ECE 200 Book

Introduction to Signals, Circuits & Systems

A Practical Introduction to Electrical Engineering

Mehmet C. Ozturk; Chapters 1 - 3, 6 - 8

Mesut Baran; Chapter 4

Joel Trussell; Chapter 5

Amir Mortazavi; Chapter 9

Edited by Mehmet C. Ozturk

North Carolina State University

Department of Electrical and Computer Engineering

Raleigh, NC

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Table of Contents

Table of Contents 3

1 Resistive Circuits 71.1 Charge, Voltage and Electric Current . . . . . . . . . . . . . . . . . . . . . . 71.2 Conductors, Resistivity and Resistance . . . . . . . . . . . . . . . . . . . . . 101.3 Resistivity and Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.4 Resistors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.5 Linear Resistors & Ohm’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . 131.6 Resistors in Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.6.1 Kirchoff’s Voltage Law . . . . . . . . . . . . . . . . . . . . . . . . . . 171.6.2 Series Equivalent Resistance . . . . . . . . . . . . . . . . . . . . . . . 181.6.3 Voltage Division Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.7 Resistors in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.7.1 Kirchoff’s Current Law . . . . . . . . . . . . . . . . . . . . . . . . . . 211.7.2 Parallel Equivalent Resistance . . . . . . . . . . . . . . . . . . . . . . 22

1.8 Circuits with Multiple Voltage Sources . . . . . . . . . . . . . . . . . . . . . 241.9 The Circuit Ground . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.10 Diode : A Non-Linear Resistor . . . . . . . . . . . . . . . . . . . . . . . . . . 28

1.10.1 I-V Characteristic of a Diode . . . . . . . . . . . . . . . . . . . . . . 281.10.2 A Simple Diode Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . 31

1.11 Other Resistive Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341.11.1 Potentiometer - Variable Resistor . . . . . . . . . . . . . . . . . . . . 341.11.2 Light Emitting Diode (LED) . . . . . . . . . . . . . . . . . . . . . . . 351.11.3 Photocell-Light Sensitive Resistor . . . . . . . . . . . . . . . . . . . . 36

2 Capacitors And RC Circuits 432.1 Capacitor Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

2.1.1 Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

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2.1.2 Charging a Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.1.3 Current-Voltage Relationship of a Capacitor . . . . . . . . . . . . . . 48

2.2 Transient Response of an RC Circuit . . . . . . . . . . . . . . . . . . . . . . 52

2.2.1 Capacitor Voltage During Charging . . . . . . . . . . . . . . . . . . . 52

2.2.2 Capacitor current during charging . . . . . . . . . . . . . . . . . . . . 55

2.3 Discharging a Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

2.3.1 Capacitor voltage during discharging . . . . . . . . . . . . . . . . . . 57

2.3.2 Capacitor Current during Discharging . . . . . . . . . . . . . . . . . 59

3 Periodic Signals in Time Domain 65

3.1 Periodic Signals: Period and Frequency . . . . . . . . . . . . . . . . . . . . . 65

3.2 Square Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3.3 Sinusoidal Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3.3.1 Phase Angle of a Sinusoidal Waveform . . . . . . . . . . . . . . . . . 70

3.4 Time-Varying Signals in Circuits with Resistive Elements . . . . . . . . . . . 72

3.4.1 Ohm’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

3.4.2 Kirchoff’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

3.5 Average or DC Value of a Periodic Signal . . . . . . . . . . . . . . . . . . . . 76

3.6 Circuits with Time-Varying Signals and Diodes . . . . . . . . . . . . . . . . 78

3.7 Half Wave Rectifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

3.8 Circuits with AC and DC Voltage Sources . . . . . . . . . . . . . . . . . . . 81

3.9 Oscilloscope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

4 Electric Power 91

4.1 Physics of Electric Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

4.2 DC Voltage Sources and Resistive Loads . . . . . . . . . . . . . . . . . . . . 93

4.3 AC Voltage Sources and Power . . . . . . . . . . . . . . . . . . . . . . . . . 96

4.3.1 Instantaneous Power . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

Square Wave and Instantaneous Power . . . . . . . . . . . . . . . . . 97

Sinusoidal Wave and Instantaneous Power . . . . . . . . . . . . . . . 100

4.3.2 Average Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

4.4 Root-Mean-Square Voltage and Current . . . . . . . . . . . . . . . . . . . . 103

4.4.1 RMS value of a ”Pure” Sinusoid . . . . . . . . . . . . . . . . . . . . . 104

4.4.2 RMS value of a Sinusoid with a DC Value . . . . . . . . . . . . . . . 104

4.5 Power Engineering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

4.5.1 Power Outlets in USA . . . . . . . . . . . . . . . . . . . . . . . . . . 106

4.5.2 Power Generation and Distribution . . . . . . . . . . . . . . . . . . . 107

4.5.3 Your Electricity Bill . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

4.5.4 Real Power with a Non-Resistive Load . . . . . . . . . . . . . . . . . 110

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5 Periodic Signals in Frequency Domain 1195.1 Adding Sinusoids and Importance of Phase . . . . . . . . . . . . . . . . . . . 1205.2 Magnitude Spectrum and Phase Spectrum . . . . . . . . . . . . . . . . . . . 1215.3 Power Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

5.3.1 Signal Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1245.3.2 Power Spectrum in Decibels . . . . . . . . . . . . . . . . . . . . . . . 125

5.4 Periodic Signals in Frequency Domain . . . . . . . . . . . . . . . . . . . . . . 1295.5 Noise in the Frequency Domain . . . . . . . . . . . . . . . . . . . . . . . . . 1315.6 Audio Signals in the Frequency Domain . . . . . . . . . . . . . . . . . . . . . 135

6 Signal Amplification 1416.1 Voltage Gain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1416.2 Transfer Characteristic of an Amplifier . . . . . . . . . . . . . . . . . . . . . 1426.3 Deviations from the Ideal Characteristic . . . . . . . . . . . . . . . . . . . . 144

6.3.1 Clipping Distortion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1446.3.2 Nonlinear Distortion . . . . . . . . . . . . . . . . . . . . . . . . . . . 1486.3.3 Total Harmonic Distortion . . . . . . . . . . . . . . . . . . . . . . . . 150

6.4 Output Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1536.5 Input Sensitivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1546.6 Power Gain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1556.7 Frequency Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1576.8 Amplifier Equivalent Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

7 Operational Amplifiers 1677.1 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1677.2 What makes an op-amp so unique? . . . . . . . . . . . . . . . . . . . . . . . 1707.3 Op-Amp as a Comparator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1717.4 Basic Amplifier Configurations . . . . . . . . . . . . . . . . . . . . . . . . . . 173

7.4.1 Non-Inverting Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . 1737.4.2 Inverting Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

7.5 Why do we need such a large Ao? . . . . . . . . . . . . . . . . . . . . . . . . 1787.5.1 Equivalent circuit of an Op-Amp . . . . . . . . . . . . . . . . . . . . 1787.5.2 Non-Inverting Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . 1787.5.3 Inverting Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181

7.6 Virtual Short Concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1837.6.1 Derivation of the Inverting Amplifier Voltage Gain . . . . . . . . . . 185

7.7 Applications of Operational Amplifiers . . . . . . . . . . . . . . . . . . . . . 1867.7.1 Audio Mixer - Summing Amplifier . . . . . . . . . . . . . . . . . . . . 1867.7.2 DC Level Shifter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1887.7.3 Differentiator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

7.8 Input Resistance of an Operational Amplifier Circuit . . . . . . . . . . . . . 190

8 Filters 197

8.1 Amplifiers and Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

8.2 Ideal Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

8.3 Realistic Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

8.4 Three dB cut-off Frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

8.4.1 Finding the Output Power Spectrum . . . . . . . . . . . . . . . . . . 205

8.5 Filtering in Time Domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

8.6 Active Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

9 Transmission and Reception of Radio Signals 211

9.1 Modulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

9.2 Amplitude Modulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

9.2.1 Analog Multiplier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

9.2.2 Multiplication of Sinusoids . . . . . . . . . . . . . . . . . . . . . . . . 214

9.2.3 Amplitude Modulation (AM) . . . . . . . . . . . . . . . . . . . . . . 216

9.2.4 AM Signal in Frequency Domain . . . . . . . . . . . . . . . . . . . . 218

9.2.5 Modulation Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221

9.2.6 Transmission of Non-Sinusoidal Signals . . . . . . . . . . . . . . . . . 222

9.3 AM Demodulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224

9.3.1 Inductor and LC Parallel Resonant Circuit . . . . . . . . . . . . . . . 227

9.4 A Simple AM Receiver (Crystal Detector) . . . . . . . . . . . . . . . . . . . 229

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Chapter 1

Resistive Circuits

In this chapter, we shall introduce the fundamental circuit laws including Ohms’ law, Kir-choff’s Voltage Law (KVL) and Kirchoff’s Current Law (KCL). These laws are applied tointroductory circuits consisting of resistors and other resistive elements, which resist theelectric current flow. Resistors are used extensively in electronic circuits whenever we mustlimit the current flow through an element to a certain maximum. They are also helpful individing up an applied voltage between different elements. Some of the resistive elements cando a lot more than just resisting the electric current. Examples include the light emittingdiode (LED), which produces light; the photocell, which can measure the intensity of lightand the thermistor, which can measure the temperature.

1.1 Charge, Voltage and Electric Current

Figure 1.1 shows a battery connected to an arbitrary load such as a light bulb, a radioor a hand-held computer. The battery, the load and the conducting wire form a complete

LOAD1.5 V

electrons

Figure 1.1: A battery connected to an arbitrary load.

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Figure 1.2: A portable water fountain, which uses an electric pump to circulate the water.

loop, which is essential for the electric current to flow. Electric current is established by thebattery, which provides the energy to move the electrons around the loop. The battery hastwo terminals, positive and negative. Electrons leave the negative terminal, make a completeloop and reenter the battery through the positive terminal. Therefore, the positive terminalof the battery serves as a sink for the electrons. It is important to note that electrons canflow only if there exists a complete loop.

The water pump used in a portable water fountain such as the one shown in Figure 1.2serves the same function as the electric battery used in an electric circuit. The water pumpcollects the water molecules from the tank and raises them to the top of the fountain, wherethey are ejected through the holes. By moving the water molecules to a greater height, thepump provides them potential energy and the molecules use this energy to drop back intothe tank. Similarly, there exists a potential difference between the positive and the negativeterminals of the battery. We refer to this potential difference as voltage. The unit of voltageis Volt, V . In Figure 1.1, the battery provides the load a voltage of 1.5 V . We can also statethat the battery results in a potential difference of 1.5 V between the terminals of the load.

From the above discussion, we understand that the term, voltage refers to an energydifference between two different physical locations (e.g. terminals of a battery) in a system(e.g. battery). You may ask why we are not using joules as our unit of energy? Whyintroduce a new unit now? The reason is that voltage refers to energy or work done per unitcharge. In the MKS system, the relationship between voltage and energy is given as

1 volt =1 joule

1 coulomb(1.1.1)

To understand the concept better, let’s consider Figure 1.3.a which shows a chargedsphere placed in an electric field. An electric force, F , is exerted on the sphere by theelectric field. If the sphere is not held in place by an external force, it will move under theinfluence of the electric force. Therefore, some work will be done by the electric electric fieldin moving the sphere. Now suppose an external force is applied on the sphere in the oppositedirection. If this force is at least equal to the electric force, the sphere will now move in the

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F

Electric Field

Fext F

d

Figure 1.3: When a positively charged ball is placed in an electric field, an electric force isapplied to the ball. The force is inversely proportional to the charge stored in the ball.

opposite direction. If the sphere moves by a distance d, then we the net work done by theexternal force can be expressed as

W = Fextd

The voltage difference between the starting and the end points is simply this work dividedby the charge stored on the ball.

Each electron carries the same charge of 1.6×10−19 Coulombs. Electric current is definedas the total charge, Q passing through any given point in the loop in one second, which canbe expressed as

I =4Q

4t(1.1.2)

The unit of current is Ampere, which is equivalent to coulomb/second. Since the charge iscarried by electrons, the magnitude of the electric current is directly proportional to how fastthe electrons circle around the loop. For example, suppose we have 1×1016 electrons enteringthe positive terminal of the battery every second. Then the current, I in the loop can becalculated by multiplying the charge carried by each electron by the number of electronsentering the battery:

1.6× 10−19 × 1× 1016 = 1.6× 10−3A = 1.6mA

which shows that we have to move a lot of electrons to create such a relatively small current.

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Figure 1.4: Benjamin Franklin

Figure 1.5: Electrons move randomly in the absence of an electric field.

1.2 Conductors, Resistivity and Resistance

We have defined the electric current as the rate of charge flow due to the flow of electrons.Majority of the electrons necessary to establish the electric current are already available inthe wire. These are free electrons, which are loosely attached to the atoms of the conductingwire. At room temperature, the free electrons are in random motion as shown in Figure 1.5and the net charge flow in any direction is zero.

When a battery is connected between the two ends of a conducting slab, an electric fieldis established in the material as shown in Figure 1.6. The direction of the field is alwaysfrom the positive to the negative polarity of the voltage applied across the slab. The electricfield, E exerts a force, F on each electron, which can be expressed as

F = qE (1.2.1)

where both F and E are vectors and q is the charge on an electron. Each electron carries afixed negative charge of 1.6 × 10−19 Coulombs. Since this charge is negative, the direction

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Electric Field

I I

Figure 1.6: An electric field is established in a material when a voltage is applied betweenthe two ends of a conducting slab. The electric field forces the free electrons to move, whichcreates the charge motion needed for the electric current.

of the force exerted on an electron is opposite to that of the electric field. Electrons movealong the direction of the applied forces and the resulting charge motion becomes the electriccurrent.

We note that the direction of current flow shown in Figures 1.6 and 1.1is clockwiseor opposite to the flow of electrons. The answer to this discrepancy lies in the historyof electricity. Scientific experiments on electricity had begun long before we knew aboutelectrons. Benjamin Franklin was the first to notice the direction of electric current flow andbecame the first to use the terms positive and negative. According to Franklin, current flowtook place from one object to another only if one of the objects had a surplus of charge. Heused the term positive to refer to the object with the excess charge. By the time electronswere discovered and the true direction of electron flow was understood, the nomenclaturehad already been so well established that no effort was made to change it.

1.3 Resistivity and Resistance

Conductors such as gold, silver and copper allow electric current to flow much better thanother materials. Most electrical wiring in our houses is made of copper because it conductswell and it is inexpensive relative to gold and silver. In computer chips, other metals suchas aluminum, copper and tungsten are also used.

Even the best conductors show some resistance to the flow of electric current. Thephysical quantity used to measure the degree of unwillingness to carry electric current iscalled resistance. The unit of resistance is ohms (Ω). The resistance of an element isdetermined by the resistivity of the material used and the dimensions of the element. The

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LI

A = area

Figure 1.7: The resistance of a rectangular prism is determined by the resistivity of thematerial used and the dimensions of the prism.

resistivity, ρ of a material is given in Ω − cm. The resistivity of a conductor is typicallyorders of magnitude lower than that of an insulator. Shown in Figure 1.7 is a rectangularprism with length, L and cross-sectional area, A. Its resistance is given by.

R = ρL

A(1.3.1)

which can be used to calculate the resistance of a cylindrical wire as well.

Example 1.1. Find the resistance of a 1 cm long copper wire with a diameter of 1 mm.

Solution:The cross-sectional area of the wire can be calculated as

A = πr2 = π × (0.05cm)2 ' 0.00785cm2

The resistivity of copper is 1.7 µΩ− cm. The resistance can be calculated from equation 1.3.1 as

R = 1.7× 10−6Ω− cm× 1cm

0.00785cm2' 2.17× 10−4Ω.

‖ X ‖

1.4 Resistors

In general, two-terminal devices that resist the electric current flow are referred to as resistiveelements. There are many different types and applications of resistive elements. For example,the filament of an ordinary light bulb or the heating element of an electric oven are bothresistive elements. In electronic circuits, we intentionally include devices called resistors,which allow us to obtain the desired voltage and current levels in different regions of an

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electric circuit. Figure 1.8 shows several resistors that can be found in electronic circuits.Such resistors come in different sizes because they are designed to handle different electricpower levels.

Figure 1.8: Resistors used in electronic circuits

1.5 Linear Resistors & Ohm’s Law

Now it is time to consider our first electric circuit shown in Figure 1.9, which consists of aresistor connected to a battery.

The battery generates a fixed voltage, say 1.5 V. The battery is a DC voltage source,where the acronym, DC stands for direct current. This implies that the current induced bythis source is constant and it always flows in the same direction. In future chapters, we willlearn about alternating current or AC voltage sources, which induce the type of current thatchanges its direction in a periodic fashion.

R

I

V s V R

Figure 1.9: A DC voltage source and a resistor in a closed loop

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I I I

Figure 1.10: A simple circuit drawn in three different ways

In circuit analysis, we assume that the conductors between different circuit elements areideal, which simply means that their resistance is zero. With this assumption, the voltageacross the resistor in Figure 1.9 should be exactly equal to the battery voltage since thereare no losses occurring on the wires connecting the voltage source to the resistor. Thebattery creates a potential (energy) difference between its terminals. Recall that voltage isjust energy (joules) per unit charge (coulombs). Because electrons do not lose any potentialenergy as they travel through ideal conductors, those electrons leaving the resistor and thoseentering the positive terminal of the battery end up having exactly the same potential energy.

The direction of the current flow is determined by the polarity of the applied voltage,and it is clockwise. Note that the current flows from the negative to the positive terminalof the battery, which is always the case with sources of electrical energy. Electrons leave thenegative terminal of the battery, travel around the closed loop until they reenter the batteryfrom the positive terminal. Since the direction of current is opposite to electron flow, thecurrent flow is clockwise.

We note that the direction of the current flow through the resistor is from positive to thenegative potential. This is always the case with passive circuit elements.

Let’s now consider Figure 1.10, which shows three seemingly different circuit diagrams.In reality, all three circuits are identical to the circuit shown in Figure 1.9. In creating acircuit diagram, we choose the arrangement that is most pleasing to the eye. In circuitanalysis, we sometimes rearrange the elements to simplify the analysis.

An ideal resistor obeys Ohms law, which states that the amount of current flowing througha resistor is equal to the voltage across the resistor divided by its resistance. If the voltageis measured in volts (V) and the resistance in ohms (Ω), the unit of current is Ampere (A).1

Ohms law is written as

I =V

R(1.5.1)

where the voltage, current and resistance are given in volts, amperes and ohms. This is astraight line with a slope equal to 1/R. Eq. 1.5.1 is plotted in Figure 1.11. Such a plot isreferred to as the current-voltage or the (I-V)2 characteristic of an ideal resistor. Because

1The units; ohm, volt and ampere are names given after their originators Georg Simon Ohm (1789- 1854),Andre-Marie Ampere (1775-1836) and Alessandro Volta (1745-1827).

2Pronounced eye-vee.

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of the linear relationship between voltage and current, a resistor is referred to as a linearcircuit element.

slope = 1/R

I

V

Figure 1.11: I-V characteristic of an ideal resistor obeying Ohm’s Law.

Example 1.2. Calculate the magnitude of the current in Figure 1.9 for a battery voltage of 9.0V and a resistor of 47 kΩ.

Solution:The voltage across the resistor is equal to the battery voltage. We write:

VR = Vs = 9.0V

Using Ohm’s law, we can calculate the current as

I =VR

R=

9.0V

47, 000Ω= 1.9× 10−4A = 0.19mA

Typical currents you will find in electronic circuits are in this range. ‖ X ‖

When a closed loop contains more than one voltage source, the direction of the currentdepends on the magnitudes and polarities of the voltage sources. In Figure 1.14, the twovoltage sources are forcing currents in opposing directions. By intuition, we can guess thatthe current flow has to be clockwise because the larger, 9 V battery on the left will force astronger, clockwise current in the loop.

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1.6 Resistors in Series

When two resistors are connected as shown in Figure 1.12, we refer to them as resistors inseries. We note that no other device is connected to the junction point of the two resistors.In a series connection, both resistors carry the same current and share the same branch. Infact, this is true for any two circuit elements in series. If this were not the case, we wouldhave electron depletion or accumulation at different points in a circuit, which we know isnot happening.

Let’s now consider Figure 1.13.a. All three resistors are in series. On the other hand, notwo resistors are in series in Figure 1.13.b. For instance, while R1 and R2 are connected toeach other at a single point, the fact that R3 is also connected to the same point means thatthe two resistors will have to carry different currents.

R1

R2

Vs

I

VR2

VR1

Figure 1.12: A DC voltage source and two series resistors in a closed loop.

R 1

R 2

R 3

R 1

R 2 R 3

(a) (b)

Figure 1.13: (a) Three series resistors connected to a DC voltage source, (b) A circuit withthree resistors. No two resistors in the circuit are in series.

19

1.6.1 Kirchoff’s Voltage Law

When a voltage is applied to two series resistors, the resistors share the applied voltage suchthat

Vs = VR1 + VR2 (1.6.1)

where VR1 and VR2 are the voltages across the resistors R1 and R2 respectively. This isanalogous to the water fountain in Figure 1.2, where the water molecules first fall into theupper tank spend some time in there and then fall into the lower tank.

The sharing of the voltages is more formally expressed using Kirchoffs Voltage Law(KVL), which states that the algebraic sum of voltages around any closed loop is zero.This is written as ∑

k

Vk = V1 + V2 + . . . + Vn = 0 (1.6.2)

where Vk is the voltage across the kth element in the loop. KVL is a universal law, whichapplies to any closed loop with any number of components. To demonstrate how KVL isapplied to a loop, lets consider our simple circuit shown in Figure 1.12 again. We have threecircuit elements in the loop and hence, three voltages to deal with. According to KVL, thesum of these voltages should be equal to zero. We write a loop equation by going around theloop adding the voltage rises and subtracting the voltage drops until we reach our startingpoint.

Applying KVL to a loop is quite straightforward if we remember the few rules and pointersgiven below.

1. First we assign current directions to different branches in the circuit. This can be donearbitrarily unless the current direction is obvious. For instance, in Figure 1.12, weknow the current direction because the single voltage source forces a clockwise currentin the loop. In more complex circuits, the current direction will not be obvious but aswe mentioned above, you must guess the current directions and it is quite alright toguess them wrong.

2. While the current directions can be assigned arbitrarily, the polarities of the voltagesassigned to the resistors must match the current directions. This means that no matterhow the current flows, it must flow from positive to negative potential.

3. We add the voltage rises and subtract the voltage drops.

4. We can start anywhere in the loop.

5. We can go around the loop clockwise or counter clockwise, it just does not matter.

Lets start tracing the loop from the negative terminal of the battery. As we pass throughthe battery, we encounter a voltage rise since we are moving from the negative to the positive

20

9 V 6 V

R

I

Figure 1.14: Two voltage sources and a resistor in a closed loop

terminal of the battery. On the other hand, as we pass through the two resistors, weencounter two voltage drops. According to KVL, the sum of these three voltages mustadd up to zero. By adding the voltage rises and subtracting the voltage drops, we obtain

∑

k

Vk = Vs − VR1 − VR2 = 0 (1.6.3)

which can be rearranged to obtain equation 1.6.1. You can try tracing the loop counter-clockwise and obtain ∑

k

Vk = −Vs + VR1 + VR2 = 0

which is equivalent to Eq. 1.6.3.

Example 1.3. A 1 kΩ resistor is used in the circuit shown in Figure 1.14. Find the magnitudeof the current flowing through the resistor.

Solution:Applying KVL to the loop, we find the voltage across the resistor as 3 V. Using Ohm’s law, thecurrent can be found as I = V/R = 3V/1kΩ = 3 mA. ‖ X ‖

1.6.2 Series Equivalent Resistance

In Figure 1.12, the same current is flowing through all three components. Expressing theindividual resistor voltages in terms of the loop current and substituting them in equation1.6.3, we find the following equation for the loop current.

I =Vs

R1 + R2

21

Therefore, the current flowing through two series resistors is equal to the applied voltagedivided by the total resistance. In a way, the series resistors are behaving as if a singleresistor with a resistance

Req = R1 + R2

is connected to the voltage source. This is called the equivalent resistance of two seriesresistors. We can easily show that this rule can be generalized for more than two resistorsin series. The series equivalent resistance of ‘n’ resistors connected in series is given by

Req =∑

k

Rk = R1 + R2 + . . . + Rn (1.6.4)

Example 1.4. Find the loop current flowing in the circuit of Figure 1.12 if Vs = 9.0V ,R1 = 47kΩand R2 = 22kΩ.

Solution:Since the resistors are connected in series, their equivalent resistance is given by Eq.1.6.4 as

Req = R1 + R2 = 47kΩ + 22kΩ = 69kΩ

The loop current can be found using Ohm’s law as

I =V

R=

9.0V

69kΩ= 0.13mA

‖ X ‖

1.6.3 Voltage Division Rule

Now lets find the individual voltages that appear across R1 and R2 in the circuit of Figure1.12. Since we already know the current flowing through the resistors, we can easily find theindividual voltages using Eq.1.5.1 as

VR1 = IR1 = VsR1

R1 + R2

VR2 = IR2 = VsR2

R1 + R2

(1.6.5)

According to these equations, the resistors share the applied voltage according to theirresistance values. If the two resistance values are identical, resistors share the applied voltageequally. If the resistances are different, a larger voltage drops across the larger resistor. Theratio of an individual resistance to the series equivalent resistance determines the fraction

22

of the applied voltage that will appear across that resistor. Eq.1.6.5 is called the voltagedivision rule and it provides a short-cut for finding individual resistor voltages when tworesistors are in series. You can easily show that voltage division is valid for more than twoseries resistors.

Example 1.5. Find the individual voltages across the two resistors of Example 1.4.

Solution:In the previous example, we found the current flowing in the loop as

I ≈ 0.13mA

We note that this current is flowing through both resistors. Thus, we can find the individualvoltages across the resistors using Ohm’s law as:

VR1 = IR1 = 0.13mA× 47kΩ ≈ 6.1V

VR2 = IR2 = 0.13mA× 22kΩ ≈ 2.9V

The two voltages can be found more quickly by applying the voltage division rule given in Eq.1.6.5. The battery voltage is divided among the two resistors according to:

VR1 = VsR1

R1 + R2= 9.0V

47kΩ47kΩ + 22kΩ

≈ 6.1V

VR2 = VsR2

R1 + R2= 9.0V

22kΩ47kΩ + 22kΩ

≈ 2.9V

Note that the two voltages add up to the battery voltage of 9.0 V.‖ X ‖

1.7 Resistors in Parallel

The two resistors in Figure 1.15 are connected in parallel. In parallel connection, bothterminals of the elements must be connected together. We note that the two resistors are alsoin parallel with the voltage source. The same voltage appears across elements connected inparallel. Hence, the three voltages in Figure 1.15 should be exactly the same: Vs = V1 = V2.

In Figure 1.15, a fraction of the current Is flows through the resistor, R1 and the remainingfraction flows through R2. In other words, the source current Is is shared between the tworesistors. This can be expressed as

Is = IR1 + IR2 (1.7.1)

Figure 1.16 shows two different versions of the circuit in Figure 1.15. While the diagramslook different, they all correspond to the same circuit. It is an excellent practice to study

23

R1

R2

Vs

I1

I2

V1

V2

Is

Figure 1.15: Two parallel resistors connected to a voltage source.

the circuit carefully before writing down equations to solve for the unknown parameters.Sometimes, a circuit may look a lot more complex than it actually is.

R1

R2

Vs

I1

I2

R1R2

Vs

I1

I2

Figure 1.16: Two different versions of the circuit shown in Figure 1.15.

1.7.1 Kirchoff’s Current Law

A circuit node is a point where more than two branches come together. In Figure 1.17 sixbranches meet to form a node. Note that some of the currents are flowing into the node whilethe others are leaving the node. According to Kirchoff’s Current Law, the algebraic sumof currents at any circuit node is equal to zero, which can be expressed as

∑

k

Ik = I1 + I2 + . . . + In = 0 (1.7.2)

In applying KCL to a node, we add the currents entering the node and subtract thoseleaving the node. For the node in Figure 1.17 KCL gives the following equation:

I1 − I2 + I3 + I4 − I5 − I6 = 0

Now lets apply KCL to the circuit in Figure 1.15. Since the source current, Is is enteringthe top node, it is positive. On the other hand, the resistor currents IR1 and IR2 are are

24

I1

I2

I3

I4

I5

I6

Figure 1.17: A circuit node

negative since they are both leaving the node. This yields,

Is − IR1 − IR2 = 0 (1.7.3)

which can be rearranged as given in Eq. 1.7.1.

Finding the individual currents, IR1 and IR2 since they both have exactly the same voltageacross them. In Figure 1.15, this is the battery voltage, Vs. Using Ohms law, we can findthe individual currents as

IR1 =VR1

R1

=Vs

R1

IR2 =VR2

R2

=Vs

R2

(1.7.4)

From these equations, we can see that a larger current will flow through the branch with thesmaller resistance. Remember that current always flows through the path of least resistance.

1.7.2 Parallel Equivalent Resistance

In the circuit shown in Figure 1.15, the current Is is shared by the two resistors. Substitutingequations 1.7.4 in 1.7.3 we obtain

Is =Vs

R1

+Vs

R2

= Vs

(1

R1

+1

R2

)=

Vs

Req

25

where Req is the equivalent resistance. Therefore, two parallel resistors behave as a singleresistor with an equivalent resistance given by

1

Req

=1

R1

+1

R2

(1.7.5)

or

Req =R1R2

R1 + R2

We can easily show that Eq.1.7.5 can be extended to parallel connections with more thantwo resistors. The equivalent resistance of n resistors in parallel can be expressed as

1

Req

=∑

k

1

Rk

=1

R1

+1

R2

+ . . . +1

Rn

(1.7.6)

Example 1.6. a) Find the equivalent resistance of three resistors in parallel whose resistancesare 10 Ω, 100 Ω and 1000 Ω b) Find the current flowing through each resistor if a voltage of 10 Vis applied to the parallel combination.

Solution:According to Eq. 1.7.6, the equivalent resistance is given by

1Req

=1

R1+

1R2

+1

R3=

110Ω

+1

100Ω+

11000Ω

= 0.111Ω−1

which yields

Req =1

0.111Ω−1≈ 9.01Ω

This result is interesting, because it demonstrates that in a parallel connection, the equivalentresistance is dominated by the smallest resistor, in this case, the 10 Ω resistor.

Realizing that all three resistors have the same applied voltage of 10 V, we can find the currentflowing through each resistor using Ohm’s law as

I1 =10V

10Ω= 1A, I2 =

10V100Ω

= 0.1A, I3 =10V

1000Ω= 0.01A

‖ X ‖Now, let’s apply what we have learned about series and parallel connections to a slightly

more complicated circuit in the following example.

26

R1

R3

R2

R4

R5

Is

Vs

Figure 1.18: A resistive network connected to a voltage source

Example 1.7. Shown in Figure 1.18 is a resistive network connected to a voltage source. Find theequivalent resistance of the resistive network and then use your result to find the current flowingthrough the voltage source. Assume R1 = R2 = R5 = 20kΩ, R3 = R4 = 10kΩ and Vs = 10V .

Solution:The resistors R3 and R4 are in series with an equivalent resistance of 20kΩ. Their equivalent re-sistor is in parallel with R2, which is also a 20kΩ resistor. The equivalent resistance of the parallelcombination is then 20‖20 = 10kΩ. This resistor is in series with R1 and R5, hence the equivalentresistance seen by the voltage source is found as

R1 + R2‖(R3 + R4) + R5 = 20 + 10 + 20 = 50KΩ

and the current flowing through the voltage source is I = Vs/Req = 10V/50kΩ = 0.2mA. ‖ X ‖

1.8 Circuits with Multiple Voltage Sources

The circuits we have considered so far contained a single voltage source. In practice, wesometimes have to work with circuits that contain multiple sources. In analyzing suchcircuits we have to remember the following tips:

1. The direction of the current flowing through a voltage source may be from the positiveto the negative terminal if there exists a second, more powerful voltage source forcingthe current in that direction.

2. The potential difference across an ideal voltage source can never change no matterwhat you connect to the voltage source.

27

I3

I2

I1

R2

R3

R1

6 V10 V

Ix

Figure 1.19: A circuit with two voltage sources and resistors.

3. The direction of the current flowing through a resistor can be assigned arbitrarily aslong as it is consistent with the polarity of the voltage.

4. If the numerical analysis of the circuit gives you a negative voltage or current value,this means that the actual direction of the current in that branch is opposite to theassigned direction.

The following two examples emphasize the fixed voltage rule for the voltage sources.

Example 1.8. In the circuit shown in Figure 1.19 assume all three resistors are 10kΩ. Find thecurrent, Ix.

Solution:The Voltage across the resistor R1 is 10 V since it is in parallel with the 10 V battery. ApplyingOhm’s law, we can calculate the current through resistor R1 as :

I1 =10V10kΩ

= 1mA

Similarly, Since R3 is in parallel with the 6 V battery, the current through resistor R3 is :

I2 =6V

10kΩ= 0.6mA

Now, applying Kirchoff’s Voltage law to the outer loop, which include the two batteries and theresistor, R2 we obtain:

VR2 = 10V − 6V = 4V

Using Ohm’s law, current flowing through R2 is I3 = 4V/10kΩ = 0.4mA. Now applying Kirchoff’scurrent law to the node connecting R1, R2 and the 10 V battery we get :

Ix = I1 + I3 = 1.4mA

28

I3

I2

I1

R1

Vs1

R2

Vs2

Figure 1.20: A resistive circuit with two voltage sources.

‖ X ‖

Example 1.9. In the circuit shown in Figure 1.20, Vs1 = 10V and Vs2 = 5V . Assume the tworesistors are both 10 kΩ. Find the current, I1.

Solution:Applying Kirchoff’s Voltage law to the loop, which includes the two voltage sources and the resistorR1, we find the voltage across R1 as:

VR1 = Vs1 − Vs2 = 5V

Using Ohm’s law, the current through R1 is I2 = 5V/10kΩ = 0.5mA. The resistor R2 is inparallel with Vs1, hence the voltage across R2 is Vs1. Using Ohm’s law, the current through R2 isI3 = 10V/10kΩ = 1mA. Applying Kirchoff’s current law to the node where the two resistors meet,we obtain:

I1 = I2 + I3 = 1.5mA

‖ X ‖

Example 1.10. Repeat the previous example with Vs1 = 2V .

