanted component of an electronic circuit that is a byproduct of fabrication, component assembly, or both

30. 1. Wires become transmission lines.

2. The length and geometry of the wires becomes an issue.

3. Bends in the circuit board layout are an issue.

4. At high frequencies you have to transition from some coaxial structure into a circuit

board to properly launch the wave.

5. Impedance match all elements in your circuit to minimizing ghosting within the circuit

board. Make sure your components are impedance matched to minimize ghosting with.

31. This can add an RF short to ground and affect the performance of the circuit.

32. 0.32” - a rule of thumb for lines on printed circuit boards is the traces should be at least four board thicknesses apart.

33. The RF chokes is used to isolate the DC circuit from the RF circuitry.

35. The output is off frequency. No input to the RF driver. Note, most RF circuits requirethat each stage be properly tuned to operate.

36. unstable transmitter operation and reduced oscillator output level or complete shutdownof the oscillator section.

37. You would observe spurious frequencies on the output of the transmitter.

38. Up conversion produces an IF signal higher in frequency than the original RF signal. This provides better image frequency suppression and less tuning range requirements.

39. The dynamic range can be improved by using a low-noise pre-amplifier with high gain. However, to maintain a high dynamic range, use only the amplification needed.

Answers to Problems - Chapter 7

237

Examples 7-7 to 7-9 demonstrate how dynamic range can be improved.

40. It is common to test a receiver's IMD using two test frequencies. Examples of this areprovided in Figures 7-12 and 7-13. IMD testing of a receiver is most critical.

41.

Answers to Problems - Chapter 8

238

1. (a) voice channel (b) voice and computer (c) computer

2. Digital signal processing uses programming techniques to process a signal while still indigital format.

3. ASCII - American Standard Code for Information InterchangeEBCDIC - Extended Binary-Coded Decimal Interchange Code

4. 5 - 0 1 1 0 1 0 1a - 1 1 0 0 0 0 1A - 1 0 0 0 0 0 1STX - 0 0 0 0 0 1 0

5. 5 - 0 1 0 1 1 1 1 1a - 0 0 0 1 0 0 1 0A - 0 0 0 1 0 0 1 1STX - 0 0 1 0 0 0 1 0

6. The gray code is based on the relationship that only one bit in a binary word changes for each binary step.

7. The Gray code is used in telemetry and systems with slowly changing data.

8. Acquisition time is the amount of time it takes the hold capacitor to reach its final value.

9. Aperture time is the time that the S/H circuit must hold the sampled voltage.

10. 1 nF

11.

Natural Sampling Flat-top Sampling

12.

13.

14

Answers to Problems - Chapter 8

239

15. Resolution refers to the accuracy of the digitizing system in representing a sampledsignal. This can be obtained by increasing the number of quantization levels (increase thenumber of binary bits) or increasing the sample frequency.

16.

17. The sampled signal is segmential into different voltage levels, each corresponding to adifferent binary number.

18. Linear PCM - each quantile interval has the same step size.Non linear PCM- each quantile interval step-size may vary in magnitude.

19. digital companding - the quantile intervals are varied to improve the weak signal's S/Nanalog amplitude companding - process of volume compression before transmission andvolume expansion after detection.

20.

21. NRZ - see Table 8-2, 0-low and 1 - highRZ - same limitations and disadvantages of NRZ, includes RZ-unipolar, RZ-bipolar

RZ-AMI (see Fig. 8-25 and Table 8-3)Phase Coded Binary - see Table 8-4Multilevel Binary - see Table 8-5

22. The plots of 1 1 0 1 0 are provided in the first five bits of Figure 8-25 for each format.

Answers to Problems - Chapter 8

240

23. The bipolar pulse +/- which reduces the DC component. Unipolar codes pulse in onedirection and have a dc component.

24. The bi-phase codes are considered to be self-synchronizing or self-clocking. The datastream transitions relative to the clock interval.

25. Dmin, also called the Hamming distance, is the distance between logical states.

26. By adding bits.

27.

28.

29. Error detected = Dmin - 1(a) 6(b) 9

30. Ethernet

31. A code in which the message and the block check code are transmitted as separate partswithin the same transmitted code.

32. n - length of the transmitted codek - length of the message

33. The code generated when creating the CRC transmit code. This code is added to themessage to create the full transmit code.

34. The feedback paths to each XOR gate are determined by the coefficient for thegenerating polynomial. A “1" indicates a feedback path, a “0" means no feedback.

Answers to Problems - Chapter 8

241

M(x)

1 x x2 x4

out

35.

36.

37. Syndrome - same as the BCC which is the value left in the CRC dividing circuit after alldata have been shifted in.

38. a) No errors detected b) errors detected

NOTE: This assumes that the registers in the CRC generating circuit were reset to alllogical zeros prior to generating the code.

39. Received data errors can be corrected.

40. CD Players. The forward error correction ability of Reed-Solomon codes allows blocksof defects (eg. a bad scratch) to be corrected.

41. recursive

42. non-recursive

43. fourth order

44. first order filter, recursive

Answers to Problems - Chapter 8

242

45.(a) 2nd-order Bandstop filter:

(b) 4th-order Bandpass filter:

46. CRC codes can effectively catch 99.95% of transmission errors. The generating anddecoding circuitry can easily be implemented in high speed applications.

47.

48.

Solutions to Problems - Chapter 9

243

1. Baseband means that the signal is transmitted at its base frequency with no modulation.

2. B. Standard analog modulation system

3. This is typical of a digital system being carried over an analog channel.

4. By using an ADC - analog-to-digital converter.

5. Binary coding systems are generally more efficient than other systems and provide betternoise immunity . A possible disadvantage is the complexity of implementing a fullydigital solution but this is becoming less of an issue.

6. Coding - transforming messages or signals in accordance with a definite set of rules.Bit - unit of information required to allow proper selection of one out of two equallyprobable events.

7.

8. The binary code cab still be recovered even if there is significant noise. For examplegiven a system with a logic 0 >> 0.0V and 1 >> 1.0V. It would take 0.5V to introduce abit error.

9.

10.

11.

12. Synchronous - implies that the transmit and receive data clocks are locked together(example computer - printer)Asynchronous - implies that the transmit and receive clocks are not locked together andthe data must provide start and stop information.

13. Protocol - a set of rules to make devices sharing a channel observe orderlycommunication procedures

Answers to Problems - Chapter 9

244

14.

15.

16. See Table 9-2

17. 24 telephone calls

18. fractional T1 - a term used to indicate that only a portion of the data bandwidth of a T1line is being used.

19. Point-of-presence - the point where the users connect their data to the communicationscarrier.

20. The CSU/DSU provides the data interface to the communications carrier providingframing and line management.

21. The data are divided into small segments called packets. They are held for short periodsof time at switching centers and therefore transmitted in near real-time.

22. Frame relay is a packet switching network designed to carry data traffic over Telco. Frame relay operates on the premise that the data channel will not introduce bit errorthereby the need for overhead bits is minimized.

23. ATM is a cell relay technique, all stations are constantly transmitting cells.

24. TDM is a technique used to transport data from multiple sources over the same datachannel.

Answers to Problems - Chapter 9

245

25. Guard times are time added to the TDMA frame to allow for the variation in data arrival.

26. See Figure 9-10

27. The encoder transmits information regarding whether the analog information increases ordecreases in amplitude.

28. The analog signal is fed into a comparator that compares the output of the integrator tothe analog signal. The comparator’s low or high output will cause the D flip-flop tooutput a low or high.

29. Slope overload occurs when the analog signal has a high rate of amplitude change. CVSD corrects the problem by increasing the step size whenever there is a longer runthan three consecutive 0's or 1's.

30. Companding is a fixed algorithm and CVSB depends on past values and adapts to these.

31. Transmitters operate on a very low duty cycle. Time intervals between pulses can befilled with samples of other messages.

32. Refer to Figure 9-14 and the related discussions.

33. A simple way to generate PWM is using a 565 PLL (see Figure 9-16)

34. See Figure 9-16

35. See Figure 9-18

36. A type of transmission where a continuous sinusoidal waveform is interrupted to conveyinformation. The carrier is turned on/off to convey information.

37. The AGC will have problems with the carrier turning on/off. In two-tone modulation, thecarrier is always being transmitted.

38. Output frequencies are 21 MHz +/- 300 Hz and 21MHz +/- 700 HzThe bandwidth required = 2 x 700 Hz = 1400 Hz.

39. RS 232 is an EIA standard. It is still used in many applications but it is slowly beingreplaced by USB.

40. RED +5 Brown - Ground Yellow - Data Blue - Data

Answers to Problems - Chapter 9

246

41. Upstream

42. Handshaking is flow control.

43. USB supports up to 480 Mbps Firewire supports up to 800 Mbps

44. See Figure 9-25, Type A is upstream, which connects to the computer. Type B isdownstream and connects to the peripheral.

45. CIR - guaranteed data rate or bandwidth to be used in the frame relay connection.CBIR - enables subscribers to exceed the CIR during times of heavy traffic

46. Red Alarm - a local equipment alarm that indicates that the incoming signal has beencorruptedYellow Alarm - indicates that a failure in the link has been detectedBlue Alarm - indicates a total loss of incoming signal

The loopback feature enables tests to be performed on the link established between theTelco and the customer.

Answers to Problems - Chapter 10

247

1. The term wireless is used today to describe telecommunication technologiesthat use radio waves, rather than cables, to carry the signal.

2. FSK is a form of FM in which the modulating wave shifts the output betweentwo pre-determined frequencies. Methods of generation include a VCO and modifyingthe oscillating frequency of a tank circuit.

3. one technique is to use a PLL and use the error signal to indicate the data change.

4. the incoming data cause the phase of the carrier to phase-shift a defined amount

5. M = 2n, number of allowable phase statesn = the number of data bits needed to specify the phase state

6. The +/- sin Tct signals are input into 1 of 2 selectors. The binary data inputis used to select the phase.

7. The received BPSK signal is input to a coherent carrier recovery circuit which produces a sin Tct signal. The recovered carrier is mixed with the BPSK inputThe output of mixer is filtered through a low-pass filter and the result is a+/- 1/2 term that indicates the phase.

8. A recovered carrier frequency is used as an input to a mixer to recover the data.

9. The recovered carrier is fed to two phase detectors circuits. One side has a 90o phase-shift. The outputs of the phase detectors are fed through low-pass filters. The I and Qdata is output from the low-pass filters.

10. See Figure 10-14 and 10-15. The QAM system uses a four-level baseband streamproviding a 16 state QAM constellation in a limited bandwidth.

11. The output signal must contain both amplitude and vector information.

12. It is used for diagnosing the performance of a digital modulation system. It allows for the immediate observation of the effects of filters, circuits,or antenna adjustments.

13. 1 1 0 1 0 0 0 1 1

-sin -sin -sin +sin +sin

Answers to Problems - Chapter 10

248

14. A PN sequence is a pseudo random sequence. It is noise-like because thesequence appears to be random like noise.

15. The RF signal is randomly spread over a range of frequencies in a noise-like manner.

16. (a) pn sequence length = 24 -1 = 15(b) pn sequence length = 29 - 1 = 511(c) pn sequence length = 223 - 1 = 8388607

17. This indicates that the PN code has a length of 2n - 1 where n is the number of shiftregisters.

18.

19. The data is transmitted by a carrier that is switched in frequency in a pseudorandomfashion.

20. The time each carrier spends at a specific frequency.

21. Shift # SR0 SR1 SR2

0 1 0 1

1 1 1 0

2 1 1 1

3 0 1 1

4 0 0 1

5 1 0 0

6 0 1 0

7 1 0 1

Answers to Problems - Chapter 10

249

22. A pseudorandom sequence to break-up the message into chips. The chips successfullymodulate fractions of the bit, typically using a phase-shift technique. The receivermultiplies the incoming signal to recover the original modulated digital signal. Thisenables many signals to be multiplexed over one communications channel.

23. When two spread spectrum transmitters momentarily transmit at the same frequency.

24. The pseudorandom digital sequence used to spread the signal.

25. See Figure 10-24.

26. spreadingxx

= ≈1 10

56 1017 863

6

.

27. See Figure 10-31

28. Flash OFDM uses a fast hopping technique to transmit each symbol over a differentfrequency.

29. D - T - VD 0 1 0 0 0 1 0 0T 0 1 0 1 0 1 0 0V 0 1 0 1 0 1 1 0

30. Both the analog and digital signals share the same channel bandwidth.

31. The digital data rate for AM HD radio is 36 kbps which produces a signal with comparable quality to the current analog FM.

32. The digital data rate for FM HD radio is 96 kbps which produces a near CD quality audiosignal.

33. a) hybrid AM – see Fig. 10-33b) hybrid FM - see Fig. 10-34

34. OFDM

35. Provides remote metering

36. Multiplexing is typically required because data must be collected from many sensors andthe sensors must share the same communications channel.

Answers to Problems - Chapter 10

250

37. The data is transmitted over fixed carrier frequencies therefore only one system cantransmit at a time.

38.Shift#

SR0 SR1 SR2

0 1 1 1

1 0 1 1

2 0 0 1

3 1 0 0

4 0 1 0

5 1 0 1

6 1 1 0

7 1 1 1

no changes with a seed value of 0 0 0.

39. The answer should include the following: unique PN codes for each transmitter andreceiver, hits cause very little loss of signal quality, other signals appear as noise

Answers to Problems - Chapter 11

251

1. Most telephone network voice connections are analog and the data speeds are limited.

2. Received voice signal ( millivolts)transmitted voice signal(1 to 2 Vrms)ringing signal (90 Vrms)

3. Subscriber lifts the handset, a switch is closed providing a dc loop current between tipand ring. Telco senses the off-hook condition and responds with a dial-tone. At thispoint, the subscriber dials the desired number.

4. Phones use pulse dialing and dual-tone multi-frequency (DTMF)

5. PBX - Private Branch Exchange, the primary function is switching one telephone line toanother.

6. Battery feeding Overvoltage protection Ringing SupervisionCoding Hybrid Testing

7. Cable resistance and transmission line effects

8. Loaded cable - cable with added inductance every 6000, 4500, or 3000 feet.

9. The “C2" line is an improved line. The limits between (500 - 2800) Hz are +1 dB to -3dB. From 300 to 500 Hz and 2800 to 300 Hz the C2 limits are +2 dB and -6 dB.

10. The signal is attenuated and distorted when carrying over long distances.

11. Attenuation distortion is the difference in gain at some frequency with respect to areference tone of 1–4 Hz.

12. Delay distortion happens when various frequency components of a signal are delayed bydifferent amounts during transmission. The delay distortion is caused by thetransmission line’s inductive and capacitive components.

13. Refer to Figure 11-6 and the related discussion.

14. Modems enable the transmission of digital data over analog phone lines. Phone lines arewidely available.

15. This maps the use of the telephone network, 9:00-11:00 AM, 2:00-4:00PM.

16. Erlangs or hundred-call seconds.

Answers to Problems - Chapter 11

252

17. A situation in a telephone switching office when calls are unable to reach theirdestination

18. Grade of service (B) =

19. The grade of service is verified using traffic scanning devices. The lower the number thehigher the grade of service.

20. Customer calling patterns are recorded. This provides data for establishing toll rates,developing contingencies for network failures, forecast future demands and projectcapital expenditures.