Solution:The first loop equation gives the the voltage across R1 as:

VR1 = Vs1 − Vs2 = 2− 5 = −3V

29

Then, the current flowing through R1 is I2 = −3V/10kΩ = −0.3mA. The negative sign impliesthat the current is flowing in the opposite direction, but its magnitude is correct. The currentflowing through R2 is I3 = 2V/10kΩ = 0.2mA. Kirchoff’s current law yields

I1 = I2 + I3 = −0.3 + 0.2 = −0.1mA

which shows that the assigned direction for I1 in figure 1.20 is also incorrect. Again, the magnitudeof I1 is indeed 0.1 mA, however, it is not flowing in the direction shown on the circuit diagram.‖ X ‖

1.9 The Circuit Ground

We calculate the potential energy of a mountain climber using the formula

E = mgh

where m is climber’s mass, g is the gravitational acceleration and h is the height the climberis at. According to this formula, the potential energy at the mountain base is zero sinceh = 0. Therefore, the climber’s potential energy is a relative quantity measured with respectto the the mountain base or the ground level.

A similar zero energy reference can be defined in electrical circuits as well. The referencepoint is commonly referred to as the circuit ground. We can calculate voltages at differentcircuit nodes by adding the voltage rises and subtracting the voltage drops as we trace thecircuit from the ground to a node of interest. The following examples illustrates the use of theground concept in a simple circuit with two DC voltage sources. Figure 1.21 shows a simpleresistive circuit with a ground terminal attached to the bottom branch. It is important tonote that the ground terminal is just a reference and its presence does not change the circuit.In other words, the voltages across the circuit elements remain the same with or without theground terminal. However, in circuits with a single DC voltage source, it is customary toconnect the ground to the negative terminal of the voltage source.

Example 1.11. Find the voltage at point X in Figure 1.21.

Solution:Using the circuit analysis methods discussed earlier in the chapter we find the voltage drops acrossthe two resistors as V1 = 3 V and V2 = 6 V . To find the voltage at point X, we can follow twodifferent paths both of which are shown on the circuit with dashed lines. Following the first path,we obtain:

Vx = 12− V1 = 12− 3 = 9 V

30

10 k

20 k12 V

X

V1

V2

50 k

3 V

Figure 1.21: A simple circuit with two voltage sources and ground.

It is important to note that the path we choose should not change the result. Following the secondpath in Figure 1.21 should therefore yield the same voltage:

Vx = V2 + 3 = 6 + 3 = 9 V

‖ X ‖

1.10 Diode : A Non-Linear Resistor

1.10.1 I-V Characteristic of a Diode

The diode is a two terminal circuit element with the non-linear I-V characteristic shown inFigure 1.22. Since we no longer have a straight line for the I-V characteristic, we can notuse Eq. 1.5.1 to relate the diode voltage to the diode current or refer to the resistance of adiode. Instead, the resistance must be calculated for a given voltage, which requires findingthe slope of the characteristic

1

R=

dI

dV|Vo (1.10.1)

at that voltage. The modern diode shown in Figure 1.23 is an electronic device madefrom a semiconducting material, most commonly silicon. However, other materials such asgermanium and gallium arsenide are also used to fabricate diodes. There are also differenttypes of diodes based on different operation principles. While the I-V characteristic of anydiode will be similar to the one shown in Figure 1.22, the exact terminal characteristics willbe determined by the diode type and the material used.

The circuit symbol of a diode is shown in Figure 1.24. Its terminals are called anode andcathode. By convention, the diode is said to have a positive voltage (V > 0) or forwardbiased when the anode potential is higher than the cathode potential. Conversely, if the

31

I

(mA)

V

50

100

150

.5 .7 1.0

Figure 1.22: Current-Voltage characteristics of a diode with a turn-on voltage around 0.7 V,which is typical for silicon PN-Junction diodes.

anode voltage is less than the cathode voltage (V < 0), the diode is said to be reversebiased . The positive direction of current flow is always from anode to cathode. Consideringthe I-V characteristic shown in Figure 1.22, we see that when the diode is reverse biased(V < 0) or forward biased (V > 0) with an applied voltage less than approximately 0.5V, the diode current is practically zero. In this region, the diode resembles a very largeresistor allowing hardly any current to pass through. Indeed, in this region, the slope ofthe characteristic is practically zero indicative of infinite resistance or an open-circuit .On the other hand, when the applied voltage is raised above 0.5 V, we see that the diodecurrent increases quickly. When the voltage is around 0.7 V, the I-V characteristic is almostperpendicular to the horizontal3 suggesting zero resistance or a short-circuit.

In circuit analysis, the diode is often viewed as a switch. Hence, when the diode resistanceis large, the diode is said to be ‘off’. Similarly, when the diode shows very little resistanceto the current flow, the diode is said to be ‘on’. The minimum voltage required to turn on adiode is naturally called the turn-on voltage, Vγ. In Figure 1.22 , this voltage is somewherebetween 0.5 V and 0.7 V. It is customary to fit a straight-line to the characteristic in the‘on’ region and extrapolate it to the horizontal axis to find the turn-on voltage. However,for simplicity in hand calculations, 0.7 V is commonly used as the standard voltage for Si

3Implies infinite slope.

32

Figure 1.23: A typical PN Junction diode used in electronic circuits.

V

I

anode cathode

+ -

Figure 1.24: Diode circuit symbol

PN Junction diodes.

We note however one important difference between a diode and a switch. While thevoltage across a closed switch is always zero, there exists a net voltage drop across a diodewhen it is ‘on’. For hand calculations, we shall assume that this voltage is equal to theturn-on voltage, Vγ.

Table 1.1 illustrates the different operation regions we have discussed above. Figure 1.25shows the I-V characteristic corresponding to the simplified model given in Table 1.1.

Reverse Bias Forward Bias Forward BiasV < 0 Vγ > V > 0 V > Vγ

I = 0 I = 0 I > 0

Table 1.1: Three regions of diode operation we shall use in simple hand calculations

33

I

V

I

VVg

Figure 1.25: a) Actual I-V characteristic of a diode, b) Simplification of the same character-istic for hand analysis.

1.10.2 A Simple Diode Circuit

Shown in Figure 1.26(a) is a resistor and a diode connected to a battery. We note that thediode is forward biased since the battery is forcing the current in normal direction of thecurrent flow in a diode. Therefore, the diode is on, it shows very little resistance to currentflow and holds a voltage equal to the turn-on voltage, Vγ across its terminals. The remainingbattery voltage should then drop across the resistor to satisfy KVL. The resistor actuallyplays a very important role in this circuit. Imagine connecting a 1.5 V battery directlyacross the terminals of a diode. This would force the diode to have 1.5 between its terminalsinstead of 0.7 V. This would in turn require a very large current flow with the possibilityof damaging the diode permanently. The series resistor eliminates this possibility by settingthe maximum current at

I =VR

R=

Vs − 0.7

R(1.10.2)

In the circuit shown in Figure 1.26(b) the voltage source is trying to force a current inthe counter-clockwise direction, which is opposite to the normal direction of current flowthrough the diode. Thus, the diode is reverse biased, it is ‘off’ and the current in the loopis zero4.

4An realistic diode allows a small leakage current to flow in the reverse direction.

34

I > 0 I = 0

(a) (b)

Figure 1.26: a) A simple circuit with a forward biased diode. A resistor is always needed inseries with a diode to limit the current. b) circuit obtained by reversing the polarity of theDC Voltage source. The diode is reverse biased, there is no current flow in the circuit.

Example 1.12. Find the diode current in the circuits of Figure 1.26a and 1.26b. AssumeVs = 5 V, Vγ = 0.7 V and R = 10kΩ

Solution:In Figure 1.26a, the diode is forward biased and there is current flow in the loop. According toequation 1.10.2, the voltage across the resistor is

VR = Vs − 0.7 = 4.3V

Ohm’s law yields a current of

I =VR

R=

4.3 V

10 kΩ= 0.43 mA

What would happen if we reduced the series resistance to 1 Ω? Our calculation would yield 4.3A, which would be too high for most diodes used in electronic circuits.

In Figure 1.26b, the diode is reverse biased and it is off. This means that there is no currentflow in the loop. It is interesting to note that from Ohm’s law, if the current is zero, there cannot be a voltage drop across the series resistor. Thus, the entire supply voltage appears across thediode. Keep in mind however that this is reverse bias and voltages larger than Vγ are perfectly safesince there is no current flow. ‖ X ‖

Example 1.13. In the circuit shown in Figure 1.27 the diode has a turn-on voltage of 1 V. As-sume Vs = 6V , R1 = 10kΩ and R2 = 5 kΩ. First determine if the diode is on and then find thecurrent flowing through the diode.

Solution:First we assume that the diode is off and then find the voltage applied across the diode. The samevoltage must appear across the 5kΩ resistor since it is in parallel with the diode. When the diode isoff, it is equivalent to an open-circuit. Removing the diode from the circuit leaves the two resistorsin series. Since both resistors now have the same current we can use the voltage division rule to

35

I3

I2I

1

R1

R2

Vs

Figure 1.27: A resistive circuit with a diode.

find the voltage across the 5kΩ resistor.

VR2 = VsR2

R1 + R2= 6V × 5

10 + 5= 2 V

Since this voltage is greater than the turn-on voltage of the diode, we conclude that the diode mustbe on. Therefore, the actual voltage across the diode has to be equal to the turn-on voltage of thediode or 1 V. Since the diode is in parallel with the 5kΩ resistor, the resistor voltage is also 1 V andthe current flowing through the resistor can be found using Ohm’s law as IR2 = 1V/5kΩ = 0.2mA.Applying Kirchoff’s voltage law to the outer loop including the diode, the voltage source and the10kΩ resistor, we can write the following equation.

VR1 = Vs − Vγ = 5 V

Then, the current flowing through the resistor is IR1 = 5V/10kΩ = 0.5 mA. The current flowingthrough the diode can be found by applying Kirchoff’s current law.

Idiode = IR1 − IR2 = 0.3 mA

‖ X ‖

Example 1.14. For the circuit shown in Figure 1.28 first find if the diode is on and then find thevoltage at the test point, X. Assume that the diode has a turn-on voltage of 0.7 V .

Solution:We first assume the diode is off and find the voltage applied to the diode. Applying Kirchoff’svoltage law to the outer loop yields,

12− VD − 3− VR = 0

36

20 k12 V

X

VD

VR

50 k

3 V

Figure 1.28: A circuit with two voltage sources and a diode

When the diode is replaced by an open circuit, current cannot flow through the outer loop resultingin VR = 0. This yields a diode voltage of 9 V , which is greater than the turn-on voltage of thediode. We conclude that the diode is on and understand that the actual diode voltage must beequal to the turn-on voltage or 0.7 V . The voltage at the test node X can be found as:

Vx = 12− 0.7 = 11.3 V

‖ X ‖

1.11 Other Resistive Elements

1.11.1 Potentiometer - Variable Resistor

A potentiometer is an ordinary resistor whose resistance can be varied by turning a knob.A typical potentiometer is shown in Figure 1.29. This is the type used to control the soundvolume in a stereo system. Figure 1.30a shows a simple resistive circuit in which we have anexternal resistor connected to a potentiometer. Figure 1.30b is the equivalent circuit, whichshows that the external resistance is in parallel with only a portion of the potentiometer re-sistance, labelled ‘RB’. The total resistance of a potentiometer is always constant and equalto RA + RB. By turning the knob the value of RB is varied between 0 and the maximumresistance, RA + RB.

There are other types of variable resistors, which are much smaller than the potentiome-ter shown in Figure 1.29. However, the operation principle demonstrated in Figure 1.30remains the same.

37

Figure 1.29: A typical potentiometer used in electronics instruments. The device has threeterminals for use as a voltage divider.

1.11.2 Light Emitting Diode (LED)

Light emitting diodes are very similar to regular diodes. The difference is they produce light!Otherwise, their I-V characteristics are exactly the same as regular diodes. However, as wenoted before, the diode turn-on voltage is determined by the properties of the semiconductormaterial. LEDs are made of materials other than silicon. Typical LED turn-on voltages arearound 2.0 V and this changes from device to device determined by the semiconductormaterial used. In a circuit, just like a regular diode, an LED requires a series resistor tolimit the amount of current flow.

LEDs come in different colors. Red, green, yellow and orange LEDs are widely available.The amount of light produced by LEDs also varies greatly. The light intensity measured inmillicandela (mcd) can vary from several hundred to several thousand. In some applications(e.g. traffic lights and tail lights of cars) many LEDs can be used together to produce largeintensities of light.

38

RA

RB

Vs

Rext

Vs

Rext

Figure 1.30: A potentiometer connected to an external resistor. Note that only a portion ofthe total resistance, labelled RB is in parallel with the external resistance.

Figure 1.31: Light emitting diode (LED)

1.11.3 Photocell-Light Sensitive Resistor

A photocell is a resistor whose resistance is determined by the intensity of the ambient light.As such, in constant lighting conditions, photocells behave as regular resistors with perfectlylinear I-V characteristics. Good photocells may have resistance values ranging from a fewhundred ohms in light to several megaohms in dark. Photocells are used in cameras, elevatordoors and many other applications where light sensing is necessary.

39

Figure 1.32: Two photocells of different sizes.

40

Problems

P 1.1 Calculate the resistance of a 1 km longelectrical wire. Assume the wire has adiameter of 3 mm and the conductorused has a resistivity of 0.5 µΩ− cm.

P 1.2 Consider the circuit shown in Figure P1.2,a) Determine the resistors connected inseries and/or in parallel, b) Assume allresistors are 10kΩ. Find the equivalentresistance between the points x and y.

R1

R2

R3x y

Figure P1.2

P 1.3 Repeat problem P1.2 for the circuit shownin Figure P1.3

R1

R2

R3

R4

x y

Figure P1.3




P 1.7 For the circuit shown in Figure P1.7,assume R1 = 1 kΩ, R2 = 2.2 kΩ, R3 =

R1

R2

R3

R4

x y

R5

Figure P1.4

R1

R2

R3

R4

x y

R5

R6

Figure P1.5

4.7 kΩ. a)Find the current flowing inthe loop, b)Find the voltage drop acrosseach resistor.

P 1.8 A 9 V battery has an internal resistanceof 10 Ω. The battery is connected to a470 Ω resistor. What is the actual volt-age that shows up across the resistor?

P 1.9 For the circuit shown in Figure P1.9,assume R1 = 1.5 kΩ, R2 = 3.3 kΩ,R3 = 4.7 kΩ, R4 = 10 kΩ. First findthe currents I1 through I4 and then findthe voltage across each resistor.

P 1.10 For the circuit shown in Figure P1.10,assume R1 = 1.5 kΩ, R2 = 3.3 kΩ, R3 =4.7 kΩ, R4 = 10 kΩ and R5 = 15 kΩ.First find the currents I1 through I5 and

41

R1

R2

R3

R4

x y

R5

R6

Figure P1.6

9 V

R1

R3

R2

I

Figure P1.7

then find the voltage across each resis-tor.

P 1.11 For the circuit shown in Figure P1.11,assume R1 = 1.5 kΩ, R2 = 3.3 kΩ, R3 =4.7 kΩ and R4 = 10 kΩ. First find thecurrents I1 through I3 and then find thevoltage across each resistor.

P 1.12 For the circuit shown in Figure P1.12,assume R1 = 1.5 kΩ, R2 = 3.3 kΩ, R3 =4.7 kΩ, R4 = 10 kΩ and R4 = 15 kΩ.First find the currents I1 through I3 andthen find the voltage across each resis-tor.

P 1.13 For the circuit shown in Figure P1.13,assume R1 = 68 kΩ, R2 = 120 kΩ, R3 =470 kΩ, R4 = 150 kΩ and R5 = 330 kΩ.

9 V

R1

R4

R2 R3

I1

I2 I3

I4

Figure P1.9

9 V

R1

R3

R2 3 V

I1

I2

I3

I4

Figure P1.10

Find the voltage at the node, X. Notethat this is the voltage measured relativeto the ground of the circuit.

P 1.14 The I-V characteristic of a non-linear re-sistor is shown in Figure P1.14. Findthe resistance of the element when theapplied voltage is a) -2 V, b) 0 V and c)2.5 V.

P 1.15 The I-V characteristic of a diode can beexpressed as

I = I0

(eV/VT − 1

)

where I0 = 1 µA and VT = 25 mV .Find the resistance of the diode whenthe diode voltage is a) -1 V, b) 0 V, c)0.2 V and d) 0.7 V

42

9 V

R1

R3

R2 3 V

I1

I2

I4

R4

Figure P1.11

9 V

R1

R3

R2 6 V

I1

I2I3 I4

R4

3 VR5

I5

Figure P1.12

P 1.16 The I-V characteristic of a particular non-linear resistor is given by

I = I0(V/V0)3

where I0 = 15 mA and V0 = 10 mV .Find and sketch the resistance of the el-ement as a function of the applied volt-age, V in the voltage range of −100 mVto +100 mV .

P 1.17 In the circuit shown in Figure P1.17 theresistor R1 is an ordinary, 10 kΩ resistorwhile R2 is a non-linear resistor. TheI-V characteristic of R2 is given by

I = I0(V/V0)3

9 V

R1

R2 6 V

3 VR4

R3

x

R5

Figure P1.13

where I0 = 1.2 mA and V0 = 1 V . Findthe current flowing in the loop and thevoltage drop across each resistor.

P 1.18 Repeat the previous problem with R1 =1 kΩ.

P 1.19 The resistance of a certain thermistorvaries with the ambient temperature ina non-linear fashion. The temperaturedependence of the resistance can be ex-pressed as

1

T=

1

T0

+1

Bln

R

R0

where T0 = 25oC, R0 = 1 kΩ and B =3oC. Suppose the ambient temperatureis 32oC. Find the current flow throughthe thermistor when the applied voltageis 9 V.

Use the piecewise linear diode modelin solving the rest of the problems.

P 1.20 The diode in Figure P1.20 has a turn-on voltage of 0.7 V. Find the resistorsuch that the current flowing through

43

1 2 3- 1- 2- 3- 4

1

2

3

- 1

- 2

- 3

V

I (mA)

Figure P1.14

9 V

R1

R2

Figure P1.17

the diode is 1 mA.P 1.21 The diode in Figure P1.21 has a turn-

on voltage of 0.7 V. Assume R = 1kΩ.First determine if the diode is on andthen find the diode current.

P 1.22 Repeat problem P1.21 with R = 100Ω.P 1.23 What is the smallest resistor, R that can

be used in Figure P1.21 without turn-ing off the diode? Assume the diode hasa turn-on voltage of 0.7 V.

P 1.24 The diode in Figure P1.24 has a turn-on voltage of 0.7 V. First determine ifthe diode is on and then find the diodecurrent.

P 1.25 Repeat problem P1.24 with a light emit-ting diode that has a turn-on voltage of2 V.

9 V

R

Figure P1.20

9 V

3 k

R

Figure P1.21

P 1.26 The diode in Figure P1.26 has a turn-on voltage of 0.7 V. First determine ifthe diode is on and then find the diodecurrent.

P 1.27 Determine the minimum voltage that thebattery on the left in Figure P1.26 mustproduce in order the keep the diode on.Assume the diode has a turn-on voltageof 2 V.

44

9 V

8 k

1 k

10 k

Figure P1.24

9 V 1 k

9 V

1 k

Figure P1.26

Chapter 2

Capacitors And RC Circuits

Until now, we have only considered circuits with resistive elements. In this chapter, weintroduce a new two-terminal element, the capacitor, which is capable of storing energy inan electric field. Capacitors have numerous applications in AC/DC conversion, filters, timingcircuits and oscillators.

The circuits constructed using resistors and capacitors are referred to as RC Circuits. Inthis chapter, we shall learn the fundamental concepts used to analyze circuits of this type.Specifically, we will focus on the transient analysis of first-order RC circuits with DC voltagesources. Examples of simple circuits with diodes will also be given.

2.1 Capacitor Fundamentals

2.1.1 Capacitance

The capacitor is a fairly simple device consisting of two metal plates separated by an insu-lator as shown in Figure P2.0. The symbols used for capacitors in circuit diagrams closelyresemble this structure. Two alternative symbols are shown in Figure P2.0.

When a capacitor is connected to a battery, electric charge accumulates on the capacitorplates. The charge stored on one of the plates has the same magnitude as the charge storedon the other plate, but its polarity is different. That is, if the charge on plate is positive,the other plate is negatively charged. This is essential to preserve the charge neutrality if weconsider the capacitor as an isolated system. Charge neutrality is a fundamental phenomenonthat is (and should be) satisfied by all real objects. Violation of this rule would result inelectric forces, which are far too powerful for the real world to handle. For a fascinatingdiscussion of the charge neutrality, and the resulting electric forces in the absence of it, thereader is referred to ”Lectures on Physics” by the Nobel prize winning physicist, RichardFeynman.

Since the two plates are separated from each other, an electric field is established between

45

46

Insulatormetal plates

(area = A)

d

Figure P2.0 A Capacitor is a passive circuit element consisting of two metal plates separatedby a insulator.

C

Figure P2.0 Two alternative symbols used for capacitors in electric and electronic circuits

47

Figure P2.0 Michael Faraday, 1791-1867

the plates, which provides the means for energy storage. The amount of charge, Q storedon each capacitor plate is given by

Q = CV (2.1.1)

where C is the capacitance and V is the potential difference between the two plates. Theunit of capacitance is Farads, which was named after Michael Faraday.

The capacitance of a parallel plate capacitor is determined by the dielectric permittivityof the insulator, ε, the cross-sectional area, A and the distance, d. The capacitance is givenby the following equation:

C = εA

d(2.1.2)

In electronic circuits, we use capacitors with capacitance values ranging from a few pi-cofarads (pF) to thousands of microfarads (µF ). Insulators with high dielectric constantsare used to make smaller capacitors. Examples of commonly used insulators include mylar,polystyrene, mica, glass and ceramic. Figure P2.0 shows examples of capacitors used inelectronic circuits.

48

(a)

(b)

Figure P2.0 Examples of capacitors used in electronic circuits. a)ceramic disc capacitorsare available at values ranging from a few picofarads to tens of nanofarads, b)electrolyticcapacitors provide larger capacitance values up to several thousand microfarads(µF )

49

Figure P2.0 A capacitor connected to a DC voltage source.

2.1.2 Charging a Capacitor

Charging a capacitor involves physically bringing new electrons to one of its plates by con-necting the capacitor to a voltage source. As shown in Figure P2.0, electrons from thenegative terminal of the battery travel to the bottom plate of the capacitor. Since eachelectron carries a fixed charge of −1.6× 10−19 coulombs, electron accumulation on the plateresults in a net negative charge. To preserve charge neutrality, the top plate must thenacquire an equal amount of positive charge. The plate accomplishes this task by losing someof its electrons to the battery. For every lost electron, a charge of +1.6 × 10−19 coulombis added to the capacitor plate. This process results in two charged plates separated by aninsulator. An electric field is established between the plates, which creates a medium forstoring electric energy.

As the charging takes place, electrons continue to travel from the negative terminal of thebattery to the bottom plate and from the top plate to the positive terminal of the battery.It is interesting to realize that, while we effectively have an open-circuit in Figure P2.0,there appears to be a continuous movement of charge around the loop. Since charge motionis the essence of current flow, there exists a net current flow in Figure P2.0 in spite of theopen-circuit between the capacitor plates. The current continues to flow in the loop untilthe capacitor is fully charged and no more electrons can be added to the negatively chargedplate. The maximum charge that can be stored in a capacitor plate for a given voltage V isgiven by equation 2.1.1.

It is now interesting to ask the following question: How long does it take to charge acapacitor? As we have seen above, this requires that we physically move the electrons toand from the battery, which takes some finite amount of time. You can imagine that thefaster we can move the electrons in the loop, the faster the capacitor will charge. In Figure

50

vc(t)V

s

t = 0 vR (t)

Figure P2.0 A capacitor charging through a resistor. The charging begins when the switchis closed at t = 0.

P2.0, the battery is connected directly across the capacitor. Thus, the only resistance tocurrent flow is the resistance of the connecting wire, which is very small. What if we inserteda resistor in series with the capacitor as in Figure P2.0. Wouldn’t you expect the resistorto slow down the electrons and delay the charging process?

We can also argue that a larger capacitor will require more time to fully charge simplybecause it can hold more electrons. Therefore, by intuition, we can conclude that if weincrease the resistance or the capacitance in Figure P2.0 we will increase the charging time.

2.1.3 Current-Voltage Relationship of a Capacitor

Resistors obey Ohm’s law, which provides a simple relationship between the resistor voltageand current. Capacitors do not obey Ohm’s law and consequently, they have a very differentI-V characteristic, which is the main subject of this section.

We have previously defined the current flow as

i(t) =dq

dt(2.1.3)

where dq is the charge moved during the time interval, dt. We can also write equation 2.1.1in the form,

dq = Cdv (2.1.4)

which states that a small change in the capacitor voltage is accompanied by a small change inthe stored charge. Substituting 2.1.4 in 2.1.3 we obtain the I-V characteristic of a capacitorgiven below.

ic(t) = Cdvc(t)

dt(2.1.5)

51

According to this equation, a larger capacitor current provides a larger change in the ca-pacitor voltage in a given time interval, which is in agreement with our discussion in theprevious section. Integrating both sides of the equation 2.1.5, we obtain another form of thisequation.

vc(t) =1

C

∫ to

0

ic(t)dt (2.1.6)

It should be obvious from our previous discussions that because it takes time to move chargearound the loop, we cannot charge a capacitor instantaneously. This is usually expressed bythe following equation:

Qc(0−) = Qc(0

+) (2.1.7)

where Qc(0−) refers to the instant of time immediately before connecting the capacitor to a

battery and Qc(0+) the time immediately after. Because charge and voltage are related to

each other by equation 2.1.1, the voltage across a capacitor can not change instantaneouslyeither. We express this by the equation below.

vc(0−) = vc(0

+) (2.1.8)

Now suppose the capacitor in Figure P2.0 is fully uncharged prior to closing the switch att = 0, hence, the capacitor voltage is zero. Equation 2.1.8 implies that at the instant weclose the switch, the capacitor voltage must remain the same. Applying Kirchoff’s voltagelaw to the loop yields

Vs − vR(0+)− vc(0+) = 0 (2.1.9)

Substituting 0 V for the capacitor voltage gives vR(0+) = Vs, which implies that the entirebattery voltage drops across the resistor at the instant we close the switch. Using Ohm’slaw, the current flowing through the resistor at t = 0+ can be found as

iR(0+) =vR(0+)

R=

Vs

R(2.1.10)

Since the resistor is in series with the capacitor, the current flowing through the capacitoris also given by equation 2.1.10. As the capacitor charges, the voltage across the resistordrops and so does the current flowing in the loop. The capacitor continues to charge until thecapacitor voltage reaches the battery voltage. At that point, the voltage across the resistorbecomes zero and the current stops flowing. We need to understand that any current flowin the loop suggests the existence of charge flow to and from the capacitor plates. This inturn suggests that the capacitor is either charging or discharging.

52

R1

C3 Vt = 0R2

Figure P2.0 When the switch is closed, the capacitor is shorted, hence, it remains un-charged. Charging begins as soon as the switch is opens.

Example 2.1. The capacitor in Figure P2.0 is completely uncharged prior to opening the switchat t = 0. Find the current flowing through the capacitor at the instant the switch is opened.Assume R1 = R2 = 1kΩ and C = 1µF

Solution:Since the capacitor cannot charge instantaneously, vc(0+) = 0. Applying Kirchoff’s voltage law tothe loop, we find that the entire battery voltage must drop across the resistor, R1 at t = 0+. FromOhm’s law, we can find the current flowing through R1 as i1(0+) = 3 V/1 kΩ = 3 mA. Since R2

is in parallel with the capacitor, its voltage must also be equal to zero, hence, no current can flowthrough this resistor. Kirchoff’s voltage law requires that the sum of the three currents flowingthrough the two resistors and the capacitor must be zero. Since the current through R2 is zero,then the entire current flowing through R1 must flow through the capacitor. ‖ X ‖

It is interesting to note that if vc(t)in Eq.2.1.5 is constant, the current through thecapacitor is zero. In other words, if a DC voltage is applied to a capacitor, the capacitorcurrent is zero. Indeed, a capacitor does not allow any DC current to flow.

Example 2.2. Find the current flowing through the capacitor in Figure P2.0 at the instant theswitch opens at t = 0. Assume Vs = 5 V , R1 = R2 = 10 kΩ, Vγ = 0.7 V and C = 1 µF .

Solution:Since the capacitor is shorted by the switch, it must be completely uncharged prior to opening theswitch at t = 0. Since the capacitor voltage can not change instantaneously, the capacitor voltagemust remain at 0 V at t = 0+, which makes the capacitor equivalent to a short-circuit.

Now we need to find if the diode is on. To do this, we assume that the diode is off and findthe voltage applied across the diode. Since both resistors are 10 kΩ, they share the battery voltageequally, hence, the voltage applied across the diode is equal to half the battery voltage or 2.5 V .

53

R1

R2

C

t = 0

Vs

Figure P2.0 An RC circuit with a diode. The capacitor begins charging as soon as theswitch openes at t = 0.

R1

CVs

t = 0

Figure P2.0 An RC circuit with an LED. The LED limits the maximum voltage the ca-pacitor can charge to.

We conclude that the diode is on, which means that the voltage across the diode has to be equalto the turn-on voltage of 0.7 V.

Applying Kirchoff’s voltage law to the loop yields the voltage across R1 as

vR1 = Vs − VD = 5− 0.7 = 4.3 V

Using Ohm’s law, the current through the resistor is IR1 = 0.43mA. Since R1 is in series with thecapacitor the same current must be flowing through the capacitor. ‖ X ‖

Example 2.3. In Figure P2.0, the switch is kept closed prior to t = 0, keeping the LED offand the capacitor uncharged. Find the capacitor current at the instant the switch opens. Assume

54

Vs = 5 V , R1 = 5 kΩ, Vγ = 1.8 V and C = 4.7 µF .

Solution:When the switch opens at t = 0, the capacitor voltage can not change instantaneously, there-fore, the diode is effectively shorted and it is off. Then, applying Kirchoff’s voltage law to theloop reveals that the entire battery voltage must drop across the resistor resulting in a current of5 V/5 kΩ = 1 mA. Since the diode is off, it is equivalent to an open-circuit and the diode current iszero. Thus, to satisfy Kirchoff’s current the capacitor current must be exactly equal to the currentflowing through the resistor. ‖ X ‖

Example 2.4. In Figure P2.0, find the maximum voltage the capacitor will charge to. Also findthe current that will flow through the LED when the capacitor reaches this maximum voltage.

Solution:The LED turns on when the capacitor voltage reaches the turn-on voltage of the LED, therefore,the capacitor voltage cannot exceed Vγ = 1.8 V . The fact that the capacitor voltage has to remainforever at 1.8 V guarantees that there can not be any charge transfer to and from the capacitor oncethe LED turns on. Since we cannot have a current without any charge movement, the capacitorcurrent has to be zero, which also means that the entire current flowing through the resistor mustflow through the diode. In other words, the capacitor behaves like an open-circuit once the LEDturns on. The voltage across the resistor is equal to the supply voltage minus the voltage acrossthe LED. The LED current can be found as

ILED =Vs − Vγ

R1=

5 V − 1.8 V

5 kΩ= 0.64 mA

‖ X ‖

2.2 Transient Response of an RC Circuit

2.2.1 Capacitor Voltage During Charging

In this section, we will derive a classic equation that describes the time dependence of thecapacitor voltage as it charges to the voltage applied to the RC circuit. The derivation willbe made for the RC circuit shown in Figure P2.0. We will assume that the capacitor iscompletely uncharged prior to closing the switch at t = 0. Closing the switch completes theloop and we can apply Kirchoff’s voltage law to the circuit.

Vs − vR(t)− vc(t) = 0 (2.2.1)

55

The above equation states that the voltage supplied by the battery is always shared by thetwo elements. Using 2.2.1 and Ohms law, the current going through the resistor can beexpressed as

iR(t) =vR(t)

R=

Vs − vc(t)

R(2.2.2)

Since the same current must be following through both elements, the resistor current mustalso satisfy the current-voltage relationship of the capacitor given in equation 2.1.5. Thisyields another equation for the resistor current given below.

iR(t) = ic(t) = Cdvc(t)

dt(2.2.3)

Equating above to equation 2.2.2 and rearranging the terms, we obtain the following differ-ential equation.

dvc(t)

dt+

vc(t)

RC− Vs

RC= 0 (2.2.4)

You will learn how to solve similar equations in future courses. For now, we will give you thesolution, which you can readily verify by substituting it back into the differential equation.The solution is the classic capacitor charging equation given below.

vc(t) = Vs(1− e−t/RC) (2.2.5)

The RC product in the exponential term is called the time constant,τ , of the circuit andit is given in units of time. We have plotted Eq. 2.2.5 in Figure P2.0 using four different RCtime constants obtained using the resistor and capacitor values given in the figure caption.We can see that in all four cases the capacitor voltage starts at 0 V and then increasesexponentially toward the voltage, Vs applied to the RC circuit. Different time constantsresult in different capacitor charging rates. In general, larger the RC time constant, longerit takes to charge the capacitor. This should make intuitive sense: a larger resistance slowsdown the charge transfer, a larger capacitance can store more charge.

You can readily predict this behavior from the charging equation. At t = 0, the exponen-tial term in equation 2.2.5 is equal to unity, hence, the capacitor voltage is zero. This satisfiesthe condition that the capacitor voltage can not change instantaneously. With increasingtime, the exponential term, e−t/RC becomes small and the capacitor voltage approaches thesupply voltage, Vs.

Example 2.5. In the RC circuit of Figure P2.0 calculate how long it would take to charge the ca-pacitor to half the battery voltage after closing the switch at t = 0. Assume Vs = 10V , C = 2.2µF ,

56

0 0.5 1 1.5 2 2.5 30

2

4

6

8

10

Time (s)

Vol

tage

(V

)

Figure P2.0 Capacitor voltage during charging plotted for four different RC time constantsobtained using R = 100 kΩ and C = 1, 3, 6 and 10 µF . The capacitor begins to charge att = 0. The voltage source used in the circuit is 10 V

57

R = 1 kΩ.

Solution:The capacitor voltage will increase according to Eq. 2.2.5. We can find the RC time constant as

τ = RC = 1 kΩ× 2.2 µF = 103 Ω× 2.2× 10−6 F = 2.2ms

Substituting Vs = 10 V and vc(t) = 5 V in 2.2.5 and solving for time, t, we find that it willtake 1.5 ms to charge the capacitor to 5 V . ‖ X ‖

Example 2.6. How long does it take the capacitor in Figure P2.0 to reach the LED turn-onvoltage after opening the switch at t = 0?