21. ISDN is physically “in band” and logically “out of band”. SS7 is both physically andlogically “out of band.

22. Layers 1-3.

23. Layers 1-7.

24. Layer 7, ISUP, communicates directly with MTP 3, SCCP, TCAP, and OAM.

25. IAM (starts the call), ACM (the phone is ringing), ANM (the called party picks up thephone), REL (either the called or calling party hangs up the phone), and RLC (the voicetrunk is available for another call).

26. A protocol analyzer.

27. A number that identifies the switch nearest to the caller.

28. A number that identifies the switch nearest to the called party.

29. A number that identifies the trunk used for the SS7 message.

30. 35, someone hangs up the phone.

Answers to Problems - Chapter 11

253

31.

32. Frequency reuse is the process of using the same carrier frequency in different cells thatare graphically separated. Cell splitting results from reducing the size of the coveragearea.

34. Mobile unit samples all received “set-up” channels and selects the strongest signal. Themobile units synchronizes the data streams and tunes to the assigned channel to place thecall. While in operation, the mobile unit monitors for a change in a cell site.

36. Rayleigh fading is the rapid variation in signal strength received by mobile units in urbanenvironments. Digital systems provide significant improvement.

37.

38.

39. The base station communicates with the mobile user. The switch collects calls from manybase stations and places the calls into the PSTN. The PSTN routs the calls to the switchnearest the called party location.

40. GSM and CDMA.

41. 30 KHz.

42. π/4 DQPSK

Answers to Problems - Chapter 11

254

43. ETSI

44. GMSK

45. 8

46. 6

47. Equalizing filter.

48. Rake receiver.

49. It is like a lighthouse. It constantly looks for users.

50. Code 0, which is the Pilot Signal.

51. It is the synchronizing signal.

52. SCH

53. TIA

54. CDMAone

55. Walsh codes.

56. PN offset.

57. The system noise level increases.

58. GPS satellites provide the frequency reference. The frequency is 1.575 GHz and thepower level is about –140 dBm.

59. Unmodulated carrier that leaks through to the output signal. The specified level is that itbe at least 25 dB below the carrier signal.

60. Signals that interfere with: GPS signal, uplink, and downlink.

61. Fout = n F1 + mF2

62. Rusty fences and piles of rusty metal.

Answers to Problems - Chapter 11

255

63. a) Span is 15 MHz, across the entire display, for both photos. b) Power for the vertical photo and time for the lower photo. c) 3 MHz d) 87

64. a) Horizontal is time and vertical is frequency.b) -3.75 MHz.c) About 1.5 MHz KHz.

65. A network of users that share computers in a limited area.

66. BUS - an older style of hardwired topology but the concept applies to modernwireless networks. TOKEN-RING - passes a token for control of the network. STAR - a central hub or switch interconnects the networking devices.

67. Ethernet is best described as carrier sense multiple access with collision detection(CSMA/CD).

68. The Ethernet frame structure is provided in Figure 11-25. The MAC address is 12hexadecimal characters made up of the vendor code and vendor assigned board ID.

69. A broadcast is defined by all 1's in the address.

70. See Figure 11-27

71. See Figure 11-28 and the related discussion.

72. See Table 11-4

73. 802.11b - 11 Mbps 802.11a - 54 Mbps 802.11g - 54 Mbps

74. LAN - Local Area Network (limited geographic area), MAN - Metropolitan AreaNetwork (metro area), WAN - Wide Area Network (regional, national, worldwide)

75. OSI - see Figure 11-23 and the related discussion

76. Bridges - use MAC addressing to interconnect LANsRouters - use layer 3 addressing to interconnect networks

77. 5.8 GHz and 2.5 GHz

Answers to Problems - Chapter 11

256

78. because of its improved NLOS characteristics

79. Frequency assignments differ, data rates differ but the main difference is the WiMax unit only has to compete once to gain entry into the network.

80. Uplink – TDMA, Downlink – TDMThe uplink is TDMA so any WiMax unit can get access to the network. The downlink is TDM to make sure time sensitive data is delivered on time.

81. the 2.4 GHz ISM band.

82. Bluetooth has three operating classes.

Bluetooth Output Power ClassesPower Class Maximum Output Power Operating Distance

1 20 dBm ~ 100 m2 4 dBm ~ 10 m3 0 dBm ~ 1 m

83. an ad hoc network of up to eight Bluetooth devices.

84. used by Bluetooth to discover other Bluetooth devices or to allow itself to be discovered.

85. it is used to establish and synchronize a connection between two Bluetooth devices.

86. Developed from ARPANET, uses TCP/IP addressing, provides extensive worldwideinter-connectivity.

87. IP addresses are divided into five classes. Classes A, B, C are used to route data overthe Internet, Class D is used for Multicasting, and Class E is an experimental range.

88. Many sites are available. Keyword for the student’s search include cellular, wireless,and PCS.

89. IP telephony is the use of routing telephone traffic over a data network.

90. QoS issues are very important when considering telephone service; key issues include

Answers to Problems - Chapter 11

257

reliability (uninterrupted service), 911 service, etc.

91. ADSL - Asymmetric DSL provides data rates up to 1.544 Mbps upstream and up to 8Mbps downstream.

92. Cable modems use the high bandwidth of the cable television systems to provide datarates from (128 kbps to 10 Mbps) upstream to (10 to 30 Mbps) downstream. Atechnique called ranging is used by the modem to determine the time it takes for datato travel to the cable head-end.

93. V.92/V.90 uses asymmetric operation, analog (customer-to Telco) and digital (ISP-Telco-customer).

94. The advantage is a standardized digital interface.

95. See Table 11-9

96. DMT is an industry standard that uses multiple subchannel frequencies to carry thedata.

97. The WAP standard has been developed to bridge the gap between mobilecommunications, the Internet, and corporate Intranets.

98. confidentiality, integrity, authentication, non-repudiation, availability of the network

99. It is a guarantee that any modification to the data set can be detected.

100. An interfering signal is transmitted over the same channel thereby disrupting or slowing down communication.

101. passive – bad guy is just listeningActive – the bad guy is disrupting the communication link.

102. internal – protection is often put in place to protect against external threats but notnecessarily against inside attacks.

103. to prevent eavesdropping

104. the secret code used in the encryption algorithm to both create and decode themessage

105. You can expect significant data loss without the equalization.

Answers to Problems - Chapter 11

258

106.. Handshaking is not actually a standard protocol but rather a way to describe that bothends of the communication line recognize each other and know how to exchangedata.

107. You need to know the point of presence.

108. Identify electrical power, wired network connections and a wireless networkcoverage plan.

Answers to Problems - Chapter 12

259

1. A transmission line is defined as the conductive connections between the systemelements that carry signal power. The effects of carrying a signal over wire are verycomplex and these effects vary greatly with frequency.

2. Two-wire, Open wire, Twisted Pair, Unshielded Twisted Pair, Shielded Twisted Pair,Coaxial lines, Balanced and Un-balanced lines (see section 12-2)

3. Flexibility of solid dielectric, expense of rigid coaxial lines

4. To keep moisture out.

5. In computer networks.

6. A category describing the performance specification for the cable.

7. NEXT - a measure of crosstalk or signal coupling.

8. PSNEXT - a measure of the total crosstalk for all four wire-pairs.

9. Unbalanced line - signal is carried in one wire and the other is at ground.Balanced line - same current flows in each wire but 180o out of phase.

10. The noise on each line of the balanced cable is 180o out of phase. The signals cancelswhen added together. Today, the common signal to both lines is removed by adifferential amplifier. The common signals are rejected.

11.

12. See Figure 12-9

13. This is the input impedance exactly matched to its characteristic impedance called Zo,(see Equation 12-1)

14.

Answers to Problems - Chapter 12

260

15.

16. Connect the transmission line to the impedance bridge and read to measurement.

17. If D increases, Zo increases (refer to Equation 12-9)

18. This problem assumes that d increases therefore Zo decreases. (Refer to Equation 12-9)

19. The surge impedance of a two-wire transmission line depends on the ratio of thedistributed inductance and capacitance in the line.

20.

21. Copper losses - at high frequencies the losses are due primarily to the skin effectDielectric losses - proportional to the voltage across the dielectric, increasing with frequency.

Radiation or induction losses - these can be greatly reduced by terminating the line with a resistive load equal to Zo and properly shielding the line.

22.

23. Surge impedance - another name for characteristic impedance.

24. This derivation is provided in equations 12-12 to 12-19.

25. Velocity .3 x 108 meters/second

Answers to Problems - Chapter 12

261

26.

27. This is the ratio of the actual velocity to the velocity in free space.

28.

29.

30.

31. Non-resonant line - one of infinite length or that is terminated with a resistive load equalin ohmic value to its characteristic impedance. The length is not critical and the voltageand current waves move in phase with one another.

32.

33. Figure 12-15 shows the wave travel. The waveform plotted at any point is a duplicate ofthe source signal.

34. Resonant line - a transmission line terminated with an impedance that is not equal to itscharacteristic impedance. Reflected waves are generated by the impedance mismatch. See Figures 12-17 and 12-18.

35. See Figure 12-17

36. See Figure 12-18

Answers to Problems - Chapter 12

262

37. Standing waves - waveforms that apparently seem to remain in one position, varying only in amplitude.SWR - another name for voltage standing wave ratio, the ratio of the maximum orminimum voltage on a line.Characteristic impedance - the input impedance of a transmission line either infinitelylong or terminated in a pure resistance equal to its characteristic impedance.Standing waves are minimized by impedance matching.

38. Refer to the text discussion for Figure 12-19 in section 12-6.

39.

40.

41.

42.

43. A mismatched line leads to a higher VSWR

44. Minimize the VSWR

45. They are 180o out of phase causing cancellation of the reflections.

Answers to Problems - Chapter 12

263

46.

47.

48. Using the Smith chart, we first normalize the load impedance (call it point A)

z ZZ

j jLL

o

= =+

= +50 75

750 666 1.

Read the location from the WTG scale at A. The load is located at 0.14258. We must travel675 degrees from the load towards the generator. The distance traveled in wavelengths is

675369

1875= . wavelengths

We add the two together to get the input, Point B on the Smith chart.

01425 1875 2 0175. . .λ λ λ+ =

The normalized impedance read at Point B is

z jin = +0 3 01. .

Thus, the input impedance is

( )Z z Z j jin in o= = + = +0 3 01 75 22 5 7 5. . / . Ω

Answers to Problems - Chapter 12

264

Answers to Problems - Chapter 12

265

j 1−:=The load impedance is ZL 50 j 75⋅+:= Ω

The characteristic impedance is Zo 75:= Ω

The line was previously found to be 1.875 λ long, compute the electrical length

βL 2 π⋅ 1.875⋅:=

Use Eq. 11-30

Zin ZoZL j Zo⋅ tan βL( )⋅+

Zo j ZL⋅ tan βL( )⋅+⋅:= Zin 22.5 7.5i+= Ω

Now work it using Eq. (12-30)

The results are the same. Ordinarily the Smith chart is not read with sufficient precision to yieldexact answers.

Answers to Problems - Chapter 12

266

z ZZ

j jLL

o

= =−

= −62 5 90

50125 18. . .

162.5 j 90⋅−

5.206 10 3−× 7.496i 10 3−

×+= S

49. Using the Smith chart, first normalize the impedance. For the purposes of this problemassume that Zo = 50 S (simply because the only Smith chart handy is a normalized one).Call the impedance location Point A. (The Smith chart is presented on the next page.)

Construct a line through the center of the chart and Point A. Extend it the same distance in theopposite direction. Read the normalized admittance at Point B.

y jL = +0 26 0 37. .

Determine the admittance by multiplying by the characteristic admittance.

Y y Y yZ

j j SL L oL

o

= = =+

= +0 26 0 37

500 0052 0 0074. . . .

Now compute the admittance directly by taking the reciprocal of the impedance (in this case wedon’t need to assume a value of characteristic impedance).

The analytical result is slightly different.

Y j SL = +0 005206 0 007496. .

Answers to Problems - Chapter 12

267

Problem 12-49

Answers to Problems - Chapter 12

268

Zo ZL

L

Zin

j 1−:=The load impedance is ZL 200 j 300⋅+:= Ω

The characteristic impedance is Zo 100:= Ω

The line is specified to be 5.35 λ long, compute the electrical length

βL 2 π⋅ 5.25⋅:=

Use Eq. 11-30

Zin ZoZL j Zo⋅ tan βL( )⋅+

Zo j ZL⋅ tan βL( )⋅+⋅:= Zin 15.385 23.077i−= Ω

50. Find the input impedance of a 100-S line, 5.35 8 long, and with ZL = 200 + j 300 S.

Answers to Problems - Chapter 12

269

Zo ZL

D

Zin

Short

L

Zo

53. The antenna load on a 150-S transmission line is 225 - j 300 S. Determine the length andposition of a short-circuited stub necessary to provide a match.

First the analytic solution

Answers to Problems - Chapter 12

270

λD2 0.129=D2if atan T2( ) atan T2( ), atan T2( ) π+,( )

2 π⋅( ):=

Second Solution

λD1 0.268=D1if atan T1( ) 0> atan T1( ), atan T1( ) π+,( )

2 π⋅:=

First Solution

T2 1.05=T2

XL− XL2 Zo RL−( )

RL2 XL

2+⎛

⎝⎞⎠

ZoRL−

⎡⎢⎢⎣

⎤⎥⎥⎦

⋅−−

Zo RL−:=

T1 9.05−=T1

XL− XL2 Zo RL−( )

RL2 XL

2+⎛

⎝⎞⎠

ZoRL−

⎡⎢⎢⎣

⎤⎥⎥⎦

⋅−+

Zo RL−:=

Now, if the load of impedance ZL is to be matched Zo line using a stub of impedance Zo. Let the distance from the load to the stub be D and the length of the stub be L.

ΩXL 300−:=ΩRL 225:=ΩZo 150:=

Answers to Problems - Chapter 12

271

λL2 0.085=L2

if atanYoB2

⎛⎜⎝

⎞

⎠0> atan

YoB2

⎛⎜⎝

⎞

⎠, atan

YoB2

⎛⎜⎝

⎞

⎠π+,

⎛⎜⎝

⎞

⎠2 π⋅( ):=

λL1 0.415=L1

if atanYoB1

⎛⎜⎝

⎞

⎠0> atan

YoB1

⎛⎜⎝

⎞

⎠, atan

YoB1

⎛⎜⎝

⎞

⎠π+,

⎛⎜⎝

⎞

⎠2 π⋅( ):=

For a short-circuited stub we find the lengths as

SB2 0.011=B2 Im YoYL j Yo⋅ tan βD2( )⋅+

Yo j YL⋅ tan βD2( )⋅+⋅

⎛⎜⎜⎝

⎞

⎠:=

βD2 2 π⋅ D2⋅:=

SB1 0.011−=B1 Im YoYL j Yo⋅ tan βD1( )⋅+

Yo j YL⋅ tan βD1( )⋅+⋅

⎛⎜⎜⎝

⎞

⎠:=

βD1 2 π⋅ D1⋅:=

The input susceptance of the line Zo without the stub attached. Solutions 1 and 2

Yo1

Zo:=YL

1RL j XL⋅+

:=

Now find the stub lengths corresponding to the two solutions.

Answers to Problems - Chapter 12

272

Now use the Smith chart. Normalize the impedance. Call it Point A.

z j jL =−

= −225 300

15015 2. .

Convert to admittance. Call it Point B. Then transform to Point C, 1 + j 1.65. This correspondsto the second analytical solution above.

From the Smith chart.

D = 0.180 - 0.055 = 0.125 8

L = 0.336 - 0.250 = 0.085 8

Answers to Problems - Chapter 12

273

Yo1

Zo:=YL

1RL j XL⋅+

:=

Now find the stub lengths corresponding to the two solutions.

λD2 0.28=D2if atan T2( ) 0> atan T2( ), atan T2( ) π+,( )

2 π⋅( ):=

Second Solution

λD1 0.396=D1if atan T1( ) 0> atan T1( ), atan T1( ) π+,( )

2 π⋅:=

First Solution

T2 5.236−=T2

XL− XL2 Zo RL−( )

RL2 XL

2+⎛

⎝⎞⎠

ZoRL−

⎡⎢⎢⎣

⎤⎥⎥⎦

⋅−−

Zo RL−:=

T1 0.764−=T1

XL− XL2 Zo RL−( )

RL2 XL

2+⎛

⎝⎞⎠

ZoRL−

⎡⎢⎢⎣

⎤⎥⎥⎦

⋅−+

Zo RL−:=

Now, if the load of impedance ZL is to be matched Zo line using a stub of impedance Zo. Let the distance from the load to the stub be D and the length of the stub be L.

ΩXL 75:=ΩRL 25:=ΩZo 50:=

54. Repeat Problem 53 for a 50-S line and an antenna of 25 + j 75 S. First the analyticalsolution

Answers to Problems - Chapter 12

274

The input susceptance of the line Zo without the stub attached. Solutions 1 and 2

βD1 2 π⋅ D1⋅:=

B1 Im YoYL j Yo⋅ tan βD1( )⋅+

Yo j YL⋅ tan βD1( )⋅+⋅

⎛⎜⎜⎝

⎞

⎠:= B1 0.045−= S

βD2 2 π⋅ D2⋅:=

B2 Im YoYL j Yo⋅ tan βD2( )⋅+

Yo j YL⋅ tan βD2( )⋅+⋅

⎛⎜⎜⎝

⎞

⎠:= B2 0.045= S

For a short-circuited stub we find the lengths as

L1

if atanYoB1

⎛⎜⎝

⎞

⎠0> atan

YoB1

⎛⎜⎝

⎞

⎠, atan

YoB1

⎛⎜⎝

⎞

⎠π+,

⎛⎜⎝

⎞

⎠2 π⋅( ):= L1 0.433= λ

L2

if atanYoB2

⎛⎜⎝

⎞

⎠0> atan

YoB2

⎛⎜⎝

⎞

⎠, atan

YoB2

⎛⎜⎝

⎞

⎠π+,

⎛⎜⎝

⎞

⎠2 π⋅( ):= L2 0.067= λ

Answers to Problems - Chapter 12

275

Now use the Smith chart. Normalize the impedance. Call it Point A.

z j jL =+

= +25 75

500 5 15. .

Convert to admittance. Call it Point B. Then transform to Point C, 1 + j 2.2. This corresponds tothe second analytical solution above.

From the Smith chart.

D = 0.088+ 0.191 = 0.279 8

L = 0.318 - 0.250 = 0.068 8

Answers to Problems - Chapter 12

276

metersd 0.012=d atanXZo

⎛⎜⎝

⎞

⎠

λ

2 π⋅⋅:=

metersλ 0.3=λ3 108⋅f

:=

Find the distance d required for the short-circuited line to look like a 2 nH inductance at 1 GHz. Equate the inductive reactance to the input impedance and solve for d. Assume the line has andielectric.

For a short-circuited line ( )Z jZ dinSC

o= tan β

ΩX 12.566=X ω L⋅:=

ω 6.283 109×=ω 2 π⋅ f⋅:=Hzf 109

:=

ΩZo 50:=HL 2 10 9−×:=55. Given

Answers to Problems - Chapter 12

277

metersd 0.288=

d if atanXZo

⎛⎜⎝

⎞

⎠0> atan

XZo

⎛⎜⎝

⎞

⎠, atan

XZo

⎛⎜⎝

⎞

⎠π+,⎛

⎜⎝

⎞

⎠

λ

2 π⋅⋅:=

metersλ 0.6=λ3 108⋅f

:=

Find the distance d required for the short-circuited line to look like a 2 nH inductance at 1 GHz. Equate the inductive reactance to the input impedance and solve for d. Assume the line has andielectric.

For a short-circuited line ( )Z jZ dinSC

o= tan β

ΩX 6.366−=X1−

ω C⋅:=

ω 3.142 109×=ω 2 π⋅ f⋅:=Hzf 500 106

⋅:=

ΩZo 50:=FC 50 10 12−×:=56. Given

64.

65. Crosstalk is often caused by placing lines close to each other when one or both of thelines are carrying strong signals. The problem can avoided by separating the lines (ifpossible) or making sure that the level on the lines is within specification.

66. AC power lines are a source of electrical noise that can be easily corrupt signals on theline.

Answers to Problems - Chapter 12

278

Zo1 ZLZoZo

0.088λ0.25λ

ZL 25 j 75⋅+:= Ω Zo 50:= Ω

First transform until the load impedance is real. The first time it is real is when the distanbetween the load and the matching section is 0.088 wavelengths. At this point the equivalent load impedance is 350 Ω, Let the characteristic impedance of the matching section be Zo1.

Zo1 350 50⋅:= Zo1 132.3= Ω

At the second point, the equivalent load impedance is 7.5 Ω (0.088+0.25 = 0.338 λ).

Zo1 7.5 50⋅:= Zo1 19.4= Ω

67. Magnetic field losses occur when the currents are induced in a nearby conductor from theinfluence of another conductor’s magnetic fields.

68. The sunlight can cause deteriation of the cable jacket or insulation.

72. Match a load of 25 + j 75 S to a 50-S line using a quarter-wavelength of matchingsection. Repeat this problem for ZL = 110 - j 50 S.

The second solution is the second time the load transforms to a real value.

Answers to Problems - Chapter 12

279

Zo1 ZLZoZo

0.338λ0.25λ

Answers to Problems - Chapter 12

280

Answers to Problems - Chapter 12

281

ZL 110 j 50⋅−:= Ω Zo 50:= Ω

First transform until the load impedance is real. The first time it is real is when the distanbetween the load and the matching section is 0.219 wavelengths. At this point the equivalent load impedance is 18.5 Ω, Let the characteristic impedance of the matching section be Zo1.

Zo1 18.5 50⋅:= Zo1 30.4= Ω

Zo1 ZLZoZo

0.219λ0.25λ

Zo1 ZLZoZo

0.469λ0.25λ

At the second point, the equivalent load impedance is 135 Ω (0.219+0.25 = 0.469 λ).

Zo1 135 50⋅:= Zo1 82.2= Ω

Now the other impedance.

The second solution is

Answers to Problems - Chapter 12

282

Answers to Problems - Chapter 13

283

1. The antenna converts electrical energy to electromagnetic energy.

2. Light waves and radio waves differ in frequency but both travel at the speed of light.

3. Electric field and magnetic field, these fields exist with any current carrying conductorand they create an electric field and are radiated.

4. These describe the direction of the electric field of an electromagnetic wave.

5. The electric field (E) and the magnetic field (H) which are perpendicular to each other.

6. Wavefront - a plane joining all points of equal phase in a wave.

7.

8.

9.

10.

Answers to Problems - Chapter 13

284

11. Inversely proportional to the square of its distance.

12. This is the degree of magnetism of a material in response to a magnetic field.

13. Radio waves are reflected by conductive material such as metal or the earth’s surface.

14. The radio wave is refracted as shown in Figure 13-4. Refractions occurs when wavespass from a medium of one density to another medium with a different density.

15. Diffraction - the phenomenon whereby waves traveling in straight paths bend around anobstacle. A shadow zone is an area following an obstacle that does not receive a wave bydiffraction.

16. It is the ratio of the reflected electric field intensity divided by the incident intensity.

17. Refraction - is the bending of a wave

18. Shadow zone - an area following an obstacle that does not receive a wave by diffraction

19. Ground wave, space wave, and sky wave

20. Ground wave - a wave that travels along the earth’s surface

21. Changes in terrain have a negative impact on ground waves, dry terrain has poorconductivity, salty water is a good conductor.

22. Attenuation of ground waves is directly related to the surface impedance of the earth. This impedance is a function of conductivity and frequency. The ground losses increaserapidly with increasing frequency.

23. ELF (30 to 300) Hz is used for submarine communications.

24. Space-wave propagation consists of direct and ground reflected space waves. A directwave travels from antenna to antenna while a reflected wave bounces off a surface suchas the earth.

25. Ghosting is when the same signal arrives at the TV receiver at two different times; thereflected signal has farther to travel and is weaker than the direct signal, resulting in adouble image.

Answers to Problems - Chapter 13

285

26.

27. Sky waves are radiated in a direction that strike the ionosphere, reflect to the ground andbounce back and so on. This produces skipping.

28. Refer to Figure 13-10.

29. Skip distance is affected by solar disturbances.

30. Critical frequency - the highest frequency that will be returned to the earth when transmitted vertically under given ionospheric conditionsCritical angle - the highest angle with respect to a vertical line at which a radio wave of a specified frequency can be propagated and still returned to the earth from the ionosphere.

Maximum Usable Frequency - the highest frequency that is returned to the earth from the ionosphere between two specific points on earth.

31. The optimum working frequency is one that provides the most consistent communicationwhile the MUF defines the maximum usable frequency.

32. Tropospheric propagation, 350 MHz to 10 GHz, and most VHS frequencies behave thesame.

33. Frequencies below 30 Mhz.

34. The skip zone occurs for a given frequency, when propagated at its critical frequency. The skip distance is the minium distance from the transmitter to the skywave that can bereturned to earth (due to the density of the atmosphere).

35. The reflection causes a 180o phase shift of the skywave.

36. See Figure 13-15. The signal is reflected beyond the earth’s horizon enabling reliablecommunications up to 400 miles. It provides for reliable links in deserts, mountainregions, and between islands.

37. Diversity is used to select the best received signal using multiple receivers to provide asummed receive signal.

Answers to Problems - Chapter 13

286

38. Space diversity - comprising two or more receiving antennas separated by 50 8 or more.Frequency diversity - transmission of the same information on different frequencies.Angle diversity - transmission of the same information at two or more more slightly different angles.

39. Skipping - the alternate refracting and reflecting of a sky-wave signal between theionosphere and the earth’s surface.

40. Fading - variation in signal strength that may occur at the receiver over a period of time.

41. Frequencies above the critical frequency will continue on out into space.

42. Satellite communications involves the use of a satellite to relay a signal from an uplinkon earth, amplify the signal and translate the frequency using a transponder and re-transmit the signal back to earth.

43. GEO - Geosynchronous orbit satellites, fixed location of 22,00 miles above the earth,these satellites have a long delay time due to the distance traveled.LEO - Low Earth Orbit satellites, small signal delay, location not fixed.

44. VSAT - see pages 647-48MSAT - a mobile satellite system

45. Frequency division multiple access - operates on different frequencies based on whichchannel is available. TDMA uses one carrier for the communication link, selectivity isaccomplished in time rather than in frequency, and is well suited for digitalcommunications.

46. This answer will vary based on geographic location.

47. Azimuth = 169.64o Elevation = 48.73

48. GPS satellites transmit two signals, a course acquisition (C/A) signal transmitted on1575.42 MHz, which is available for civilian use, and a precision code (P-code),transmitted on 1227.6 MHz and 1575.42 MHz, which are for military use only.

49. using the on-line calculator, distance . 38298 km, round trip delay . .25549 seconds.

Answers to Problems - Chapter 13

287

50. refer to figure 13-18, perigee - closest distance of the orbit to earth and apogee - farthestdistance of the orbit from earth.

51. altitude – 485 miles orbital period – 100 minutes and 28 seconds66 satellite using near polar-orbit

52. NF(dB) = 10 log (100/290 + 1) = 1.29 dB

53. 28.65 K (using http://web.nmsu.edu/~jbeasley/Satellite/)

54. 204.4 dB (using http://web.nmsu.edu/~jbeasley/Satellite/)

55. The uplink transmit power is 4.5 W. Both the uplink and downlink C/N are well withinlimits specified in the problem.

56. EMI causes distortion to the audio and causes vertical bands of dots moving on thescreen.

57. Install a high-pass filter between the antenna and receiver input on the lead-in wire or, ifpossible, eliminate the source of the interference.

58. The “ghost” signal adds noise to the desired received signal, the noise reduces the systemperformance.

59. Try changing the orientation of the receive antenna.

60. EMI typically shows up as vertical bands. RFI displays itself as several bars or wavylines and strong RFI will cause a complete loss of the TV picture or it will cause garbledsound in FM.

61. Via the antenna or lead-in wire, power-line, sources include microwave ovens andmotors.

62. EMI has differing symptoms, see the problem 56 answer.

Answers to Problems - Chapter 13

288

63.

64. The skip distance is how far away a received signal is obtained. The skip zone definesthe region where no signal is received, found or reflected.

65. This varies with frequency. You can expect variations from day to night and with solaractivity.

Answers to Problems - Chapter 14

289

1. The transmit antenna must be vertically polarized. The receive antenna must also bevertically polarized.

2.

3.

4.

5.

6. Field intensity - this is measured in terms of the field strength which is inverselyproportional to the distance from the transmitter.

7. Polarization - the direction of the electric field of a given electromagnetic radiated signal.

8. This refers to the fact that an antenna will work equally well for transmit and receivegiven adequate power boundaries.

9. Refer to Figures 14-2 and 14-3. The half-wavelength dipole is comprised of two quarter-wave sections. The total length of the two sections is ½ 8.

10. See Figure 14-2.

Answers to Problems - Chapter 14

290

12. The impedance value for the half-wave dipole varies from about 2500 S at the open endsto 73 S at the source ends.

13. The reception of the induction field requires that the receive antenna be close to thetransmit antenna. For practical purposes, it does not radiate a field.

14. See Figure 14-5.

15. The angular separation between the half-power points on an antenna’s radiation pattern.

16.

17.

18.An antenna with a gain of 4.7 dBi has a gain of (4.7 - 2.15) dBd = 2.55 dBd, which is lessthan 2.6 dBd.

19. The size of the antenna below 2 MHz is physically too large.

20. A half-wave dipole has an approximate physical length of one-half wavelength of theapplied frequency

21.

Answers to Problems - Chapter 14

291

22.

23. near field region - less than 20/8 from the antennafar-field region - region greater than 2D/8 from the antenna.

24. Radiation resistance - the portion of an antenna’s input impedance that results in powerradiated into space.

25.

26. For a given power, if antenna current increases , the effect is as if the radiation resistancedecreases.

27. The value falls steadily to a minimum value of 70S at a height of 8/2 above ground. Thevalue rises and falls by several ohms.

28. The ground has an effect on the radiation resistance

29.

30. (a) Physical length - this affects the radiation resistance, also the physical length ~ 85%of the electrical length.(b) electrical length - determined from 8 = speed of light / frequency(c) polarization - the direction of the electric field of a given electromagnetic field(d) diversity reception - using multiple antennas with different heights and angles to provide a choice for obtaining the best signal(e) corona discharge - luminous discharge of energy by an antenna from ionization of the air around the surface of the conductor

Answers to Problems - Chapter 14

292

31. The physical length can be approximated to be about 95% of the calculated electricallength.

32. Antenna length is an issue and the resulting impedance. A properly matched antenna willlook resistive.

33.

Electrical length = 0.95 x 240 = 2238 metersphysical length = 123.45/228 =~ .54 8

34.