Solution:When the capacitor begins to charge at t = 0 the LED is off and it has to remain off untilthe capacitor voltage is equal to the turn-on voltage of the diode. During this time, the diode isequivalent to an open-circuit, hence, the capacitor begins to charge with a time constant of

τ = RC = 5000 Ω× 4.7× 10−6 F = 0.0235 s

The capacitor voltage during charging is given by equation 2.2.5. We note that this equation wasobtained for the simple RC circuit in Figure P2.0. The only reason we can use the same equationin this problem is that the diode is off, hence, it is equivalent to an open circuit. Substituting theknown parameters in 2.2.5 we obtain:

vc(t) = Vs

[1− e−t/RC

]= 5

[1− e−t/0.0235

]V

When vc = 1.7 V , the diode turns on. Since the LED voltage has to remain constant at the turn-onvoltage, the capacitor charging stops when the LED is on. Thus the charging time is the timerequired to charge the capacitor to only 1.7 V instead of the battery voltage of 5 V. Substituting1.7 V for the capacitor voltage in the above equation and solving for time,

t = −0.0235 ln[1− 1.7 V

5 V

]' 9.8 ms

‖ X ‖

2.2.2 Capacitor current during charging

Since we know how the capacitor voltage is changing with time, it is easy to find the timedependence of the capacitor current. Substituting the capacitor voltage, vc(t) from equation2.2.5 into the capacitor I-V characteristic given in 2.1.5 we obtain the following equation,

58

0 0.5 1 1.5 2 2.5 3

1

2

3

4

5

6

7

8

9

10

11x 10

−5

Time (s)

Cur

rent

(A

)

Figure P2.0 Capacitor current during charging plotted for four different RC time constantsobtained using R = 100 kΩ and C = 1, 3, 6 and 10 µF . The capacitor begins to charge att = 0. The voltage source used in the circuit is 10 V

ic(t) = Cd

dt

[Vs(1− e−t/RC)

]=

Vs

Re−t/RC (2.2.6)

which is plotted in Figure P2.0. We can see that the maximum current flows through thecapacitor at t = 0, the instant the switch is closed and the capacitor begins to charge. Thiscurrent can be found by substituting t = 0 in equation 2.2.6. This yields,

iR(0) =Vs

R

which we have already found in the previous section by using vc(0−) = vc(0

+) = 0 leavingthe entire battery voltage drop across the resistor.

With time, the capacitor current decreases exponentially approaching zero at t = ∞.Equation 2.2.6 indicates that when the capacitor becomes fully charged, that is, when vc(t) =Vs, the capacitor acts like an open circuit and current can no longer flow in the loop.

59

t = 0

vc (t) v

R (t)

ic (t)

Figure P2.0 The capacitor is charged to a voltage Vo before closing the switch at t = 0.When the switch is closed, the capacitor starts discharging through the resistor.

2.3 Discharging a Capacitor

Now lets consider how a charged capacitor discharges when it is connected to a resistor.Suppose the capacitor in Figure P2.0 is fully charged to a voltage Vo before it is connected tothe circuit. When the switch is closed at t = 0, a current starts flowing in the loop dischargingthe capacitor. During this process, electrons flow in the counter-clockwise direction untilno excess electrons are left on the negatively charged capacitor plate. At that point, thepotential difference and the electric field across the capacitor are both zero. It is interestingto note that the capacitor in Figure P2.0 closely resembles a voltage source providing energyto the circuit. The difference is that the capacitor is able to provide this energy only for alimited time until it is completely discharged. The energy stored in the capacitor is convertedto heat as the current travels through the resistor. We will return to this subject when wediscuss electric power in Chapter 4.

2.3.1 Capacitor voltage during discharging

We will now derive equations for the capacitor voltage and current during the dischargecycle. First, we note that our simple circuit in Figure P2.0 has a unique property. Sincethe capacitor and the resistor form a closed loop, the same current must be flowing throughboth elements. This yields the equation,

Cdvc(t)

dt=

vR(t)

R(2.3.1)

where we used Ohm’s law and equation 2.1.5for the resistor and capacitor current respec-tively. Since both terminals of the elements are connected to each other, they must also havethe same voltage, that is, vc(t) = vR(t). Substituting above and rearranging the terms wefind the differential equation given below.

60

0 0.5 1 1.5 2 2.5 3

1

2

3

4

5

6

7

8

9

10

11

Time (s)

Vol

tage

(V

)

Figure P2.0 Capacitor voltage during discharging plotted for four different RC time con-stants obtained using R = 100 kΩ and C = 1, 3, 6 and 10 µF . The capacitor begins todischarge at t = 0 from an initial voltage of 10 V .

61

dvc(t)

dt− vR(t)

RC= 0 (2.3.2)

which we need to solve for the capacitor voltage. The solution yields,

vc(t) = Voe−t/RC (2.3.3)

where Vo is the capacitor voltage at the instant the switch is closed. You can verify that Eq.2.3.3 is indeed a solution by substituting it into the differential equation. Equation 2.3.3 isplotted in Figure P2.0. Similar to the charging process, the rate of change of the capacitorvoltage is highest at the instant the switch is closed. As time passes, the voltage continuesto decrease reaching o V at t = ∞.

Example 2.7. A 1 µF capacitor is charged to 5 V before it is connected to a 22 kΩ resistor. Howlong will it take for the capacitor voltage to drop to 2.5 V?

Solution:The capacitor will discharge through the resistor according to equation 2.3.3 with a time constantof

RC = 22000 Ω× 10−6 F = 0.022 s

Substituting Vo = 5 V and vc(t) = 2.5 V in equation 2.3.3 we obtain,

2.5 = 5e−t/0.022

which yields,

t = −0.022 ln2.55' 15.2 ms

‖ X ‖

2.3.2 Capacitor Current during Discharging

The capacitor current during discharge can be readily found by substituting equation 2.3.3in the capacitor I-V characteristic given in equation 2.1.5. This yields,

ic(t) = Cd

dt

[Voe

−t/RC]

= −Vo

Re−t/RC (2.3.4)

Therefore, as the capacitor voltage decreases exponentially, the capacitor current alsodecreases at the same rate with a time constant equal to RC. The largest current flowsthrough the capacitor at the instant the capacitor begins to discharge. This current can be

62

found by substituting t = 0 in the above equation. Alternatively, since the capacitor voltageis always equal to the resistor voltage, at the instant the switch in Figure P2.0 is closed, theresistor voltage is equal to VR.

63

Problems

P 2.1 The heart of a typical MOS transistorused in a CMOS integrated circuit is acapacitor, which consists of a thin layerof silicon dioxide sandwiched between metaland silicon plates. Suppose the surfacearea of a capacitor plate is 50 nm ×50 nm and the thickness of the oxide is2 nm. Find the capacitance of the MOScapacitor.

P 2.2 A 220 µF capacitor is charged to 9 V.Calculate the net charge stored on oneof its plates.

P 2.3 In Figure P2.3 the switch S1 is closedand the switch S2 is open for t < 0.Suppose S1 is opened and S2 is closedat t = 0. Find the voltage on each ca-pacitor in terms of the battery voltage,Vs.

Vs C1 C2

S1 S2

Figure P2.3

P 2.4 In Figure P2.3 Vs = 9 V , C1 = 100 µFand C2 = 220 µF . The switch S1 isclosed and the switch S2 is open for t <0. Suppose S1 is opened and S2 is closedat t = 0. Find the electric charge storedin each capacitor.

P 2.5 In Figure P2.5 assume the switches S1

and S2 are both closed for t < 0 and S2

is opened at t = 0. Find the voltage oneach capacitor in terms of the batteryvoltage, Vs for t > 0.

Vs

C1

C2

S1

S2

Figure P2.5

P 2.6 In Figure P2.5 Vs = 9 V , C1 = 330 µFand C2 = 47 µF . Assume the switchesS1 and S2 are both closed for t < 0 andS2 is opened at t = 0. Find the electriccharge stored in each capacitor for t > 0.

P 2.7 Find the time constant of the RC circuitshown in Figure P2.7.

9 V

5.6 k

1 F

Figure P2.7

P 2.8 In the RC circuit shown in Figure P2.7,the capacitor is uncharged prior to clos-ing the switch at t = 0. Find how muchtime the capacitor needs to charge toone third of the battery voltage. Findthe current flowing in the loop at thatinstant.

P 2.9 The time constant of the basic RC cir-cuit in Figure P2.7 can be reduced byusing a smaller capacitor or resistor. Howmuch would the capacitor need to charge

64

to one third of the battery voltage if thetime constant is reduced to one half ofits original value in P2.7.

P 2.10 The switch in Figure P2.10 is kept closedprior to t = 0. a) Find the capacitorvoltage at the instant the switch is openedat t = 0+, b) Find the resistor voltageand loop current at the same instant.

9 V

5.6 k

1 F

Figure P2.10

P 2.11 In Figure P2.10, what is the maximumvoltage the capacitor can charge to andhow long does it take for the capacitor toreach 90% of this voltage after openingthe switch at t = 0?

P 2.12 In Figure P2.10, we want to double thetime constant by using a larger capaci-tor. Find the standard capacitance valuethat is nearest to your desired value. Hint:Visit an online electronic store for thestandard capacitance values.

P 2.13 In Figure P2.10, the switch is openedat t = 0. Calculate and plot the timeneeded for the capacitor to reach 6 Vfor capacitance values ranging from 1 to10 µF.

P 2.14 The capacitor in Figure P2.14 is un-charged prior to closing the switch at t =0. a) Find the capacitor voltage and cur-rent immediately after closing the switch;b) Find the maximum voltage the capac-itor can charge to if the switch is kept

closed for a very long time.

9 V

10 k

2.2 F

3.3 k

Figure P2.14

P 2.15 Assume the capacitor is not charged priorto closing the switch at t = 0 in Fig-ure P2.14. Find the time constant forcharging the capacitor.

P 2.16 The switch in Figure P2.16 is openedat t = 0. a) Find the capacitor voltageand current immediately after openingthe switch; b) Find the maximum volt-age the capacitor can charge to if theswitch is left open for a very long time.

9 V

4.7 k

10 F 2.2 k

Figure P2.16

P 2.17 In Figure P2.16, determine how muchtime is needed for the capacitor to reach2 V after opening the switch at t = 0.

P 2.18 The switch in Figure P2.18 is openedat t = 0. Obtain a differential equationfor the capacitor voltage, vc(t) for t ≥ 0.Solve the differential equation.

P 2.19 Find the time constant for charging thecapacitor in Figure P2.18. Assume R1 =

65

Vs

R1

C

R3

R2

Figure P2.18

220 Ω, R2 = 150 Ω, R3 = 680 Ω andC = 0.033 nF .

P 2.20 Assume Vs = 9 V in Figure P2.18. Usethe component values given in the previ-ous problem to find the current flowingthrough R3, a) immediately after open-ing the switch at t = 0 and b) 1 ms afteropening the switch.

P 2.21 Assume the switch in Figure P2.21 iskept closed for a long time to charge thecapacitor and then opened at t = 0 dis-connecting the battery from the circuit.a) Construct and equivalent circuit tofind the maximum voltage the capaci-tor can charge to and use your circuit tofind this voltage; b) What is the capac-itor voltage immediately after openingthe switch? c) Construct an equivalentcircuit to find the capacitor current im-mediately after opening the switch anduse your circuit to find this current.

P 2.22 When the switch in Figure P2.21 is opened,the capacitor begins to discharge becauseit is disconnected from the battery. a)Obtain the differential equation that de-scribes the circuit for t ≥ 0; b) Solve thedifferential equation to determine how

9 V

10 k

220 F

47

Figure P2.21

the capacitor voltage decreases with time;c) Use your result to find the expressionfor the capacitor current for t ≥ 0; d)Use the current expression found in part(c) to find the current immediately afterclosing the switch, i.e. at t = 0+ Ver-ify that your result matches the currentfound using the equivalent circuit in theprevious problem.

P 2.23 Repeat the previous two problems forthe circuit shown in Figure P2.23.

9 V

1.5 k

1.0 F

4.7 k

2.2 k

Figure P2.23

P 2.24 The capacitor in Figure P2.24 is un-charged prior to closing the switch. TheLED in parallel with the capacitor hasa turn-on voltage of 1.7 V. a) The LEDdoes not light up when the switch is closed.Explain why; b) Derive an expression

66

describing how the LED voltage changeswith time; c) Use your expression to de-termine how much time has to pass toturn on the LED.

9 V

10 k

220 F

Figure P2.24

Chapter 3

Periodic Signals in Time Domain

The voltage sources you have seen in the first two chapters were constant or DC voltagesources. In this chapter, we will discuss characteristics of time-varying sources and we willuse them in simple circuits with resistive elements. A commonly used term for time-varyingsignals is Alternating Current or AC signals.

3.1 Periodic Signals: Period and Frequency

A periodic signal is a time-varying voltage waveform that repeats itself continuously. Anexample is shown in Figure 3.1 , which is very similar to the waveforms produced by musicalinstruments. A periodic voltage waveform v(t) satisfies the equation,

v(t) = v(t + T ) (3.1.1)

where T is referred to as the period of the signal. Note that we used a lower-case v for thevoltage in the above equation. In this book, lower case letters will be used for time-varyingsignals whether they are periodic or not.

The frequency, f , of a periodic signal is the number of periods (or cycles) in one second.While cycles/second is occasionally used as the unit of frequency, its standard unit is Hertz(Hz)1. The frequency of a waveform can be determined from the period using,

f =1

T(3.1.2)

1After Heinrich Hertz, Figure 3.2

67

68

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04

−20

−15

−10

−5

0

5

10

15

20

Time (s)

Vol

tage

(V

)

Figure 3.1 : A periodic waveform similar to those produced by musical instruments.

Figure 3.2 : Heinrich Hertz (1857-1894), German scientist, did his early work on the natureof electromagnetic waves

69

1 2 3 4

5 V

t (ms)

vs (t)

5 6

T = 2 ms

Vp-p = 5 V

Figure 3.3 : A square wave with a duty cycle of 50%

3.2 Square Wave

A square wave is a periodic waveform that cycles between two discrete voltage levels. Anexample is shown in Figure 3.3 . The percentage of time a square wave spends at the highvoltage level is called the duty cycle of the square wave. The waveform in Figure 3.3 has aduty cycle of 50%.

Let’s now consider the square wave in Figure 3.4 , which cycles between 0 and 5 V ata frequency of 100 Hz. The duty cycle of the square wave is 25 %. Suppose we apply thisvoltage waveform to a 10 kΩ resistor. Equivalent circuits can be constructed using constantvoltage sources to analyze the circuit behavior in different regions of the square wave. Asshown, when the voltage level is high, the circuit is equivalent to a resistor connected to a 5V battery and the current flowing through the resistor can be found using Ohm’s law as 0.5mA. When the voltage level is zero, the battery can be replaced by a short-circuit and thecurrent in the loop is zero.

One of the discrete voltage levels of a square wave can be negative as shown in Figure3.5 . In this example, when the square wave switches to - 5 V, the current flow becomesnegative, which implies that the current flow is counter-clockwise or opposite to the currentdirection shown on the equivalent circuit. The duty cycle of this square wave is also 25 %.

It should be clear from the above discussion that when a square wave is applied to aresistor, the resulting current waveform is also a square wave and the two waveforms arerelated to each other by Ohm’s Law.

3.3 Sinusoidal Wave

A sinusoidal voltage waveform is a periodic signal of the form

v(t) = Vo cos(2πft) (3.3.1)

where Vo is the amplitude, f is the frequency. Sine wave and sinusoid are other namescommonly used to refer to sinusoidal waveforms. A 100 Hz sinusoid with an amplitude of 1V is shown in Figure 3.6 . The peak-to-peak value, Vp−p of a sinusoid is the voltage differencebetween the two peaks and it is equal to twice the amplitude, Vo.

70

10 20 30 40

5 V

t (ms)

vs (t)

vR (t) = 5 V

5 V

i > 0

vR (t) = 0 V

i = 0

vR (t)v

R (t)

Figure 3.4 : When a square wave is applied to a resistor, the current can be found by usingequivalent circuits for different regions of the input waveform. When the square wave voltageis zero, the voltage source is replaced by a short-circuit.

10 20 30 40

5 V

t (ms)

vs (t)

vR (t) = 5 V

5 V

i > 0

vR (t) = - 5 V

- 5 V

5 V

i < 0

vR (t)v

R (t)

Figure 3.5 : When the square wave voltage is negative, the polarity of the battery in theequivalent circuit is reversed such that it forces a counter-clockwise current through the loop.

71

0 0.5 1 1.5 2

x 10−5

−1

−0.5

0

0.5

1

Time (s)

Vol

tage

(V

)

Period

Amplitude

Figure 3.6 : A sinusoid

Sinusoids are extremely important in electrical engineering. The 120V/60 Hz voltageavailable at electrical outlets is a sinusoid 2, a fact that placed the signal on the forefrontof all waveforms since the beginning of electrical engineering. There is however a far moreimportant reason for giving the sinusoid a special place amongst all other waveforms. InChapter 5, you will learn that any periodic signal, even the square wave with its sharp cornerscan be represented as an infinite sum of sinusoids. This is a subject of great importance andit provides the key to signal analysis in frequency domain.

Example 3.1. Find the amplitude and frequency of the signal v(t) = (10mV ) cos(200t)

Solution:Comparing v(t) with the equation 3.3.1, the amplitude is found as 10mV . The argument of thecosine is 2πf = 200, from which we can find the frequency as

f =2002π

= 31.83 Hz

The period can be found as T = 1f = 0.03141s ‖ X ‖

2We should make a quick note here that 120 V is not the amplitude but the root-mean-square value ofthe sinusoid, which will be covered in Chapter 4.

72

0 0.5 1 1.5 2

x 10−5

−1

−0.5

0

0.5

1

Time (s)

Vol

tage

(V

)

θ = 0

θ = − π/3

Figure 3.7 : Two sinusoids with phase angles of θ = 0 and θ = −π/3 radians.

3.3.1 Phase Angle of a Sinusoidal Waveform

The more general formula for a sinusoidal waveform is given by

v(t) = Vo cos[2πf(t + to)] (3.3.2)

where to is an offset or time delay, which can be positive or negative. A positive to shifts thewaveform to the left. The above equation is commonly expressed in the form

v(t) = Vo cos(2πft + θ) (3.3.3)

where θ is the phase angle in radians given by

θ = 2πfto (3.3.4)

The phase angle, θ is required to be in the interval −π < θ < π. If θ is positive, the waveformis shifted to the left, if it is negative, it is shifted to the right. This is demonstrated in Figure3.7 , which shows two sinusoids with the same amplitude and frequency but different phaseangles of θ = 0 and θ = −π/3 radians. The phase angle is negative because the waveformcan be obtained by shifting the cosine function to the right by π/3 radians.

Example 3.2. Find the amplitude, frequency and the phase angle of the sinusoid shown in Figure3.8 and write a mathematical expression for the waveform.

Solution:By inspection, we determine the amplitude and the period as 2 V and 10 µs respectively. From

73

−5 0 5 10 15

x 10−6

−3

−2

−1

0

1

2

3

Time (s)

Vol

tage

(V

)

θ

Figure 3.8 : A sinusoid with a non-zero phase angle. We can see that the waveform can beobtained by shifting the cos(2πft) function to the left.

Eq. 5.1.1, the frequency is found as 100 kHz. Thus, we know that the sinusoid is in the form

v(t) = 2V cos(2π105t + θ)

where θ is the non-zero phase angle to be determined. First, we see that the waveform has a non-zero phase angle. We can also see that the waveform can be obtained by shifting (2V ) cos(2π105t)to the left, so the phase angle must be positive. To determine the phase angle, we first determinethe amount of shift on the time axis, which we find as T/4 seconds or 2.5 µs. Then, the phaseangle can be determined as

θ = 2πfto

Another approach to finding the phase angle is to remember that one period corresponds to 360

or 2π radians, T/4 seconds corresponds to 90 or π/2 radians. Therefore, the sinusoid can beexpressed as

v(t) = (2V ) cos(2π105t + π/2)

Using the trigonometric identity

cos(α + π/2) = − sin(α)

the above waveform can also be written as

74

vR (t)

iR (t)

vs (t)

ground

Figure 3.9 : A simple circuit with an AC voltage source and a resistor.

v(t) = −(2V ) sin(2π105t)

‖ X ‖

Example 3.3. Find the phase angle of the waveform, v(t) = (10mV ) sin(200t)

Solution:To find the phase angle we have to express the waveform as a cosine function. Using the trigono-metric identity

cos(α− π/2) = sin(α)

the waveform can be written as

v(t) = (10mV ) cos(200t− π/2)

which gives us the phase angle as −90 or −π/2 radians. ‖ X ‖

3.4 Time-Varying Signals in Circuits with Resistive El-

ements

The fundamental laws you have learned in the previous chapter are applicable to time-varyingsignals as well. In this section, we will go over several examples to demonstrate how you cananalyze circuits with time-varying voltage and current signals.

75

3.4.1 Ohm’s Law

When a periodic voltage waveform is applied to a resistive element as in Figure 3.9 , thecurrent through the resistor can be found using Ohm’s law as

i(t) =v(t)

R(3.4.1)

which indicates that the functional form of the current waveform is identical to the voltagewaveform. For instance, for a sinusoidal voltage waveform, v(t) = Vo cos(2πft + θ) thecurrent, i(t) is given by

i(t) =Vo

Rcos(2πft + θ)

Thus, the current through the element is also a sinusoid with the same frequency and phaseangle.

The ”+” signs next to the top terminal of the voltage source has a special meaning. Itindicates that an equation used to describe the characteristics of the waveform gives thevoltage at that node relative to ground level.

3.4.2 Kirchoff’s Laws

When the voltages across the elements of a closed loop change with time, Kirchoff’s voltagecan be expressed as

n∑

k=1

vk(t) = v1(t) + v2(t) + . . . + vn(t) = 0 (3.4.2)

The ”+” sign next to a time-varying voltage source becomes important when there are twoor more time-varying sources in the circuit. For example, consider the circuit in Figure 3.10with two time-varying voltage sources and a resistor. Both sources have their ”+” signs

next to their top terminals. Now suppose both voltage sources generate identical sinusoids(amplitude, frequency and phase). In other words, when the voltage level at one of thoseterminals is positive and rising, the voltage at the other other top terminal must behaveexactly the same way. This in turn means that when the left source is forcing a clockwisecurrent, the right source must be forcing a current in the opposite direction. ApplyingKirchoff’s voltage law to the loop results in the following equation:

v1(t)− vR(t)− v2(t) = 0

If the voltage waveforms produced by the two sources are identical, i.e. v1(t) = v2(t), theycancel each other leaving zero volt across the resistor, which implies that there cannot beany current flowing in the loop at any time.

76

iR (t)

v1 (t) v

2 (t)

Figure 3.10 : A circuit with two time-varying sources and a resistor

i2 (t)

v1 (t)

R2

v2 (t)R

1i1 (t)

ix (t)

Figure 3.11 : A circuit with two time-varying voltage sources.

When we have time varying currents entering or leaving a circuit node, Kirchoff’s currentlaw can be applied to the node according to

n∑

k=1

ik(t) = i1(t) + i2(t) + . . . + in(t) = 0 (3.4.3)

Example 3.4. Find the current, ix(t) in Figure 3.11 . The two voltage sources generate thewaveforms v1(t) = (10 V ) cos(2π100t + π/3) and v2(t) = (20 V ) cos(2π50t − π/4). AssumeR1 = R2 = 10kΩ.

Solution:Since R1 is in parallel with v1(t), the voltage across R1 can be written as

vR1(t) = v1(t) = (10 V ) cos(2π100t + π/3)

The current, i1(t) can then be found using Ohm’s law as

77

i1(t) =(10 V ) cos(2π100t + π/3)

10kΩ= (1 mA) cos(2π100t + π/3)

Applying Kirchoff’s voltage law to the outermost loop including v1, v2 and R2 the voltage acrossthe resistor can be found as

vR2(t) = v1(t)− v2(t) = (10 V ) cos(2π100t + π/3)− (20 V ) cos(2π50t− π/4)

Using Ohm’s law, we can find i2(t) as

i2(t) = (1 mA) cos(2π100t + π/3)− (2 mA) cos(2π50t− π/4)

Applying Kirchoff’s current law to the top node we obtain,

ix(t) = i1(t) + i2(t)

which yields,

ix(t) = (2 mA) [cos(2π100t + π/3)− cos(2π50t− π/4)]

The resulting current waveform is shown in Figure 3.12 . It is important to note that whilethe two added current waveforms were both sinusoids, the resulting waveform is not. You canpredict from this example that by adding different sinusoids we can create infinitely many differentwaveforms. This forms the basis for Fourier Series representation of periodic signals, which weshall discuss in Chapter 5.

0 0.02 0.04 0.06 0.08−5

0

5

Time (s)

Cur

rent

(m

A)

Figure 3.12 : The current waveform, ix(t) is obtained by adding sinusoidal current waveformsat two different frequencies.

‖ X ‖

78

3.5 Average or DC Value of a Periodic Signal

The time average of a periodic signal over a single period is referred to as its average or DCvalue. The average value of a periodic signal can be found by computing the time averageaccording to

VDC =1

T

∫ T

0

v(t)dt (3.5.1)

which is equivalent to finding the area under the signal within one period and dividing thearea by the period. The DC value of a sinusoid is zero because within one period we alwayshave a positive area corresponding to the portion of the signal above the time axis andanother equal but negative area under. Remembering this way of thinking will come veryhandy when we need to find the average value of simple signals such as square waves.

vs (t)

VDC

vR (t)

Figure 3.13 : A resistor connected to a DC and AC voltage source in series.

Now consider Figure 3.13 , which shows a circuit with a battery connected in series withan AC voltage source. According to Kirchoff’s voltage law, the voltage across the resistor isgiven by

vR(t) = vs(t) + VDC (3.5.2)

As an example, vs(t) = 0.5 cos(2π105t) V and VDC = 2 V . Then, the resistor voltage is

vR(t) = 0.5 cos(2π105t) + 2 V

and it is plotted in Figure 3.14 . Note that adding a DC value to a sinusoid is equivalent toshifting the signal up by an amount equal to the DC value. The voltage across the resistorhas a DC value of 2 V.

Finding the DC value of a sinusoid is rather straightforward since the DC value appears asa constant added to the cosine function. For more complex signals, we have to use equation3.5.1. The following example demonstrates the use of this equation.

79

0 0.5 1 1.5 2 2.5 3

x 10−5

0

0.5

1

1.5

2

2.5

3

Time (s)

Vol

tage

(V

)

DC Value = 2 V

Figure 3.14 : A sinusoidal voltage waveform with a DC value of 2 V. The amplitude of thewaveform is 0.5 V.

2 t (s)6 8 104

v = t2

vs (t) = (t - a) 2

Figure 3.15 : A periodic waveform consisting of repeating parabolas

Example 3.5. Find the average value of the waveform shown in Figure 3.15 , which consists ofrepeating parabolic segments of the form,

vs(t) = (t− a)2

where a = 0, 2, 4, · · ·, 2n.

Solution:The period of the waveform is 2 s. To apply equation 3.5.1 to the waveform, consider the firstperiod for 0 ≤ t < 2. The equation that represents the waveform in this time interval is vs(t) = t2.Substituting this function in equation 3.5.1 the DC value can be found as

80

5 V

- 5 V

1 2 3 4 ms

4 V

1 2 3 4 ms

vin(t)

vR(t)

10 k

vin(t)

vR(t)+ -

(a)

(b)

(c)

Figure 3.16 : a)The voltage waveform of the AC source b)Circuit with a AC source andDiode c)Output voltage waveform across Resistor.

VDC =1T

∫ T

0vs(t)dt =

12

∫ 2

0t2dt =

12

t3

3|20 ' 1.33 V

‖ X ‖

3.6 Circuits with Time-Varying Signals and Diodes

When a time-varying voltage source is used in a circuit including a diode we analyze thecircuit for different regions of the voltage source since the state of the diode may changedepending on the polarity and the magnitude of the voltage source. We will follow the sameprocedure introduced in chapter one to determine the state of the diode, which involvesfinding the voltage across the diode assuming it is off.

Example 3.6. The square wave shown in Figure 3.16 (a) is applied to the series connection ofa resistor and a diode. First show that the voltage waveform across the resistor is the waveformgiven in Figure 3.16 (c) and then find the average value of the waveform. Assume the diode has aturn-on voltage of 1.0 V.

Solution:When vin(t) = 5 V the diode is forward biased and current flows in the loop. The voltage across

81

R2

R1

6 V

- 6 V

0.1 0.2 0.3 0.4 ms

vin (t)

(a)vin(t)

1 V

- 2 V0.1 0.2 0.3 0.4 ms

vD (t)

(b) (c)

iD (t)i

1 (t)

i2 (t)

Figure 3.17 : a)Circuit with a AC source, Diode and Two Resistors. b)The voltage waveformof the AC source c)Output voltage waveform across diode.

the resistor can be found using Kirchoff’s voltage law as

vR(t) = vin(t)− Vγ = 5− 1 = 4 V

When vin(t) = −5 V , the diode is reverse biased and there is no current flow in the loop. Hence,the voltage drop across the resistor is zero. Therefore, the resistor voltage cycles between 0 and 4V as shown in Figure 3.16 (c).Since the waveform spends half of its period at 4 V, its average or DC value is 2 V. Alternatively,

VDC =1T

∫ T

0vin(t)dt =

area

T=

4 V × 1 ms

2 ms= 2 V

‖ X ‖

Example 3.7. Consider the circuit shown in Figure 3.17 .a. The square wave applied to thecircuit is given in Figure 3.17 .b. First show that the voltage waveform across the diode looks asshown in figure 3.17 .c and then find its average value. Assume that R1 = 10kΩ, R2 = 5kΩ andvγ = 1V for the diode.

Solution:

82

First we need to determine if the diode is on when vin(t) = 6 V . We begin by assuming that thediode is off and find the voltage across the terminals of the diode, which is the voltage across the5kΩ resistor. When the diode is off, it is equivalent to an open-circuit. Removing the diode fromthe circuit puts the two resistors in series allowing us to use voltage division to find the voltageacross the 5kΩ resistor. Since both resistors are 10 kΩ, they share the source voltage equally.Therefore, the voltage across R2 is 2 V. Since this voltage is greater than the turn-on voltage ofthe diode, we conclude that the diode is on. Therefore, the voltage across the diode is equal to theturn-on voltage of the diode or 1 V.

When the input voltage is - 2 V, the diode is reverse biased and it is off. Replacing the diode byan open-circuit and following the above procedure we find the voltage across R2 as - 2 V. Combiningthe two cases we arrive at the waveform shown in figure 3.17 (c).To find the DC value of the waveform we will follow the simple procedure used in the previous ex-ample. Since each period includes a positive and a negative rectangle, the area under the waveformwithin one period is given by,

Area = 1 V × 0.1 ms + (−2 V )× 0.1 ms = −0.1 V.ms

We can find the DC value by diving this area by the period according to equation 3.5.1:

VDC =−0.1 V.ms

0.2 ms= −0.5 V

‖ X ‖

Example 3.8. Find the DC value of the current flowing through R2 in the previous example.

Solution:Since we know the DC value of the voltage across R2 we do not need to find the current waveformand then find its average value. Instead, we can find the DC value of the current by diving the DCvalue of the voltage by 10kΩ according to Ohm’s law:

I2,DC =−0.5 V

10kΩ= −0.05 mA

The minus sign indicates that half the time the current through R2 is flowing in a directionopposite to that shown on the circuit. ‖ X ‖

3.7 Half Wave Rectifier

One of the most useful applications of diodes is AC/DC conversion. The first stage of anAC/DC converter is a full or half-wave rectifier. In this section, we will discuss the half-waverectifier and leave the operation of the full-wave rectifier to more advanced classes.

83

vin(t) v

out(t)

Figure 3.18 : Half-Wave Rectifier

The circuit is shown in Figure 3.18 . The AC voltage input to the circuit is usuallysupplied by a transformer, which reduces the magnitude of the standard 120 V/60 Hz ACoutlet voltage to a desired, smaller value.

Theoretically, the circuit will work with any AC signal, however, the term half-waverectifier is mostly associated with sinusoidal waveforms.

During the positive cycle of the input sinusoid, a clockwise current is forced around theloop. The diode turns on only if the input voltage is greater than the turn-on voltage of thediode. Since the loop has to satisfy Kirchoff’s Voltage Law, the output voltage across theresistor is equal to the input voltage minus the voltage drop across the diode,

vout(t) = A cos(2πft + θ)− Vγ when vin(t) > Vγ (3.7.1)

where we used the fact that the voltage across the diode is equal to Vγ.When the input voltage is less than the turn-on voltage, the diode is off even if the input

voltage is positive. Under these conditions, the diode acts as an open-circuit and the outputvoltage remains zero.

When the input voltage is negative, the diode is reverse biased. Since the diode will notallow any current flow in the reverse direction, it acts like an open-circuit resulting in 0 Vacross the resistor.

The resulting output signal shown in 3.19 is a half-wave rectified sinusoid with the sameperiodicity as the input signal. The signal is not exactly equivalent to the constant voltagegenerated by a battery, however, since it is always positive it has a finite DC value. In thenext chapter, we will learn that by adding a capacitor a much flatter output signal can beobtained.

3.8 Circuits with AC and DC Voltage Sources

In this section, we combine AC and DC voltage sources in the same circuit. There is nothingreally new that we have to learn to analyze these circuits so we will go ahead and dive intoexamples. We must however remember the fact that an AC voltage source is represented by

84

Voltage

A - Vg

A

Time

Figure 3.19 : Input sinusoid (dashed line) with an amplitude, A and the half-wave rectifiedsinusoid with a peak voltage of A− Vγ

V2

R2

R1 R3v1(t)

i1(t) i2(t)A B

iR1 (t)

iR2 (t)

iR3 (t)

Figure 3.20 : Circuit with a sinusoidal voltage source and a DC Source.

a time dependent equation while we just have a voltage value for a DC source.

Example 3.9. Consider the circuit shown in Figure 3.20 . Assume v1(t) = 3 V cos(2π1000t) andV2 = 2 V . Find the currents i1(t) and i2(t).Assume that all three resistors are 10 kΩ.

Solution:The resistor R1 is parallel with the sinusoidal voltage source and therefore the voltage across R1 isv1(t). The current flowing through R1 can be found using Ohms’ law as

iR1(t) =3 V cos(2π1000t)

10 kΩ= 0.3 cos(2π1000t) mA.

85

Similarly, since R3 is in parallel with the DC Voltage source V2, the voltage across R3 is equal toV2. The current flowing through R3 can be found using Ohms’ law as

iR3(t) =2V

10kΩ= 0.2mA.

Using Kirchoff’s Voltage law in the outermost loop including both sources and the resistor R2,weobtain the voltage across R2 as

vR2(t) = v1(t)− V2 = 3 cos(2π1000t)− 2 V

Using Ohm’s law, the current through R2 can be found as

iR2 = vR2/10kΩ = 0.3 cos(2π1000t)− 0.2 mA.