35. The antenna feed line is a transmission line. A resonant feed line is not widely used; it isinefficient and length is critical, however impedance is not an issue.

36. A non-resonant feed line is more widely used, it has negligible standing waves and itsoperation is practically independent of its length.

37. Delta-match - impedance matching device that spreads the transmission line as itapproaches the antenna. It is convenient to use when the transmission line does not havea characteristic impedance sufficiently low to match a center-fed dipole.

39. See Figure 14-10(b)

40.

41. See Figure 14-10(c) and the related discussion.

42. The monopole (vertical) antenna.

Answers to Problems - Chapter 14

293

43. P = I2R

44. The monopole (vertical) type antenna requires a conducting path to ground and the half-wave dipole does not.

45. See Figure 14-13. Yes the reciprocity theorem applies.

46. Image antenna - the simulated 8/4 antenna resulting from the earth’s conductivity with avertical antenna.

47. The ground plane would be the roof of the car (assuming the roof top is a conductivesurface). If the whip antenna is placed on the bumper, it distorts the radiation patterns andincreases the directivity of the antenna.

48. They provide the ground plane if the actual earth ground can not be used. Bad reflectionswill result in poor radiation.

49. The series inductor tunes out the capacitive appearance of an antenna so the antennsappears resistive.

50. The antenna becomes highly capacitive.

51. By adding a loading coil to the antenna.

52. The resonant frequency would increase.

53. Top loading enables maximum possible radiation.

54. The loading coil can be used to compensate for the fact that the vertical antenna is not afull quarter wavelength.

55. A parasitic element is located 8/4 behind the dipole. The parasitic element reflects the

waves doubling the signal propagated in that direction.

56. (a) driver element - an antenna element that is excited through a transmission line.(b) parasitic elements - not electrically connected(c) reflector - the parasitics that effectively reflects energy from the driven element(d) parasitic element - effectively directs energy in the desired direction.

57.

Answers to Problems - Chapter 14

294

58. (a) F/B = 7 - (-3) = 10 dB (b) F/B = 18 - (5) = 13 dB

59. See Figure 14-17

60. See Figure 14-18, any combination of half-wave elements are excited by a connectedtransmission line. This results in a more directed antenna.

61. See Figure 14-19, a group of half-wave elements is mounted vertically. Thisarrangement provides greater directivity in both the vertical and horizontal planes.

62. Refer to Figure 13-20 and the drawing at the intersection of the phase difference = 0 andthe spacing = ½ 8.

63. This is accomplished using a vertical array.

64. The changes in the ionized layers of the atmosphere will change the sky-wave coverage(refer to Figure 13-10).

65. Phased array - combination of antennas in which there is control of the phase and powerof the signal applied to each antenna resulting in a wide variety of possible radiationpatterns.

66. A parasitic array is developed when one or more of the elements in an antenna is notdriven.

67. A log-periodic antenna is a special case of a driven array, good gain over an extremelywide range of frequencies. The longest and shortest dipoles are a half-wavelength for theupper and lower frequencies.

68. (a) horizontal half-wave dipole (see Figure 14-4)(b) vertical half-wave dipole (see Figure 14-5)(c) monopole (vertical) loop antenna

(d) horizontal loop antenna (see Figure 14-11)(e) vertical antenna (see Figure (14-22)

69. It is bi-directional. see Figure 14-22

70. Ferrite loop antennas are found in most AM receivers. The large number of loops woundaround a ferrite core serves to greatly increase the effective diameter of the loops. Itenables the reception of lower frequencies using a small antenna.

Answers to Problems - Chapter 14

295

71. The radiated resistance is 288S, it offers relatively broadband operation. The foldeddipoles provide a high input impedance for Yagi-Uda antennas.

72. A smooth surface can be obtained (see Figure 14-24). Phase arrays can be built into thewings of the aircraft, the phase arrays provide for a directive radiation pattern.

73

74. Bandwidth specifies the usable frequencies, beamwidth describes the radiation pattern.

75. VSWR is a measure of the standing waves on the transmission line. A high VSWR canindicate a transmission line or antenna problem.

76. Check VSWR, it should be low, visually inspect the transmission line and the antenna,look for weather damage or ice build-up.

77. Grid-dip meter - a device that measures the resonant frequency of tuned circuits andantenna.

78. The SWR meter displays the standing wave ratio, see Figure (14-29).

79. A high VSWR indicates some type of problem. Excessive VSWR can damage thetransmitters output power amplifier or cause heating and eventual failure of either thetransmission line or amplifier.

80. Anechoic chamber - a large enclosed room that prevents reflected electromagnetic wavesand shields out interfering waves. The room has to be large enough so that the testmeasurement receiver is at least 2 wavelengths away from the antenna.

81.

82 You need to know the TX output power and the antenna gain.

Answers to Problems - Chapter 14

296

84.

Answers to Problems - Chapter 15

297

1. Factors should include:(1) initial cost and long-term maintenance(2) frequency band(3) selectivity and privacy(4) reliability and noise characteristics(5) power level

2. Waveguide - a medium that guides a wave, generally it is a transmission line, hollowmetal tube or pipe, that conducts electromagnetic waves through its interior.

3. The coaxial transmission line has high attenuation at high frequencies, the waveguide hasminimal loss at the same frequencies.

4. TE - Transverse electric, TM - transverse magnetic, the wave that is propagated by awaveguide is electromagnetic and therefore has an E-field and an H-field. If nocomponent of the E-field is in the direction of propagation, we say that it is the TE mode. TM is the mode whereby the magnetic field has no component in the direction ofpropagation.

5. TE - Transverse electric, TM - transverse magnetic, the wave that is propagated by awaveguide is electromagnetic and therefore has an E-field and an H-field.

6. TE10, the most “natural” one for operation. It has the lowest cutoff frequency.

7. This defines the range of sizes of waveguide that can be used for a given length.

8.

9. Refer to question 4.

10. a = 8/2 (TE10) a for TE20 = 2 half-wavelengths

Answers to Problems - Chapter 15

298

11.

12. The wave that travels in a waveguide propagates down the guide at less than the velocityof light. This is called the group velocity. In Smith chart calculations, the group velocityshould be used when making moves, not the free-space wavelength.

13. The cross-sectional area of a circular waveguide must be more than double that of arectangular guide.

14. The cross-sectional area of a circular waveguide must be more than double that of arectangular guide. Circular waveguide can be rotationally symmetrical which means itcan be rotated with no electrical disturbance.

15. ridged-waveguide: Advantage - allows operation at lower frequencies for a given set ofoutside dimensions; Disadvantage - expensive to manufacture

16. See Figure 15-10, outside section is covered with a soft dielectric such as rubber, insidehas spiral wound ribbons of brass or copper.

17. The main advantage is in applications where space is a premium.

18. Operating frequency and the choice of dielectric. At high frequencies the current tends toflow only on the surface of the conductor . Coaxial cable will have a significant loss athigh frequencies which reduces its power handling capacity.

Answers to Problems - Chapter 15

299

19. The purpose of the waveguide is to carry the guided wave from point A to point B. Becareful of dents, dings, twists, or severe bends. Also keep the waveguide pressurized tokeep out contaminants.

20. Abrupt changes in the waveguide path will cause disruptions in the waves.

21. See Figure 15-12 and the related discussion.

22. Slide - screw tuner [see Figure 15-14(a)]. The effect of the protruding object is toproduce shunting reactance across the guide.Double-slug tuner [see Figure 15-14(b)]. This tuner provides for the adjustment of thelongitudinal position of the slugs aand the spacing between them.

23.

24.

25. See Figure 15-15 (a) fill the end of the waveguide with graphited sand; (b) highresistance rod; (c) use a wedge of resistive material

26. Flap attenuator - attenuation is accomplished by inserting a thin card of resistive materialthrough a slot in the top of the guide [see Figure 15-16(a)]vane attenuator - attenuation is provided by changes in the sides, minimum (vanes areclose to the side walls), maximum (vanes are in the center)

27. See Figure 15-17, a directional coupler consists of two pieces of waveguide with one sidecommon to both guides and with two holes in the common side.

Answers to Problems - Chapter 15

300

28.

29. Capacitive coupling (Probe - see Figure 15-18) the probe is excited by an RF signalsetting up an electric field. The probe is in the center of the waveguide and 8/4 from theshort-circuited end.

30. Inductive (loop) coupling [see Figure 15-19], the current flow in the loop sets upmagnetic field inside the guide.

31. Slot (aperture) coupling [see Figure 5-20], slot A is at an area of the maximum E-field,slot B is at an area of the maximum H-field, slot C is at an area of the maximum E and H-field and is a form of electromagnetic coupling.

32. (a) as frequency increases, the size decreases(b) TE - Transverse electric, TM - transverse magnetic, the wave that is propagated bya waveguide is electromagnetic and therefore has an E-field and an H-field. If nocomponent of the E-field is in the direction of propagation, we say that it is the TE mode. TM is the mode whereby the magnetic field has no component in the direction ofpropagation.(c) Probe, loop, and aperture coupling is used(d) The wave that travels in a waveguide propagates down the guide at less than thevelocity of light. This is called the group velocity. In Smith chart calculations, the groupvelocity should be used when making moves, not the free-space wavelength.

33. Cavity resonators are metal-walled chambers filled with devices for admitting andextracting electromagnetic energy.

34. Resonant cavity walls are made of highly conductive material [see Figure 15-21].Resonant mode occurs at frequencies for which the distance between end plates is a half-wavelength or multiples of half-wavelengths.

35. Waveguide - a medium that guides a wave, generally it is a transmission line, hollow

metal tube or pipe that conducts electromagnetic waves through its interior

36. See Figure 15-22, the movement of the disk may be calibrated in terms of frequency.

37. Cavity volume, cavity inductance, cavity capacitance.

38. A radar uses reflected waves to determine the direction and distance of a target.

Answers to Problems - Chapter 15

301

39. Radars typically use high power and operate at high frequencies, coaxial cable high lossat high frequencies.

40. (a) Target - the object struck by the radar’s radio waves(b) Echo - part of the returning radar energy collected by the antenna and sent to thereceiver

(c) Pulse repetition rate - number of pulses transmitted per second(d) Pulse repetition time - the time from the beginning of one pulse to the beginning ofthe next(e) Pulse Width - the duration of the time in transmitting energy(f) Rest Time - the time between pulses(g) Range - the time it takes for a pulse of energy to travel to a target and return

41.

42.

43. Double Range Echoes are produced when the reflected beam makes a second round trip.

44. A target too close will return an echo before the transmitter turns off, masking the echo.

45. Long duty cycles allow the transmitter output components to be physically much smaller. A short pulse is advantageous for seeing closely spaced objects.

46.

Answers to Problems - Chapter 15

302

47. Refer back to the basic radar block diagram section

48.

49. Doppler Effect is the phenomenon whereby the frequency of a reflected signal is shiftedif there is a relative motion between the source and the reflecting object. A common useis for measuring the speed of objects in sports.

50. Doppler radars are always on, hence the term continuous wave (CW).

51. refers to the reflection of the radio waves striking the RFID tag and reflecting back to thetransmitter source.

52.1. means of powering the tag2. frequency of operation3. communication protocol

53. power is provided by rectifying the RF energy transmitted by the reader that strikes theRF tag antenna.

54. the two dipoles of the dual dipole antenna are oriented at 90o angles to each other whichmeans that tag orientation is not critical

55.1. can incorporate wireless Ethernet connectivity2. can incorporate location capability3. the unit is always turned “on”

56. LF – 125/134 kHz, HF - 13.56 MHz, UHF – 860-960 MHZ and 2.4 GHz

57. Refer to Figure 15-31 and the related discussion.

58. A dielectric waveguide is a waveguide with just a dielectric (no conductors) used toguide electromagnetic waves. Advantage - easy to manufacture; Disadvantage - higherloss

Answers to Problems - Chapter 15

303

59.

60. See Figure 15.32

61. Check that the line is pressurized; check flanges; look for evidence of arcing; inspect forworn components, cracking, and corrosion.

62. Make sure joints are properly fitted.

63. The most likely problems are with joints or other flanges.

64. Refer to Figure 14-35, connect the power meter to the reflected coupler.

65.

Answers to Problems - Chapter 15

304

66. See Figure 15-13, yes, the receive signal can be sent to the transmitter without affectingthe transmit signal.

67.

68. See Figure 15-36 and the related discussion.

Answers to Problems - Chapter 16

305

1. The shape of the different types of antennas focus the RF signal and provide gain.

2. See Figure 16-1.

3. These antennae are highly directive. See Figure 16-12, equation 16-1, and Example 16-1.

4. Refer to Figure 16-2.

5.

6.

7. Radome - a low-loss dielectric material used as a cover for a microwave antenna. It isused for protecting the antenna from the environment.

8. Zoning - a fabrication process that allows a dielectric to change a spherical wavefrontinto a plane wave. See figure 16-4.

9. fr = 1.3 Ghz BW . 10% of fr = 130 Mhz.

10.

Answers to Problems - Chapter 16

306

11. focal length = D / 16h = 5 / (16)(1.2) = .26 meters

12.

13. .

The beamwidth is narrower as the wavelength gets smaller.

14.

15. A magnetron produces a microwave frequency output within its enclosure without theuse of external components such as crystals, inductors, capacitors, etc. Basically it is adiode without the grid.

16. See Figure 15-7

17. The open space between the plate and the cathode where the electric and magnetic fieldsexert force on the electrons.

18. negative resistance - a static negative resistance between its electrodeselectron-resonance - the oscillation is dependent on the transit time for the electrons totravel from the cathode to the plate. Hint: go on-line and search magnetron class

Answers to Problems - Chapter 16

307

19. See Figure 16-11 and the related discussion.

20. - apply the modulating signal to a modulation grid to amplitude modulate the signal- TWT mixer, provides gain and mixing- TWT oscillator- TWT amplifier

21. Helix of the TWT amplifier is longer than that of the amplifier, no input connection,lossy wire attenuator is eliminated, this allows forward and backward waves. Theoperating frequency is determined by the pitch of the tubes helix.

22. TWTs are widely used in wideband communication repeater links and communicationsatellites. Advantages: long life, reliable, and high output power.

23. BWO - backward-wave oscillator, TWT that allows both forward and backward wavesand can be used as an oscillator.

24. Velocity modulation - an electron beam moving along in bursts of electrons, the bunchesof waves resemble ocean waves.

25. There are more solid-state devices for microwave operation. The advantage is typicallysimplicity, the disadvantage is that solid-state devices are low power devices.

26. The discovery made was that microwaves could be generated by applying a steadyvoltage across a chip of n-type gallium arsenide crystal. A common application is thehand-held radar guns used by the police. The Gunn oscillator is small, rugged, and has alow manufacture cost, good efficiency, and a lack of vacuum or filaments.

27. IMPATT - impact ionization avalanche transit time, operates in the avalanche break-down region.

28. See Figure 16-16, as the voltage rises above the breakdown during the positive half-cycle, a sharply peaked charge lags the RF voltage by 90o. Maximum negative resistanceis obtained when Tt . 0.74B (the transit angle).

29. The important difference is in their modes and in thermal design. IMPATT diodesoperate in the avalanche breakdown region. Varactor diodes normally operate in areverse bias mode.

Answers to Problems - Chapter 16

308

30. See Figure 16-17.

31. Ferrites are compounds of iron, zinc, manganese, magnesium, cobalt, aluminum, andnickel oxides. They are manufactured by pressing into shape the required mixture offinely divided metallic oxide powders.

32. The ferrite allows energy to travel in one direction but absorbs energy traveling in theopposite direction. Refer to Figure 16-21 and the related discussion.

33. Faraday Rotation Shift - when microwaves are passed through a piece of ferrite in amagnetic field and their frequency is above the resonant frequency of the ferriteelectrons, the plane of polarization of the waves is rotated.

34. circulator - a ferrite device that provides isolation between the input and output power,see Figure 16-25 and the related discussion.

35. ferrite bead - small bead of ferrite material that can be threaded onto a wire to form adevice that offers no impedance to dc and low frequencies, but a high impedance at RF.

36. parametric amplifier - provides low-noise amplification at microwave frequencies via thevariation of reactance. The distance between capacitor plates is varied. This is called thepump force (see Figure 16-27).

37. degenerate mode - parametric amplifiers with pump frequencies double the signalfrequencynon-degenerate mode - parametric amplifiers with reduced gain with frequencies otherthan two-times the signal frequency. This allows the parametric amplifier to functiondirectly as a mixer, with its output being the first IF frequency.

38. See Figure 16-29 and the related discussion.

39. maser - microwave amplification by stimulated emission of radiation, ammonia iscapable of being used to amplify one frequency,

40. laser - low-noise light wave amplifier, it differs from a maser in terms of the frequenciesinvolved. Lasers have multiple uses; examples include, medicine, communications,military, engineering, CD player.

41. See Figure 16-31 and the related discussion. A temperature change can halt the lasingaction.

42. It is limited to line of sight and the beam can easily be attenuated or disrupted by debris

Answers to Problems - Chapter 16

309

or weather.

43. Transphasor - an optical switch using a laser beam, see Figure 16-32. Its importance tocomputers is the expected operating speed.

44. Linear (see Figure 16-34)switching (see Figure 16-35)

45. Bad active device, failure in the regulated power supply.

46. This breaks the feedback path which will shut down or cause the switching power supplyto operate poorly.

47. Visually inspect the system, check power supplies, check for proper input drive level.

48. This can be caused by cable problems or input drive levels.

49.

50. GaAS (gallium arsenide) devices are preferred above 5 Ghz, BJTs are used below 5 Ghz. Tubes are preferred above 20 GHZ and for higher power applications.

51. See Figure 16-21, this arrangement attenuates frequencies at the resonant frequency ofthe electrons in the ferrite. Changing the strength of the dc field produces a change in thefrequency that is attenuated. A ferrite vane extending into the waveguide can be used toprovide variable attenuation.

Answers to Problems - Chapter 16

310

52.

Answers to Problems - Chapter 17

311

1. The 4:2:2 represents the sample rate for the luminance (Y), the R-Y (R = red), and B-Y(B = blue) components, see Table 17-1.

2. MPEG2 uses statistical data redundancy, psychvisual redundancy, entropy,luminance/chrominance contrast, spatial redundancies, and temporal redundancies.

3. The AC-3 audio channel has 5 audio channels [left, right, center, surround left and right]plus one low frequency enhancement channel.

4. See Figure 17-2

5. See Figure 17-3; it is an 8-level vestigial sideband modulator.

6. The first byte in each MPEG2 packet is used to synchronize its circuits. ATSC Pilot Carrier - provides a clock reference for the 8VSB receiver to lock onto.Segment Sync - a repetitive 1-byte pulse that is added to the front-end of the datasegmentFrame Sync - data pulse repeated once every 313 data segments to identify that a frame has been completed

7. Frame synchronizer Data Randomizer, Read Solomon Encoder, Data Interleave, TrellisEncoder, Pilot Insertion. Refer to Figure 17-2 and related discussion on pages803 and804.

8. Trellis coding breaks up an 8 bit byte into foot 2 bit words. A 3-bit code is generatedfrom the present and past 2-bit words, hence 2/3 encoding.

9. 19.39mbps

10. 1. Video data stream (MPE62 encoding)2. Audio data stream (AC-3 encoding)3. PSIP, program, Information, control, ect

11. This is the HDTV signal for channel 7

12. It IDs the channel

13. It is used by the digital receiver to lock onto the digital data stream so the multiplexeddata can be recovered.

14. It is the power ratio of the pilot signal to the average channel power level.

Answers to Problems - Chapter 17

312

15. It is used by test equipment to synchronize itself to the TX PN-23 code sequenceenabling a bit by bit comparison to determine the batteries.

16. See figure 17-3

17. A perfectly compacted 8VSB signal

18. The point where there is a lot of multiple echo and you have reached the threshold ofvisibility.

19. Sound (Aural) FM

20. Both are frequency modulated, maximum intelligence frequency = 15 kHz for both, DR-TV = 1.67, DR-FM = 5

21. Video (Visual) AM

22. Bandwidth savings

23. TV camera - converts the visual information into an electrical signalmicrophone - converts the voice into an electrical signal speaker - provides an output device for the amplified audiopicture tube - provides a visual output deviceTransmit antenna - used to broadcast the visual and aural signalReceive antenna - used to receive the RF signal

24. The diplexer allows for both the visual and aural carrier to be fed into the same antenna.

25. Photosensitive cells are arranged in a two-dimensional array. Light, striking the cells inthe array, is converted to an electronic analog and stored in the cells. The electronicinformation is serially shifted out of the device. This is called a “bucket brigade.”

26. See Figure 17-18(b); it is an array of pixels.

Answers to Problems - Chapter 17

313

27. Refer to Figure 17-18

28. It is the ratio between the frame width to the frame height.

29. NTSC (4:3) HDTV (16:9)

30. Provides for horizontal and vertical synchronization of the video signal for the electronbeam.

31. When reading you horizontally scan a line (left-to-right) then return to the left and beginreading the next line.

32. The scanning lines remain the same.

33. 30 frames per second (actually 29.97)

34. See Figure 17-19; interleaving fields of 242.5 horizontal lines (even/odd lines)

35. To reduce flicker.

36. They are used to keep the horizontal and vertical signals locked.

37. The human eye perceives the decay of the TV screen phosphor as flicker. This problemis minimized by using 60 fields.

Answers to Problems - Chapter 17

314

Horizontal sync pulses must occur once for each of the 525 lines that occur in 1/30 sec.

From 428 lines to 3.5 MHz × 53.5 µsec × 2 = 375 lines

38. blanking pulses, see Figure 17-20 and 17-21. This is used to blank out the electron beamduring retrace.

39.

40. field frequency = 59.94 Hz

41. To minimize interference with the electric company’s 60 Hz power line frequency. Itallows for good stability of the receiver’s oscillator.

42. See Figure 17-20, 17-21 and 17-22.

43. Amplifier with bandpass characteristics from dc up to 4.2 MHz.

44. Resolution - ability to resolve detailed elements in a TV picturevertical resolution - number of horizontal lines that actually make up a TV displayhorizontal resolution - number of vertical lines that can be resolved in a TV display

45.

46.

47.

Answers to Problems - Chapter 17

315

48. 87.75 MHz

49. 100% modulation implies +/- 25 MHz deviation of the aural carrier.

50. 6 aural carrier (87.75 MHz) is directly below the start of the FM band

51. Vestigial sideband operation is a form of amplitude modulation in which one of the sidebands is partially attenuated.

52. See Figure 17-23 and the related discussion.

53. approx 1 kV/each diagonal inch, therefore 14" x 1 kV/in = 14 kV.

54. Front end (1) provides amplification, (2) prevents the local oscillator from being driveninto the antenna and thus producing RFI, (3) steps the received RF signal down to thefrequency required for the IF stages, and (4) provides proper impedance matchingbetween the antenna feed-line into the tuner.

55. See Figure 17-24

56. Stagger tuning is cascading a number of tuned bandpass filters, each having a slightlyoffset bandpass frequency. This provides a wider flat bandpass with steep high-0 andlow-frequency roll-off skirts.

57.

58. This is done to reverse the vestigial sideband characteristics generated at the transmitter. If all the receiver IF responses were equal for all video frequencies, the lower ones wouldhave excessive output levels.

59. wavetrap - high-Q bandstop circuit that attenuates a narrow band of frequencies. Thewavetrap eliminates interference from the adjacent channel carrier frequencies.

60. No, the signal is obtained from the common IF and through the video detector.

61. Restores the dc portion of the video signal that is often removed by amplifier coupling.

Answers to Problems - Chapter 17

316

62. Separates the horizontal sync pulses, a low-pass filter removes the 59.94 Hz sync signaland a high-pas filter removes the 15.75 kHz sync signal.

63. Clipper

64. See Figure 17-31, horizontal sync pulses drive the horizontal oscillator. These pulses areamplified to a powerful level and applied to a high voltage transformer (flyback). Thisdrives the horizontal yoke windings and provides high voltage to the CRT. Itscomplexity and power level make it prone to failures.

65. See Figure 17-32

66. See the horizontal deflection circuit section (pgs. 832-834)

67. The damper shuts out unwanted damped oscillations during the flyback period.

68. See Figure 17-34, Interleaving

69. See Figure 17-37 and the related discussion

70. The compatible color transmission can be reproduced in black-and-white shades by amonochrome receiver. This requires the use of the same 6 Mhz bandwidth.

71. The color subcarrier is 3.579545 MHz. The color information is clustered around theblack and white signals using a technique called interleaving.

72. See Figure 17-38 and the related discussion.

73. The color crystal frequency is used to allow proper signal detection. The color burst isused to keep the receiver properly phase locked.

74. The chroma killer prevents the output from the chroma circuits to be transmitted with amonochrome signal. A defective color killer results in a colored noise, called confetti.

75. See Figure 17-40 and 17-41, static convergence - proper beam convergence at the centerof a CRT, dynamic convergence - beam convergency away from the center of a CRT.

76. See Figure 17-42, the TV stereo signals remain noise free well beyond that of the FMbroadcast band. This is partly possible because of the increased amplitude of the (L-R)signal.

Answers to Problems - Chapter 17

317

Channel is 192-198 MHz (from Table 16-1).Sound carrier is 0.25 MHz below upper channel frequency.f sound carrier = 198 MHz-0.25 MHz= 197.75 MHzf video carrier =lowest channel fequecy + 1.25 MHz

= 192 MHz + 1.25 MHz = 193.25 MHz

LO frequency is 41 MHz above upper channel frequency = 198 MHz = 41 MHz= 239 MHzˆIF freq. sound carrier = 239 MHz-197.25 MHz = 41.25 MHz And IF freq. Picture carrier = 239 MHz - 193.25 MHz = 45.75 MHz

77.

78. HDTV (High Definition TeleVision), the HDTV signal can carry multiple SDTVprograms over a single television channel or it can be used to transmit one channel withgreater quality (resolution).

79. Check antenna, tuner, and IF amps, the raster is the illuminated area on the picture of aTV receiver when no signal is being received.

80. Possible tuner problem, check sound IF amplifiers, detector, audio amplifier, and speakerconnection.

81. No 45 MHz signal to the RF amplifier, therefore no reception for UHF signals.

82. You will see the raster.

83. You won’t have gain control of the previous amplifying stages.

84. The overall picture will be black or very dark.

87. The .7 is an estimation of the maximum detail that is visible. You have to be able todistinguish space between the black and white lines.

88.

Answers to Problems - Chapter 17

318

89. Contrast - adjusts the level of the video signalBrightness - varies the dc level applied to the control grid or cathode of the CRT. It controls overall picture brightness.

90. See Figure 17-34. The color information is centered in clusters between the black andwhite signals.

91. Poor SNR will result in frozen pictures and repeated pxels (called pixelate). There mayalso be adjacent channel problems with digital and NTSC transmissions.

Answers to Problems - Chapter 18

319

1. See Figure 18-1, modulator, connector, optical fiber, light detector.

2. (1) bandwidth (2) immunity to electrostatic interference (3) elimination of crosstalk (4)security (5) lower signal attenuation

3. Refractive index - ratio of the speed of light in free space to its speed in a given material.

4.

5. 6.

7. Infrared light - the electromagnetic waves just below the frequencies in the visiblespectrum.

8. See Table 18-1

9. See Table 18-1

10. See Figure 18-5

11. Pulse dispersion - a broadening of received pulse width because of the multiple paths taken by the light. This limits the maximum distance and data rates.

12. Multimode - fibers with cores of about 50 to 100 :m that support many waveguidemodes; light takes many paths.

13. Graded-index fiber was developed to overcome the pulse dispersion problem. Core sizes of 50 and 62.5 :m are commonly used and both have 125 :m cladding.

14. Single mode fibers are used in high data rate and/or long distance systems.

15. Core diameters of 7 to 10 :m are typical. The cladding size is 125 :m.

16. Mode field diameter is the actual guided optical power distribution diameter.

Answers to Problems - Chapter 18

320

17. The wavelength at which the material dispersion and waveguide dispersion cancel oneanother.

18. short-range sensors (on the manufacturing floor) and in short data lengths.

19. attenuation and dispersion

20. scattering, absorption, macro-bending, micro-bending

21. dispersion - broadening of a light pulse as it propagates through a fiber strand

22.

23. chromatic, polarization, and modal

24. acts like an equalizer, canceling dispersion effects and yielding close to zero dispersionin the 1550 nm region

25. Refer to Table 18-6

26. The diode laser is basically an optical oscillator. When forward biased, a large numberof free holes and electrons are created near the junction. The collision of holes andelectrons produce a photon of light. At some point, a density level is reached where therelease of one photon can trigger several more. The wavelength is determined by thematerials used.

27. Dense Wavelength Division Multiplex incorporates the propagation of severalwavelengths in the 1550 nm range of a single fiber.

28. A laser in which the fundamental wavelength can be shifted a few nanometers, ideal fortraffic routing in DWDM systems.

29. An in-line passive device that allows optical power to flow in one direction only.

30. reduce the signal level

31. isolators, attenuators, branching devices, splitters, couplers

32. They are used to convert the transmitted light back into an electrical signal.

Answers to Problems - Chapter 18

321

33. DFB - a more stable laser suitable for use in DWDM systems

34. VCSEL - lasers with the simplicity of LEDs and the performance of lasers.

35. axial misalignment, angular misalignment, air gap, rough surface, numerical aperturedifferences, core-size differences, core concentricity or offset, core ellipticity

36. Fusion splicing is a long-term method and should always be used if the cost is justified.

37. SC, ST, MT-RJ

38. The ends on mechanical splices must be polished. Connections with air gaps require theuse of index matching gel.

39. GENERAL RULE - do not splice single mode and multimode together. In anemergency, small-to-larger core splicing results in minimal insertion loss, larger-to-smaller core splicing results in significant insertion loss and an increase in reflectedpower.

40. link distance, bit-rate, received signal level

41. Long-haul - intercity or interoffice class of system, high bit-rate, high channel density,high reliability, redundant

42. RSL - received signal level

43. Accounts for the system degradation due to addition of link splices, added losses due towear and tear, misalignment, and repairs.

44. The fiber length can exceed the cable run by .5% to 3% due to the construction of theloosely enclosed fiber in the buffer tubes.

45.

46. cable losses, splice losses, connector losses, losses due to splitters, couplers, WDMdevices, patch panels, etc.

47. environment, exposure factors, pulling tensile, bend radius

Answers to Problems - Chapter 18

322

48. OTDR - optical time-domain reflectometer, sends a light pulse down the fiber andmeasures the reflected light which provides some measure of performance for the fiber.