Applying Kirchoff’s current law to node A and solving for i1(t) we obtain:

i1(t) = −iR2(t)− iR1(t) = −0.6 cos(2π1000t) + 0.2 mA

Now applying Kirchoff’s current law to node B and solving for i2(t) we obtain

i2(t) = iR2(t)− iR3(t) = 0.3 cos(2π1000t)− 0.4 mA

It is interesting to note that while V2 is a DC Voltage source the current flowing through thesource has a sinusoidal component forced by v1(t). Similarly, even though v1(t) is a purely sinusoidalwaveform it has a DC value of 0.2 mA due to the DC voltage source in the circuit. ‖ X ‖

3.9 Oscilloscope

An Oscilloscope is one of the most helpful test instruments used by electrical and computerengineers. The primary function of an oscilloscope is to display time varying signals on acalibrated screen from which the signal parameters can be graphically obtained. Modernfeatures available on digital oscilloscopes such as the one shown in Figure 3.21 can makethe measurements much quicker and easier.

Any time varying voltage waveform can be displayed by an oscilloscope. For instance, youcan connect a microphone and view the speech waveforms as you speak into the microphone.Oscilloscopes are widely used in testing of analog and digital circuits.

Figure 3.22 shows a square wave displayed by a digital oscilloscope made by Agilent.Signal parameters can be measured graphically using the vertical and horizontal grid lines.As an example, let’s measure the peak-to-peak value and period of the square wave shown.This particular display was obtained using a vertical calibration of 2 Volts/division and ahorizontal calibration of 200 µs/division. You can find these numbers displayed usually atthe top of the screen. To measure the peak-to-peak value of the square wave, we countthe number of vertical divisions the square wave spans, which we find as three. Since each

86

Figure 3.21 : A digital oscilloscope made by Textronix.

division corresponds to 2 V , the peak-to-peak value of the square wave has to be 6 V . Tofind the period, we count the number of horizontal divisions, which in this case is equal totwo. Since each horizontal division corresponds to 200 µs, the period is determined as 400µs. We can also use the same display to find the average value of the signal. To do this, weneed to locate the zero volt reference, which is indicated by an arrow connected to a groundsign on the far left side of the screen. Using this reference, we can determine the high andlow values of the square wave as + 5 V and - 1 V respectively.

While the graphical method described above applies to any oscilloscope, digital oscil-loscopes can make these measurements automatically and display the numerical results onthe screen as shown in Figure 3.22 . The buttons under the screen display the duty cycle,frequency and period of the square wave.

87

Figure 3.22 : A square wave with a duty cycle of 70% displayed on an oscilloscope screen.Digital oscilloscopes can measure the parameters of a periodic signal. The buttons underthe screen indicate the duty cycle, frequency and period of the square wave.

88

Problems

P 3.1 Find the amplitude and peak-to-peak-value of the sinusoidal waveforms listedbelow.

(a) v(t) = 10 cos(2π100t) V

(b) v(t) = −10 cos(2π100t) V

(c) v(t) = 10 cos(2π100t− π/4) V

(d) v(t) = −10 cos(2π100t− π/4) V

P 3.2 Find the frequency and period of thewaveforms listed below.

(a) v(t) = 10 cos(2π100t) V

(b) v(t) = 10 sin(2π100t) V

(c) v(t) = 10 cos(4π100t) V

(d) v(t) = 10 cos(100t) V

P 3.3 Find the DC value of the signals listedbelow. Hint: Use trigonometric identi-ties to expand the waveforms where nec-essary.

(a) v(t) = 10 cos(2π100t) V

(b) v(t) = 20 + 10 cos(2π100t) V

(c) v(t) = −20 + 10 cos(2π100t) V

(d) v(t) = −20− 10 cos(2π100t) V

P 3.4 Find the phase angle of the sinusoidalwaveforms listed below.

(a) v(t) = 10 cos(2π100t) V

(b) v(t) = 10 cos(2π100t + π/4) V

(c) v(t) = −10 cos(2π100t + π/4) V

(d) v(t) = −10 cos(2π100t− π/4) V

P 3.5 Construct a time-domain equation de-scribing the waveform shown in FigureP3.5.

t (ms)105 15 20

T = 20 ms10

5

-5

-10

V

Figure P3.5

P 3.6 Suppose the sinusoid shown in FigureP3.5 is applied to a 9.1 kΩ resistor. Con-struct a time-domain equation for thecurrent flowing through the resistor.

P 3.7 Construct a time-domain equation de-scribing the waveform shown in FigureP3.7.

P 3.8 Suppose the sinusoid shown in FigureP3.5 is applied to a 9.1 kΩ resistor. Con-struct a time-domain equation for thecurrent flowing through the resistor.

P 3.9 Find the frequency and the DC valueof the signal,v(t) = 10 cos2(2π100t) V .Hint: Use a trigonometric identity to ex-pand the waveform.

P 3.10 Find the phase angle of the following si-nusoidal waveforms. Hint: Convert thewaveforms to cosine functions.

(a) v(t) = 10 sin(2π100t) V

(b) v(t) = 10 sin(2π100t + π/4) V

89

t (ms)105 15 20

10

5

-5

-10

V

15

T = 20 ms

Figure P3.7

(c) v(t) = 10 sin2(2π100t) V

P 3.11 Given the square waveform shown in theoscilloscope display of Figure P3.11, finda) peak-to-peak value, b) DC value c)frequency and d) duty cycle of the wave-form.

Vertical = 2 V/division

Horizontal = 100 s/division

Figure P3.11

P 3.12 Given the square waveform shown in theoscilloscope display of Figure P3.12, finda) peak-to-peak value, b) DC value c)frequency and d) duty cycle of the wave-form.



Figure P3.12

P 3.13 Given the triangular waveform shown inthe oscilloscope display of Figure P3.13,find a) peak-to-peak value, b) DC valueand c) frequency of the waveform.



Figure P3.13

P 3.14 Given the sawtooth waveform shown inthe oscilloscope display of Figure P3.14,find a) peak-to-peak value, b) DC valueand c) frequency of the waveform.

P 3.15 The triangular waveform shown in Fig-ure P3.13 is applied to a 1 kΩ resis-tor. Find the average value of the cur-rent flowing through the resistor.

P 3.16 The sawtooth waveform shown in FigureP3.14 is applied to two resistors (10 kΩ

90

Vertical = 0.1 V/division

Horizontal = 1 ns/division

Figure P3.14

and 1 kΩ) connected in series. Find thepeak-to-peak value and DC value of thesawtooth waveform on the 1 kΩ resistor.

P 3.17 A 2 V battery is connected in series withthe function generator supplying the rect-angular waveform shown in Figure P3.11.What is the DC value and the peak-to-peak value of the resulting signal?

P 3.18 The AC signal in Figure P4.10 is a 100Hz sinusoidal waveform with a DC valueof zero and an amplitude of 2 V. Findthe currents i1, i2, i3 and the voltageacross the 3.9 kΩ resistor.

vs(t)

5 V

100

3.9 k

i1(t) i2(t)

i3(t)

Figure P3.18

P 3.19 Repeat P4.10for the circuit shown inFigure P4.2.

vs(t) 5 V

100 47

3.9 k

i1(t) i2(t)

i3(t)

Figure P3.19

P 3.20 The AC signal in Figure P3.20 is a 60Hz sinusoidal waveform with a DC valueof zero and an amplitude of 10 V. Thediode is a standard silicon diode (e.g.1N4001) with a turn-on voltage of 0.7V. Determine the peak-to-peak to valueof the output signal.

vs(t)vR(t)

Figure P3.20

P 3.21 Determine what fraction of the periodthe diode remains on in P3.20. Hint:To keep the diode ’on’, the input voltagehas to be at least equal to the turn-onvoltage.

91

P 3.22 Determine the DC value of the voltageacross the resistor in P3.20. Assumethe resistor is 1 kΩ and find the DCvalue of the current flowing through theresistor.

P 3.23 Determine the DC value of the voltageacross the resistor in P3.20 if the si-nusoidal input source is replaced by thesquare wave shown in Figure P3.11.

P 3.24 Assume the sinusoidal input signal inthe half-wave rectifier of P3.20 is re-placed by a triangular waveform with apeak-to-peak value of 12 V and zero DCoffset. Find the DC value of the voltageapplied to the resistor.

92

Chapter 4

Electric Power

You must be already familiar with the daily use of the term, electric power and its unit Watt,W. We know that a 100 W light bulb is much brighter than a 40 W bulb. A typical, goodquality home system may produce a music power 100 W of music power. The transmissionpower of a radio station determines how far the electromagnetic waves transmitted by thestation’s antenna will travel. A typical FM station produces several kilowatts of outputpower. These are just a few examples from many practical uses of the quantity, electricpower.

The chapter begins with the physical origins of electric power and consider power dissi-pation with DC and AC voltage sources. These concepts are then applied to simple circuitswith resistive elements. The chapter concludes by an overview of power engineering conceptsrelevant to the standard household voltage.

4.1 Physics of Electric Power

By definition, power is the work done per unit time, therefore, it is a measure of how fastthe work is done. A powerful machine is the one that can produce the most work in theshortest possible time. Let’s consider the following example from classical mechanics. Whenyou climb the stairs, the energy stored in your body is used to move your legs, which carryyour weight up the stairs by doing work against the gravitational force. By doing so, youburn calories and this body energy is converted to potential energy. The mechanical poweryou dissipate is determined by how fast you climb the stairs.

In theory, electric power is no different than the mechanical power but the work is per-formed by negatively charged electrons moving in an electric field. As such, the forcesinvolved in the process are electric forces and the power is determined by how fast theelectrons move, the intensity of the electric field and the distance they travel.

In Chapter I, we have learned that work, W is done in moving a charged particle frompoint a to point b in an electric field. The work done per unit charge (i.e. 1 Coulomb) is the

93

94

potential difference or voltage, Vab, which can be expressed as

Vab =Wab

Q(4.1.1)

Therefore, if 10 joules of work is to be done to carry a charge of 5 Coulombs from a tob, then, the voltage Vab is equal to 2 volts. The following example applies this concept toelectrons moving in an electric field.

Example 4.1. A battery applies 10 V to an ordinary resistor. Calculate the work done in moving1000 electrons between the two end of the resistor.

Solution:The potential difference between the two ends of the resistor is 10 V. This difference establishesan electric field in the resistor and the work associated with moving the electrons is done by thisfield. Every electron carries a net charge of 1.6× 10−19 C. To find the work done in moving 1000electrons, we can use equation 4.1.1.

Wab = QVab

= 10 V × 1000× 1.6× 10−19 C

= 1.6× 10−15 Joules

‖ X ‖To find the electric power associated with moving charges, we need to know how fast thecharges move or the electric current, I in amperes. Recall from Chapter I that the electriccurrent, I is expressed as

I =Q

∆t(4.1.2)

where Q is the net charge passing through a surface in a conductor in time, ∆t. The productof the equations, 4.1.1 and 4.1.2 yields the work done per unit time or the electric power, P.

P =W

Q× Q

∆t=

W

∆t

Therefore, electric power is expressed as

P = V I (4.1.3)

which is the main subject of this chapter. In the remaining sections, we will apply equation4.1.3 to practical circuits with DC and AC voltage sources. We will also look at essentials

95

R V

I

Figure 4.1 : A simple circuit with a DC voltage source and a resistor.

of power distribution as two subjects that are important to electrical engineers specializingin power systems or power electronics.

4.2 DC Voltage Sources and Resistive Loads

Let’s now consider the simple circuit of Figure 4.1 consisting of a DC voltage source anda resistor. The voltage, V produced by the voltage source is applied to the resistor anda current, I, flows in the loop. The electric power dissipated on the resistor is given byequation 4.1.3. According to this equation, if the voltage applied to the resistor is 1 V andthe current flowing in the loop is 1 A, the power dissipated on the resistor is 1 W.

Using Ohm’s law, we can rewrite equation 4.1.3 in two different forms. SubstitutingV = IR in 4.1.3 for the voltage, we obtain,

P = V I = (IR)I = I2R (4.2.1)

Alternatively, if we substitute I = V/R for the current, we obtain,

P = V I = VV

R=

V 2

R(4.2.2)

All three forms of the electric power equation are commonly used.

In the circuit of Figure 4.1 , as a result of the current flowing in the loop, the resistorheats up and begins to radiate heat. In this process, electrical energy supplied by the voltagesource is first used to move the electrons, converted to heat as the electrons travel throughthe sea of atoms and electrons and finally lost to ambient via radiation. It may help to thinkthat heat is produced as a results of electrons’ huffing and puffing as they travel throughthe resistor, which resists electrons motion. Therefore, we can view this simple circuit asan energy converter since the circuit receives electrical energy from the voltage source and

96

Figure 4.2 : Resistors with different power ratings.

produces heat in return. In practice, all electric heaters used in ovens, hair dryers etc. areresistors operating on this principle.

The resistors used in electric circuits also come with their power ratings. Figure 4.2shows resistors with different power ratings. As the power rating increases, the resistor

becomes larger. This allows heat to radiate from a larger surface area to achieve moreeffective cooling.

Example 4.2. A 12 V battery is connected to two resistors in series. Assume the resistances areR1 = 100 Ω and R2 = 500 Ω. Find the power dissipated on each resistor.

Solution:Applying the voltage division rule, the voltage across each resistor can be found as

V1 =100 Ω

100 Ω + 500 Ω12 V = 2 V

V2 =500 Ω

100 Ω + 500 Ω12 V = 10 V

We can now apply equation 4.2.2 to find the power dissipated on each resistor.

P1 =V 2

1

R1=

(2 V )2

100 Ω= 0.04 W = 40 mW

P2 =V 2

2

R2=

(10 V )2

500 Ω= 0.2 W = 200 mW

97

I

VD

VR

Vs

Figure 4.3 : A simple circuit with a resistor and a diode.

‖ X ‖

Let’s apply our new knowledge to a slightly more complex circuit shown in Figure 4.3 .In this circuit, the voltage produced by the DC voltage source is shared between a resistorand a diode. Our goal is to calculate the power dissipated on both elements. We will assumethat the voltage source is able to provide the voltage needed to turn on the diode. Otherwise,current can not flow in the loop and the power dissipation on both elements is zero accordingto equation 4.1.3. When the diode is on, the voltage across the diode is equal to the turn-onvoltage, Vγ, which leaves VR = Vs − Vγ on the resistor according to Kirchoff’s voltage law.The power dissipated across the resistor can then be found using equation 4.2.2 as

PR =V 2

R

R=

(Vs − Vγ)2

R

To find the power dissipated on the diode, we can use neither 4.2.2 nor 4.2.1 because theseequations are only valid for linear resistors. To use them, we need a resistance, R, indepen-dent of the voltage across the element and we can not define such a resistance for the diode.On the other hand, equation 4.1.3 is valid for any element, linear or non-linear. To use thisequation, we must know both the diode voltage and current. Since we know the voltageacross the resistor, the loop current can be found as

I =VR

R=

(Vs − Vγ)

R

The power dissipated on the diode is then given by

PD = VDID = Vγ(Vs − Vγ)

R

Similar to resistors, diodes have different power ratings as well. Figure 4.4 shows different

98

Figure 4.4 : Power diodes can be mounted on heat sinks for efficient cooling.

silicon power diodes with different power ratings. These diodes are designed so they can bemounted on heat sinks for efficient cooling via radiation.

Example 4.3. Suppose Vs = 9 V , Vγ = 0.7 V and R = 1 kΩ in the circuit of Figure 4.3 . Findthe power dissipated on the resistor and the diode.

Solution:The voltage across the resistor is VR = 9 V − 0.7 V = 8.3 V . The power dissipated on theresistor is then equal to PR = (8.3 V )2/1 kΩ ≈ 68.9 mW . The current flowing in the loopis given by I = 8.3 V/1 kΩ = 8.3 mA. The power dissipated on the diode is then equal toPD = 8.3 mA× 0.7 V = 5.81 mW‖ X ‖

4.3 AC Voltage Sources and Power

How can we calculate the power dissipation on a resistive element when the voltage appliedto the circuit is time varying? Will the power dissipation also change with time? If so, canwe calculate an average power? These are the questions we shall try to answer in this section.We will introduce two new terms, instantaneous power and average (or real) power, which

99

R

i(t)

v(t)

Figure 4.5 : A simple circuit with an AC voltage source and a resistor.

will lead us to the root-mean-square (or RMS) voltage, which is commonly used quantity inreferring to the strength of AC signals.

4.3.1 Instantaneous Power

Consider the simple circuit shown in Figure 4.5 , which consists of a time varying voltagesource, v(t) connected to an ordinary resistor. Since both the resistor voltage and currentare time-varying signals, their product, i.e. the power dissipated on the resistor must alsobe a time varying waveform. We define the instantaneous power, p(to) as the product of thevoltage v(t) and current, i(t) at the instant, t = to. Therefore, the instantaneous power, p(t)is given by,

p(t) = v(t)i(t) (4.3.1)

Since the load in Figure 4.5 is a simple linear resistor obeying Ohm’s law, we can write theequivalents of the equations 4.2.1 and 4.2.2 for time-varying signals given as

p(t) =v2(t)

R(4.3.2)

and

p(t) = i2(t)R (4.3.3)

We will now apply these concepts to a few standard signals.

Square Wave and Instantaneous Power

Suppose the square wave shown in Figure 4.6 .a is applied to a 100 Ω resistor. The signal hastwo discrete voltage levels, VH = 3 V and VL = −2 V . We can use equation 4.3.2 to calculatethe instantaneous power dissipated on the resistor at each voltage level, which yields the p(t)

100

vR(t)

t

VH = 3 V

p(t)

t

90 mW

VL = - 2 V

40 mW

(a)

(b)

Figure 4.6 : a) A square wave applied to a 100 Ω resistor, b) Resulting instantaneous powerwaveform.

101

0 0.005 0.01 0.015 0.024

2

0

2

4

time (s)

Voltage (V)

0 0.005 0.01 0.015 0.020

0.01

0.02

0.03

0.04

0.05

time (s)

Power (W)

(a)

(b)

Figure 4.7 : a) A sinusoidal wave applied to a 100 Ω resistor, b) Resulting instantaneouspower waveform.

waveform shown in Figure 4.6 .b. Note that instead of using equation 4.3.2, we could havedivided the voltage waveform, v(t) by 100 Ω to find the current waveform, i(t) and then useequation 4.3.1 to find p(t).

It is important to understand that the dissipated power is positive even when the inputvoltage is negative. If this result hits you with surprise, you must realize that when the inputvoltage switches from 3 V to −2 V , all that is happening is that the direction of the loopcurrent is changing from clockwise to counter-clockwise. If you are comfortable with thisconcept, we can raise the following critical question: Since power dissipation is the energylost per unit time, can the current direction have anything to do with power? The answer is’no’ because the electrons will be huffing and puffing no matter how they flow. As a resultof the current flow, the resistor will warm up and radiate heat to the ambient. This willhappen regardless of the current direction or the polarity of the input voltage.

102

Sinusoidal Wave and Instantaneous Power

Now let’s consider the case of a pure sinusoid, v(t) = Vp cos(2πft) applied to a resistor.Since the load is a resistor obeying Ohm’s law, we shall again use equation 4.3.2 to find theinstantaneous power. This yields,

p(t) =v2(t)

R=

V 2p cos2(2πft)

R(4.3.4)

which can be further simplified using the trigonometric identity,

cos2 θ =cos 2θ + 1

2(4.3.5)

Substituting the above identity in the p(t) equation we obtain,

p(t) =V 2

p

2R[1 + cos(2π2ft)] (4.3.6)

Figure 4.7 shows an example voltage waveform, v(t) = 2 cos(2π100t) applied to a 100Ωresistor and the resulting instantaneous power waveform, p(t). First, we note that p(t) isalways positive as it should be with a resistive load. The frequency of the p(t) waveform istwice the frequency of the applied voltage waveform. The waveform is raised to a DC off-setsuch that at the minimum point of the sinusoidal waveform the instantaneous power is zero.Even though the current direction is constantly switching between clockwise and counterclockwise, the power dissipation is always positive, i.e., regardless of the current direction,electrical energy is always converted to heat.

Example 4.4. The voltage waveform measured across a 1 µF capacitor is v(t) = 10 cos(2π100t) V .Find the instantaneous power dissipated on the capacitor.

Solution:Since the capacitor is not a resistive element, we can not use equations 4.3.2 or 4.3.3, however,equation 4.3.1 is valid for any load resistive or not. To use this equation, we must first find thecapacitor current. Using the I-V characteristic of a capacitor, we can find the capacitor current as

ic(t) = Cdvc(t)

dt

= 1 µF × d

dt[10 cos(2π100t) V ]

= −0.002π sin(2π100t) A

103

We can now find the instantaneous power using equation 4.3.1.

p(t) = v(t)i(t)= 10 cos(2π100t) V ×−0.002π sin(2π100t) A

= −0.02π cos(2π100t) sin(2π100t)

Using the trigonometric identity, sin 2θ = 2 sin θ cos θ, the above equation can be written as

p(t) = −0.01π sin(2π200t)

To simplify the result even further, the minus sign can be absorbed in the sinusoid by inserting aphase angle of ±π in its argument.

p(t) = 0.01π sin(2π200t± π)

Therefore, the instantaneous power is a sinusoid with a frequency of 200 Hz, i.e. twice the frequencyof the original voltage waveform. Therefore, unlike the resistive loads we have considered before,the power dissipated on the capacitor is not always positive. Is this an anomaly? Not really! Ifpositive power is indicative of energy lost, negative power is indicative of energy gained. As such,a capacitor does not really dissipate power like a resistor. Instead, it stores and releases energyfollowing the input waveform. ‖ X ‖

4.3.2 Average Power

In the previous section, we have seen that the instantaneous power, p(t), is a time varyingsignal. Just like finding the average value of a voltage waveform, we can find the averagevalue of the instantaneous power. Assuming that p(t) is a periodic waveform, the averagepower can be determined from

P =1

T

∫ T

0

p(t)dt (4.3.7)

The above equation first finds the area under p(t) within a period, T, and then, dividesthe area by the period to find the average. Because the waveform is periodic, it sufficesto evaluate the average for a single period of the waveform. If p(t) is a simple waveform(e.g. a square wave), we may be able to find the area under the signal without integration.Therefore, it is convenient to express equation 4.3.7 as

P =Area under p(t) within T

T(4.3.8)

Example 4.5. Find the average power dissipated on the resistor in Figure 4.6 .

104

Vdc v(t)

(a) (b)

Figure 4.8 : The two circuits will dissipate the same average power if the RMS value of theAC voltage source is equal to Vdc.

Solution:Since the instantaneous power has already been found, all we have to do is find its average valueusing equation 4.3.8

P =0.09T

2 + 0.04T2

T= 0.065 W

‖ X ‖

Example 4.6. Find the average power dissipated on the resistor in Example 4.7 .

Solution:The instantaneous power dissipated on the resistor was found as

p(t) = 20 + 20 cos(2π200t) mW

To find the average value of this equation, we can use equation 4.3.7, however, there is a mucheasier solution than integrating the cosine function. Let’s consider each term separately becausethe average value of the whole function is equal to the sum of the average values of its components.The first term is a constant and the average value of the constant is the constant itself. The secondterm is a pure cosine function and the area under a pure sinusoid is zero. Therefore, the averagepower is equal to 20 mW. ‖ X ‖

In power engineering, the average power given by equation 4.3.7 is commonly referredto as real power. In this book, we will stick to the term, average power except when wecalculate the power in problems related to the standard household voltage, power generationand distribution, i.e. concepts relevant to power engineering.

105

4.4 Root-Mean-Square Voltage and Current

Consider the two circuits shown in Figure 4.8 . They are very similar other than theirvoltage sources. We note that we impose no restrictions on the AC voltage source. It canbe generating any arbitrary waveform.

With the DC source, the power dissipated on the resistor is given by

P =V 2

dc

R

With the AC voltage source, we can calculate an average power dissipation if we know whatthe waveform looks like. To do this, we first find the instantaneous power, p(t) using equation?? and then compute its average value using equation 4.3.7. This yields,

P =1

T

∫ T

0

v2(t)

Rdt

Suppose now that the average power found above is identical to V 2dc/R. Equating the two

power equations and solving for the voltage, Vdc, we obtain,

VRMS =

√1

T

∫ T

0

v2(t)dt (4.4.1)

where we have switched the voltage Vdc with VRMS, the root-mean-square voltage becausethis is the name given to this special voltage. Therefore, if Vdc = VRMS, the two circuitsare equivalent in terms of the average power dissipated on their resistors. This equivalencyprovides us a new tool, a convenient standard in quantifying the strength of AC signals. Weneed such a tool because not all AC signals can be fit into a few categories.

We can also refer to the RMS value of an AC current. For instance, in Figure 4.8 .b,there will be a current, i(t) flowing in the loop due to the AC voltage source. The RMSvalue of the current can be found from

IRMS =

√1

T

∫ T

0

i2(t)dt (4.4.2)

We note that equations 4.4.1 and 4.4.2 have exactly the same form. Indeed, this isa mathematical form that applies to any function. To find the RMS value of a periodicfunction, f(·), we execute the following procedure:

1. Find the square of the function, f(·) −→ f 2(·)

2. Find the average value of f 2(·) −→ 1T

∫ T

0f 2(·)dt

106

3. Find the square root of the above average −→√

1T

∫ T

0f 2(·)dt

4.4.1 RMS value of a ”Pure” Sinusoid

We will now apply equation 4.4.1 to the pure sinusoidal voltage waveform, v(t) = Vp cos(2πft+θ). Substituting v(t) in equation 4.4.1 we obtain,

VRMS =

√V 2

p

T

∫ T

0

cos2(2πft + θ)dt

Application of the trigonometric identity 4.3.5 to the cosine squared term yields,

VRMS =

√V 2

p

T

∫ T

0

1 + cos[2(2πft + θ)]

2dt

Rearranging the terms,

VRMS =

√V 2

p

2T

[∫ T

0

dt +

∫ T

0

cos[2(2πft + θ)] dt

]

In the above equation, the second integration computes the area under the sinusoid, cos[2(πft+θ)], which has to be equal to zero. It is important to note that the area under a sinusoidis always zero unless the integration includes a partial cycle of the waveform. In our case,the frequency of the sinusoid is 2f = 2/T and the limits of the integration are from 0 toT , hence, the integration computes the area under two full cycles of the sinusoid. Afterevaluating the first integral and substituting zero for the second one we find the RMS valueof a pure sinusoid as

VRMS =Vp√

2(4.4.3)

We must remember the above equation because the sinusoid is such a basic waveform that itsRMS value will be frequently needed throughout the rest of this book. Before we concludethis section, we need to emphasize again the fact that equation 4.4.3 is valid only for puresinusoids. This means that if the sinusoid has a DC offset, its RMS Value will be entirelydifferent. We will discuss this special case in the next section.

4.4.2 RMS value of a Sinusoid with a DC Value

Now we consider the sinusoid, v(t) = Vdc + Vp cos(2πft) shown in Figure 4.9 . Our goal isto find the RMS value of this waveform or underline the fact that we can not use equation4.4.3 when we do not have a pure sinusoid. Even the presence of a DC value will change the

107

v(t)

t

Vdc

Vp

Figure 4.9 : A sinusoidal voltage with a DC off-set.

calculation substantially. At this point, we have no choice but to employ equation 4.4.1, inthe next chapter we will learn to look at this problem from a different angle.

Substituting the waveform in equation 4.4.1, we obtain

VRMS =

√1

T

∫ T

0

[Vdc + Vp cos(2πft)]2 dt

Expanding the argument of the integral yields,

VRMS =

√V 2

dc

T

∫ T

0

dt +2VdcVp

T

∫ T

0

cos(2πft) dt +V 2

p

T

∫ T

0

cos2(2πft) dt

The above equation is not as complex as it may seem at a first glance. The first integrationyields just the constant, V 2

dc. The second integral finds the area under the sinusoid, cos(2πft),hence, it has to be equal to zero. The argument of the second integral can be expandedusing the trigonometric identity 4.3.5 as we have done in the previous section. After thesesimplifications we obtain the expression,

VRMS =

√V 2

dc +V 2

p

2T

∫ T

0

[1 + cos(4πft)] dt

Following the procedure used in the previous section, we can show that

V 2p

2T

∫ T

0

[1 + cos(4πft)] dt =V 2

p

2

where, we have again used the fact that the area under a sinusoidal waveform is zero.Therefore, the RMS value of the sinusoid is given by

108

VRMS =

√V 2

dc +V 2

p

2(4.4.4)

4.5 Power Engineering

This entire section is dedicated to a specific sinusoidal waveform. As you may have alreadyguessed, we are referring to the standard household voltage available from the AC outlets inour homes. The waveform is a pure 60 Hz sinusoid with an RMS value of 120 V. Since thewaveform does not have a DC value, its amplitude can be found using equation 4.4.3 as

Vp = 120√

2 ≈ 170 V

Therefore, we can express the waveform as

v(t) = 170 cos(2π60t + θ) (4.5.1)

where θ is an arbitrary phase angle. In practice, the waveform is slightly distorted and itsRMS value may differ slightly from 120 V. That is why, the appliances are usually designedto operate in a voltage range of 110 - 125 V.

In different parts of the world, different standards are used. For instance, in Europe, 220V/50 Hz is used in many countries as the standard household voltage.

4.5.1 Power Outlets in USA

The standard power outlet used in USA is shown in Figure 4.10 . As shown, it has threeterminals referred to as Line, Neutral, and Ground. The AC voltage is supplied through theLine and Neutral terminals. The Ground terminal provides a return path for the current ifthe insulation between the electrical wires and the equipment breaks down.

Example 4.7. The standard 120V/60Hz AC voltage from a power outlet is applied to a 100 Ωresistor. Find the amplitude of the current flowing through the resistor.

Solution:The RMS value of the current flowing through the resistor can be found as,

IRMS =VRMS

R=

120 V

100 Ω= 1.2 A

109

Figure 4.10 : Standard power outlet used in USA.

Figure 4.11 : Sources of electricity in the U.S. in 2005.

Since the current is purely sinusoidal, its amplitude can be found from its RMS value using 4.4.3.

Ip = IRMS

√2 = 1.2

√2 =

‖ X ‖

4.5.2 Power Generation and Distribution

Let’s now consider how the standard AC voltage is generated and distributed to our homesand factories.

Power plants are huge engineering marvels that convert one form of energy to electricalenergy. They became feasible after realizing that AC power could be transported great

110

distances at very low cost with the help of transformers used at different points on the lineto raise and lower the voltage levels according to the needs of the distribution system.

The source of energy for the first power plants was either coal or flowing water. Lately,nuclear power plants have gained considerable popularity despite the fact that their long-term safety is still controversial in nature. Interest in other sources of energy including solar,thermoelectric and wind is also growing. Figure 4.11 shows different sources of electricityin the U.S. in 20051.

Figure 4.12 illustrates the main components of a power generation/distribution system.The voltage generated by the power plant is first raised to a very high voltage within 69- 700 kV by step-up transformers. The voltage level is determined by factors such as theamount of power to be delivered and the length of the transmission line. This voltage travelsfor miles and miles to distribution centers in residential areas, where step-down transformersreduce it down to smaller yet still fairly high distribution voltages. In US, typical levels usedat this stage are 8.66, 15, 25, 34.5 kV determined by the power demands of an individualneighborhood and how the electricity will be distributed. This voltage is carried to the polesnear the houses, where it is further reduced down to 120 V by step-down transformers onthe poles before it is delivered to individual houses.

4.5.3 Your Electricity Bill

Your local electric power company charges you for the total energy consumed in your homeor business over a period of time, to, typically a month. A modern electric meter such as theone shown in Figure 4.13 measures the instantaneous voltage and current first to determinethe instantaneous power consumption, p(t), and then integrates it real-time to find the totalenergy consumption during the entire billing period . Mathematically speaking, the metercomputes the area under p(t) over a period, to:

E =

∫ to

0

p(t)dt

The above equation gives the energy consumption in Joules or watt-seconds. It is muchmore common to measure the energy consumption in kilowatt-hours or kWh. Realizing that1 kW = 1000 W and we have 3600 seconds in an hour, the above equation can be written as

E =1

3.6× 106

∫ to

0

p(t)dt kWh (4.5.2)

which yields the consumption in kWh and that is the number we have on our electricity bill.

1Reproduced from Wikimedia Commons, based on Net Generation by Energy Source by Type of Producer,(c. 2006), Washington: U.S. Dept. of Energy, Energy Information Administration

111

69 – 700 kV

Power Plant

step-up

transformer

step-down

transformer

8.66 – 34.5 kV

120 V

Figure 4.12 : Main components of a power system.

112

Figure 4.13 : A residential power meter.

Example 4.8. Calculate the monthly cost of using a 2000 W heater for 1.5 hours everyday. As-sume the power company charges 10 cents per kWh.

Solution:First we need to realize that the power rating of any electrical load such as the heater in thisexample represents the average (i.e. real) power dissipation on the load. Since we already knowthe average value of p(t), we can multiply this average power by the duration of use to find theenergy consumption. This yields,

E = 2000 W × 1.5 h = 3 kWh

Therefore, the total energy consumption in 30 days is 90 kWh. The cost of using the heater canthen be calculated as

Cost = 10cents

kWh× 90 kWh = $9.00

‖ X ‖

4.5.4 Real Power with a Non-Resistive Load

When a sinusoidal voltage, v(t) = Vp cos(2πft) is applied to an ordinary resistor, the currentflowing through the resistor can be determined using Ohm’s law. The resulting currentwaveform is another sinusoid, i(t) = Ip cos(2πft) where Ip = Vp/R. Since both waveforms

113

v(t)

t

i(t)

t

to

Figure 4.14 : If the load is not purely resistive, there exists a time delay between the voltageand current waveforms.

have the same frequency and phase, they line up perfectly without a time delay betweenthem.