49. A- dead zone B- begins useful trace info

C- spliceD- termination at the fiber endE- end of the fiber

50. speed, decreasing cost of fiber, increasing data capacity

51. FTTC - fiber to the curb FTTH - fiber to the home

52. OC-192 is a high speed (~10 Mbps) connection

53. The fiber uses high-speed dedicated connections that provide high bit-rates over increaseddistances.

54. FDDI - fiber distributed data interface uses two counter-rotating rings to increase datathroughput.

55. The average distance between stations can not exceed 200 meters, therefore the distance tothe last stage can not exceed 208 meters.

56 .

57.

58.

Answers to Problems - Chapter 18

323

59.

324

Chapter 1 - Troubleshooting with EWB Multisim - Solution UNDERSTANDING THE FREQUENCY SPECTRA

Exercise 1. seventh harmonic – 7.016 kHZ, ninth harmonic – 9.005 kHz The ideal harmonic values should be 7 kHz and 9 kHz. There is some small measurement error with the spectrum analyzer. Exercise 2. The most obvious change is that frequencies above 500 Hz are severely attenuated compared to a square wave spectral content. Exercise 3. The multisim 9 solution is provided in the Lab CD (E1-2-solution). EWB lab exercises on CD-ROM Fig1-33.ms9 FigE1-1.ms9 FigE1-2.ms9 FigE1-2-solutions.ms9

Chapter 2 - Troubleshooting with EWB Multisim – Solution AM Measurements

Exercise 1. Modulation index = 1 Exercise 2. Using the cursors, T2 – T1 = 66.582 µs. Frequency = 1/T = 15.019 kHz which is approximately 15 kHz Exercise 3.

AM signal Input Signals EWB lab exercises on CD-ROM Fig2-33.ms9 FigE2-1.ms9 FigE2-2.ms9 FigE2-3.ms9

325

Chapter 3 - Troubleshooting with EWB Multisim – Solution

AM Demodulation Exercise 1. The output shows a rectified signal but it still contains the high frequency RF carrier. The problem is because capacitor C2 is open. Exercise 2. The output signal does not show any rectification. This problem can be attributed to a shorted rectifier diode. Exercise 3. C1 = 50%, C3 = 70% EWB lab exercises on CD-ROM Fig3-32.ms9 FigE3-1.ms9 FigE3-2.ms9 FigE3-3.ms9 FigE3-4.ms9

Chapter 4 - Troubleshooting with EWB Multisim – Solution Single-Sideband Generation

Exercise 1. L1 = 35%, L2 = 55%

Changing the inductance values changes the frequency response (shape) of the filter.

Exercise 2. The output signal has changed significantly indicating a possible

problem with one of the passive components. In this case, resistor R2, is open.

Exercise 3. There isn’t a signal going to the spectrum analyzer. This indicates a possible problem with a series component, C1, C2, or C3. In this case, the problem is C1 is open. EWB lab exercises on CD-ROM Fig4-27.ms9 FigE4-1.ms9 FigE4-2.ms9 FigE4-3.ms9 FigE4-4.ms9

326

Chapter 5 - Troubleshooting with EWB Multisim – Solution Generating and Analyzing the FM Signal

Exercise 1. The image produced on the spectrum analyzer is very noisy but a

crude approximation shows that the bandwidth is ~120 kHz. Using Carson’s rule the expected bandwidth :

∆max = 5 (10 kHz) = 50 kHz

BW = 2(50 kHz + 10 kHz) = 120 kHz

Exercise 2. The image produced on the spectrum analyzer is very noisy but a

crude approximation shows that the bandwidth is ~53 kHz. Using Carson’s rule the expected bandwidth :

∆max = 1.67 (10 kHz) = 16.7 kHz

BW = 2(16.7 kHz + 10 kHz) = 53.4 kHz

EWB lab exercises on CD-ROM Fig5-33.ms9 Fig5-1.ms9 FigE5-2.ms9 FigE5-3.ms9

Chapter 6 - Troubleshooting with EWB Multisim – Solution Generating and Analyzing the FM Signal

Exercise 1. There is no signal coming from the output amplifier. There is no signal coming off the mixer. Inductor L1 is open. Exercise 2. The output signal is limited to +/- 5 V by diodes D3 and D4 and V1 and V2. If the output signal tries to exceed 4.3V + 0.7 (5V), diode D3 turns on limiting the output voltage to +5V. If the output signal tries to exceed -4.3V - 0.7 (-5V), diode D4 turns on limiting the output voltage to –5V. Exercise 3. The circuit is a mixer and is used to down convert the RF signal of interest to the IF frequency of 10.7 MHz. EWB lab exercises on CD-ROM Fig6-27.ms9 FigE6-1.ms9 FigE6-2.ms9 FigE6-3.ms9 FigE6-4.ms9 FigE6-5.ms9

327

Chapter 7 - Troubleshooting with EWB Multisim – Solution

Mixer and Squelch Circuits

Exercise 1. There is a signal appearing on the input to transistor Q1 but no signal is coming off the collector of Q1. The power supply voltages are good so the problem is either transistor Q1 or resistors R3 or R4. In this case, transistor Q1 has a short from collector to emitter. Exercise 2. The circuit seems to work fine when the input is 100 mV but when the the input signal falls to 10 mV the signal is poor. It turns out that amplifier A1 is not working properly. Exercise 3. The signal quality is bad off Q1 but it is strong off amplifier A1. A screwdriver was dropped onto the circuit and most likely damaged a component after amplifier A1. The power supply voltage is correct and further checks reveal that D1, R6, and C2 have been damaged. EWB lab exercises on CD-ROM Fig7-33.ms9 Fig7-36.ms9 FigE7-1.ms9 FigE7-2.ms9 FigE7-3.ms9

Chapter 8 - Troubleshooting with EWB Multisim – Solution Sampling the Audio Signal

Exercise 1. The 3 kHz signal comes from subtracting the 5 kHz input signal from the 8 kHz sample frequency. 8 kHz – 5 kHz = 3 kHz.

Exercise 2. Diode D1 is open Exercise 3. The output signal is a flat line. The power supply voltages are good. It turns out that the operational amplifier U1 is bad. EWB lab exercises on CD-ROM Fig8-45.ms9 FigE8-1.ms9 FigE-2.ms9 FigE8-3.ms9 FigE8-4.ms9

328

Chapter 9 - Troubleshooting with EWB Multisim – Solution Sequence Detector

Exercise 1. Another sequence of three consecutive ones appears at data sequence 0025 – 0027. Exercise 2 The is no change on the output of the sequence detector. Use the additional inputs on the logic analyzer or use an oscilloscope to isolate the fault. There is a fault with U2. Exercise 3 There is no change on the output of the sequence detector. Use the additional inputs on the logic analyzer or use the oscilloscope to isolate the fault. There is a fault with U3B. EWB lab exercises on CD-ROM Fig9-34.ms9 FigE9-1.ms9 FigE9-2.ms9

Chapter 10 - Troubleshooting with EWB Multisim – Solution BPSK Transmit-Receive Circuit

Exercise 1. The BPSK received signal is distorted. However the input to the coherent receive circuit is good. Capacitor C1 is bad. Exercise 2. The signal at the input of the coherent carrier receive circuit is not good. Device A3 is bad. Exercise 3. The solution is provided in FigT10-1-solution.ms9 EWB lab exercises on CD-ROM Fig10-42.ms9 FigE10-1.ms9 FigE10-2.ms9 Fig10-42-solution.ms9

329

Chapter 11 - Troubleshooting with EWB Multisim – Solution Analog Signal Measurements

Exercise 1. 8 dB of attenuation Exercise 2. It is approximately 1.5% Exercise 3. 5.76% EWB lab exercises on CD-ROM Fig11-44.ms9 FigE11-1.ms9 FigE11-2.ms9 FigE11-3.ms9 FigE11-4.ms9

Chapter 12 - Troubleshooting with EWB Multisim – Solution Network Analyzer

Exercise 1. Remember to multiply displayed on the network analyzer at 10 MHz by the characteristic impedance of 50Ω. (0.5 – j3.183x10-5) Z= 25 – j 1.59x10-3 Exercise 2. @1 GHz Z = (0.5 + j1.257 x 103) X 50 basically an open circuit @1 GHz Z = 25 + j6.285 x 104 basically an open circuit @10 GHz Z = 0.516 + j1.257 x 104 basically an open circuit @10 GHz Z = 25.8 + j 62.85 x 104 basically an open circuit

Exercise 3. F = 50.11 kHz EWB lab exercises on CD-ROM Fig12-35.ms9 FigE12-1.ms9 FigE12-2.ms9 FigE12-3.ms9 FigE12-4.ms9 FigE12-5.ms9

330

Chapter 13 - Troubleshooting with EWB Multisim – Solution Crystals and Crystal Oscillators

Exercise 1. The resonant frequency is about 3 MHz. Exercise 2. Check the power supply voltages. There is a problem with the +12V supply. Exercise 3. The supply voltage is good but the JFET transistor is bad. EWB lab exercises on CD-ROM Fig13-25.ms9 FigE13-1.ms9 FigE13-2.ms9 FigE13-3.ms9 FigE13-4.ms9

Chapter 14 - Troubleshooting with EWB Multisim – Solution Dipole Antenna Simulation and Measurements

Exercise 1. L1 = 1.198 µH, C1 = 2.5 pF Exercise 2. L1 = 50 nH, C1 = 2.5 pF Exercise 3. The length of the adjustable plunger needs to be changed to approximately 7.3e-1 EWB lab exercises on CD-ROM Fig14-32.ms9 Fig14-34.ms9 FigE14-1.ms9

331

Chapter 15 - Troubleshooting with EWB Multisim – Solution Lossy Transmission Lines and Low-Loss Waveguide

Exercise 1. Microstrip B has the greatest loss. Exercise 2. The first resonant frequency is 1.123 GHz. Exercise 3. The normalized impedance at 1.122 GHz = 0.0100 – j0.0081. The normalized impedance at 226.4 MHz = 0.074 – j3.487. EWB lab exercises on CD-ROM Fig15-37.ms9 Fig15-40.ms9 FigE15-1.ms9 FigE15-2.ms9 FigE15-3.ms9

Chapter 16 - Troubleshooting with EWB Multisim – Solution Characteristics of High-Frequency Devices

Exercise 1. L1 – 3dB frequency ~ 160 kHz, L2 – 3dB frequency ~1.5 GHz Exercise 2. The resonant frequency ~ 1.071 GHz Exercise 3. There is no output signal on the oscilloscope. The power supply voltage is good. The problem is a base-collector short in Q1. EWB lab exercises on CD-ROM Fig16-38.ms9 Fig16-41.ms9 FigE16-1.ms9 FigE16-2.ms9 FigE16-3.ms9

332

Chapter 17 - Troubleshooting with EWB Multisim – Solution The NTSC Television Spectra

Exercise 1. This is VHF channel 4 frequencies, 67.25 MHz (visual carrier), 71.75 MHZ aural carrier.

Exercise 2. C1 is open Exercise 3. C1 is shorted EWB lab exercises on CD-ROM Fig17-43.ms9 Fig17-45.ms9 FigE17-1.ms9 FigE17-2.ms9 FigE17-3.ms9 FigE17-4.ms9

Chapter 18 - Troubleshooting with EWB Multisim – Solution Light Budget Simulation

Exercise 1. Resistor R2 is open. Exercise 2. The sine-wave circuit is shorted. Exercise 3. The problem with this circuit is the voltage level coming from the function generator is too low which is causing the system to not function. Setting the levels within proper operational specifications will fix the problem. EWB lab exercises on CD-ROM Fig18-30.ms9 FigE18-1.ms9 FigE18-2.ms9 FigE18-3.ms9

333

334

335

336

337

338

339

340

341

342

343

344

345

346

347

348

349

350

351

352

353

354

355

356

357

358

33 359

360

361

362

363

364

365

366

367

368

369

370

371

372

373

374

375

376

377

378

379

380

381

382

383

384

385

386

387

388

389

390

391

392

393

394

395

396

397

398

399

400

401

402

403

404

405

406

407

408

409

410

411

412

413

414

415

416

417

418

419

420

421

422

423

424

425

426

427

428

429

430

431

432

433

434

435

436

437

438

439

6 440

441

442

443

444

445

446

Experiment 22 – Solution dB Measurements in Communications

VIRTUAL TEST EQUIPMENT Multim eter AC Voltage Source Virtua l resistors, capacitors Part 1: Measuring the Insertion Loss Provided by Passive Resistive Attenuators Circuits

1. __-0.017 dB ( ~ 0 dB)____ ___773.5 mV____ 2. 0 dBm (600: ) yields 0.774 V see section 1-2 of the text for a

derivation. 3.

dB level measured at the attenuator input ___~ 0 dB___ dB level measured at the attenuator output ___-20 dB___ insertion loss (input – output) ____20 dB___

4. Repeat step 3 for the resistor values provided for the T-type attenuator

in Table 22-1. Table 22-1

R2 R3 R4 Input level (dB)

Output level (dB)

Insertion loss (dB)

230 230 685 ~ 0 -6.95 6.95 69 69 258 * 4.5 -8.18 8.18 588 588 12 0 -40 40 312 312 422 0 -10 10 563 563 38 0 -30 30

* Ask the students why the input level does not equal to 0 dBm. This is not a 600: attenuator, therefore 0.774 V does not equal to 0 dBm(600)

447

5. Repeat step 5 for the resistor values provided in Table 22-2. Make sure that the T-type attenuator is terminated with a 600: resistor. The resistor R5 is the termination resistor in Figure 22-3.

Table 22-2 R2 R3 R4 Input Resistance (: ) 230 230 685 605.28 69 69 258 * 255.19 588 588 12 599.88 312 312 422 600.5 563 563 38 599.8

* Note: The resistant of the attenuator does not equal 600: which explains why the input level did not equal 0 dBm..

Part 2: Setting System Wide dB Levels 5. Set the levels according to Table 22-3.

Table 22-3 Device Output Level Level Control Key Tone 0 dBm A increase, a decrease Production Audio 0 dBm B increase, b decrease VCR1 0 dBm C increase, c decrease Satellite Feed 0 dBm D increase, d decrease Rotary Switch 0 dBm None STL + 8 dBm S increase, s decrease 4. Does the position of the rotary switch make any difference when setting the audio levels? Explain your answer. The feed input resistance to the STL is 600: .. The levels for each source must be set with the source terminated therefore yes, the position of the switch is important.

5. Do you need to add a 600: termination resistor when measuring the audio level from the output of the rotary switch? Explain your answer.

No, the input to the STL feed is already terminated at 600: .

448

6. Do you need to add a 600: termination resistor when measuring the

audio level feeding the STL? Explain your answer.

Yes, the +8 dBm level is obtained relative to a 600: load, therefore a 600: must be used when setting the level.

Part 3: EWB Exercises on CD-ROM

The Multisim files for laboratory 22 are provided on your laboratory CD-ROM so that you can gain additional experience simulating electronic circuits with EWB and gain more insight into the characteristics of the dB measurement and system wide level settings. The file names and their corresponding figure number are listed.

FILE NAME Lab22-Fig22-1 Lab22-Fig22-2 Lab22-Fig22-3 Lab22-Part 2

449

Experiment 23 – Solution Smith Chart Measurements using the EWB Network Analyzer

VIRTUAL TEST EQUIPMENT Multim eter AC Voltage Source Virtual resistors, capacitors, inductors Virtual Network Analyzer

Part 1: Using the Network Analyzer

3. Change the characteristic impedance of the network analyzer to 75: . This will

require changing the characteristic impedance setting for the network analyzer and will also require that the resistor connected to port 2 (P2) be changed to 75: . Change the value of resistor R1 to 75: . Restart the simulation and record the normalized value (Z11) and convert the normalized value to actual resistance. Repeat this for the resistor R1 values provided in Table 23-1. Record the measured normalized network impedance and calculate the actual resistance.