The above argument can not be true if the load is not a pure resistor. In such a case, eitherthe current or the voltage waveform may fall behind, resulting in a phase difference betweenthe two waveforms as shown in Figure 4.14 . This turns out to be an important subjectin power engineering because it creates a challenge for the power company in measuringthe energy consumption of certain businesses that rely on heavy machinery equipped withelectric motors and similar non-resistive loads. It is important to note that both waveformsin Figure 4.14 have the same period, therefore, the time delay between the two correspondsto a phase angle of

θ =

(toT

)2π

radians.A good example for a non-resistive load is the capacitor. In example 4.4, we have shown

that when a sinusoidal voltage is applied to a capacitor, the instantaneous power dissipatedon the capacitor is also a sinusoid with zero DC offset. Since the standard power metermeasures the area under p(t) and since the area under a pure sinusoid is zero, the electriccompany can not charge you anything for the energy you provide to a purely capacitive load.Electric motors behave in a manner very similar to capacitors. In the case of a capacitor,the phase difference between the voltage and current waveforms is always π/2 radians. Morecomplex devices may combine elements of both resistive and capacitive loads. Considerfor example the series combination of a capacitor and a resistor as shown in Figure 4.15 .This load is neither resistive nor capacitive, therefore, we should expect some arbitrary timedelay between the two waveforms but it will never be equal to half the period as it is for apure capacitor. Suppose this arbitrary time delay corresponds to a phase angle of θ radians

114

v(t)

Load

Figure 4.15 : The equivalent circuit of a non-resistive load might be the series combinationof a capacitor and a resistor.

between the two waveforms. Then, when the voltage, v(t) = Vp cos(2πft) is applied to sucha non-resistive load, the resulting current will be of the form,

i(t) = Ip cos(2πft + θ)

We note that the current waveform is again a perfect sinusoid with the same frequency asthe voltage waveform. The instantaneous power dissipated on the load can now be writtenas

p(t) = v(t)i(t) = VpIp cos(2πft) cos(2πft + θ)

where θ is the phase angle corresponding to the time delay, to between the two waveformsillustrated in Figure 4.14 . Using the trigonometric identity

cos A cos B =1

2[cos(A + B) + cos(A−B)]

we can simplify the instantaneous power expression as

p(t) =1

2VpIp cos(4πft + θ) +

1

2VpIp cos(−θ) (4.5.3)

To find the average i.e. the real power, we need to evaluate the average value of the expressionabove. We note that the first term is a pure sinusoid, hence, its average value has to be zero.The second term looks like a sinusoid, but it actually is not because the argument of thecosine function, the phase angle, θ is a constant. Therefore, the entire second term is also aconstant. Since the average value of a constant is the constant itself, the real power is given

115

by

P =1

T

∫ T

0

p(t)dt =1

2VpIp cos(−θ)

We can further simplify the above equation by using the RMS values of the voltage andthe current waveforms instead of using their amplitudes. Since both voltage and currentwaveforms are sinusoidal, the RMS values of the waveforms are given by

VRMS =Vp√

2

IRMS =Ip√2

as derived in section 4.4.1. Using these relationships, we can simplify the real power foundabove as:

P = VRMSIRMS cos(−θ) (4.5.4)

which indicates that the real power has two components, the so-called apparent power givenby

S = VRMSIRMS (4.5.5)

and the power factor, cos(θ). The unit of apparent power is volt-ampere or simply VAinstead of watts. The apparent power, S is a useful quantity because it determines the peakpower required, which determines the capacity required by energy companies.

Example 4.9. Calculate how much real and apparent power an electric fan will consume whenit is supplied from an ideal AC source of 120 V. Assume that the electric motor that drives the fanwill draw a sinusoidal current with an amplitude of Ip = 5.0A and there exists a phase angle of−32 between the voltage and current waveforms.

Solution:To calculate the power consumed by the fan, we first need to determine the current it draws fromthe AC source. The peak value of the load current is 5 A. The RMS value of the current can becalculated as:

IRMS =Ip√2

=5√2≈ 3.53 A

The apparent power can be calculated as

S = VRMSIRMS = 120× 3.53 = 423 V A

The power factor is given by

116

cos(−θ) = cos(32o)

The real power can then be found as

P = 423 V A× cos(32) = 359 W

‖ X ‖

117

R1

R2

5 V

Figure P4.2

Problems

P 4.1 A 9 V battery is connected to two re-sistors in series. Suppose the resistancevalues are 1 Ω and 100 Ω. Find thepower dissipated on each resistor.

P 4.2 In the circuit shown in Figure P4.2, theresistor, R1 is a non-linear resistor withan I-V characteristic of I = I0(V/V0−1)3

with I0 = 100 mA and V0 = 200 mV .The resistor, R2 is an ordinary, linearresistor with a resistance of 1.5 kΩ. a)Find the voltage drop across each resis-tor; b) Find the power dissipated acrosseach resistor.

P 4.3 The electric power dissipated by an in-candescent light bulb is 100 W when itis powered by the standard, 120 V / 60Hz AC voltage obtained from a wall out-let. a) Calculate the internal resistanceof the light bulb; b) Calculate the powerdissipated on the light bulb when theRMS value of the applied voltage is re-duced to 90 V.

P 4.4 A 50 Ω resistor is connected in serieswith a 100 W light bulb. The series com-bination is then connected to a standard



Figure P4.6

AC wall outlet. Find the power dissi-pated on the light bulb and the seriesresistor.

P 4.5 Find the DC and RMS values of the fol-lowing signals.

(a) v(t) = (10V ) cos(2000t)

(b) v(t) = −(10V ) cos(2000t)

(c) v(t) = (10V ) cos(106t)

(d) v(t) = (10V ) cos(106t + π/3)

(e) v(t) = (2V ) + (10V ) cos(106t)

(f) v(t) = 2V

P 4.6 The square wave shown in Figure P4.6is applied to a 100 Ω resistor. a) Findthe DC value of the voltage applied tothe resistor; b) Find the RMS value ofthe voltage applied to the resistor; c)Find the average (real) power dissipatedon the resistor.

P 4.7 Repeat the previous problem for the sig-nal shown in Figure P4.7.

P 4.8 A periodic signal, v(t) has been createdby adding two sinusoidal waveforms, v1(t)and v2(t) at different frequencies such

118

Vertical = 0.1 V/division

Horizontal = 1 ns/division

Figure P4.7

that v(t) = v1(t) + v2(t). Show that theRMS value of the resulting voltage wave-form is given by Vrms =

√V 2

rms1 + V 2rms2.

P 4.9 A sinusoidal voltage waveform with anamplitude of 2 V and a DC value of 1 Vis applied to a 100 Ω resistor. a) Findthe RMS value of the voltage applied tothe resistor; b) Find the instantaneouspower dissipated on the resistor; c) Findthe average power dissipated on the re-sistor.

P 4.10 The AC voltage source in the circuit shownin Figure P4.10 generates a sinusoidalwaveform with an RMS value of 2 V. a)Find the RMS value of the current ix(t);b) Find the average power dissipated onthe 10 kΩ resistor.

P 4.11 A sinusoidal voltage obtained from a trans-former is half-wave rectified by a diodewith a negligibly small turn-on voltage.The RMS value of the voltage producedby the transformer is 12 V and the loadhas a resistance of 250 Ω. Find the fol-lowing:

(a) DC value of the voltage across theload.

vs(t)

22 k

1.5 V

10 k ix(t)

Figure P4.10

(b) RMS value of the voltage acrossthe load.

P 4.12 The 120 V/60 Hz sinusoidal supply volt-age is applied to a non-resistive load,which results in a sinusoidal current withan amplitude of 2 A. The voltage andcurrent signals have a phase differenceof π/4 radians (45). Find the follow-ing:

(a) RMS value of the voltage appliedto the load.

(b) RMS value of the load current.

(c) Power factor.

(d) Apparent power.

(e) Real power.

P 4.13 An average power of 2 kW is deliveredto a non-resistive load. The apparentpower is 2.5 kVA. Calculate the powerfactor.

P 4.14 The voltage signal v(t) = (4V ) cos(200t)is applied to a non-resistive load, which

119

results in the current waveform, i(t) =(0.2A) cos(200t + π/4). Find the follow-ing:

(a) Power factor.

(b) Apparent power.

(c) Real power.

P 4.15 An inductor is a two terminal passivecircuit element, which consists of manyturns of insulated electrical wire woundaround a cylindrical core. The voltage,vL(t) across an inductor is given by

vL(t) = LdiL(t)

dt

where L is the inductance in Henry (H)and iL(t) is the current through the in-ductor. Suppose we have an inductorwith an inductance of 100 H. There isa sinusoidal current flowing through theinductor given by iL(t) = (100mA) cos(1200t+6). Find the following:

(a) Voltage across the inductor.

(b) RMS values of inductor voltage andcurrent.

(c) Power factor.

(d) Apparent power.

(e) Real power.

P 4.16 A standard, 60 W light bulb is pow-ered by a commercial light dimmer. Thedimmer knob is set to a position, whichleaves the TRIAC open for pi/ 4 radiansafter each zero crossing. Find the powerdissipated on the bulb.

120

Chapter 5

Periodic Signals in Frequency Domain

The sinusoid is the most common periodic function that is encountered in electrical engi-neering, as well as everyday life. However, there are a great many more types of periodicfunctions and signals that surround us. Some of these are perfectly periodic and some arenearly periodic. The square wave you have seen in previous chapters is periodic. Many partsof speech, as we will see later in this chapter, are nearly periodic. In previous chapters, wehave focused on the properties of periodic signals in time domain. In this chapter, we willlearn how to describe and analyze periodic signals in the frequency domain.

Everyday, we use the term frequency to describe aspects of common devices and physicalphenomena. The term is used to denote a specified number of events that occur within aspecific time, usually one second. The clock frequency of PCs are in excess of 500 Mega-hertz (MHz). This refers to the number of elementary operations (500 million) that can beperformed in one second. Personal computer modems have data rates ranging from 19.6Kilobits per second (Kbs) to 56 Kbs. High-speed lines promise rates in excess of 10 Mbs.This refers to the number of bits that can be transmitted between computers in one second.Broadcast radio stations use frequencies between 550 Kilohertz (kHz) and 1650 kHz for AM,and between 88 MHz and 108 MHz for FM. This refers to the number of oscillations ofan electromagnetic wave in one second. We see these frequencies on the dials (or digitaldisplays) and are more aware of them. We have heard the debate about selling the broad-cast spectrum of satellite communication systems. Companies are buying the right to usespecified frequencies for commercial purposes.

In the audio range, we are aware that touch-tone phones use specific frequencies toencode the digits of the number that we “dial.” Actually, each digit requires a combinationof two frequencies to make the system more robust. Stereo systems are often equipped withgraphic equalizers and analyzers. The equalizers allow the listener to adjust the amplificationof various frequency bands to suit the acoustics of the room or the preferences of the listener.The analyzer is a device qualitatively similar to the spectrum analyzer that will be used inthis lab. It allows the listener to visually track the power of the output signal of the stereo

121

122

in various frequency bands.

5.1 Adding Sinusoids and Importance of Phase

In this chapter, we shall learn that virtually all periodic waveforms that are encountered inelectrical engineering can be represented as the sum of sinusoids of different frequencies.

In Chapter Periodic Signals in Time Domain, we learned that the phase difference canbe thought of as a time delay or lag. An example of two signals with a 90 degree (π/2) phasedifference is shown in Figure 5.1 . The second signal has exactly the same waveform as thefirst but appears to be delayed in time.

0 10 20 30 40 50 60 70 80

-1.5

-1

-0.5

0

0.5

1

1.5

volts

cos(2p60t)cos(2p60t - p/2)

time (ms)

Figure 5.1 : Two 60 Hz signals with 90 degree phase difference.

For most cases where we are concerned with only a single frequency, the phase is not ofconsequence. However, when dealing with multiple sinusoids, their relative phases can bevery important. Previously, we have seen how the phase difference between the voltage andcurrent signals had an impact on the real power consumed by the load.

As another example, consider the case of the sum of a 60 Hz signal and a 180 Hz signalof different magnitudes and different phase angles

v(t) = cos(2π60t + θ1) + 0.5 cos(2π180t + θ2) (5.1.1)

The two component signals and their sum are shown in Figure 5.2 . For this exampleθ1 = θ2 = 0. To illustrate the effect of a relative phase difference, we let θ1 = 0 and θ2 = 180and present the results in Figure 5.3 . Since time is not an absolute quantity, it is therelative phase that is always of interest. Not only does the waveform appear very different,the magnitudes of the two summations are significantly different.

123

10 20 30 40 50 60 70 80

-1

0

1

60 Hz signal

volt

s

10 20 30 40 50 60 70 80

-1

0

1

180 Hz signal

signal obtained by adding the above sinusoids

vo

lts

10 20 30 40 50 60 70 80

-1

0

1

vo

lts

time (ms)

Figure 5.2 : Sum of two sinusoids with zero relative phase.

5.2 Magnitude Spectrum and Phase Spectrum

The examples of Figure 5.2 and Figure 5.3 look similar to what would be seen on anoscilloscope. In this section, we shall learn that we can also represent such signals in frequencydomain. Any sinusoid is fully defined by its magnitude and phase, that is, the value of signal,

v(t) = V cos(2πft + θ)

can be determined at any time by knowing its f and θ. The sum of two sinusoids

v(t) = V1 cos(2πf1t + θ1) + V2 cos(2πf2t + θ2)

requires two frequencies and two phases for complete definition. If the magnitude at thefrequency f is denoted V (f) and the phase is denoted θ(f), the sum of two sinusoids inFigure 5.3 can be expressed in the frequency domain as

V (f) =

1 if f = 600.5 if f = 180

0 else(5.2.1)

124

10 20 30 40 50 60 70 80

-1

0

1

60 Hz signal

vo

lts

10 20 30 40 50 60 70 80

-1

0

1

180 Hz signal

vo

lts

10 20 30 40 50 60 70 80

-1

0

1

volt

s

time (ms)

signal obtained by adding the above sinusoids

Figure 5.3 : Sum of sinusoids with 180 degree relative phase.

θ(f) =

θ1 if f = 60θ2 if f = 1800 else

(5.2.2)

This functional representation may appear cumbersome. Its purpose is to get you to thinkof the magnitude and phase as functions of frequency, not as isolated constants. A visual-ization of the frequency representation requires a graph of both the magnitude and phasecomponents. For the summation signal of Figure 5.3 , the graphs are shown in Figure 5.4 ,for θ1 = 0 and θ2 = 180.

Complicated periodic signals are composed of the sums of a large, sometimes infinite,number of sinusoids. Each requires two real numbers to define the value. To express thisrepresentation, we will generalize the two functions above. The function, V (f) is called themagnitude spectrum and θ(f) is called the phase spectrum. Note that in the figures, thefrequency axis is shown only to 1 kHz. Since the functions are zero beyond this frequency,this is sufficient. In general, you must choose a limited frequency range to display.

Example 5.1. Sketch the magnitude spectrum of the voltage signal,

v(t) = (5V ) + (10V ) cos(2π1000t + 3π/4).

125

100 200 300 400 500 600 700 800 900 1000

0

0.2

0.4

0.6

0.8

1.0

Magnitude in Frequency Domain

Frequency (Hz)

100 200 300 400 500 600 700 800 900 1000

-200

-100

0

100

200

Phase in Frequency Domain

Frequency (Hz)

Am

pli

tude (

V )

Ph

ase

( d

egre

es

)

Figure 5.4 : Frequency domain representation of the summation signal shown in Figure 5.3: a) magnitude spectrum and b) phase spectrum.

Solution:This voltage signal has two terms. The first term is the constant, DC term. In frequency domain,we like to think of the signals DC component as another sinusoid with zero frequency. This ofcourse relies on the fact that cos(0) = 1 . The second term in v(t) is an ordinary sinusoid with anamplitude of 10 V. Thus, the magnitude spectrum consists of two components at f = 0 and f =1000Hz with magnitudes of 5 and 10 V respectively. The resulting magnitude spectrum is shownin Figure 5.5 . It is important to note that the individual phases of the signal components have no

500 1000 1500 20000

0

5

10

f requency (Hz)

Am

pli

tud

e (

V )

Figure 5.5 : Magnitude Spectrum

126

effect on the magnitude spectrum. ‖ X ‖

In some applications, we are less concerned with the relative phase. The most commonexample is audio analysis. The ear does not hear relative phases well. On a monophonicspeaker, the two summation signals of Figure 5.2 and Figure 5.3 would sound indistinguish-able. However, we do use relative phase in our ears (stereo receivers) to locate the source ofa sound.

5.3 Power Spectrum

In many cases, we are interested in the power of a signal as a function of frequency. Incommunications systems, it is the power at various frequencies that is measured to determineif broadcasters are in compliance, that is, their electromagnetic transmission stays withintheir allocated frequency band. It is also the power as a function of frequency that is used todetect radio and cellular communications signals. This is one way that the cellular systemknows when to change the user from one cell (local area) to another.

5.3.1 Signal Power

In the previous chapter, we have learned that if a time-varying signal is applied to a resistor,R, the instantaneous power dissipated on the resistor is given by

p(t) = v(t)i(t) =v2(t)

R

The average value of p(t) is the real power, P, which can be expressed as

P =1

T

∫ T

0

p(t)dt =V 2

RMS

R

We define the Signal Power as the real power dissipated on a resistance of 1 Ω. Hence, theabove equation reduces to

P =V 2

RMS

1 Ω= V 2

RMS (5.3.1)

given in Watt. In creating the magnitude spectrum of a periodic signal, we were interestedin the amplitudes of sinusoids at different frequencies. In this section, we are interested infinding the signal power of the individual sinusoids. In the previous chapter, we have seenthat the RMS voltage of the sinusoid, Vo cos(2πft + θ) is given by

VRMS =Vo√2

(5.3.2)

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From equations 5.3.1 and 5.3.2, the signal power of a sinusoid can be found as

P =V 2

o

2(5.3.3)

Therefore, we can construct the power spectrum of a signal using equation 5.3.1 to computethe signal power at each frequency.

Example 5.2. Sketch the power spectrum of the signal, v(t) = (5V )+(10V ) cos(2π1000t+3π/4)whose magnitude spectrum was found in Example 5.1

Solution:To compute the power spectrum, we need to calculate the signal power at each frequency. Thisyields,

P (f) =

25 if f = 0Hz50 if f = 1000Hz0 else

and the corresponding power spectrum is shown in Figure 5.6 . It is important to note that

500 1000 1500 20000

0

50

25

f requency (Hz)

Watt

Figure 5.6 : Power Spectrum

the signal power in the DC term is calculated as V 2DC whereas the signal power at 1000 Hz is

V 2rms = V 2

o /2 according to Eq. 5.3.1 ‖ X ‖

5.3.2 Power Spectrum in Decibels

When plotting the power spectra of periodic signals it is generally preferred to use a logarith-mic axis for the signal power. This is done by expressing the signal power in decibel-watts ordBW, which is nothing more than the logarithm of the signal power multiplied by a factorof ten. Therefore, conversion of signal power from W to dBW can be done using the formula

PdBW = 10 log P (5.3.4)

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where P is the power in Watts. While this is a simple expression that we can accept andremember, it is important to understand its origins. This is one of those basic concepts,which we hope you will not forget as long as you work as an engineer.

Strictly speaking, the unit decibel (dB) is used to compare the signal intensity of twodifferent signals. For instance, dB can be used to describe the difference between the musictransmitted by a radio station and the background noise. It is defined as,

dB = 10 logP2

P1

(5.3.5)

where P1 and P2 are the signal powers of the two signals compared. That is, we first evaluatethe logarithm of the power ratio and multiply the result by a factor of ten. For example,suppose you are at a rock concert running an experiment on the impact of high sound levelson human hearing. You measure the sound level when the band is playing and determinethe average power level to be roughly equal to 1000 W. When the music stops, you measurethe background noise level as 10 W. To express the difference between the music and noiselevels in dB, we write,

dB = 10 logPmusic

Pnoise

= 10 log1000

10= 20dB

We explain our measurement by stating that ”there is a difference of 20 dB between themusic and noise levels”.

Expanding the logarithm in equation 5.3.5 yields the following expression

dB = 10 log P2 − 10 log P1 (5.3.6)

where 10 log P1 and 10 log P2 give us the signal power in dBW for the signals ”1” and ”2”respectively. Therefore, dB is the difference between the two signal powers measured indBW.

Now let’s go back to the origin of dBW and explain how it relates to dB. The inevitablequestion is the following: ”If dB needs two quantities to compare, how can we use the conceptto refer to the power of a specific signal?” The honest answer is that we cheat and definethe signal power in dBW as

PdBW = 10 logPsignal

1 W(5.3.7)

where we have taken the signal power, Psignal as our P2 and assigned 1 W to P1. By doing so,we are comparing our signal power to a reference of 1 W . This happens to be a convenientreference since 1 W corresponds to 0 dBW according to equation 5.3.4. In a sense, when werefer to the signal power in dBW, we are implicitly stating that the signal power is so muchdB above the reference level of 0 dBW.

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Source of Sound dBW (approximate)Threshold of Hearing (TOH) 0

Rustling Leaves 10Whispering 20

Normal conversation 60Busy Street Traffic 70Vacuum Cleaner 80Large Orchestra 90

Walkman at Maximum Level 100Front Rows of Rock Concert 110

Threshold of Pain 120Military Jet Takeoff 140

Instant Perforation of Eardrum 160

Table 5.1: Acoustic Loudness in Decibels

Another very popular unit used in radio communications is dBm, which uses 1 mW asthe reference power. Therefore, the unit dBm is given by

PdBm = 10 logP

1 mW(5.3.8)

The most common use of the decibel scale is in audio signal processing where scales onstereo systems are calibrated in dB. This is meaningful since the ear can be modeled as alogarithmic detector. In other words, we hear in dB. A sound that is twice as powerful asanother does not seem to us to be twice as loud. It seems only slightly louder. In the dBscale, we would compute the difference as

dB = 10 logP2

P1

= 10 log2P1

P1

= 10 log(2) ≈ 3 dB

Therefore, increasing the signal power by a factor of two, increases the apparent sound byonly 3 dB. For a sound to seem twice as loud, the magnitude of the signal must increase bya factor of ten. This implies a factor of 100 in power, or two orders of magnitude (20 dB).There are sound theoretical reasons for this relationship. In fact, Alexander Graham Bell, oftelephone fame and for whom the decibel is named, helped work out this theory. Examplesof loudness are given in Table 5.1.

Example 5.3. Sketch the power spectrum of the signal in Examples 5.1 and 5.2 using the decibel

130

scale.

Solution:To solve the problem, we need to calculate the power at each frequency in decibels using Eq. 5.3.4.This yields,

PdB =

10 log(25) = 13.98dB if f = 0Hz10 log(50) = 16.99dB if f = 1000Hz

0 else

and the corresponding power spectrum is shown in Figure 5.7 . ‖ X ‖

500 1000 200015000

0

20

10

f requency (Hz)

Pow

er

( dB

W )

Figure 5.7 : Power Spectrum in dB

Example 5.4. What is the amplitude in volts of a sinusoid represented by a 0 dB spike in apower spectrum graph?

Solution:According to Eq. 5.3.4, the power in dB is given by

PdBW = 10 log(P )

Solving the above equation for the signal power we obtain,

P = 10PdBW /10

Substituting 0 dBW for signal power yields, P = 1 W . The signal power at any frequency is givenby

P =V 2

p

2

which yields Vp =√

2 V for the amplitude of the sinusoid. ‖ X ‖

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5.4 Periodic Signals in Frequency Domain

The real power of frequency domain analysis lies in the fact that any periodic signal1 canbe represented by the sum of sinusoids. The mathematical term for this representation isthe Fourier Series2. If v(t) is any periodic signal with period T, it can be expressed as aninfinite sum of sinusoids according to

v(t) =∞∑

n=0

Vn cos [2πfnt + θn] (5.4.1)

where Vn is the magnitude of the sinusoid at frequency, fn = nfo Hz and θn is the corre-sponding phase3. According to Fourier Series, the only possible frequency components lieat multiples of the fundamental frequency, fo. The fundamental frequency is the frequencyassociated with the smallest period of a periodic signal. We have to say smallest since aperiodic signal with a period of T is also periodic with a period of 2T, 3T, etc. In Eq. 5.4.1,the fundamental frequency is fo = 1/T . Frequencies that are multiples of the fundamentalfrequency are called harmonics. The fundamental tone of a periodic signal is also the firstharmonic obtained with n = 1.

It is also important to note that the fundamental frequency, fo of a periodic signal is thefrequency that we would measure from a display of the signal in time domain. As such, thefundamental sets the main oscillations and the rest of the harmonics provide the delicatefeatures of the signal.

For a powerful demonstration of Eq. 5.4.1, consider the square wave shown in Figure 5.8a, which has a frequency of 100 Hz and a duty cycle of 50%. The spectrum of this signal isshown in Figure 5.8 b. From the graph, we see that the only nonzero frequency componentsexist at the odd multiples of 100 Hz (i.e., 100, 300, 500 etc.). This implies that Vk = 0 for k= 2,4,6 etc.

It requires an infinite number of harmonics in order to represent a square wave.Thegraph of the spectrum (or spectrum analyzer) shows only a finite number of these terms. Itis interesting and instructive to see the effect of partial summations. The first term in thesummation corresponds to k = 0 . This term represents the average or the DC value of thesignal with f = 0Hz . The DC value of the square wave shown in Figure 5.8 a is 0.5 V. Thesecond term is the fundamental, which corresponds to k = 1 and has the frequency of 100Hz. For k = 3 and k = 5 we obtain the first two harmonics at 300 and 500 Hz respectively.

1There are some mathematical constraints on this statement but it holds true for any realizable signalthat you will encounter.

2Jean Baptiste Joseph Fourier (1768-1830) travelled with Napoleon on several occasions, an early exampleof military funded research.

3The theory on the computation of the magnitude and phase is covered in a junior level course on signalsand systems. It is also noted that periodic signals can be represented in the frequency domain by continuousfunctions of frequency. These functions are obtained using the Fourier transform.

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10 20 30 40 50 60 70 80 90

-1.5

-1.0

-0.5

0

0.5

1.0

1.5Pulse train with 50% duty cycle

Time (ms)

100 200 300 400 500 600 700 800 900 1000

-80

-60

-40

-20

0Power Spectrum of the above Pulse train

Frequency (Hz)

Vo

lta

ge (

V )

Po

wer

( d

BW

)

Figure 5.8 : Pulse train with 50% duty cycle and its power spectrum

Let us begin a series of graphs of successive summations with this constant function.The first term is shown in the top graph of Figure 5.9 . The next graph shows the result ofsumming the DC term and the fundamental frequency term ( k = 0 and k = 1). The resultis

v0(t) + v1(t) = 0.5V + V1 cos(2π100t + θ1)

which is nothing but a cosine with a DC value. The next graphs show the effect of summingthe next two terms. These signals can be written as

v0 + v1 + v2 = 0.5V + V1 cos(2π100t + θ1) + V2 cos(2π300t + θ2)

and

v0 + v1 + v2 + v3 = 0.5V + V1 cos(2π100t + θ1) + V2 cos(2π300t + θ2) + V3 cos(2π500t + θ3)

From this, it is apparent that the result of adding more terms is an increasingly betterapproximation to the pulse train.

133

10 20 30 40 50 60 70 80 90

-1

0

1

volt

s

DC term

10 20 30 40 50 60 70 80 90

-1

0

1

volt

s

10 20 30 40 50 60 70 80 90

-1

0

1

volt

s

10 20 30 40 50 60 70 80 90

-1

0

1

vo

lts

Time (ms)

DC term and 1st harmonic (fundamental)

DC term, 1st and 2nd harmonics

DC term, 1st, 2nd and 3rd harmonics

Figure 5.9 : Just like any other periodic signal, a pulse train can be represented by a Fourierseries, which is an infinite sum of sinusoids. As shown, a reasonably good approximation to asquare wave can be obtained by a DC term and only three frequency components (sinusoids).

5.5 Noise in the Frequency Domain

Noise is a major factor in communication systems and signal processing applications. Oneof the reasons digital systems are favored is their ability to continue functioning over noisychannels. An important characteristic of noise is its power spectrum. White noise is calledwhite because, like white light, it has roughly equal power at all frequencies (or wavelengths).A sample of white noise and its spectrum are shown in Figure 5.10 . As expected the spectrumis relatively flat. Since the noise is random, it is only the average or expected value that isactually flat. The power of a sample of a random process will show random variations. Thenoise is added to the square wave signal and shown in Figure 5.11 . The quantity of interestin a noisy system is not the magnitude of the noise power but its relation to the signal power.This is formally quantified by the signal-to-noise ratio or SNR, which is defined by

SNR =Ps

Pn

(5.5.1)

134

0.5 1 1.5 2 2.5 3 3.5 4 4.5

-2

-1

0

1

2

Time (ms)

0 2 4 6 8 10 12 14 16 18 20-80

-60

-40

-20

0

Frequency (kHz)

Volt

ag

e (

V )

Pow

er

( dB

W )

Figure 5.10 : White noise and its power spectrum.

where Ps is the average power of the signal and Pn is the average power of the noise. Bothare computed in an analogous way to Eq. 5.3.3, for example,

Ps =1

T

∫ T/2

−T/2

s2(t)dt (5.5.2)

The SNR in decibels is given by

SNRdB = 10 log10

(Ps

Pn

)(5.5.3)

The signal-to-noise ratio for the noisy signal in Figure 5.11 is 20 dB. This means thatthe ratio of the signal power to noise power is 100. For this case, the square wave signalis still easy to see. The edges of the pulses are still readily identified. Let us consider anoisier communication channel. The noise has been increased to yield an SNR of 0 dB forthe signal in Figure 5.12 . This means that the noise is just as powerful as the signal. Itis difficult to determine the pulses in the time domain graph. Certainly, the edges cannotbe determined with any accuracy. However, because of the different distribution of powerin the frequency domain, the signal is easily detected, that is, the peaks associated withthe square wave spectrum are easily identified. This means that the square wave can berecovered by appropriate processing. Exactly what processing is appropriate is a subject of

135

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

-2

-1

0

1

2

Time (ms)

0 2 4 6 8 10 12 14 16 18 20

-80

-60

-40

-20

0

Frequency (kHz)

Vo

lta

ge (

V )

Pow

er

( dB

W )

Figure 5.11 : a) Signal plus white noise with a signal-to-noise ratio of 20 dB, b)Powerspectrum of the signal.

more advanced courses in communications.

Example 5.5. The AM section of a radio receiver has a Signal-to-Noise Ratio of 40 dB. If thevolume level of the receiver is set to produce a music power of 100 W, calculate the noise power.

Solution:According to Eq. 5.5.3, the signal-to-noise ratio is given by

SNRdB = 10 log10

(Ps

Pn

)

Solving for Pn , we obtain

Pn = Ps.10−SNR/10 = 100W × 10−40/10 = 0.01W = 10mW

‖ X ‖

136

0.5 1 1.5 2 2.5 3 3.5 4 4.5

-2

-1

0

1

2

2 4 6 8 10 12 14 16 18 20-80

-60

-40

-20

0

Time (ms)

Frequency (kHz)

Volt

ag

e (

V )

Po

wer

( d

BW

)

Figure 5.12 : a) Signal plus white noise with a signal-to-noise ratio of 0 dB, b) Powerspectrum of the signal.

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5.6 Audio Signals in the Frequency Domain

As audio signals are one of the most common signals encountered, most students are familiarwith many of their characteristics. The graphic spectrum analyzers and equalizers have arange between 20 Hz and 20 kHz. From this, one may safely assume that the limits onhearing are in this range. Most humans produce little audio (voice) power above 3 kHz.This permits the telephone system to use a much smaller frequency range than CD qualitystereo. Let us consider an audio signal. Figure 5.13 shows a segment of the long ‘oh’ soundin the word ‘oak.’

5 10 15 20 25 30 35-0.2

-0.1

0

0.1

0.2

500 1000 1500 2000 2500 3000 3500 4000-80

-60

-40

-20

0

Time (ms)

Frequency (kHz)

Volt

ag

e (

V )

Po

wer

( d

BW

)

Figure 5.13 : a) human voice: long ‘O’ sound in the word ‘oak’, b) power spectrum of thelong ’O’ sound.

This signal is not quite periodic but it is close enough for practical purposes. From thegraph, the period is seen to be about 4 ms, which implies the fundamental frequency shouldbe about 250 Hz. This agrees with the frequency peaks of the power spectrum graph. Ifyou were to consider encoding4 the long ‘oh’ signal, you could reproduce a recognizable ‘oh’sound with very few frequencies. There are only about five or six major peaks. Note howthe signal bears a qualitative resemblance to the sum of two sinusoids of Figures 5.2 and

4Data is encoded by changing its representation. Many times the data is also compressed by reducingthe number of symbols in the representation.

138

5.3 . The sound of such a coded voice sounds mechanical to us. It is the deviations fromperfect periodicity that give the human voice its distinctive quality. For comparison, Figure5.14 shows the same ‘O’ sound from a different speaker.

0 5 10 15 20 25 30 35-0.4

-0.2

0

0.2

0.4

Time (ms)

Volt

ag

e (

V )

Figure 5.14 : Human voice: long ‘O’ sound from a different speaker

We can also see that a frequency range of 4 kHz is quite adequate to represent this voicedsound. The power in the 3-4 kHz range is down over 40 dB from the peak power of the signal.With typical telephone line signal-to-noise ratios, the signal power in this range would belost in the noise of the system. The laboratory will give you some practice at generatingsignals and getting a ‘feel’ for the relative power in the frequency domain.

While the voice has a limited frequency range, musical instruments can generate muchhigher frequencies. The characteristic sound of most instruments is a combinations of a fewdistinct lower frequency tones and a large number of their harmonics. To adequately displaythese spectra, we must extend the frequency axis of the graph to 20 kHz. A single notefrom a recorder (a wooden flute commonly used in baroque music) is shown in the time andfrequency domain in Figure 5.15 . Significant harmonics are evident to at least 10 kHz. Bysignificant, we mean that these harmonics are obviously identified above the noise level.

An interesting comparison is done by plucking a single string of an acoustic guitar. Thespectrum is shown in Figure 5.16 . Note that there are significant harmonics almost outto 20 kHz. One might expect that the single string would produce only harmonics of asingle fundamental frequency. However, we note several lower harmonics, which are criticalin giving the guitar its characteristic sound.

It is important to select the frequency range and resolution that is appropriate for theproblem. In the laboratory, we will learn how to do this on the spectrum analyzer. Asusual, there are engineering trade-offs to consider. Higher resolution and larger frequencyranges require more processing power and higher processing speeds, which, in turn, meanshigher cost. Furthermore, frequency resolution is dependent on the number of samples usedto compute the spectrum and the sampling frequency. A higher sampling frequency implies

139

0 5 10 15 20 25-0.20

-0.15

-0.10

-0.05

0.00

0.05

0.10

0.15

0 5 10 15 20-60

-50

-40

-30

-20

-10

0

10

Time (ms)

Frequency (kHz)

Vo

lta

ge (

V )

Pow

er

( d

BW

)

Figure 5.15 : a) single note from a wooden recorder, b) power spectrum of the signal

a wider total range. In fact, the range is always one half of the sampling frequency 5. Theresolution of the spectrum is proportional to the reciprocal of the number of samples. Thus,high resolution in the frequency domain implies a large number of samples6.

5This will be explained in a junior level signal and systems course.6In fact, the resolution is proportional to ∆f =sampling rate/number of samples.

140

0 5 10 15 20 25-0.6

-0.4

-0.2

0

0.2

0.4

0 5 10 15 20-60

-40

-20

0

Time (ms)

Frequency (kHz)

Vo

lta

ge (

V )

Po

wer

( d

BW

)

Figure 5.16 : a) single note from an acoustic guitar, b) power spectrum of the signal.

141

Problems

P 5.1 Sketch the magnitude and phase spectraof the following waveforms:

(a) v(t) = 2 + 3 cos(2π50t)

(b) v(t) = 2− 3 cos(2π50t)

(c) v(t) = 2 + 3 cos(2π50t + π/4)

(d) v(t) = 2− 3 cos(2π50t + π/4)

(e) v(t) = −2− 3 cos(2π50t + π/4)

P 5.2 Sketch the power spectra of the wave-forms given in the previous problem. Use”Watts” for the vertical axis.

P 5.3 Sketch the power spectra of the wave-forms given in the previous problem. Use”dbW” for the vertical axis.

P 5.4 Fourier series coefficients of the voltagewaveform,

v(t) =∞∑

n=1

An cos(2π50nt + θn)

v(t) are given by An = 120n−2 V andθn = nπ/8.

(a) If we were to apply this signal to aDC voltmeter, what would be thereading on the voltmeter?

(b) Find the signal power of the first,second and third harmonics in dBWand sketch the power spectrum.

(c) Use the first five harmonics of thesignal to create an approximate sketchof the signal in the time domain.Suggestion: Create this plot on yourcomputer.

P 5.5 Given the signal,

v(t) =3∑

n=1

12

ncos(2π1000nt) V

(a) Find the signal power of v(t).

(b) Find the RMS value of v(t).

(c) The same signal is applied to a1 kΩ resistor. Find the averagepower dissipated on the resistor.

P 5.6 An AM radio station is broadcasting atan average signal power of 1000 W. Ifsignal-to-noise ratio of the transmissionsystem is 60 dB, find the signal power ofthe background noise.

142

Chapter 6

Signal Amplification

Transducers such as microphones and electric guitar pick-ups produce tiny electrical signalsin the millivolt range. Compact disc and tape players might produce larger signals butthey too are too small for the audio speakers to produce strong sound waves that can fillconcert halls. We use special electronic circuits called amplifiers that increase the intensityof electrical signals and we refer to this process as signal amplification.

Amplifiers constitute the most important class of electronic circuits. Not only they arewidely used in audio and video systems but they are also fundamental to all electroniccircuits, analog and digital combined. For this reason, a standard introductory course onelectronic circuits typically focuses just on designing simple amplifier circuits using transis-tors.

In this chapter, we shall focus on amplification as a fundamental analog signal processingtechnique and review the basic properties of amplifiers.

6.1 Voltage Gain

The voltage gain, Av of an amplifier can be loosely defined as the ratio between the outputand input voltages and it can be expressed as

Av =Vout

Vin

(6.1.1)

This definition assumes that the output signal is a larger replica of the input signal withoutany distortion.

In the laboratory, we can use this simple definition to measure the voltage gain of an am-plifier. To do this, we apply a sinusoidal test signal to the amplifier, measure the amplitudesor the peak-to-peak values of the input and output waveforms and determine the voltagegain as the ratio of the two numbers. This procedure assumes that the output signal is alarger replica of the input signal without any obvious distortion when the amplifier output

143

144

is viewed in time domain. In other words, if the input is a sinusoid, the output should be asinusoid as well.

The following example demonstrates this measurement procedure.

Example 6.1. A sinusoidal test signal is applied to an audio amplifier. The input and outputwaveforms are measured by an oscilloscope and the the resulting display is shown in Figure 6.1 .Determine the voltage gain of the amplifier.

Solution:

First, we realize that the output signal displayed on Channel 2 is indeed an amplified sinusoidwithout any obvious distortion. Interestingly however, it is inverted relative to the input signal,thatis, when the input is positive, the output is negative. We can also look at signal inversion as atime delay between the input and output signals. The time delay is exactly equal to half theperiod, which corresponds to a phase difference of π radians. Signal inversion is the norm forbasic amplifier stages and typically, we have to do extra work to avoid it. Fortunately, for manyapplications including audio, it does not matter if the signal is inverted. In this problem, it does notinterfere with our gain measurement. The important thing is that the output is a larger sinusoid,whether it is inverted or not.

We note that the two oscilloscope channels are using different vertical scales: 50 mV/divisionfor Channel 1 and 1 V/division for Channel 2. Using these scales, we can measure the peak-to-peakvalues of the input and output waveforms as 100 mV and 2.0 V respectively. Then, the voltagegain is the ratio of the two numbers, i.e. 2.0/0.1 = 20.

‖ X ‖

6.2 Transfer Characteristic of an Amplifier

According to equation 6.1.1, the amplifier output voltage is given by

Vout = AvVin

which is a straight line with a slope, Av as shown in Figure 6.2 . We refer to this plot asthe transfer characteristic of the amplifier. If the characteristic is just a straight line as inFigure 6.2 , we use the adjectives ideal or linear to describe the amplifier behavior.

The transfer characteristic can be very helpful in visualizing the amplification process.Figure 6.3 illustrates a graphical method for obtaining the amplifier output waveform usingthe transfer characteristic. Using this method, the output signal can be constructed for anyinput signal by going through the characteristic point by point. In this example, we notethat the output signal is indeed a larger replica of the input signal.

145

Figure 6.1 : Input and output waveforms of an amplifier displayed by an oscilloscope. Theoutput signal is inverted during amplification. For sinusoidal signals, inversion is equivalentto a phase angle of π radians.

Vin

Vout

Av = Slope

Figure 6.2 : Transfer characteristic of an ideal amplifier. The slope of the straight line isequal to the voltage gain, Av. A linear amplifier such as this one applies the same voltagegain to the input signal regardless of the input voltage level.

146

Obviously, for a perfectly linear characteristic the graphical method described in Figure6.3 may seem quite unnecessary. Let’s just say that we will soon be discussing amplifierswith nonlinear features, hopefully then, you will appreciate the usefulness of the transfercharacteristic concept and the graphical method presented in this section.

6.3 Deviations from the Ideal Characteristic

Suppose we apply a larger test signal to the amplifier of Figure 6.1 , say a sinusoid withan amplitude of 300 mV. Will the amplifier still produce an output signal 20 times largerthan the input signal? You are probably thinking, ”why wouldn’t it?” After all, the voltagegain of an amplifier should have nothing to do with the intensity of the input signal. Smallor large, an amplifier has to perform the same algebraic operation, that is, multiply theinput signal by the same voltage gain. Now suppose we use an even larger input signal, thistime, a sinusoid with an amplitude of 50 V. Can the output signal have an amplitude of20 × 50 = 1000 V? Probably not! At this point, we should begin feeling a bit suspiciousabout the simple Av equation and the ideal transfer characteristic we have been using. Afterall, if this were true, our energy problem could be solved rather easily. A few batteries andan amplifier would be sufficient to produce thousands of volts!

6.3.1 Clipping Distortion

An amplifier is an electronic circuit, which requires an external energy source, a DC powersupply or a battery, to function. This external source determines the maximum voltagethat can be measured anywhere on the circuit, including the amplifier output terminals. Tounderstand this concept, you can think of any circuit constructed using resistors and a DCbattery. Can you measure a voltage greater than the battery voltage anywhere in the circuit?Your answer should be a firm ”no”. An amplifier is not much different than the resistivecircuits we have seen before. Therefore, if the output signal of an amplifier attempts toexceed the limit set by the amplifier’s energy source, the DC battery voltage, it simply can’t- it is clipped. This means that if the output signal is outside a certain voltage range, theamplifier does not function properly and its behavior can not be described by equation 6.1.1.Therefore, the transfer characteristic shown in Figure 6.2 must be improved to take intoaccount this limitation imposed by the power supply. The resulting transfer characteristic isshown in Figure 6.4 . For this characteristic, we have assumed that the amplifier is poweredby two voltage sources, one producing a positive and the other producing a negative supplyvoltage relative to the ground.

We can see that as long as the input signal is within the limits set by the power supply,the transfer characteristic shown in Figure 6.4 is perfectly linear, hence, the voltage gainof the amplifier can still be determined from the slope of the straight line. On the otherhand, if the input signal is larger than Vin,max the output voltage is independent of the

147

Vin

Vout vout(t)

vin (t)

t

t

Figure 6.3 : A graphical method can be used to construct the output signal using thetransfer characteristic of the amplifier. In this example, an arbitrary input signal is appliedto an amplifier with a perfectly linear characteristic.

148

Vin

Vout

Av = Slope

+ Vs

- Vs

Vin, max-Vin, max

Figure 6.4 : The maximum output voltage range of a practical amplifier is determinedby the positive and negative power supply voltages. The transfer characteristic shown isperfectly linear between the two limits.

input signal and it is equal to the positive supply voltage. Similarly, if the input goes below−Vin,max, then the output voltage is equal to the negative supply voltage. The slope of thecharacteristic in the saturated regions is zero indicative of no amplification.

Figure 6.5 illustrates how the output signal is distorted when it attempts to go beyondthe linear range of the amplifier. The transfer characteristic used in this example belongs toan amplifier powered by a +/−15V dual power supply. From the slope of the characteristic,we can determine the voltage gain as 15/0.1 = 150. If the input signal is within −100mV <Vin < +100mV , the output signal is an amplified replica of the input signal. Outside thelinear range, the peaks of the signal are clipped.

An oscilloscope display demonstrating clipping of a sinusoidal input signal is shown inFigure 6.6 . The input signal is displayed on Channel 1. We can see that the output signalis clipped both on top and bottom, however, it is interesting to note that clipping on top ofthe signal does not result in a fixed voltage level as in Figure 6.5 . We really can not explainthis behavior without seeing the actual transfer characteristic of the amplifier. As we shallsee in the next section, the transfer characteristic may be quite nonlinear between the twopower supply limits, which may lead to the clipping shown in Figure 6.6 .

Clipping is a common cause of distortion that plagues practical amplifiers, however, thereis at least one application we are aware of in which clipping is put into good use. Electricguitar players use various signal conditioning techniques to distort the natural sound of theguitar. One of these techniques is hard clipping, which was made popular by the heavy metalbands of the early eighties. To create this type of distortion, all we need to do is boost thevoltage gain such that the output signal is pushed outside the linear range of the amplifier.

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Vin

Vout

+ 15 V

- 15V

- 100 mV

+ 100 mV

vout(t)

vin (t)

t

t

Peaks outside

the linear range

clipping

Figure 6.5 : When the input signal goes beyond the linear region of the amplifier, the outputsignal is clipped resulting in ”clipping distortion”. The graphical method shown can be usedto predict the output signal.

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Figure 6.6 : Oscilloscope display showing the input and output waveforms of an amplifier.The output signal displayed on channel 2 is clipped.

Example 6.2. An audio amplifier with a voltage gain of 200 is powered by a +/ − 12 V dualpower supply. Find the peak-to-peak value of the largest input signal the amplifier can processwithout causing the output signal to be clipped?

Solution:The peak-to-peak value of the largest output signal the amplifier can produce is 24V equal to thedifference between the positive and negative power supply voltages. The peak-to-peak value of theinput signal can be found by dividing this voltage by the voltage gain, Av:

Vin, p−p =Vout, p−p

Av=

24V

200= 0.12 V

‖ X ‖

6.3.2 Nonlinear Distortion

Another deviation from equation 6.1.1 arises from the fact that amplifiers use transistorswith highly non-linear current - voltage characteristics. We must emphasize the fact thattransistors are the essential semiconductor devices that make the amplification happen. This

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Vin

Vout

+ Vs

- Vs“Almost“

linear range of the

amplifier

Figure 6.7 : A non-linear transfer characteristic. There is a small section of the transfercharacteristic where some linearity is achieved. To minimize distortion, only a tiny signal,which can stay within this linear range should be allowed.

non-linear transistor behavior shows up in the transfer characteristics of virtually all am-plifiers. Consequently, the voltage gain of a realistic amplifier varies with the input voltagelevel, which becomes another major source of distortion. An example characteristic is shownin Figure 6.7 . We note that the slope of the characteristic is different at different input volt-age levels, hence, we need a mathematically more rigorous definition to describe the voltagegain. Our improved expression is given below:

Av =dvout(t)

dvin(t)(6.3.1)

This equation looks at the small variations in the output signal, dvout that result in responseto small variations, dvin in the input signal. Because it is a derivative, we can determine thevoltage gain around a specific voltage level.

Clearly, an amplifier with such a nonlinear characteristic will also suffer from clippingdistortion, however, the distortion caused by the non-linearity may be just as bad when youlisten to the music produced by such an amplifier. We note that the characteristic has atiny ”almost” linear region in the middle. To minimize distortion, the input signal shouldbe kept within the bounds of this region. Extending the range of this ”almost” linear regionis a challenging task for circuit designers.

Now, let’s examine Figure 6.8 , which shows the impact of a non-linear transfer charac-teristic on the output signal. Also shown for comparison is the output signal from an idealamplifier for the same input signal. We note that the signal is not really clipped but it

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does not quite have the same general shape as the signal produced by the ideal amplifier.In other words, it is distorted in a non-linear fashion. Therefore, unlike in the case of hardclipping, the output signal still resembles the input signal, yet it is a different signal and itwill definitely sound distorted.

6.3.3 Total Harmonic Distortion

Total harmonic distortion is not a different type of distortion, rather it is a generally acceptedway of quantifying the distortion caused by clipping or the nonlinear transfer characteristicof an amplifier. This discussion requires examining the impact of distortion in the frequencydomain.

Suppose a sinusoidal test signal of amplitude Vp and frequency fo is applied to the inputport of an amplifier. Since it is a pure sinusoid, the test signal is represented by a single peakat f = fo in the frequency domain. Let us now suppose that the amplifier produces a slightlydistorted output signal due to clipping or due to a nonlinearity in its transfer characteristic.Whatever the cause of the distortion might be, the output signal will no longer be a puresinusoid, instead it will still be a periodic signal consisting of infinitely many sinusoids. Notehowever that the distorted output signal is still organically tied to the input signal - afterall, it is just a distorted version of the same signal. As such, the frequency of the outputsignal will be fo just like the input signal. We shall use this opportunity to once again stressthe fact that the fundamental frequency of the distorted output signal is the frequency ofthe signal that we would measure in time domain from an oscilloscope screen. Therefore,the distorted output signal can be expressed as a Fourier series given by,

vout(t) =∞∑

n=1

Vn cos(2πnfo + θn) (6.3.2)

The power spectrum of the above signal will consist of infinitely many peaks occurring atmultiples of fo. The signal power of the nth harmonic, Pn will then be

Pn =

(Vn√

2

)2

=V 2

n

2(6.3.3)

The total harmonic distortion of an amplifier is defined as the ratio of the total signal powerof all harmonics above the fundamental frequency to the signal power of the fundamental.It can be written as

THD =P2 + P3 + P4 + . . . + Pn

P1

× 100% (6.3.4)

Figure 6.9 shows the measured output spectra of an amplifier with and without dis-tortion. For both spectra, the input signal is a pure, 2.5 kHz sinusoid. The top spectrum

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Vin

Vout

vout(t)

vin (t)

t

t

Vs

-Vs

output signal

of an ideal amplifier

output signal of a

non-linear amplifier

Figure 6.8 : Amplification of an arbitrary input signal using two different transfer charac-teristics. One of the characteristics is perfectly linear and it results in a larger, undistortedreplica of the input signal. The second characteristic is quite nonlinear between the powersupply limits and it results in nonlinear distortion. The output signal is not clipped but itdoes not have the same general shape of the input signal.

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Figure 6.9 : Output power spectra of an amplifier displayed by a spectrum analyzer: withoutclipping (top) and with clipping (bottom). The frequency span is from 0 to 50 kHz. Thecenter of the vertical scale is at - 20 dB and each vertical division corresponds to 10 dB.

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was obtained by setting the amplitude of the sinusoid to a sufficiently small value to avoidclipping. As expected, the spectrum consists of a single peak, corresponding to a sinusoidat 2.5 kHz. This shows that the amplifier did not cause any distortion. The input was asinusoid, the output is a sinusoid as well. The bottom spectrum was obtained by increasingthe amplitude of the input signal to force the output into clipping. Now we can see that thepower spectrum includes many new distortion harmonics which do not exist in the spectrumof the input signal. In other words, the amplifier is producing these distortion harmonicsout of nowhere. As expected, the harmonics exist at multiples of 2.5 kHz, the fundamentalfrequency of the distorted sinusoid. From theory, we know that the harmonics should ex-tend all the way to infinity, however, in Figure 6.9 , it is difficult to discern them from thenoisy background beyond the 9th harmonic. We note that the vertical scale is logarithmic,hence, including nine harmonics in our calculation is probably more than enough to obtaina fairly accurate THD measurement. The signal power of the first nine harmonics is listedin the table given below. The measurements were made directly from the bottom spectrumanalyzer display in Figure 6.9 and then they were converted from dBW to Watt using

P = 10dBW/10 (6.3.5)

The signal power of the fundamental is 10 W. Summing up the signal power of the distortionharmonics, n = 2 through n = 9 we obtain the distortion power as 5.96 mW. Substitutingthese values in equation 6.3.4, we obtain a total harmonic distortion, THD of 0.06%. Whilethis seems like a fairly small distortion, it really is not because our hearing is surprisinglysensitive to this type of distortion. It is not uncommon to find THD figures as low as 0.003%in high quality audio systems.

Frequency (kHz) Power (dBW) Power (Watt)2.5 10 105.0 -29 1.26× 10−3

7.5 -26 2.51× 10−3

10.0 -30 1.00× 10−3

12.5 -32.5 5.62× 10−4

15.0 -34 3.98× 10−4

17.5 -38 1.58× 10−4

20.0 -43 5.01× 10−5

22.5 -48 1.58× 10−5

6.4 Output Power

The output power, Pout of an amplifier is the real power delivered to the load resistanceconnected to the amplifier’s output port, hence, it can be expressed as:

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Pout =V 2

out

Rload

(6.4.1)

where Vout is the RMS value of the output voltage. For an audio amplifier, the load is anaudio speaker, with a typical impedance of 4 Ω or 8 Ω. Impedance refers to the resistance ofthe speaker at a specific frequency. We expect our speakers to respond the same way to allaudible frequencies from 20 Hz to 20 kHz; Unfortunately, this is only partially true even forthe best speakers.

Example 6.3. An audio amplifier can produce a maximum output power of 100W , when it isconnected to an audio speaker with an impedance of 8Ω. If this power rating corresponds to thelargest output signal the amplifier can produce without clipping, determine the power supply usedby the amplifier.

Solution:Using equation 6.4.1 with P = 100 W and Rload = 8 Ω, the RMS value of the output signal can befound as

Vout =√

100× 8 = 28.28 V

Assuming that this is the RMS value of a pure sinusoidal waveform used to test the amplifier, wecan find its amplitude by multiplying RMS value by

√2. This yields:

Vp = Vout

√2 ≈ 40 V

Therefore, the amplifier is able to produce an output signal with a peak-to-peak value of 2×40V =80V without clipping. To achieve this, the amplifier must employ a dual power supply of ±40 V .‖ X ‖

6.5 Input Sensitivity

In audio terms, input sensitivity is the minimum RMS input voltage required to drive anamplifier to its rated output level. In other words, this is the minimum voltage that causesthe output to clip. Input sensitivity is commonly listed as one of the key specifications ofa high-fidelity amplifier. In essence, it is a fairly simple concept, which provides anotherway of referring to the voltage gain. The following example demonstrates the relationshipbetween voltage gain and input sensitivity.

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Example 6.4. The input sensitivity of an audio amplifier is 500mV . The amplifier produces arated (maximum) output power of 100W when it is connected to a speaker with an impedance of8Ω. Find the voltage gain of the amplifier.

Solution:Using equation 6.4.1 the RMS output voltage producing the rated output power can be found as

Vout =√

PoutRload =√

800V

The input sensitivity is the RMS input voltage leading to the rated output power. Therefore, thevoltage gain can be found as the ratio between the output and input RMS voltages. This yields,

Av =√

8000.5

≈ 56.57 V

‖ X ‖

6.6 Power Gain

The power gain of an amplifier is measured in decibels and it is defined as

G = 10 logPout

Pin

(6.6.1)

where Pin and Pout are both defined for the standard load resistance of 1 Ω. Expanding thelogarithm in the above equation we obtain

G = 10 log Pout − 10 log Pin (6.6.2)

where 10 log Pin and 10 log Pout correspond to input and output signal power levels expressedin dBW respectively. Therefore, if the input power is given in dBW, the output power canbe found by simply adding G to it.

Example 6.5. The signal power applied to an amplifier is 1W . The amplifier has a power gainof 25 dB. First find the output signal power in dBW and then convert it to watts.

Solution:First we have to express the input signal power in dBW.

Pin = 10 log 1W = 0 dBW

Using equation 6.6.2, the output power can be found by adding the power gain, G to the input

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power expressed in dBW. This yields:

Pout = G + Pin = 25 + 0 = 25 dBW

Since dBW = 10 log P , The resulting output power can be expressed in watts as

Pout = 1025/10 ≈ 316.23 W

‖ X ‖

Example 6.6. The amplifier of the previous example is connected to a 4Ω speaker. Find theoutput power.

Solution:We need to remember that the power gain is defined for the standard load resistance of 1 Ω. TheRMS voltage resulting in this power is

Vout =√

316.23W × 1ΩV

If this voltage is applied to a load resistance of 4 Ω, the resulting power is given by

Pout =316.23V 2

4Ω≈ 79 W

‖ X ‖

It is also possible to express G in terms of Av. Using the relationship between powerRMS value and signal power, the power gain can be written as,

G = 10 logV 2

out

1ΩV 2

in

1Ω

= 10 logV 2

out

V 2in

which can be simplified as:

G = 20 log Av (6.6.3)

This is a convenient expression that allows us to go back and forth between the voltage gainand the power gain. Amplifier manufacturers prefer the power gain in dB to the unitlessvoltage gain, Av in their list of specifications.

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Example 6.7. Find the voltage gain of an amplifier with a power gain of 30 dB.

Solution:Using equation 6.6.3, the voltage gain can be found as

Av = 10G/20 = 1030/20 ≈ 31.62

‖ X ‖

6.7 Frequency Response

The voltage gain of an amplifier varies with frequency, which creates yet another challengefor the circuit designers. In general, the goal is to achieve a fairly constant voltage gain in alimited frequency range for a given application. For instance, audio amplifiers are expectedto provide constant gain within 20 Hz to 20 kHz. Outside this range, we do not really careif the gain shows any major fluctuations. Figure 6.10 shows a typical frequency responsefor an audio amplifier. The vertical axis gives the power gain of the amplifier in dB. Alogarithmic frequency axis is generally preferred. We can see that this particular amplifierdoes not have an entirely flat frequency response. A manufacturer may make the followingstatement referring to their amplifier: ”The frequency response is flat within 20 Hz to 20kHz within ± 3 dB”. This sentence implies that the power gain is really not constant butthe variations are contained within a hopefully acceptable range. For high fidelity amplifiers,even 3 dB is considered too much of a variation. As the price of the amplifier goes up, it isnot uncommon to find amplifiers frequency responses that are almost perfectly flat within± 0.03 dB!

2

4

6

Po

wer

Ga

in (dB)

Frequency (Hz)

10 10 10 102 3 41

8

10

12

14

Figure 6.10 : Frequency response of a typical amplifier.

While the impact of a varying frequency response on music is not considered as distortion,

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it can still have a profound impact on the music we hear. For instance, suppose we havean amplifier whose gain appreciably drops beyond 10 kHz. You might argue that therearen’t many musical instruments that produce their fundamental tones above this frequency,hence, the amplifier should have at best a minimal effect on music. Unfortunately, this way ofthinking would not be correct because it completely ignores the importance of the harmonics.We must remember that it is those harmonics and their relative magnitudes that determinewhy a piano sounds very different than a guitar. Therefore, an amplifier that can’t amplifythe guitar sound above 10 kHz would suppress the higher harmonics above this frequency,ultimately changing the instrument’s sound in a subtle way. The change perhaps will not beas obvious as the one caused by hard clipping, however, the overall character of the musicwill be different even if we don’t recognize it.

As it turns out, music is not a very demanding application considering the state of theamplifier electronics today. There are however, many other applications for which the signalbandwidth is much wider than the modest, 20 Hz - 20 kHz bandwidth of audio signals. Ingeneral, applications that involve signals with sharp corners (e.g. square wave) will requirebetter amplifiers because their finer features come with strong higher order harmonics.

An example is provided in Figure 6.11 , which shows the input and output waveforms ofa real amplifier. The input signal displayed on Channel 1 is a square wave and the amplifiedsignal is displayed on Channel 2. We note that the output waveform is inverted similar to theamplifier output shown in Figure 6.1 . Not surprisingly, the output signal does not possessthe sharp corners of the input square wave. This is due to the fact that the amplifier is notable to process the higher order harmonics as mentioned above.

6.8 Amplifier Equivalent Circuit

Figure 7.10 shows the equivalent circuit of an amplifier. An interesting point is that thereis not an internal connection between the input and output ports, nevertheless, they areconnected externally through the common ground.

The resistor, Rin is the input resistance of the amplifier. This is an important parameterbecause any two-port electronic circuit has a characteristic input resistance. We expect thisresistance to be as high as possible in a good voltage amplifier to avoid any voltage divisionfrom taking place at the input port of the amplifier.

Possibly, the most interesting feature of the equivalent circuit is that it includes a newelement, a voltage controlled voltage source. Controlled voltage sources are indispensablecircuit elements in equivalent circuits constructed to describe the operation of complex elec-tronic circuits. The source included in our equivalent circuit somehow senses the voltageacross the input resistance of the amplifier and generates a voltage equal to the product ofthe input voltage and the voltage gain, Av. Other than the fact that the voltage generatedby a controlled voltage source depends on the voltage somewhere else in the circuit, it is

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Figure 6.11 : Channel 1 displays the input signal of an amplifier. The amplified outputsignal is displayed on Channel 2, which shows that all the sharp corners of the square waveare lost due to the limited frequency response of the amplifier.

still a voltage source, which behaves just like the independent voltage sources we have seenbefore.

The resistor, Rout is the output resistance of the amplifier and it has to be as small aspossible, again to avoid any voltage division with the load resistance connected to the outputport of the amplifier. As such, its impact on the circuit performance is identical to that ofthe output resistance of an independent voltage source.

Example 6.8. A microphone with an output resistance of 1000Ω is connected to the input portof an amplifier with a voltage gain of 100. The amplifier has the input and output resistances of4000Ω and 32 Ω respectively. The output port of the amplifier is connected to an audio speakerwith an equivalent resistance of 8Ω. If the microphone is generating an input signal with an RMSvalue of 100mV , what is the output power of the amplifier?

Solution:The equivalent circuit for the entire circuit with all the resistance values is shown in Figure 6.13 .The microphone signal, vs(t) is divided between the 1 kΩ output resistance of the microphone andthe 4 kΩ input resistance of the amplifier. Applying the voltage division rule to the two resistors,the true input voltage of the amplifier can be found:

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Rin

Rout

Avvinvin(t) vout(t)

Figure 6.12 : Equivalent circuit of an amplifier.

vin(t) =4000

1000 + 4000vs(t) = 0.8vs(t)

The voltage generated by the voltage controlled voltage source is then equal to

100vin(t) = 80vs(t)

The voltage generated by the voltage controlled voltage source is divided between the 32Ω outputresistance of the amplifier and the 8Ω audio speaker. Applying the voltage division rule, the outputvoltage can be found as follows:

vout(t) =8

8 + 3280vs(t) = 16vs(t)

Assuming a linear amplifier, the RMS value of the output signal should be 16 times larger thanthat of the input signal, i.e. 100mV × 16 = 1.6V

‖ X ‖

In the example above, we have seen that even though the amplifier has a voltage gain of100, the usable voltage gain is only 16 due to the voltage divisions at the input and outputports. This is why a good voltage amplifier must have the largest possible input resistanceand the smallest possible output resistance. These are important design considerations forthe circuit designer.

Example 6.9. An amplifier with Rin ≈ ∞ and Rout ≈ 0 has a resistor, Rf connected between theinput and output ports. Because the resistor creates a link between the two ports it is referred toas the feedback resistor. The voltage source providing the input signal has a finite output resistanceRs and the amplifier is connected to a load resistance, RL. Derive an equation for the output

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4000

32

100vinvin(t) vout(t)vs(t)

1000

8

Figure 6.13 : Equivalent circuit of a microphone amplifier. The microphone and the loadare both included in the circuit.

voltage in terms of the voltage source, vs(t).

Solution:

The equivalent circuit of the amplifier is shown in Figure 6.14 . Since Rin ≈ ∞, there can notbe any current flowing into the input port of the amplifier, hence, iin = 0. To satisfy Kirchoff’scurrent law at the input node, we must have is = if . Applying Kirchoff’s voltage law to the firstloop yields:

vs − isRs − vin = 0

Considering the third loop, we can see that that the voltage controlled voltage source is in parallelwith the load resistor, hence, we must have vout = Avvin. We can then write vin as vin = vout/Av

and substitute it in the equation above. Solving the resulting equation for is yields,

is =vs − vout

Av

Rs

Applying Kirchoff’s voltage law to the second loop yields,

vin − isRf − vout = 0

Substituting vin and is in the equation above, we obtain,

vout

Av− vs − vout

Av

RsRf − vout = 0

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Solving this equation for vout yields,

vout =vs

Rf

Rs

1Av

(1 + Rf

Rs

)− 1

We can see that both the feedback and the source resistors impact the output voltage. It isinteresting to consider a special case in which the voltage gain of the amplifier is very very large.We can find the output voltage for this case by evaluating the limit of the above expression as thevoltage gain approaches infinity. This yields, the following result

vout ≈ −vsRf

Rs

which has two interesting features. First, the result is negative, which indicates that the amplifierwill invert the input signal as in Figure 6.1 . Second, the output voltage is independent of thevoltage gain of the amplifier and it is only determined by the ratio of the resistors Rf and Rs.Therefore, if the voltage gain of the amplifier is sufficiently large, its impact on the output voltagewill be negligibly small. We shall return to this very concept when we begin designing our amplifiersusing operational amplifiers. This is the subject of another chapter in this book. ‖ X ‖

Avvin

voutvs(t)

Rs

RL

Rf

is

i f

iin iout

if

iL

Loop 1

Loop 2

Loop 3vin

Figure 6.14 : An amplifier with a feedback resistor, Rf between the input and output ports.Rin ≈ ∞ and Rout ≈ 0

Example 6.10. Figure 6.15 shows the equivalent circuit of an amplifier with the feedback re-sistor of the previous problem but with a finite input resistance and a non-zero output resistance.

165

Use Kirchoff’s laws to come up with five independent equations that can be solved to find the fiveunknown currents, is, iin, if , iout and iL.

Solution:We can write the following three independent loop equations:

vs − isRs − iinRin = 0 · · · KV L1

AviinRin − ioutRout − iLRL = 0 · · · KV L2

iinRin − ifRf + ioutRout −AviinRin = 0 · · · KV L3

where we have used vin = iinRin. We can also write the following independent node equations:

is − if − iin = 0 · · · KCL1

if + iout − iL = 0 · · · KCL2

The five equations can be solved simultaneously to find the five unknown currents. Once the loadcurrent is known, an equation can be derived for the output voltage in terms of the source voltage,vs. ‖ X ‖

Rin

Rout

Avvin

voutvs(t)

Rs

RL

Rf

is

i f

iin iout

if

iL

Loop 1 Loop 2

Loop 3

vin

Figure 6.15 : Equivalent circuit with the feedback resistor, Rf , finite Rin and non-zero Rout

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Problems

P 6.1 A sinusoidal test signal, which has anamplitude of 12 mV is applied to the in-put port of a perfectly linear amplifier.The amplifier has a voltage gain of 100.Assume the input signal is sufficientlysmall to avoid any clipping of the out-put signal. a) Find the amplitude of theoutput signal; b) Find the RMS value ofthe output signal.

P 6.2 An amplifier is producing an output sig-nal with an RMS value of 2 V. If theamplifier is connected to an 8 Ω speaker,what is the output power of the ampli-fier?

P 6.3 Suppose the total signal power at theoutput port of an audio amplifier is 10W. Find the output power if the speakerhas an impedance of 4 Ω. Hint: The sig-nal power is the electric power dissipatedon a load of 1 Ω.

P 6.4 A microphone is connected to the inputport of an audio amplifier. The amplifierhas a voltage gain of 4000. The outputport of the amplifier is connected to anaudio speaker with an impedance of 8Ω. a) Find the amplifier output powerif the RMS value of the signal generatedby the microphone is 10 mV ; b) Use yourresult to calculate the signal-to-noise ra-tio of the amplifier (in dB) if the RMSvalue of the noise at the output port is400 mV .

P 6.5 A microphone amplifier used in a largeconcert hall is capable of producing 400W with a 4 Ω audio speaker when theRMS value of the signal at its micro-phone input is 10 mV. Find the voltagegain of the amplifier.

Figure P6.6

P 6.6 A sinusoidal test signal is applied to theinput port of a mediocre amplifier. Fig-ure P6.6 shows the output signal and itspower spectrum. The output waveformappears to be a sinusoid without anysign of clipping, however, its spectrumincludes several equally spaced harmon-ics indicative of distortion. a) What isthe frequency of the original signal? Hint:You can find this in the time-domain;b) What is the separation between dif-ferent harmonics in Hz? c) Determinethe center frequency and the frequencyspan of the power spectrum display; d)Determine the total harmonic distortionat the amplifier output. Note that thevertical axis is given in dBW.

P 6.7 An amplifier, which has a voltage gain of200 is powered by a dual ±12 V powersupply. An 8 Ω speaker is connected tothe output port. a) What is the largestoutput power the amplifier can producewithout clipping distortion? b) Find theRMS value of the largest input signal wecan apply without causing the output toclip?

P 6.8 An audio amplifier has the input andoutput resistances of 100 kΩ and 25 Ω

167

4Vx5 V

Vx

Figure P6.9

respectively. The amplifier has a voltagegain of 100 and its output port is con-nected to an 8 Ω audio speaker. Assumethe signal source is a CD player with asource resistance of 100 Ω. Determinethe amplifier output power when the sig-nal power of the music generated by theCD player is 100 mW.

P 6.9 Find the voltage, Vx and the current flow-ing in the loop shown in Figure P6.9.Assume the resistor is 1 kΩ.

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Chapter 7

Operational Amplifiers

We have discussed the concept of signal amplification in Chapter 6. We have also learnedthat this fundamental signal processing operation is performed by special electronic circuitscalled amplifiers. In this chapter, we introduce an actual amplifier, available in the formof an integrated circuit (IC) or a chip that you can buy from your neighborhood electronicstore for less than a dollar! The term IC refers to a complete circuit, which may require onlya few external components to do its job. ICs come in plastic or metal casings such as theones shown in Figure 7.1 .

This entire chapter is dedicated to a particular IC called the operational amplifier or inshort, op-amp. Because of their low cost and unique properties, operational amplifiers arewidely used in a variety of analog signal processing applications. Furthermore, they are soeasy to use that you do not have to be a circuit expert to design circuits with op-amps,which is yet another fundamental reason behind their popularity.

7.1 Basic Properties

The circuit symbol of an operational amplifier is the triangle shown in Figure 7.2 . Thesymbol shows two independent input terminals labeled + and - and a single output terminal.Here, we need to stress the fact that the two input terminals are not identical to each otherand we will explain below why this is the case. Just as any other electronic circuit, an op-amprequires a power supply to operate. A standard op-amp employs a dual (±) power supply,

Figure 7.1 : Dual-in-line package is a standard integrated circuit package.