Table 23-1

R1 (: ) Network Impedance Z11 Resistance (: ) 75 1 + j 0.0 75 50 0.667 + j 0.0 50 100 1.3333 + j 0.0 100 600 8.000 + j 0.0 600 300 4.00 + j 0.0 300

Part 2: Measuring Complex Impedances With the Network Analyzer

1. Calculate the capacitive reactance (Xc) of the 6.4 nF capacitor at 1 MHz. Record your result.

Xc = ___24.87_: ____

2. Repeat step 1 to measure the impedance of the RC networks specified

in Table 23-2 for the frequencies listed. Table 23-2

R1 (: ) C1 Frequency Network Impedance, Z11 25 6.4 nF 100 kHz 0.5 – j 4.9736 25 6.4 nF 100 MHz 0.5 – j 0.005 10 10 pF 1 MHz 0.2 – j318.31 10 10 pF 100 MHz 0.2 – j 3.1831 50 50 nF 100 kHz 1.0 – j 0.6366

450

3. Calculate XL of the 4PH inductor at 1 MHz. Record your value.

XL _ = 25.133 :

4. Measure the impedance of the RL networks specified in Table 23-2 for

the frequencies listed.

Table 23-3 R1 (: ) L1 Frequency Network Impedance, Z11 25 4 PH 10 kHz 0.5 + j 0.0050 25 4 PH 100 kHz 0.5 + j 0.0503 25 4 PH 10 MHz 0.5 + j 5.0265 25 4 PH 100 MHz 0.5 + j 50.265 25 4 PH 1 GHz 0.5 + j 502.65

5. What observation can you make about the RL network of step 4 at low and high frequencies? (Hint: Express your observation in terms of the changes in the inductance value.

low frequencies? This is showing basically a short..

high frequencies? This is showing an open condition.

Part 3: Transmission and Waveguide Impedance Measurements

2.

Waveguide Section

Network Impedance Z11

Frequency

A 0.0035 – j 0.0081 1.122 GHz B 0.0148 – j 0.0082 1.122 GHz C 0.0315 – j 0.0084 1.122 GHz E 0.0638 – j 0.0092 1.122 GHz F 0.0502 – j 0.0066 1.1263 GHz G 0.0345 + j 0.0014 1.2626 GHz

3. Based on the measurements made in Step 2, which waveguide has the greatest loss? Record your answer.

_____E_____

451

3. Based on your measurements made in Step 2, what frequency are the

waveguide sections A-E designed to carry? Record your answer.

___1.122 GHz____

5. Based on your measurements, what frequency is waveguide G designed to carry? Record your answer. (Hint: This is the frequency of the waveguide is when the resistance and reactance are at a minimum.)

___1.2626 GHz___

6. Explain what the Smith Chart is showing when the frequency is set to 1.12 GHz.

Your measurement should indicate an open condition. Z11 = 38.6852 + j 31.0611

7. Connect the network analyzer to transmission line I and sweep it from 1.1 to 1.3 GHz. Explain what result the Smith Chart is showing. What is the impedance of this waveguide at 1.119 GHz. How does the loss of this transmission line compare to the sample waveguide sections? Z11 = 0.3994 – j 0.0648, showing almost a short. This is a very lossy waveguide.

Part 4: EWB Exercises on CD-ROM

FILE NAME

Lab23-Fig23-1 Lab23-Fig23-3 Lab23-Fig23-5 Lab23-Fig23-7 Lab23-Part 1_Step3_75_ohms Lab23-Part 3_6

452

453

454

EXPERIMENT 32 Solutions

Using a Spectrum Analyzer

TABLE 32-1. Sine Wave Harmonics Frequency 1 MHz 2 MHz 3 MHz 4 MHz 5 MHz 6 MHz 7 MHz Power -10 dBm -36 dBm -49 dBm -58 dBm -58 dBm -58 dBm -59 dBm

TABLE 32-2. Odd Square Wave Harmonics Frequency 1 MHz 3 MHz 5 MHz 7 MHz 9 MHz 11 MHz Meas. DBm

-9 dBm -20 dBm -30 dBm -38 dBm -48 dBm -62 dBm

DB, Rel to 1 MHz

0 dB, ref -11dB -21dB -29dB -39dB -53dB

Calculate dB, Rel to 1 MHz

0 dB, ref -9.55 dB -13.9 dB -16.9 dB -19 dB -20.8dB

TABLE 32-3. Even Square Wave Harmonics Frequency 2 MHz 4 MHz 6 MHz 8 MHz 10 MHz 12 MHz Meas. DBm

-20 dBm -20 dBm -20 dBm -20 dBm -20 dBm -20 dBm

NOTES: 1. Table 32-1 could show the sine wave harmonics to be <-60 dBm if the

function generator is a higher quality source. 2. Table 32-2:

a. The calculations in the third row are made from the following equation: dB = 20 log (V2/V1).

b. At 3 MHz, V2/V1 =3. Therefore the relative voltage in dB is –9.55. c. At 5 MHz, V2/V1 =5. Therefore the relative voltage in dB is –13.9.

3. Table 32-2. The numbers in row 2 do not match the numbers in row 3,

especially at 9 MHz and 11MHz. This is a measure of the “lack of quality” of the square wave produced by the function generator. A Multisim version of this experiment would show better agreement between rows 2 and 3.

4. Table 32-3 would show the even square wave harmonics to be <-60 dBm if

the function generator were a higher quality source.

455

EXPERIMENT 33 Solutions

Using Capacitors for Impedance Matching

A. Circuit Description 2. Compute wavelength in centimeters.

C = 3 X 10 exp 10 cm per second. __30 cm_____________(a) B. Zi along the Transmission Line

3. What is the VSWR for all points on this circle?____2:1_________(b) 4. Compute the number of wavelengths for a distance of 3

cm.___0.1_____(c) 5. From the point identified in step B1, move along the circle (towards

the generator) by 3 cm. Using a ruler measure Zi / Z 0 = __0.67_________ + j __0.48_____ (d)

6. Calculate Zi = ____33.5_______ + j __24.0_____ (e)

C. Impedance Matching with a series capacitor. 1. Find the number of wavelengths from the point in step B1 required to

make Re (Zi) = 50 ohms. _____0.15___________(f). 2. Compute the corresponding physical distance._____4.5 cm_______(g) 3. What is the magnitude and phase of the reflection coefficient at this

point?___.34, 70 degrees___________(h). 4. What is Im (Zi)? _____.7 X 50 = 35 ohms_________________(i) 5. Compute the amount of series capacitive reactance required to make

the reflection coefficient equal to zero._____ 35 ohms __________(j) 6. Compute the corresponding capacitance for step C5. ____4.5 pF___(k)

D. Impedance Matching with a shunt capacitor. 1. Start at a point defined in ZL = 15 + j 15 ohms. (0.3 + j 0.3) 4. Note the wavelength towards the generator and record ____.30______(l) 6. Note the wavelength towards the generator and record

_____.326_______(m) 7. What is the physical distance between D4 and D6? _.026X30=.78 cm______________(n) 8. What is Im (Yi / Y 0)?________1.4______________(o) 9. Compute Yi __0.02_____ - j___.028________(p) 10. Compute the value of shunt capacitance is required to make the

reflection coefficient equal to zero___4.456 pF_______________(q).

456

EXPERIMENT 33 Solutions Using Capacitors for Impedance Matching 11. Compute the value of inductance, for a reactance of 15 ohms at 1

GHz______ E. Change Frequency and Dielectric Constant. 1. For the series matching capacitor:

a. Calculate the capacitance is the frequency is changed from 1 GHz to 100 MHz:_______45 pF_____________________________. (r)

b. Calculate the length for the matching element if the dielectric constant is 2.1 (teflon):____45 cm /sq root of 2.1 = 31 cm_________. (s) HINT: See page 572, equation 12-21. The velocity equals “c” divided by the square root of the dielectric constant.

2. For the parallel matching circuit start with a 25-ohm load. Change the frequency to 100 MHz and the dielectric constant to 2.1. a. Calculate the capacitive reactance and capacitance: b. Capacitive reactance = 1/.014 = 71.42 ohms (t)

Capacitance = 22 pF (u) c. Calculate the # wavelengths and length for the matching element:

(iii) Wavelengths __.10______ (v) (iv) Length ___20.7 cm_______ (w)

457

EXPERIMENT 34 - Solutions

Electronics Workbench Multisim Impedance Matching

Part II: Measurements for the Series Capacitor 1 a. Record the minimum value of S11: (.007) b. Record the frequency when S11 is minimum: 1016 MHz

6. Calculate the bandwidth for each value of VSWR.

Table 34-1. Results for Series Capacitor Circuit

# S11 Frequency (MHz)

VSWR Bandwidth (MHz)

1 Min: ( .007 ) 1016.0 1.01 -------- 2 .05L 966.0 1.11 -------- 3 .10L 918.3 1.22 -------- 4 .15L 868.9 1.35 -------- 5 .05R 1079.0 1.11 113 6 .10R 1140.0 1.22 221.7 7 .15R 1211.0 1.35 342.1

Part III: Measurements for A Shunt Capacitor

4. Make the measurements for the shunt capacitor circuit in a manner similar to that used for the series capacitor. Enter the results into Table 34-2. Table 34-2. Results for Shunt Capacitor Circuit

# S11 Frequency (MHz)

VSWR Bandwidth (MHz)

1 Min: ( .013 ) 1045.0 1.025 -------- 2 .15L 909.8 1.35 -------- 3 .15R 1161.0 1.35 251.2

Part V: Optimizing the Shunt Capacitor Design – Table 34-4

# C1 S11 Phase (Degrees)

Frequency (MHz)

1 22.0 pF .012 42.3 99.76 2 22.5 pF .007 41.9 98.90 3 23.0 pF .002 42.8 98.05 4 23.5 pF .003 -140.2 97.21 5 24.0 pF .008 -140.2 96.38 6 23.1 pF .0009 98.8 97.77 7 23.2 pF .0006 -104.4 97.77 8 23.3 pF .002 -94.1 97.77

458

Part VI: Optimizing the Series Capacitor Design

Table 34-5. Optimizing Results for Series Capacitor Circuit # C1 S11 Phase

(Degrees) Frequency

(MHz) 1 43.7 pF .003 -83.2 101.2 2 43.8 pF .002 -84.9 101.2 3 43.9 pF .001 -79.5 101.2 4 44.0 pF .0006 -66.5 101.5 5 44.1 pF .0003 39.4 101.5 6 44.2 pF .001 75.5 101.5 7 44.3 pF .002 176 101.5 8 44.4 pF .002 152 101.5

459

EXPERIMENT 35 - Solutions AM Generation Using an Electronic Attenuator

TABLE 35-1: Measurement Limits

Oscilloscope Spectrum Analyzer Minimum Carrier Frequency

< 60 Hz 50 KHz, dynamic range reduced by start spur

Maximum Carrier Frequency

50 MHz, typical 1 GHz

Minimum Modulation Frequency

< 60 Hz 10 KHz, 3 times the minimum RBW

Maximum Modulation Frequency

10 MHz, typical 100 MHz

Minimum Modulation Level (Maximum Dynamic Range)

20 dB 60 dB

460

EXPERIMENT 36 - Solutions

Generating FM from a VCO

11. Measure and record the rms voltage of the 100 KHz signal at the first J0 null (x=2.4) : 18 mv

12. Measure and record the rms voltage of the 100 KHz signal at the 2nd J0 null (x=5.5) : 29 mv

13. Measure and record the rms voltage of the 100 KHz signal at the 3nd J0 null (x=8.7) : 46 mv

14. Calculate the AC voltage for x= 5.0: 26 mv

J # Voltage, from

page 213 of textbook

Calculated (dB)

Measured (dBm)

Measured minus calculated

values Ref JN 1 (Reference) 0 (Reference) -20 dBm No modulation 1 J0 -.18 -14.9 dB -35 dBm -0.1 2 J1 -.33 -9.63 dB -30 dBm -0.37 3 J2 .05 -26 dB -45 dBm 1.0 4 J3 .36 -8.87 dB -30 dBm -1.13 5 J4 .39 -8.17 dB -30 dBm -1.83 6 J5 .26 -11.7 dB -32 dBm -0.3 7 J6 .13 -17.7 dB -38 dBm -0.3

Table 5A-1. Calculations and Measurements for VCO Project. (Bessel Functions for x=5)

461

EXPERIMENT 37 – Solutions Upconversion and Downconversion

# Frequency (MHz)

m n Order * Calc. Power (dBm)

-20 dBm input

Measured with

-2 dBm input

at 21 MHz 1 1 1 1 2 -26 -8 2 2 2 2 4 -38 -55 3 20 1 0 1 --- -55 4 21 0 1 1 -20 -41 5 22 1 2 3 -32 -44 6 23 2 3 5 -44 <-60 7 19 2 1 3 -32 -51 8 39 3 1 4 -38 -20 9 41 1 1 2 -26 -8 10 43 1 3 4 -38 -40 11 59 4 1 5 -44 -49 12 61 2 1 3 -32 -50 13 62 1 2 3 -32 -44 14 64 1 4 5 -44 -52 15 81 3 1 4 -38 -20 16 82 2 2 4 -38 <-60 17 83 1 3 4 -38 -54 18 101 4 1 5 -44 -47 19 102 3 2 5 -44 -55 20 104 1 4 5 -44 -51

Table 37-1. Calculated Signals Generated For a Mixer, When FLO = 20 MHz, and Fi = 21 MHz (-20dBm).

462

EWB Experiment 1 – Solution Simulation of Active Filter Networks

VIRTUAL TEST EQUIPMENT : Dual-trace oscilloscope AC Voltage Source Bode Plotter Power Supplies Virtual resistors, capacitors Simulation model of the 741 operational amplifier Part I

Note: Make sure the students change the initial (I) Horizontal setting on the Bode Plotter to 500 mHz as shown.

3. Use the slider on the Bode Plotter to measure the critical frequency. This will be the point where the plot drops from 0dB to -3dB. Record the value..

___ 478 Hz_____

4. Move the slider to 1 kHz and record the dB level. ____-13.108 dB__

5. Move the cursor to 10 kHz and record the dB level. _____~ -53 dB__

6. Calculate and record the dB level difference for 1 kHz and 10 kHz.

____~ 40 dB____

This is the amount in dB that the magnitude of the output voltage is being attenuated. The students should recognize that the measurement of 1 kHz to 10 kHz is one decade and for a second order filter, the attenuation is 40 dB/decade.

Part II

2. Move the slider to the critical frequency you measured in step 3 of part 1. Record the amount of phase shift, in degrees.

____~ 90 o__

At the critical (or 3 dB frequency`) the phase shift of a second order Butterworth filter should be -90o.

463

EWB Experiment 1 – Solution (continued) Simulation of Active Filter Networks

3. Move the cursor to 10 kHz and record the phase. _(0)_- 180 o__

Your answer should be close to 0 for a total phase shift of -180o. This is the amount of phase shift for an ideal 2nd order Butterworth active filter. Double click on the Bode Plotter and change the final frequency to 500 kHz. Restart the simulation and you will see that the phase shift actually exceeds a total of 180o. This is due to the additional contributions in phase shift provided from the LM741. The simulation for the LM 741 models its behavior on real operating characteristics including the non-ideal characteristics such as the increased phase shift.

Part III: Additional EWB exercise on CD-ROM The active filter networks for EWB Experiment 1 are provided on the laboratory CD-ROM so that the student can gain additional experience simulating electronic circuits with EWB and gain more insight into the characteristics of active filter networks. The file names and their corresponding figure number are listed. You might be able to give the students the task of simulating one of these circuits in addition to the low-pass filter active filter. The students will benefit from the additional exercise with EWB and simulation.

FILE NAME Lab1-low_pass_filter Lab1-high_pass_filter Lab1-bandpass_filter Lab1-notch_filter

464

EWB Experiment 2 – Solution Using The Spectrum Analyzer in the Simulation and Analysis of

Complex Waveforms VIRTUAL TEST EQUIPMENT :

Dual-trace oscilloscope Function generator Spectrum Analyzer Virtual Resistor

8. Calculate the minimum resolution frequency for the following given an end frequency of:

end frequency Minimum resolution frequency 100 kHz 97.656 Hz 455 kHz 444.336 Hz 108 MHz 105468.75 Hz 433 MHz 422821.156 Hz

9. How many frequency components are displayed by the spectrum analyzer shown in Figure E2-1? One, two, three, ..? Use the slider to determine the frequency of the displayed signal. Is this what you expected for a sine wave? Record and justify your answer. One Frequency _____1 KHz___

The spectral content of a sine wave is just the fundamental frequency therefore only one frequency is expected. The function generator is set to 1 kHz therefore only a 1 KHz component is expected.

Part II. Spectral Analysis of a Square Wave

2. Identify the frequency of the components (harmonics) being displayed, their

dB levels and harmonic number. Do this by placing the slider directly over each harmonic. You should be seeing the 1st through 13th harmonics. The first harmonic is 1 kHz and the 3rd harmonic is 3 kHz and so on..

Frequency ___1__ ___3__ __5___ ___7__ ___9__ __11 _ __13__

(kHz) harmonic ___1__ ___3__ __5___ ___7__ ___9__ __11 _ __13__

dB value _27.86_ _16.17 _12.5__ __11__ ___7__ __5.6_ __5.5_

Note: the dB values will vary depending on the exact point where the measurement is made on the spectrum analyzer. Make sure the students sets the cursor close to each harmonic frequency when making a measurement. It might be easier for the student to measure the first harmonic with the cursor and approximate the 3rd – 13th harmonic measurements relative to the first. The spectrum analyzer has a dB/Div setting specified on the control panel to help with the measurement.. The answers that are provided are only an approximation. Make sure the dB value for each harmonic is decreasing.

465

EWB Experiment 2 – Solution (continued)

3. Which harmonic frequency has the greatest amplitude (dB value)?

____1st____

4. How many dB down is the 7th harmonic relative to the 1st harmonic? __16.86 dB_

5. How many dB down is the 3rd harmonic relative to the 1st harmonic?

___11.69dB__

Part 3: Spectral Analysis of a Triangle Wave

2. Identify the frequency of the components (harmonics) being displayed, their

dB levels and harmonic number. Do this by placing the slider directly over each harmonic. You should be seeing the 1st through 13th harmonics. The first harmonic is 1 kHz and the 3rd harmonic is 3 kHz and so on..

Frequency ___1__ ___3__ __5___ ___7__ ___9__ __11 _ __13__

(kHz) harmonic ___1__ ___3__ __5___ ___7__ ___9__ __11 _ __13__

dB value 23.1__ ___5__ __-7___ __-8___ __-10_ __-13_ _-14__

3. Which harmonic frequency has the greatest amplitude (dB value)?

____1st____ The fundamental frequency will have the strongest level.

4. How many dB down is the 11th harmonic relative to the 1st harmonic?

____36.1___ 5. How many dB down is the 5th harmonic relative to the 1st harmonic?

___30.1____ Note: the dB values will vary depending on the exact point where the measurement is made on the spectrum analyzer. Make sure the students set the cursor close to each harmonic frequency when making a measurement. It might be easier for the student to measure the first harmonic with the cursor and approximate the 3rd – 13th harmonic measurements relative to the first. The spectrum analyzer has a dB/Div setting specified on the control panel to help with the measurement.. The answers provided are only an approximation.

466

Part 3: EWB Simulation Files on CD-ROM

This EWB laboratory has demonstrated that complex waveforms, such as a square wave or a triangle wave, generate multi-frequency components, called harmonics. The laboratory exercise demonstrated how the spectrum analyzer can be used to observe and analyze the spectral content of a square-wave. The simulation files for this exercise have been provided on the Lab CD-ROM.

FILE NAME

Lab2-sine_wave Lab2-square_wave Lab2-triangle_wave

467

EWB Experiment 3 – Solution Simulation of Class C Amplifiers and Frequency Multipliers

VIRTUAL TEST EQUIPMENT :

Dual-trace oscilloscope AC Voltage Source

Bode Plotter Virtutal 2N2222A Transistor Virtual resistor Virtual capacitor Virtual inductor

Part 1: EWB Analysis of a class C amplifier

2. Calculate the resonant frequency of the tank circuit made up of L1 and C3. Use the equation

LCfr π2

1= . Record your answer.

_27.694 kHz

3. The cursor on the Bode Plotter to determine the resonant frequency of the class C amplifier. The resonant frequency, fr, is the point on the plot where the amplitude is maximum. Record your answer and compare the test results obtained with the Bode Plotter to the answer calculated in part 2.

_27.608 kHz The calculated and measured values are very close in frequency as expected.

4. Use the Bode Plotter to determine the bandwidth of the tank circuit. To determine the bandwidth, move the cursor so that it is sitting directly over the peak magnitude. Record the resonant frequency and amplitude value in dB in the table provided. Next, move the marker to the left of the peak until the magnitude drops 3 dB of the peak value. Record the –3 dB frequency which will be indicated as f(-). Repeat this procedure to determine the -3 dB frequency to the right of the peak magnitude and record your measurement. This point will be indicated as f(+). Use the –3 dB measured frequencies to determine the bandwidth BW using the equation, BW = f(+) – f(-).

fr f(+) F(-) BW 27.608 kHz ~27.7 kHz ~ 27.5 kHz 200 Hz

5. Use the measured values obtained from the Bode Plotter to determine the Q of

the amplifier. Record your answer.

Q = fr/BW __138.04__

468

Part 2: Frequency Multiplication

4. Use the values of L1 and C1 to calculate the resonant frequency of the tank.

LCfr π2

1= __5032 Hz__

5. Use the cursors on the oscilloscope to measure the frequency of oscillation.

This can be done by setting one cursor at the beginning of the cycle for a sine wave and set the other cursor at the end of the cycle. The difference, in time

(T), is displayed as T2 – T1 on the oscilloscope. Use the equation T

f 1= to

calculate the frequency. Record your answer and compare the measured frequency to the calculated frequency. They should be close in value.

__4878 Hz__

The frequency of oscillation can also be made using the EWB frequency counter. The measured value should be close to the calculated value of 5032 Hz.

Part 3: Additional EWB exercise on CD-ROM

This EWB laboratory has demonstrated that analysis and simulation of a class C amplifier the development of a frequency multiplier. The EWB instruments were used to confirm theoretical results. Your EWB Lab CD-ROM contains the simulation files for this exercise and the full implementation of the frequency doubler.

FILE NAME Lab3-classC_amplifier Lab3-tank_circuit Lab3-frequency_doubler

469

EWB Experiment 5 – Solution Simulation of a Phase-Shift Oscillator

VIRTUAL TEST EQUIPMENT :

Dual-trace oscilloscope AC Voltage Source Virtutal 741 op-amp Virtual resistor

Virtual capacitor DC Voltage Source

Part 1: Analyzing and Testing the components of the

Phase-Shift Oscillator.

3. Start the simulation and verify that U1 and U2 are working properly. Use the oscilloscope to measure and record the peak to peak voltage level of the signals at the output of U1 pin 6 (U1-6) and U2 pin 6 (U2-6).

U1-6 ___400 mV____ U2-6 ____400 mV____

4. Table E5-1 Test Results on the Phase Shift Circuit.

Frequency (Hz) T1(Vin) T2(Vf) T2 – T1 Phase Shift500 20.5 ms 22.3 ms 1.8 ms 324o 1000 10.3 ms 11.3 ms 1 ms 360o 1500 880.3 µs 930.3 µs 50 µs 27o 2000 628.9 µs 678.9 µs 50 µs 36o

5. at 500o 324o (considerable phase shift) at 1000o 360 o (in phase – Barkhausen criteria met) at 1500o 27o (small phase shift) at 2000o 36o (slightly larger phase shift) 6.

Record the frequency of the peak value in step 6. ___1 kHz___ Record the phase of the peak frequency. -2.39 o ( ~ 0 o )

Is the phase result consistent with the Barkhausen criteria? Why? Yes, the total phase shift is ~ 360 degrees and the gain ~ 1.

470

Part 2: EWB Simulation of the Phase-Shift Oscillator.

1. Determine the oscillating frequency? Is this the frequency you expected, why?

T = 1 ms, f = 1 kHz which is the resonant frequency of the circuit as tested in part I – 6.

Note: To help start the oscillations, press the space bar to toggle the feedback switch

2. Why does the circuit no longer maintain oscillation?

The gain is now less than 1 therefore the circuit does not meet the Barkhausen criteria to maintain oscillation.

Part 3: EWB lab exercises on CD-ROM This EWB laboratory has demonstrated the operation of the phase shift oscillator. The EWB Lab CD-ROM contains the simulation files for this exercise.

FILE NAME

Lab5-phase_shift_oscillator Lab5-phase_shift_test Lab5-damped_osc Lab5-phase_shift_Bode

471

EWB Experiment 6 – Solution Simulation of an LC Feedback Oscillator

VIRTUAL TEST EQUIPMENT : Dual-trace oscilloscope AC Voltage Source Virtual 741 op-amp Virtual resistor

Virtual capacitor DC Voltage Source Virtual inductor Bode Plotter Virtual switch

Part 1: Analyzing and Testing the Components of the LC Oscillator.

2. Calculate the resonant frequency of the LC oscillator circuit shown in Figure

E6-1 using the equation and record the value.

1121

CLf

π=

___33.884 kHz__

4. Record the peak frequency value measured with the Bode Plotter and compare

your result with the frequency calculated in step 2.

__33.859 kHz___

Table E6-1 Test Results on the Phase Shift Circuit. Frequency (kHz) T1(Vin) T2(Vf) T2 – T1 Phase Shift 10 7.925 ms 8.003 ms 77.8 µs 280o 25 5.9 ms 5.9 ms 34.5 µ 310.5o 33.9 2.2 ms 2.2 ms ~ 0 ~ 0o 65 2.3 ms 2.3 ms 4.6 µs 107.64o 100 822.3µs 826.3µs 40 µs 144o

Note: Some of the T1 and T2 values look equal but notice that the T2 -= T1 values show a small difference. This is due to the resolution of the T1 and T2 values.

472

Part 2: EWB Simulation of the LC Oscillator.

1. Change the feedback switch so that the loop for the oscillator is closed. Re-start the simulation and observe the output of the oscillator. It may take up to .02s of simulation time for the oscillation to start. Determine the oscillating frequency? Is this the frequency you expected, why?

You can kick start the simulation by pressing the spacebar twice to open and close the feedback loop. T = 29.8 µs f = 33.557 kHz. Yes, the resonant frequency should be close to 33.884 kHz. You may see some difference in the value displayed on the frequency counter. This is most likely due to a setting, sensitivity, trigger level, coupling.

2. Change R3 to 300 Ω and restart the simulation. What change do you observe with the oscillator? Why do you think this happened?

The feedback path no longer is sufficient to maintain the oscillation. 3. Change R3 to 470 Ω and restart the simulation. You will need to help start

the oscillator by momentarily switching the input which opens and closes the feedback path. Change the time base on the oscilloscope to 100µs/div to better see the output signal. What change do you observe with the oscillator? Why do you think this happened?

The feedback path no longer is sufficient to maintain the oscillation. Part 3: EWB lab exercises on CD-ROM

FILE NAME Lab6-LC_oscillator Lab6-LC_R4_300 Lab6-LC_R4_480 (note: this is part II-part 3, R3 = 470Ω)

473

EWB Experiment 11 – Solution Percentage of Modulation Measurement of an Amplitude

Modulated Waveform VIRTUAL TEST EQUIPMENT Dual-trace oscilloscope AC Voltage Source

Virtual resistor AM Source

Part 1: Trapezoidal Measurements of the AM Waveform

The equation for calculating the percentage of modulation is

%100% xABABm

+−

=

The equation for calculating power is

)2/1( 2mPcPt +=

Where Pc = (10 * .707)2 / 50 = 1 W.

Table E11-1 File Name (from EWB Lab CD-ROM)

Percentage of Modulation (%m)

Output Power (Pt) Calculated

Output Power (Pt) Measured

Lab11-AM-1 0 1 W 1 W Lab11-AM-2 100 1.5 W 1.523 W * Lab11-AM-3 50 1.25 W 1.249W * Lab11-AM-4 75 1.28125 W 1.265 W * Lab11-AM-5 66 1.2178 W 1.203 W * Lab11-AM-6 90 1.405 W 1.427 W *

* The wattmeter value will vary and can only be used to provide an estimate of output power.

Lab11-AM-7 is showing an example of over modulation.

474

Notes to the Instructor In this laboratory instructor’s manual, typical data and answers for the laboratory experiments found in the Laboratory Manual to Accompany Modern Electronic Communications, Ninth Edition. A few of the experiments are prone to oscillation and noise interference. This includes experiment 14. In experiment 14, several of the DSB-SC scope displays will appear a little fuzzy due to RF noise. Changes in board layout should remedy this problem. A few of the experiments use components that are not commonly found at electronics parts distributors. Hard to get parts can be obtained from:

Part Source CFM-455D Ceramic Filter Milgray Electronics XR-2206, XR-2211 Marshall Electronics Toroid T-106 Mix 2 Amidon Associates AMP Fiber Optic Cable and Connectors Pioneer - Standard

Milgray Electronics Marshall Electronics 97 Monroe Avenue 1280 Scottsville Rd. Pitts, NY 14534 Rochester,. NY 14624 716-385-9330 716-235-7620 Amidon Associates Pioneer-Standard Electronics 12033 Ostego St. 840 Fairport Park North Hollywood, CA 91607 Fairport, NY 14450 716-381-7070

The EWB labs and Troubleshooting exercises have been tested using EWB Multisim 9 and are easy for the student to follow. It is best if the students build their own circuit from the schematic provided in the exercise. The textbook does include a CD-ROM containing all of the EWB multisim (.ms9) files used in the lab manual. The instructor can use these files to verify student work or to demonstrate what the simulation should look like. The student can use the files to continue their investigations. Feel free to contact us if you have any questions about any of the experiments. Mark Oliver, Associate Professor Jeff Beasley, Professor Electrical Engineering Technology Department of Engineering Technology Monroe Community College New Mexico State University 1000 E. Henrietta Rd. Box 30001 MSC 3566 Rochester, NY 14623 Las Cruces, NM 88003 email: MOLIVER@MONROECC.EDU email: jbeasley@nmsu.edu David Shores, Keely Communications email: davidshores@comcast.net