169

170

Output

Input 1

Input 2

Figure 7.2 : The circuit symbol of an operational amplifier

1(t) vs2(t)RLvout(t)

Figure 7.3 : Operational amplifier is designed to amplify the signal applied between theinput terminals. Ao is the voltage gain of the operational amplifier

hence, two terminals of a typical op-amp IC are reserved for the power supply connections.These terminals are normally omitted from the op-amp symbol to avoid unnecessary clutterin circuit diagrams.

Figure 7.3 shows a basic amplifier circuit, which includes an op-amp, two independentvoltage sources, vs1 and vs2 and a load resistor, RL. We shall refer to the voltages applied tothe two input terminals as v+ and v− such that v+ = vs1 and v− = vs2. The output voltageof this amplifier is given by,

vout = Ao

(v+ − v−

)= Ao (vs1 − vs2) (7.1.1)

where Ao is the open-loop voltage gain of the operational amplifier. According to thisequation, an op-amp multiplies the difference between the two input voltages by the open-loop voltage gain, Ao.

While we have the option of using two input sources, there is no rule that says we can

171

vs(t) RL

vs(t)RL

(a)

(b)vout

vout

Figure 7.4 : In both circuits, one of the input terminals is grounded and the input signal isapplied to the other terminal.

not use a single source and ground the other terminal. After all, when we ground one of theinput terminals, we are still applying a voltage, it is just 0 V ! With a single source, we havetwo options; which input terminal should we connect the signal source to? The terminallabeled with a plus or a minus sign? The two circuit choices are shown in Figure 7.4 .

In the first circuit, the input signal, vs is applied to the terminal labeled with a plus sign,that is, v+ = vs and the other input is grounded. In this case, the output voltage is given by

vout = Ao (vs − 0) = Aovs

In the second circuit, the signal is applied to the terminal labeled with a minus sign, thatis, v− = vin. The output voltage is then given by

vout = Ao (0− vs) = −Aovs

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As we can see, the only difference between the two equations is the minus sign in thesecond equation, which implies inversion of the input signal during amplification. Becausethis happens only when the signal is applied to the input with the minus sign, that input iscalled the inverting input of the op-amp. Not surprisingly, the other input is referred to asthe non-inverting input.

7.2 What makes an op-amp so unique?

What separates operational amplifiers from other amplifiers? What is unique about them?Why the name operational? These are important questions that we need to answer. Firstand foremost, operational amplifiers are amplifiers just like all other amplifier circuits. Thedifference is, op-amps have some extraordinary properties. Actually, there are four keyrequirements that must be satisfied to qualify an amplifier as an operational amplifier. Allfour of these requirements must be satisfied for this qualification. These are:

1. An operational amplifier is a difference amplifier, that is, it must have two inputterminals and it should be able to provide an output voltage proportional to the thevoltage difference between them.

2. The open-loop voltage gain, Ao of an op-amp must be very large. Practical op-ampshave voltage gains as high as 200,000. For an ideal op-amp, Ao = ∞.

3. The input resistance of an op-amp must also be very large. For an ideal op-amp,Rin = ∞. This guarantees that the currents flowing in and out of the input terminalsare negligibly small.

4. The output resistance of an op-amp must be negligibly small. For an ideal op-amp,Rout = 0.

Any good amplifier should have a high input resistance and low output resistance butordinary amplifiers do not have infinitely large gains, no matter how good they are. If youare wondering, ”why would I need an amplifier with such a high gain? Would clipping not bea problem with almost any input signal?” you are absolutely right! With an infinitely largevoltage gain, it is almost certain that the output voltage will be clipped for any practicalinput level. Just imagine having an input voltage as small as 1 mV . With a gain of 200000,the output voltage will want to reach 200 V ! Unless the power supply voltage is that high,the output will be badly clipped. Clearly, we can’t have much use for an amplifier with sucha large gain. It turns out that except the comparator application that we shall introduce inthe next section, we never use op-amps without external resistors, which serve the purposeof reducing the gain to reasonable levels.

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vin(t)RL vout(t)Vref

Figure 7.5 : An op-amp used as a comparator. When the input signal is larger than thereference voltage, Vref , the output voltage is equal to the positive supply voltage. Otherwise,it is equal to the negative supply voltage.

7.3 Op-Amp as a Comparator

In this section, we show how an op-amp can serve as a comparator, which happens to be theonly op-amp application that does not rely on external components.

The circuit diagram is shown in Figure 7.5 . You can see that all we have in this circuitis an operational amplifier with two input sources. An AC voltage source is connected tothe non-inverting input of the operational amplifier. We will refer to this signal as the inputsignal of the comparator. The DC voltage source connected to the inverting input is thereference voltage, Vref . Using equation 7.1.1, the output voltage can be written as.

vout = Ao (vs − Vref )

Suppose the operational amplifier is powered by a dual power supply with voltages, ±Vs.According to the equation above, if vs > Vref , the output will try to reach a very largepositive value but it will never get there due to clipping at the positive supply voltage level.Alternatively, if Vref > vs the output will try to go to a very large negative value but itwon’t be able go below the negative supply voltage, −Vs. In short, the circuit can have onlytwo output voltage levels, ±Vs.

This circuit is called a comparator because it compares the input signal, vin to the DCreference voltage, Vref . A comparator circuit can be used to convert any periodic signal to asquare pulse train. This is demonstrated in Figure 7.6 for an arbitrary periodic input signal.We can see that the output is a square wave oscillating between the positive and negativepower supply levels.

174

Vref

time

time

Input Signal

Comparator Output Signal+Vs

-Vs

Figure 7.6 : The input and output signals of a comparator. When the input signal largerthan the reference voltage, the output is equal to the positive supply voltage. If smaller, theoutput is equal to the negative supply voltage. This is identical to hard clipping discussedin the amplification chapter. The gain of the amplifier is so high that the only voltage levelswe can get out of this circuit are the clipped output levels.

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7.4 Basic Amplifier Configurations

In this section, we introduce two basic amplifier configurations, which can both be realizedwith a single operational amplifier and a few external resistors. The circuits, which arereferred to as the inverting and non-inverting amplifier stages are shown in Figure 7.7 .We note that both amplifiers include a resistor, Rf connected between the output terminaland the inverting input of the op-amp. Because Rf forms a link between the input andoutput ports of the amplifier, it is referred to as the feedback resistor, hence, we use thesubscript, ”f ”. By the word ”link,” we imply that some information from the output (inthis case, the output voltage) is sent back to the input port and the amplifier somehow usesthis information to fine tune its operation. This is the essence of feedback and it is a veryimportant concept in electrical engineering.

We must also stress the fact that the choice of the inverting input for Rf is not arbitrary.In fact, we will go ahead and state that we can not construct an amplifier by connectinga resistor between the output and the non-inverting input of an op-amp. We will try toexplain this later in the chapter. For now, just remember that the feedback resistor isalways connected between the output and the inverting input of the amplifier.

Figure 7.8 shows the same amplifier stages drawn in a slightly different way. Thedifference is that a separate ground symbol is used at every point on the circuit where thereis a connection to the ground line in Figure 7.7 . This is how we normally show our figuresto reduce the clutter as much as possible. Therefore, from now on, we shall rely on you toconnect the ground points in your head. The usefulness of this drawing style may not be soobvious to you with these relatively simple amplifier configurations but we assure you thatit will be when we gradually move to more complex circuits.

7.4.1 Non-Inverting Amplifier

In the non-inverting amplifier configuration shown in Figure 7.8 .a, the input signal is applieddirectly to the non-inverting input. As noted above, the feedback resistor is connectedbetween the output and the inverting input of the operational amplifier. A second resistor,Rn is used between the inverting input and the ground. The voltage gain of this amplifier isgiven by the following simple equation:

Av = 1 +Rf

Rn

(7.4.1)

It is important to note that that the infinitely large open-loop voltage gain of the op-amp,Ao, is not even in the final voltage gain equation of the amplifier. This implies that we canuse op-amps from two different manufacturers with two different voltage gains in the circuitof Figure 7.8 .a. The overall gain of the amplifier, Av will remain the same because it isindependent of Ao.

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Another missing item in equation 7.4.1 is the load resistor. Therefore, it just does notmatter if the load resistance is infinitely large or only 4 Ω! It turns out that this can be trueonly if the output resistance is very small compared to the load resistance. Remember thatto have a negligibly small output resistance is one of the four requirements that an amplifiermust satisfy to be qualified as an op-amp.

It is indeed remarkably simple to design a fully functional amplifier using two externalresistors. To a large extent, even the actual values of the resistors do not matter. The onlything that concerns us is the ratio of Rf and Rn. This simplicity is a key advantage ofoperational amplifiers. Not that we want to discourage you from pursuing a career path incircuit design, but you really do not have to be a circuit expert to use operational amplifiersin your circuits.

The transfer characteristic of the non-inverting amplifier is shown in Figure 7.9 .a. Asalways, the slope of the characteristic is the voltage gain of the amplifier and it is givenby equation 7.4.1. We must stress the fact that in designing an op-amp, much attentionis given to achieving an almost perfectly linear transfer characteristic between the powersupply limits. Other than that, the transfer characteristics shown in Figure 7.9 are identicalto those we have seen in the chapter on signal amplification.

7.4.2 Inverting Amplifier

The inverting amplifier configuration is shown in Figure 7.7 .b. Again, we have the feedbackresistor Rf between the output and the inverting input of the operational amplifier. Thistime, the non-inverting input of the op-amp is grounded. The voltage gain of this amplifierstage is given by:

Av =vout

vin

= −Rf

Rn

(7.4.2)

The negative voltage gain implies that the output signal is a larger but inverted version ofthe input signal. If the input signal is a sinusoid, the input and output signals will have aphase difference of π radians or 180 degrees.

The transfer characteristic of the inverting amplifier is shown in Figure 7.9 .b. Besidesthe fact that it has a negative slope, it does not have any unusual features. Again, the slopeof the characteristic is equal to the voltage gain and it is given by equation 7.4.2.

Example 7.1. The signal v(t) = (10mV )cos(100t + π/4) is amplified by an inverting amplifier.The resistors used in the circuit are Rf = 100kΩ and Rn = 1kΩ. Find the output signal

Solution:First we need to find the voltage gain of the amplifier using Eq.7.4.2:

177

Rf

Rn

vin(t) RL vout(t)

Rf

Rn

vin(t)RL vout(t)

(a)

(b)

Figure 7.7 : Circuit diagrams of two basic op-amp amplifier stages: a) non-inverting am-plifier and b) inverting amplifier.

178

Rf

Rn

vin(t)RL vout(t)

Rf

Rn

vin(t)RL vout(t)

(a)

(b)

Figure 7.8 : Circuit diagrams of the a) non-inverting and b) inverting amplifier stages drawnusing the ground symbol at multiple points on the circuit. The circuits are identical to thoseshown in Figure 7.7

179

Vin

Vout

Slope = 1 + Rf /Rn

+ Vs

- Vs

Vin, max-Vin, max Vin

Vout

Slope = - Rf /Rn

+ Vs

- Vs

Vin, max-Vin, max

(a) (b)

Figure 7.9 : Transfer characteristic of a) non-inverting and b) inverting amplifier stages.

Av = −Rf/Rn = −100kΩ/1kΩ = −100

We then find the output signal by multiplying the input signal with the voltage gain:

vout(t) = −100× vin(t) = −1000 cos(100t + π/4) mV

Recall that the minus sign can be absorbed in a phase angle of ±π/2. If we use +π/2, this yields,

vout(t) = 1000 cos(100t + π/4 + π/2) mV = 1000 cos(100t + 3π/4) mV

‖ X ‖

Example 7.2. What is the peak-to-peak voltage of the largest input signal that a non-invertingstage can amplify without clipping with Rf = 100 kΩ and Rn = 10 kΩ if the amplifier is poweredby a dual ±9 V power supply?

Solution:From equation 7.4.1, the voltage gain of the amplifier can be found as

Av = 1 +Rf

Rn= 11

Since a dual, ±9 V power supply is used, the peak-to-peak value of the largest output the

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amplifier can produce without clipping is 2 × 9 V = 18 V . We can find the corresponding peak-to-peak input voltage by dividing this maximum output voltage by the voltage gain, which yields18/11 ' 1.64 V . ‖ X ‖

7.5 Why do we need such a large Ao?

Now that we know how the amplifier circuits that utilize op-amps look like, we can goback and explain some of the arguments we have previously made in this chapter. Mostimportantly, we need to understand the need for that huge open loop gain of op-amps. Inthis pursuit, we will first learn how to construct the equivalent circuit of an op-amp and thenuse this circuit to derive the equations 7.4.1 and 7.4.2.

7.5.1 Equivalent circuit of an Op-Amp

We have learned how to work with amplifier equivalent circuits in the chapter on signalamplification. Figure 7.10 shows two equivalent circuits used for operational amplifiers.The top circuit includes the input and output resistances. The bottom circuit assumesRin = ∞ and Rout = 0. The open-loop voltage gain, Ao has been used to create the voltagecontrolled voltage source.

Comparing these circuits to those we have seen in the chapter on signal amplification,we realize that the only difference is the fact that op-amp has two inputs as opposed tothe single input of a standard amplifier. The input voltage that appears across the inputresistance is the difference between the voltages applied to the non-inverting and invertinginput terminals as we have seen before.

7.5.2 Non-Inverting Amplifier

Figure 7.11 shows the equivalent circuit of the non-inverting amplifier configuration. Youshould compare this circuit with the non-inverting amplifier in Figure 7.8 .a. Our goal inthis section is to use the equivalent circuit to derive an equation for the output voltage.Let’s begin our analysis from the output port. Since the output resistance of the op-ampis zero, the voltage controlled voltage source is connected directly across the load resistor.This yields,

vout = Aov′in

wherev′in = v+ − v−

Therefore, we need to express the op-amp input voltage v′in in terms of the source voltage,vin. First we note that because the input resistance of the op-amp is infinitely large, the

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Rin Aovin

Rout

voutvin

Aovin voutvin

Realistic Op-Amp

Ideal Op-Amp

Figure 7.10 : Equivalent circuits used for operational amplifiers. The top circuit includesboth input and output resistances. The bottom circuit assumes Rin ≈ ∞ and Rout ≈ 0.

182

currents flowing into the input terminals of the op-amp are both zero. Applying Kirchoff’scurrent law to the meeting node of the resistors Rf and Rn, we find,

if = in

Now, we shall write two loop equations. The first one will include the source voltage, theinput port and the resistor, Rn:

vin − v′in − inRn = 0

which can be solved for vin to yield,

v′in = vin − inRn

The second loop equation will include both ports and it will go through the feedback resistor:

vout − in (Rf + Rn) = 0

where we have used if = in. This equation can be solved for in to yield

in =vout

(Rf + Rn)

Combining the two equations, we obtain:

v′in = vin − Rn

Rf + Rn

vout

By substituting this equation in vout = Aov′in and solving the equation for the voltage gain,

Av = vout/vin, we obtain:

Av =

[1

Ao

+Rn

Rf + Rn

]−1

(7.5.1)

We can see that the final equation is a function of the open-loop gain, Ao. Figure 7.12shows Av plotted as a function of Ao. This particular plot was created for Rf/Rn = 100.

We can see that if Ao is sufficiently large, the voltage gain of the non-inverting amplifier isindependent of Ao. Indeed, if we take the limit of the above gain equation as Ao approachesinfinity, the resulting voltage gain equation is identical to the non-inverting gain given inEquation 7.4.1.

Av = 1 +Rf

Rn

Therefore, it really does not matter if an op-amp has an open-loop voltage gain, Ao offifty thousand or one million! This is a great advantage considering a practical but a very

183

Aovin’ voutvin’

Rf

Rn

RL

vin=vs(t)

i fi n

if

iL

iout

i = 0

i = 0

if

Figure 7.11 : Equivalent circuit of the non-inverting amplifier shown in Figure 7.8 .a.

real problem. Integrated circuits do show variations in their specifications arising fromfabrication related variations, hence, it is virtually impossible to find two identical ICs.The clear advantage of the very large open-loop gain is that the variations in the op-ampspecifications do not influence the performance of the non-inverting amplifier.

7.5.3 Inverting Amplifier

We will now repeat the same analysis for the inverting amplifier stage shown in Figure 7.7.b. The equivalent circuit is shown in Figure 7.13 . The output voltage is again determinedby the voltage controlled voltage source:

vout = Aovin

where vin is the voltage between the input terminals of the op-amp. It is important to notethat the equivalent circuit does not include the input and output resistances since we assumeRin ≈ ∞ and Rout ≈ 0. We begin our derivation by applying Kirchoff’s voltage law to theinput loop and obtain the following equation:

vin − inRn + v′in = 0

184

1 10 102 103 104 105 106

120

100

80

60

40

20

0

Open-Loop Gain (Ao)

Voltage Gain (Av)

Rf / Rn = 100

Figure 7.12 :

185

Solving the above expression for vin and then substituting it in the equation for vout, weobtain:

vout = Ao (inRn − vin)

We need to express in in terms of the source voltage and the independent parameters. To dothis, we apply Kirchoff’s voltage law to the loop that includes both input and output portsas well as the feedback resistor. This yields,

vout − ifRf + inRn − vin = 0

Applying Kirchoff’s current law to the node of the inverting input we obtain:

if = −in

The two equations above can be combined to yield

in =vin − vout

Rf + Rn

We can now substitute the above equation for in in the equation for the output voltage.Solving the equation for the voltage gain, Av = vout/vin yields,

Av =vout

vin

= −Ao

Rf

Rf+Rn

1 + AoRn

Rf+Rn

(7.5.2)

Similar to the equation derived for the non-inverting amplifier, the voltage gain is a functionof the open-loop gain and the external resistors, Rf and Rn. If we take the limit of the aboveequation as Ao approaches infinity, we obtain the inverting amplifier gain given in equation7.4.2.

Av = −Rf

Rn

It is also important to note that the load resistance, RL does not appear in either 7.4.2 or7.5.2. This is again due to the fact that the output resistance of the op-amp, Rout shown inFigure 7.10 is not included in the analysis assuming that it is negligibly small.

7.6 Virtual Short Concept

In this section, we introduce the virtual short concept, which greatly simplifies the analysis ofoperational amplifier circuits. In the previous section, we have used the op-amp equivalentcircuit to derive the voltage gain expressions, 7.5.1 and 7.5.2 for the non-inverting and

186

Aovin’ voutvin’

Rf

Rn

RL

i f

in

if

iL

iout

i = 0

i = 0

if

vin = vs(t)

Figure 7.13 : Equivalent circuit of the inverting amplifier stage shown in Figure 7.8 .b.

inverting amplifier stages respectively. In both expressions, the voltage gain was expressedin terms of a finite open loop gain, Ao. We then asked the question, ”what if Ao is very verylarge?”. To find the answer, we evaluated the limits of both expressions as Ao was allowedto approach ∞, which yielded the simple expressions 7.4.1 and 7.4.2.

The virtual short concept upfront assumes that the open loop gain, Ao is infinitely largein addition to the assumptions Rin ≈ ∞ and Rout ≈ 0. Using the op-amp equivalent circuitin Figure 7.10 , we can write

v+ − v− = v′in =vout(t)

Ao

(7.6.1)

If we now make the assumption, Ao ≈ ∞, the fraction yields zero no matter how large theoutput voltage is as long as it is finite. Therefore, by assuming an infinitely large voltagegain for the operational amplifier, we are implicitly assuming

v+ = v− (7.6.2)

which is exactly what we would have if the two input terminals were shorted. We know thatthey are not shorted, hence, we say that ”there exists a virtual short” between the inputterminals.

It turns out that virtual short is a very powerful tool in analyzing op-amp circuits. In

187

essence, by making the infinitely large open-loop gain assumption upfront, we are avoidingany complex expressions in our derivations. Because the two gain stages we have seen so farare very basic, the need for an analysis short-cut may not be so obvious at this point. Justbe patient, before the chapter is over we will be analyzing far more complex circuits, whichwill be very difficult to analyze without the virtual short concept. Before we move on, weshall repeat ourselves one more time and derive the expression for the inverting amplifiergain using the virtual short concept. We shall leave derivation of the non-inverting amplifiergain given in equation 7.4.1 to you as an exercise.

7.6.1 Derivation of the Inverting Amplifier Voltage Gain

As a first example of this technique, we will derive the inverting amplifier gain equation. Wewill again refer to Figure 7.8 . Since the non-inverting input is grounded, v+ = 0. Using thevirtual short concept,

v− = v+ = 0

We can find the current, in as,

in =v− − vin

Rn

= −vin

Rn

The current if is given by,

if =vout − v−

Rf

=vout

Rf

Since the current flowing into the inverting input of the operational amplifier is zero, Kir-choff’s current law yields,

if = in

Substituting the expressions for the two currents, we obtain

vout

Rf

= −vin

Rn

Solving for the ratio vout/vin, we obtain the voltage gain of the inverting amplifier that wehave found before.

Av =vout

vin

= −Rf

Rn

We hope you realize how simple this analysis is relative to the one given in the previoussection.

188

7.7 Applications of Operational Amplifiers

In this section, we will introduce a few applications of op-amps in analog signal processingto give you a basic idea about the things you can do with them. These are real circuits usedin real hardware.

7.7.1 Audio Mixer - Summing Amplifier

In live audio recording, one of the indispensable tools of a recording engineer is an audiomixer. The instrument provides inputs for a number of microphones with an individual levelcontrol for each input. The output signal can then be sent to a recording device. The blockdiagram of a typical recording set-up is shown in Figure 7.14 . Slider potentiometers areoften used on mixers to set the voltage gain for each microphone input as shown in Figure7.16 . A high quality audio mixer can be realized using a single operational amplifier asshown in Figure 7.15 . Microphones are replaced by the voltage sources and the operationalamplifier is used in the inverting configuration. The circuit is commonly referred to as asumming amplifier since it can be used to add any number of voltage sources.

Using the virtual short concept, we can analyze the circuit and find the output voltagevery quickly and easily. Since the non-inverting input terminal is grounded, using the virtualshort concept we can claim that the voltage at the inverting input terminal is also zero.

v− = v+ = 0

The current flowing through each input resistor can be written as,

in =vn − v−

Rn

=vn

Rn

AUDIO MIXER RECORDING

DEVICE

MICROPHONES

Figure 7.14 : Set up used to record signals from four microphones

189

Rf

R4

R3

R2

R1

i1

i2

i3

i4

iT

if

v1

v2

v3

v4

vout

Figure 7.15 : Summing Amplifier

Figure 7.16 : A commercial audio mixer.

190

Rf

R2

R1

i1

i2

iT

if

vs

VDC

vout

Figure 7.17 : An op-amp circuit used to add a DC off-set to a time-varying signal.

where n = 1,2,3 or 4. The current if is given by,

if =vout − v−

Rf

=vout

Rf

Applying Kirchoff’s current law to the node of the inverting input terminal yields,

iT + if = 0

where we assumed that the current flowing into the inverting input of the operational am-plifier is zero. Since the current iT is equal to the sum of the currents flowing through theinput resistors, we have,

i1 + i2 + i3 + i4 + if = 0

Substituting the current expressions and solving for vout we obtain,

vout = −Rf

(v1

R1

+v2

R2

+v3

R3

+v4

R4

)(7.7.1)

which gives the output signal in terms of the four input voltages. We can see that a differentgain can be applied to each input source simply by changing the resistors, Rn. In an audiomixer, potentiometers are used in place of the fixed resistors.

7.7.2 DC Level Shifter

Sometimes, it is necessary to add a DC offset to a time varying signal, which can be easilyaccomplished using a summing amplifier. The circuit is shown in Figure 7.17 . We can followthe steps used in the previous section to find the output voltage as given below.

191

Rf

C

vin(t)RL

Figure 7.18 : Differentiator circuit diagram. The output signal is proportional to thederivative of the input signal.

vout = −Rf

(vin

R1

+vDC

R2

)(7.7.2)

By making the resistor R2 variable, the signal vin can be raised to the DC offset desired.

7.7.3 Differentiator

Now we shall consider another classic application of operational amplifiers. The circuitdiagram is given in Figure 7.18 . The circuit is referred as a differentiator because theoutput signal is proportional to the derivative of the input signal.

In analyzing this circuit, we have to remember and use the I-V characteristic of a capacitorgiven by,

ic = Cdvc

dt

We note that the non-inverting input of the operational amplifier is grounded. Using thevirtual short between the two input terminals, we conclude that the inverting input is alsoat the ground potential.

v− = v+ = 0

We can then write the equations given below for the currents if and ic.

if =vout − v−

Rf

=vout

Rf

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ic = Cdvc

dt= C

d(vin − v−)

dt= C

dvin

dt

Applying Kirchoff’s current law to the node where Rf and the capacitor meet,

if = −ic

Substituting the expressions found for these two currents, we obtain the final expression forthe output voltage.

vout = −CRfdvin

dt

7.8 Input Resistance of an Operational Amplifier Cir-

cuit

As we have seen in the previous chapter, the input resistance of a voltage amplifier hasto be as high as possible. In this section, we shall learn how to find the input resistanceof an actual amplifier circuit employing an operational amplifier. In the chapter on signalamplification, we have defined the input resistance of an amplifier, Rin as

Rin =vin

iin(7.8.1)

Let’s apply this definition to the inverting amplifier circuit. The amplifier input currentis the current flowing through the input signal source, which is the same current flowingthrough the resistor, Rn. The input current can be expressed as:

iin =vin − v−

Rn

=vin

Rn

where we used v− = v+ = 0 using the virtual short concept. Rearranging the equation, wecan find the input resistance of the amplifier circuit as

Rin = Rn (7.8.2)

Therefore, the resistor Rn must be as large as possible in a good amplifier design. Sincethe voltage gain is equal to −Rf/Rn, the feedback resistor can then be chosen to obtain thedesired voltage gain.

Let’s now consider the input resistance of the non-inverting amplifier. The input voltagesource is connected directly to the non-inverting input of the operational amplifier, hence theinput current is supposed to flow directly into the operational amplifier. But we know that

193

in a good operational amplifier, this current is practically zero, which yields an infinitelylarge input resistance according to

Rin =vin

iin≈ vin

0= ∞ (7.8.3)

Clearly, this gives a major advantage to the non-inverting amplifier stage when the applica-tion requires an amplifier with a high input resistance.

194

1 k

10 k

vin(t) vout

iin

Figure P7.1

Problems

P 7.1 Given the amplifier shown in Figure P7.1,a) Determine if the amplifier is an in-verting or non-inverting stage; b) Findthe voltage gain of the amplifier; c) Findthe RMS value of the output signal if theinput signal is a 10 kHz sinusoid with apeak-to-peak value of 200 mV. Assumethe output signal is not clipped.

P 7.2 Find the input resistance of the amplifiershown in Figure P7.1.

P 7.3 Given the amplifier shown in Figure P7.3,a) Determine if the amplifier is an in-verting or non-inverting stage; b) Findthe voltage gain of the amplifier; c) Findthe input resistance of the amplifier.

P 7.4 A 100 Ω load resistor is connected tothe output port of the operational am-plifier circuit shown in Figure P7.4. a)Find the voltage gain of the amplifier; b)What is the impact of the load resistoron the voltage gain?

P 7.5 A 1.5 kΩ resistor has been connected be-tween the non-inverting input of the op-erational amplifier and the ground in theinverting amplifier stage shown in Fig-ure P7.5. a) Find the voltage gain of

1 k

10 k

voutiin

vin(t)

Figure P7.3

1 k

10 k

100 voutvin(t)

Figure P7.4

195

1 k

10 k

100 1 k voutvin(t)

Figure P7.5

Rn

R1 R2

Rf

vout

if

vs(t)

in

i2i1

Figure P7.6

the amplifier; b) What is the impact ofthe extra resistor on the voltage gain?

P 7.6 Given the op-amp circuit shown in Fig-ure P7.6, a) Find the currents, i1, i2,if , in in terms of the resistors and thesource voltage, vs. Your answers shouldagree with the current directions shownon the circuit diagram; b) Derive an equa-tion for the output voltage; c) Find theinput resistance of the circuit.

P 7.7 A 1 kΩ resistor has been connected be-tween the non-inverting input of the op-erational amplifier and the input sourcein the non-inverting amplifier stage shownin Figure P7.7. a) Find the voltage gainof the amplifier; b) What is the impact

10 k

1 k

vout

1 k

vin(t)

iin

Figure P7.7

of the extra resistor on the voltage gain?;c) Does the extra resistor have an impacton the input resistance? Explain why orwhy not.

P 7.8 Derive an equation for the output volt-age in terms of the input sources, v1 andv2.

P 7.9 All three operational amplifiers shown inFigure P7.8 are powered by a dual±9 Vpower supply. What is the largest peak-to-peak value each input source can haveto avoid clipping distortion?

P 7.10 The input sources v1 and v2 in FigureP7.8 are generating sinusoids at two dif-ferent frequencies but with the same peak-to-peak value of 1 V . Suppose the am-plifier is connected to an audio speakerwith an impedance of 4 Ω. Find the av-erage power dissipated on the speaker.

P 7.11 Derive an equation for the output volt-age of the circuit shown in Figure P7.11in terms of the source voltage, vs and theDC voltage source, VDC .

P 7.12 Derive an equation for the output volt-age of the operational amplifier circuitshown in Figure P7.12 in terms of the

196

10 k

3.3 k

47 k

vout

v1(t)

v2(t)

Figure P7.8

22 k

1 k

vs(t) vout

VDC

Figure P7.11

Rn

Rf

voutv1(t)

Rn

Rfv2(t)

Figure P7.12

source voltages v1 and v2.P 7.13 Suppose the AC voltage source in Figure

P7.11 is generating a sinusoid, whichhas an RMS value of 100 mV and VDC =50 mV . Find the signal power of theoutput signal.

P 7.14 The feedback resistor in the op-amp cir-cuit shown in Figure P7.14 is replacedby a non-linear element with an IV char-acteristic given by ix = v2

x. Derive anequation for the output voltage in termsof the source voltage, vs. Note that yourfinal equation should not include eithervx or ix.

P 7.15 Given the circuit shown in Figure P7.15,derive a differential equation that relatesthe source voltage to the output voltagein terms of the resistors, R1, R2, Rf andthe capacitor, C.

197

10 k

vs(t) vout

vx

ix

Figure P7.14

R1

Rf

vout

R2

Cvs(t)

Figure P7.15

198

Chapter 8

Filters

We can find many different types of filters in our daily lives. A water filter can removeundesirable chemicals such as lead from the drinking water. Sunglasses reduce the intensityof the ultraviolet radiation entering our eyes. A photographer uses color filters to enhance orattenuate different wavelengths in the visible spectrum. In electronics, a filter is used whenwe need to amplify or attenuate the signal intensity in a certain frequency range.

There are many practical applications of filters in electronics. Here are just a few examplesfrom the audio world:

• A radio receiver makes use of a filter to tune to a radio station broadcasting in a givenfrequency range.

• A music signal from a stereo system may contain some high-frequency noise generatedby the electronic circuits used in signal processing. A filter can be used to remove thenoise without altering the rest of the music.

• The treble control on your stereo system relies on a filter to change the intensity of thehigh frequency content of the music played.

• A typical high-end audio speaker box houses three speakers tuned to different rangesof the audible spectrum. A sub-woofer is used for low-frequencies, a tweeter is used forhigh frequencies and a good general purpose speaker is used for the mid-range. Filtersin the speaker box divide the music into three different frequency ranges such that onlycertain frequencies can be applied to each speaker.

A filter can be constructed from just resistors and capacitors or it can be an electroniccircuit employing op-amps or individual transistors. Nowadays, high quality filters are avail-able in integrated circuit (IC) form requiring a few external components to operate. In thischapter we will discuss fundamentals of filtering as an analog signal processing technique.

199

200

8.1 Amplifiers and Filters

There are great similarities between amplifiers and filters. First and foremost, just likeamplifiers, filters are two-port circuits. An electrical signal is applied to the input port of afilter and the filtered signal is taken from the output port.

A filter attenuates the harmonics of an input signal that fall outside a certain frequencyrange, often referred to as the pass-band of the filter. The harmonics that are within thepass-band, pass through the filter without any attenuation. This is equivalent to having anamplifier with a voltage gain of unity (or zero power gain since G = 20 log Av) in a certainfrequency range. Outside the pass-band, the signal is attenuated or suppressed, that is,Av < 1 or G < 0.

As shown in Figure 8.1 , the equivalent circuit of an amplifier is identical to that of anamplifier. Just like any two-port circuit, a filter has an input resistance and an output re-sistance. Typically, a filter is not expected to amplify the signal in the pass-band (Av ≈ 1),hence, the voltage waveform generated by the voltage controlled voltage source is identicalto the waveform applied to the input port of the filter. We should add however that ampli-fication, (i.e. Av > 1) is a possibility with electronic filters. As long as the input signal issuppressed outside the pass-band, we have a filter.

Rin

Rout

Avvinvin(t) vout(t)

Figure 8.1 : Equivalent circuit of a filter is identical to that of an amplifier.

8.2 Ideal Filters

An ideal filter completely removes the harmonics of the input signal that fall outside thepass-band of the filter. Figure 8.2 illustrates the frequency response of three commonlyused filters referred to as low-pass, high-pass and band-pass filters.

For an ideal filter, Av = 1 in the pass-band, which means that the input signal passes

201

fc fc fc1 fc2

Av Av AvLow-Pass High-Pass Band-Pass

Figure 8.2 : Frequency response of ideal low-pass, high-pass and pass-band filters.

through the filter without any attenuation. Outside the pass-band (referred to as the stop-band), Av = 0 or G = −∞, hence, harmonics of the signal in this range are completelysuppressed.

The transition from the pass-band to the stop-band occurs at a specific frequency calledthe cut-off frequency, fc. A band-pass filter has two cut-off frequencies, fc1 and fc2 as shownin Figure 8.2 . The frequency response of an ideal filter is characterized by an abrupt dropof the voltage gain from unity to zero.

As an example, consider Figure 8.3 , which demonstrates the behavior of an ideal low-pass filter. The amplitude spectrum of the input signal consists of harmonics occurringat multiples of the fundamental frequency. Within the pass-band of the filter, the outputspectrum is identical to the input spectrum. In the stop-band, the harmonics are completelysuppressed.

8.3 Realistic Filters

Realistic filters can not make the abrupt transitions between the pass-band and the stop-band shown in Figure 8.2 for ideal filters. Figure 8.4 illustrates the frequency response of arealistic band-pass filter. In this example, we have plotted the power gain, P as a function offrequency, however, it is also possible to plot the voltage gain, Av because the two quantitiesare related to each other as we have shown in Chapter 6:

G = 20 log Av (8.3.1)

We can see that within the pass-band, the power gain is 0 dB and it continually dropsas we move away from this range. It is not obvious how we would define the stop-band inFigure 8.4 because the voltage gain never drops to zero with an abrupt transition as inFigure 8.2 .

In designing an amplifier, a grand challenge is to keep the voltage gain constant in thedesired frequency range. For instance, for audio applications, this range would be from 20Hz to 20 kHz. Obviously, it is desirable to have a fixed power gain of 0 dB in the entirepass-band of a filter as in Figure 8.4 . Interestingly however, what happens outside the

202

V

Input Signal

Amplitude Spectrum

Hz

Av

Ideal Low-Pass Filter

Frequency Response

Hz

1

0

V

Output Signal

Amplitude Spectrum

Hz

cut-off

frequency

Figure 8.3 : With a voltage gain of zero beyond the cut-off frequency an ideal low-pass filterremoves all harmonics outside its pass-band.

203

Power Gain (dB)

Frequency (Hz)

Pass-Band

0

- 60

Figure 8.4 : Frequency response of a filter.

Low-Pass Filter

vin (t) vout (t)

Figure 8.5 : A simple low-pass filter can be created using a capacitor and a resistor.

pass-band is more important for filters. Most importantly, it is desirable to have the voltageor power gain to drop as abruptly as possible once we leave the pass-band of the filter. Thisis the single most important property that determines the complexity and the cost of a filter.We shall return to this subject later when we discuss realistic filters in section 8.3.

A simple, inexpensive low-pass filter can be constructed using a resistor and a capacitoras shown in Figure 8.5 . This is a 1st order filter because it employs a single capacitor. Thefrequency response of the RC filter is given by:

Av(f) =1

[1 + (2πf)2R2C2]1/2(8.3.2)

According to the above equation, the voltage gain can be unity only when f = 0 and itcontinually decreases at higher frequencies, hence, the frequency response of this filter doesnot have a pass-band where the voltage gain is unity like the ideal filters shown in Figure8.2 .

Better filters can be constructed by repeating the RC section as illustrated in Figure8.6 , which shows a fourth order RC filter. The filter order, n is equal to the number of

204

4th order RC low-pass filter

vin (t) vout (t)

Figure 8.6 : Higher quality RC filters can be constructed by repeating the RC section. Thefilter shown is a 4th order filter because it has four capacitors.

RC sections and the quality improves as the filter order increases. Figure 8.7 shows thefrequency responses of three low-pass filters with the same cut-off frequency of 1000 Hz butdifferent filter orders of n = 1, 4 and 20. Considering the frequency response of the 1st orderfilter, we can see that it does not have a very well defined pass-band. We had successfullypredicted this behavior above using the equation for its frequency response. On the otherhand, the fourth order filter displays a reasonably flat pass-band and a relatively abrupttransition into the stop-band. These characteristics further improve when the filter order israised to twenty approaching the box-like frequency response of an ideal low-pass filter.

The responses shown in Figure 8.7 belong to Butterworth type filters. The name comesfrom its inventor, Stephen Butterworth, a British engineer who described his filter in a paperpublished in 1930. The voltage gain of an nth order low-pass Butterworth filter is describedby

Av(f) =Av(0)√

1 +(

ffc

)2n(8.3.3)

There are other types of filters including Chebychev and Bessel filters also named aftertheir inventors. These filters differ in characteristics such as flatness of the pass-band orthe abruptness of the roll-off above the cut-off frequency. Among all known filter types,Butterworth filters provide the best flatness in the pass-band but they do not have thesharpest transition into the stop-band.

All three filters shown in Figure 8.7 have the same cut-off frequency of 1000 Hz. Notethat the three curves cross each other at this frequency. How can we define the cut-offfrequency when the transition into the stop-band is gradual as it is the case for the 1st orderfilter? We need a criterion, which we shall define later in this chapter.

The frequency response of real filters typically plot the power gain as a function offrequency as shown in Figure 8.8 for three Butterworth filters. These are the same filters

205

0 1000 2000 3000 4000 50000

0.2

0.4

0.6

0.8

1n = 1

n = 4

n = 20

Frequency (Hz)

Vo

ltag

e G

ain

(A

v)

Figure 8.7 : Frequency response of three low-pass filters. The orders of the filters are 1, 4and 20. All three filters have the same cut-off frequency of 1000 Hz

we have shown before in Figure 8.7 . In the new plot, the vertical axis is given in decibelsand a logarithmic axis is used for the frequency.

A key observation we can make from Figure 8.8 is that transition into the stop-band islinear, which is always the case for realistic filters. In fact, the slope of the filter roll-off isused as one of the key filter specifications. The slope has the units of dB/decade, where onedecade corresponds to a factor of 10 increase in frequency. This means that if the power gaindrops by a certain dB amount from 1 to 10 kHz it will drop by the same amount within thenext decade, from 10 to 100 kHz.

A filter may also have a positive power gain (i.e. Av > 1) in the pass-band. The treblecontrol on your home stereo is a high-pass filter with zero power gain at the center positionof the knob. By turning the knob clockwise or counter-clockwise we can apply positive ornegative power gain to the music above the cut-off frequency of the filter.

Example 8.1. Find the filter roll-of in dB/decade for the 1st order RC filter whose frequencyresponse is shown in Figure 8.8 .

Solution:We measure the slope of the straight line in the transition region of the filter. Considering the

206

1 10 100 1 103

1 104

1 105

1 106

40

30

20

10

0n = 1

n = 4

n = 20

Frequency (Hz)

Po

wer

Gai

n (

dB

)

Figure 8.8 : Frequency response of three Butterworth filters (n = 1, 4 and 20) plotted usingthe logarithmic dB scale. Note that the frequency axis is also logarithmic.

frequency range from 10 to 100 kHz, we measure a drop of 20 dB in power gain. Hence, the filterroll-off is 20 dB/decade. ‖ X ‖

8.4 Three dB cut-off Frequency

In Figure 8.2 , it was easy to define the cut-off frequency of an ideal filter. How can we dothis when the transition from the pass-band to the stop-band is not so abrupt? We need acriterion or convention: The cut-off frequency of a realistic filter is defined as the frequencyat which the output power is reduced to 50% of the input power. Using the definition ofpower gain, we can show that this corresponds to a power gain (or loss) of

G(fc) = 10 logPin/2

Pin

= 10 log1

2= −3dB (8.4.1)

Since the power gain is negative, the signal is attenuated by 3 dB at the cut-off frequency.It is important to realize that while 3 dB does not appear to be much of an attenuation, itcorresponds to losing 50% of the original power at that frequency so it is not negligible, atleast electrically!

The voltage gain of a butterworth filter is given by equation 8.3.3. Assuming that voltagegain in the pass-band is unity, we can determine the voltage gain at the cut-off frequency as:

207

Av(fc) =1√2

(8.4.2)

Using equation 8.3.1, this corresponds to a power gain of

G = 20 log Av(fc) = −20 log√

2 = −3dB

in agreement with the definition of the cut-off frequency given above.

Example 8.2. Derive an equation for the 3-dB cut-off frequency of the RC filter whose frequencyresponse is given by equation 8.4.3.

Solution:Substituting 8.4.2 in the equation for the filter frequency response we obtain:

Av =1√2

=1

[1 + (2πf)2R2C2]1/2(8.4.3)

Solving this equation for the frequency, f yields:

f3−dB =1

2πRC

as the 3-dB cut-off frequency of a first order RC filter. ‖ X ‖

8.4.1 Finding the Output Power Spectrum

Now let’s look at how we can use this information to obtain the power spectrum of thefiltered signal. Using the definition of power gain, the signal power at the filter output canbe found as

10 log Pout = 10 log Pin + G (8.4.4)

which can be written as

Pout,dBW = Pin,dBW + G (8.4.5)

According to this equation, we just add the filter power to the input signal power at thefrequency of interest.

208

Example 8.3. The input signal applied to a filter is given by vin = 2 cos(2π1000t) V . Find the sig-nal power of the output signal in dBW assuming that the filter has a voltage gain of 0.1 at 1000 Hz.

Solution:We can solve this problem using two different approaches. We shall follow both to demonstratetheir equivalency.

Since the voltage gain of the filter is 0.1, the amplitude of the output sinusoid has to be2 V × 0.1 = 0.2 V . Also, since the output signal is a perfect sinusoid, its RMS value is given by0.2/

√2 V . The signal power can then be found as P = V 2

rms = 0.02 W . We can easily convert thisvalue to dBW using PdBW = 10 log(0.02) ≈ −17 dBW .

Alternatively, we can first find the signal power of the input signal and then find the outputpower in dBW by adding the power gain at 1000 Hz. The RMS value of the input signal is2 V/

√2 =

√2 V , which corresponds to a signal power of P = V 2

rms = 2 W . This yields a signalpower PdBW = 10 log(2) ≈ 3 dBW . The power gain of the filter can be found from equation ??as G = 20 log(0.1) = −20 dB. Using equation 8.4.5, we obtain the same output power of -17dBW.‖ X ‖

Example 8.4. Figure 8.9 shows the frequency response of a band-pass filter and the powerspectrum of the input signal applied to its input port. The signal consists of a series of harmonics,which have exactly the same power. The output spectrum shown was simply found by adding thepower gain at different frequencies to the signal power of the harmonics according to equation 8.4.5.Note that because the vertical axis is logarithmic, it does not have an origin. The power spectrumonly shows the harmonics that have a signal power greater than -30 dBW.

‖ X ‖

8.5 Filtering in Time Domain

So far we have discussed filtering only in the frequency domain. It is important to understandthe impact of filtering on the time domain signal. Imagine we apply a square wave to a low-pass filter. Suppose the square wave has a frequency of 1 kHz and the filter has a cut-offfrequency of 1.5 kHz. If the filter is ideal, it will only allow the fundamental to pass, hence,the output will be a single sinusoid at the frequency of the square wave. On the otherhand, with a realistic filter several more harmonics will pass through the filter with someattenuation. The effect of losing the higher frequency harmonics amounts to losing the sharpfeatures of a signal such as the corners of a square wave.

209

Hz

Hz

0

-10

-20

-30

G (dB)

dBW

-30

-20

-10

0

10

10.5 1.5 2.0

10.5 1.5 2.0

Input Spectrum

Hz-30

-20

-10

0

10

10.5 1.5 2.0

Output Spectrum

Frequency Response

dBW

Figure 8.9 : If the frequency response is given in decibels, the signal power of the harmonicsat the filter output can be readily found by adding the filter power gain to the signal powerof the harmonic.

210

8.6 Active Filters

The RC filters we have discussed in the previous section are great when a low-cost design ispreferred. An important limitation of the RC filters is that to obtain the desired frequencyresponse, precision resistors and capacitors must be used, which is certainly not an easytask considering the typical tolerances of these elements. Surely, low-tolerance resistorsand capacitors are available, but they are considerably more expensive than run-of-the-millcomponents. Another important limitation of these filters is that it is practically impossibleto make the cut-off frequency variable (e.g. by turning a single knob) because changing thecut-off frequency requires changing all of the filter components simultaneously.

Active filters were developed to address the above limitations. They are called activebecause they are electronic circuits requiring an external voltage source. In return, activefilters can provide power gain in the pass-band.

Another advantage of active filters is that they are very easy to use because they areavailable as integrated circuits. An example is given in Figure 8.10 , which shows the firstpage from the data sheet of a commercial low-pass filter. From the data sheet, we understandthat the filter order is eight and it can be configured to simulate both Butterworth and Besseltype low-pass filters. At the lower left corner of the page you can see the circuit diagramillustrating the connections. Note that the IC does not require any external componentssuch as resistors or capacitors. However, it requires a ±8V split power supply. The cut-offfrequency of the filter is determined by the frequency of a clock signal (i.e. square wave)applied to pin 11. Therefore, we can easily change the cut-off frequency by changing theclock frequency.

211

Figure 8.10 : First page from the data sheet of a commercial active filter.

212

dBW

0

- 4

- 6

- 2

0 10 20 30

dB

0

- 20

- 30

- 10

0 10 20 30

40kHz

40kHz

input

filter

Figure P8.4

Problems

P 8.1 A 10 kHz sinusoidal voltage with an RMSvalue of 1 V is applied to the input portof a Butterworth filter, which has thecut-off frequency of 10 kHz. Find thesignal power at the filter output in Watts.

P 8.2 Find the voltage gain at the cut-off fre-quency of an 8th order Butterworth low-pass filter. Assume Av = 1 at 0 Hz.

P 8.3 Given the power spectrum of the filterinput signal and the filter frequency re-sponse shown in Figure P8.4, find thesignal power at the filter output at 10kHz. Give your result in Watts.

P 8.4 Find the total signal power at the filteroutput in Figure P8.4.

P 8.5 The signal applied to the input port of afirst order Butterworth filter, which hasa cut-off frequency of 1000 Hz has theharmonics at 1, 2 and 3 kHz with the

amplitudes of 10, 8 and 6 V respectively.a) Find the amplitudes of the harmonicsat the filter output; b) Find the totalsignal power at the filter output in Wattsand dBW.

Chapter 9

Transmission and Reception of RadioSignals

In this chapter, we introduce fundamental concepts concerning transmission and receptionof radio frequency (RF) signals. The most familiar application of these concepts is that ofbroadcast radio. Other applications include satellite communication, TV, radar, and cellulartelephones.

A radio transmitter uses a high frequency sinusoidal signal called the carrier, which car-ries the information (e.g. voice or music) to be transmitted. The signal to be transmittedmodifies the carrier in such a way that once the radio signal reaches a radio receiver, theoriginal signal can then be extracted from the carrier. The process of modifying the carrieris called modulation. There are different modulation schemes such as amplitude modulation(AM), frequency modulation (FM) and phase modulation (PM). The process of extract-ing the information from a modulated signal by a radio receiver is called demodulation.Common consumer radios receive commercial broadcast signals transmitted using amplitudemodulation (AM) or frequency modulation (FM). In this chapter, we will learn about AMtransmission and reception.

9.1 Modulation

There are two important reasons for using modulation in broadcasting radio signals. The firstreason has to do with the ability to tune our radios to the station of our choice. Suppose wewant to set up several radio stations broadcasting audio. As we have seen in earlier chapters,audio signals cover a frequency band from 20 Hz to 20 kHz with an effective bandwidth ofabout 20 kHz. If all of our stations attempted to broadcast such low frequency signalssimultaneously using the same frequency band, the result would be similar to a noisy partywhere everyone is talking at once. Modulation avoids this problem by mapping the signalsfrom each station to a different frequency band. Figure 9.1 illustrates this concept with the

213

214

dBW

20 kHz

dBW

20 kHz

dBW

1000 1020 1030 1050 kHz

Station 1

Station 2

modulation

Guard

Bandmodulation

Figure 9.1 : Two radio stations broadcasting audio signals. Modulation is used to mapsignals in a low frequency band to a much higher frequency band.

power spectra of the audio signals transmitted by two radio stations. As shown, a differentfrequency band is assigned to each station and these bands are separated by a guard bandof unused frequencies to avoid interference.

The second reason for using modulation is that if we wanted to broadcast audio signals,the antennas that could efficiently convert such low frequencies (audio frequencies) to elec-tromagnetic energy would be kilometers long. This is due to fact that the optimum antennalength is on the order of λ/2 where λ is the wavelength of the broadcasted signal. Forexample, suppose we want to transmit a 20 kHz sinusoid. The wavelength of the signal is,

λ =c

f=

3× 108m/s

20, 000Hz= 15, 000m = 15km

where c is the speed of light. A typical antenna needed for this wavelength would then be onthe order of λ/2 = 7.5km which of course is highly impractical. Electromagnetic signals inthe radio-frequency (RF) spectrum have frequencies ranging roughly from several hundredkHz to several hundred MHz. These frequencies can be detected using antennas of reasonablelengths. You will study these phenomena in future electromagnetics classes and others onantennas and radio frequency engineering.

215

v1 (t) v

2 (t)

v1 (t) . v

2 (t)

Figure 9.2 : An analog multiplier with two input signals.

9.2 Amplitude Modulation

While amplitude modulation may not be the most efficient modulation method, it is stillwidely used, it is relatively easy to explain and it beautifully demonstrates the essence oftransmitting signals in air. Before we begin our discussion on this subject, we need to look atmultiplication of periodic signals and learn how to analyze the resulting waveforms in bothtime and frequency domain. We will begin our discussion with multiplying two sinusoidsand later we will extend our scope to more complex waveforms.

9.2.1 Analog Multiplier

A multiplier is an electronic circuit, which produces a signal proportional to the product oftwo input signals as shown in Figure 9.2 . The output signal of a multiplier can be expressedas

vout(t) ∝ v1(t)v2(t) (9.2.1)

which is a voltage waveform. It is important to note that we have used the proportionalitysign to avoid a unit conflict between the two sides of the equation. We can also write 9.2.1as

vout(t) = ηv1(t)v2(t) (9.2.2)

where η is a constant with a unit of V −1. In this chapter we will assume that η is alwaysunity.

Multiplication requires a complex electronic circuit with many transistors, resistors andcapacitors. Fortunately, practical multipliers are available in integrated circuit form andthey are very easy to use requiring perhaps a few external components.

216

0 0.005 0.01 0.015 0.02 0.025 0.03

−1

−0.5

0

0.5

1

Time (s)

Vol

tage

(V

)

Figure 9.3 : Signal obtained by multiplying two sinusoids.

9.2.2 Multiplication of Sinusoids

Let’s first consider the special case of multiplying two sinusoidal waveforms,

v1(t) = A1 cos(2πf1t + θ1)

and

v2(t) = A2 cos(2πf2t + θ2)

The multiplier output signal can be expressed as

v1(t)v2(t) = A1A2 cos(2πf1t + θ1) cos(2πf2t + θ2) (9.2.3)

Figure 9.3 shows the output signal for f1 = 100 Hz, f2 = 1000 Hz, A1 = A2 = 1 V andθ1 = θ2 = π/3. We can look at this signal as a 1000 Hz sinusoid with a variable amplitude.We can also see that the variations are sinusoidal and occur at the rate of 100 Hz. In otherwords, we are modulating the amplitude of the 1000 Hz sinusoid with the lower frequency,100 Hz sinusoid. Thus, the product carries elements of both sinusoids.

Multiplication is indeed a form of modulation. To verify that it can indeed accomplishthe frequency mapping shown in Figure 9.1 , we have to consider the multiplier output signalin frequency domain. To create the magnitude spectrum of v1(t)v2(t), we need to break itdown to its individual sinusoidal components. To do this, we can apply the trigonometricidentity,

cos α cos β =1

2cos(α + β) +

1

2cos(α− β) (9.2.4)

217

500 1000 Hz

0.5 V

Figure 9.4 : Magnitude spectrum of the signal shown in Figure 9.3 .

to equation 9.2.3, which yields,

v1(t)v2(t) =A1A2

2cos[2π(f1 + f2)t + (θ1 + θ2)] +

A1A2

2cos[2π(f1 − f2)t + (θ1 − θ2)]

The magnitude spectrum resulting from multiplication of two sinusoids is shown in Figure9.4 , where we have used the same parameters of the signal shown in Figure 9.3 . As shown,the product consists of two sinusoids one at 900 Hz and the other at 1100 Hz. We also notethat the phase information is irrelevant since we have no way of using it in creating themagnitude spectrum.

As a second example, let’s consider the product of the signals,

v1(t) = 2 cos(2π100t) + 3 cos(2π200t)

and

v2(t) = 10 cos(2π1000t)

The resulting signal is shown in Figure 9.5 . We can see that similar to the previous example,the amplitude of the high frequency sinusoid is modulated by a signal at a lower frequency.

The magnitude spectrum of the signal can again be obtained by breaking the productv1v2 into its sinusoidal components. Applying equation 9.2.4 yields

v3(t) = 10 cos(2π1100t) + 10 cos(2π900t) +

15 cos(2π1200t) + 15 cos(2π800t)

218

0 0.005 0.01 0.015 0.02 0.025 0.03

−40

−20

0

20

40

60

Time (s)

Vol

tage

(V

)

Figure 9.5 : Product of two signals. The first signal was obtained by adding two sinusoidsat 100 Hz and 200 Hz. The second is a 1000 Hz sinusoid.

The magnitude spectra of v1(t) before multiplication and the signal at the multiplier outputare both shown in Figure 9.6 . Similar to the previous example, the multiplier outputspectrum consists of peaks centered around 1000 Hz, which happens to be the frequencyof the higher frequency sinusoid, v2(t). We can see in Figure 9.6 that the peaks of theoriginal spectrum are shifted to higher frequencies, which is exactly what we want to dowith modulation.We also note that during this process, a mirror image of the spectrum isalso created, which can be filtered out if desired.

The high frequency sinusoid, v2 is referred to as the carrier signal, vc(t). The low fre-quency signal is the modulating signal, vm(t) and it can be of any shape consisting of manyharmonics in the frequency domain.

9.2.3 Amplitude Modulation (AM)

We have seen in the previous section that multiplication is a form of modulation. In am-plitude modulation, the modulating signal, vm(t) is raised to a DC offset, VDC , before itis multiplied with the carrier as shown in Figure ??. The high frequency carrier signal isalways the sinusoid,

vc(t) = Vc cos(2πfct)

while the modulating signal, vm(t) can be music or any other analog signal. The signal atthe multiplier output can then be expressed as

vam(t) = Vc cos(2πfct)[VDC + vm(t)] (9.2.5)

Suppose we use the signal shown in Figure 9.7 as the modulating signal, vm(t) to modulate

219

500 1000 Hz

10 V

500 1000 Hz

2 V

3 V

15 V

Figure 9.6 : Product of two signals. The first signal was obtained by adding two sinusoidsat 100 Hz and 200 Hz. The second is a 1000 Hz sinusoid.

0 0.01 0.02 0.03 0.04−20

−10

0

10

20

Time (s)

Vol

tage

(V

)

Figure 9.7 : A 100 Hz periodic signal.

220

the carrier sinusoid, 10 cos(2π2000t). Shown in Figure 9.8 are four different AM signalsobtained with different DC values added to the modulating signal before it is multipliedwith the carrier according to 9.2.5. We can see that all four signals have the same envelopeidentical to the modulating signal shown in Figure 9.7 . The fact that the modulating signalis preserved in this envelope is the only advantage of AM over other modulation methods.Later in this chapter we will see how we can demodulate the AM signal using an envelopedetector, which was exactly how the early radios worked.

In Figure 9.8 , we can see that the DC offset determines the separation between theupper and lower envelope. Now the question is, can we choose the DC value randomly oris there an optimum value for it? We note that when the DC offset is 4 V, the envelopesare actually butting against each other. Therefore, anything less than 4 V will result inoverlapping envelopes, which will then make it impossible to demodulate the signal usingan envelope detector. Therefore, the DC offset should be chosen to avoid this undesirablesituation. Considering the modulating signal shown in Figure 9.7 , its minimum voltage levelis about - 4 V. Hence, by raising the signal to a DC offset of 4 V, we make sure that thesignal is always positive before it is multiplied with the carrier sinusoid.

The above discussion explains why the DC offset should not be below a certain level.Does this mean that we can choose any DC offset provided it is large enough to preserve theenvelope? To answer this question, we need to look at the AM signal in frequency domain.

9.2.4 AM Signal in Frequency Domain

We begin by assuming that the modulating signal is also a sinusoid, hence, the AM signal isgiven by equation 9.2.5, which can be expanded as

vam(t) = (VDCAc) cos(2πfct) +(AmAc

2

)cos[2π(fc + fm)t] +

(AmAc

2

)cos[2π(fc − fm)t]

using the trigonometric identity in 9.2.4. The signal is composed of three sinusoids. The firstsinusoid is the carrier and it can be shown by a peak at the carrier frequency, fc. The othertwo sinusoids exist at fc ± fm and they are referred to as the upper and lower sidebands.To construct the power spectrum, we find the signal power at each frequency using

P = V 2rms =

[Vp√

2

]2

=V 2

p

2

which yields

221

0 0.01 0.02 0.03 0.04−200

−100

0

100

200

Time (s)

Vol

tage

(V

)

0 0.01 0.02 0.03 0.04−200

−100

0

100

200

Time (s)

Vol

tage

(V

)

0 0.01 0.02 0.03 0.04−200

−100

0

100

200

Time (s)

Vol

tage

(V

)

0 0.01 0.02 0.03 0.04−200

−100

0

100

200

Time (s)

Vol

tage

(V

)

V = 10 VDC

V = 8 VDC

V = 6 VDC

V = 4 VDC

Figure 9.8 : The periodic signal shown in Figure 9.7 is used to modulate the carrier sinusoid,vc(t) = 10 cos(2π2000t). Plots correspond to modulation using different DC offsets.

222

volts

fc - fm fc + fmfc

Frequency (Hz)

Figure 9.9 : Magnitude spectrum of a carrier modulated by a sinusoid of frequency fm.

Pcarrier =

(VDCAc√

2

)2

(9.2.6)

and

Pupper sideband = Plower sideband =

(AcAm

2√

2

)2

(9.2.7)

It is important to note that the DC value only affects the signal power of the carriergiven by 9.2.6, which does not carry any information about the original signal. Therefore,it makes sense to use the minimum possible DC offset for the modulating signal. However,as we have seen in the previous section, there exists a minimum acceptable DC level belowwhich, the AM envelope is not identical to the modulating signal.

Example 9.1. The carrier sinusoid, vc(t) = 10 cos(2π106t) V is modulated by the signal 10 cos(2π100t) V .The modulating signal is given a DC offset of 8 V before it is multiplied with the carrier. Find thetotal power of the signal broadcasted by the radio station. What percentage of this power is usedto transmit the carrier?

Solution:The AM signal can be found using equation 9.2.5 as

vam(t) = 10 cos(2π106t) [8 + 10 cos(2π100t)]

which can be expanded using 9.2.4 as

223

vam(t) = 80 cos(2π106t) + 50 cos[2π(106 + 100)t] + 50 cos[2π(106 − 100)t]

The total transmitted power is given by

PT = Pcarrier + Pupper sideband + Plower sideband

We can find the signal power of each sinusoid using P = V 2RMS . This yields,

PT =802

2+

502

2+

502

2= 3200 + 1250 + 1250 = 5700 W

The percentage of the station power used to broadcast the carrier is given by,

% =32005700

× 100 = 56.14%

which shows that more than half of the station power is used to broadcast the carrier sinusoid,which does not carry any information about the modulating signal.

‖ X ‖

9.2.5 Modulation Index

Modulation index, m is a measure of the efficiency of the modulation. By efficiency, we referto signal power wasted to broadcast the carrier sinusoid. For simplicity, lets assume that themodulating signal is also a sinusoid. The resulting AM signal is shown in Figure 9.10 . Themodulation index is defined as

m =Vmax − Vmin

Vmax + Vmin

(9.2.8)

where Vmax and Vmin are the maximum and minimum voltage levels of the envelope asillustrated in Figure 9.10 . Modulation index is usually given as a percentage. The aboveformula is valid for any signal shape.

In Figure 9.10 , the minimum acceptable DC value has to be equal to Vm yielding a Vmin

level of zero. This results in a modulation index of m = 1 or 100% modulation. From thediscussion above, this is the best modulating index we can have since the power wasted onthe carrier is minimized. Any DC offset less than this would result in overlapping envelopesand it must be avoided to be able to use an envelope detector to demodulate the signal.

Example 9.2. The AM signal shown in Figure 9.11 was obtained using a DC offset of 5 V. Findthe modulation index if vc(t) = 10 cos(2π104t) V and vm(t) = 3 cos(2π1000t) V .

224

Time

vam(t)

Vmin

Vmax

VDC

Figure 9.10 : Determining the Modulation Index

Solution:The AM signal is given by,

vam(t) = 10 cos(2π104t) [5 + 3 cos(2π1000t)]

The AM signal envelope reaches its maximum voltage level, Vmax when both cosines are equal tounity, hence Vmax = 80 V . The minimum voltage level is reached when the carrier cosine is +1and the modulating signal cosine is -1. This yields, Vmin = 20 V . The modulation index can becalculated from 9.2.8 as

m =80− 2080 + 20

× 100 = 60%

‖ X ‖

9.2.6 Transmission of Non-Sinusoidal Signals

In AM, the carrier signal is always a sinusoid, however, the modulating signal can be of anyshape. In this section, our objective is to demonstrate that the method we have learned inthe previous section can be extended to complex signals with many harmonics.

We begin our discussion by considering a modulating signal consisting of two sinusoidsas given below.

225

0 1 2 3 4

x 10−3

−100

−50

0

50

100

Time (s)

Vol

tage

(V

)

Figure 9.11 : The AM signal was obtained by raising the modulating signal, vm(t) =3 cos(2π1000t) V to a DC offset of 5 V . The carrier was 10 cos(2π104t)

vm(t) = A1 cos(2πf1t + θ1) + A2 cos(2πf2t + θ2)

The resulting signal can be expressed as

vam(t) = Vc cos(2πfct)[VDC + A1 cos(2πf1t + θ1) + A2 cos(2πf2t + θ2)]

Expanding the above equation using the trigonometric identity in 9.2.4, we can show thatthe spectrum will consist of peaks at the frequencies of fc , fc ± f1 and fc ± f2 . In otherwords, for every peak in the original signal, we expect to have two new peaks centered aroundthe carrier frequency.

It should be clear at this point that no matter how complicated the modulating signal is,it can be broken down to its harmonics and each harmonic results in a peak at fc +fn, wherefn is the frequency of the nth harmonic. This way, the entire spectrum can be duplicated inupper and lower sidebands. Figure 9.12 shows spectra of three stations. All three stationsbroadcast the carrier, upper and lower sidebands.

Example 9.3. An AM station is broadcasting at 1.2 MHz. If a guardband of 20 kHz is leftbetween two stations, what is the broadcasting frequency of the nearest station broadcasting at ahigher frequency? Assume the radio stations are broadcasting audio signals within 20 Hz to 20 kHz.

Solution:The broadcasting frequency of the station can be found by sketching the power spectra of thestations as shown in Figure 9.13 . The separation between the frequencies of the two stations can

226

dBW

fc1

fc2

fc3

Figure 9.12 : Three AM Stations

dBW

1200 1220 1240 1260 12801180 kHz

Guard

band

Figure 9.13 : Power spectra of the radio stations.

be found by adding the upper sideband of the first station, guardband between the two stationsand the lower sideband of the second station as 60 kHz. Since the first station is operating at 1200kHz, the next station must have the broadcasting frequency of 1260 kHz. ‖ X ‖

9.3 AM Demodulation

A simple AM radio can be constructed using a technique called “envelope” detection. Thisdetector is a simple combination of circuit elements that we have already used, a diode, aresistor and a capacitor. In order to explain how the circuit works, we will consider thecircuit in parts.

Consider Figure 9.14 , which shows a half wave rectifier. The input is the modulatedsignal received by the antenna and it is shown in Figure 9.15 (a). We know that whena diode is forward biased, current can pass through the diode. However, a small voltageapproximately equal to the turn-on voltage drops across the diode. In this application, weshall choose a diode with a small turn-on voltage of approximately 0.2 volts. This voltage isso small that in the following discussion, we shall ignore this small voltage drop and assumethat the diode is “on” if the input signal is positive and “off” if the input signal is negative.That is, we shall assume that a forward biased diode is just like an ideal switch with zerovoltage drop across its terminals. Thus, the positive part of the input signal is replicatedexactly, while the negative part is set to zero. The signal across the load resistor is shownin Figure 9.15 (b).

227

vam (t) vR (t)

+

-

Figure 9.14 : Half-wave rectifier studied in earlier Chapters.

Time

Time

(a)

(b)

vam(t)

vR(t)

Figure 9.15 : a) AM signal and b) half-wave rectified signal

228

vam (t) vR (t)

+

-

Figure 9.16 : Half-wave rectifier with a filter capacitor. This circuit was called the crystaldetector and used by many radio enthusiasts.

Time

vRC(t)

Figure 9.17 : When a capacitor is connected in parallel with the resistor, the output signalis the envelope of the AM signal, which is just the original signal with a DC off-set.

Now let’s add a capacitor in parallel with the resistor. The resulting circuit is shown inFigure 9.16 . The effect of capacitor on the output signal is shown in Figure 9.17 . Duringthe positive cycle of the input signal the diode turns on and the capacitor charges up witha very short time constant until the input waveform begins to decrease. At this time, thediode turns off and the capacitor discharges through the load resistance according to

vL(t) = Vpe−t/RC

where Vp is the peak value of the input signal, and the diode opens at time t = 0. The RCtime constant determines the rate at which the capacitor discharges. Hence, the larger thevalue of capacitance used, the smaller will be the output ripple.

We note that in Figure 9.17 , the output signal obtained with the capacitor is very closeto the envelope of the AM signal. You can see that the original signal can be obtained fromthe envelope by filtering out the DC value of the signal. The ripple will inevitably result

229

Cylindrical

Shaft

Insulated

Wire

Circuit

Symbol

Figure 9.18 : a) A typical inductor is constructed by wrapping an insulated wire around acylindrical shaft, b) circuit symbol of an inductor.

in some distortion, however, considering the fact that the frequency of the carrier is muchlarger than the modulating signal the ripple can be very small.

9.3.1 Inductor and LC Parallel Resonant Circuit

The crystal detector shown in Figure 9.16 does not have the capability to tune to a desiredradio station. Consequently, this simple detector receives all AM signals from many differentstations in the air. To avoid this, each radio has a tuning circuit which enables the receiverto respond to the signal that has the desired carrier frequency. A simple tuning device can beconstructed using an inductor and a capacitor. You learned about capacitors and RC circuitsin earlier Chapter . In this section, we shall briefly review inductors and the principles of anLC parallel resonant circuit.

The inductor is a passive element that stores energy in the form of magnetic field. Asshown in Figure 9.18 , a typical inductor consists of a cylindrical shaft with insulated wirewrapped around it. The circuit symbol for the inductor is shown in Figure 9.18 .

The voltage, vL(t) across the terminals of an inductor is proportional to the time deriva-tive of the current, iL(t) and is expressed as

vL(t) = LdiL(t)

dt(9.3.1)

where the proportionality constant is called the inductance with units of Henry (H)1. Recallfrom Chapter “Capacitors and RC circuits” that Eq.9.3.1 is very similar to the current-voltage relationship of a capacitor. The difference is for a capacitor, current is proportionalto the derivative of the voltage.

1After Joseph Henry, 1797-1878.

230

Similar to capacitors, inductors store energy. While a capacitor stores energy as an elec-tric charge, as mentioned, an inductor stores energy as a magnetic field. You will learn moreabout this in a later course on electromagnetics. Energy storage in inductors is associatedwith the current through the inductor and is given by

E = LI2

2(9.3.2)

A sudden change in the current through the inductor ( diL/dt ) can cause a large voltageacross its terminals. This is used in car ignition systems to make a spark inside the cylindersto ignite the gas-air mixture. In a car, the 12 volts from the battery is converted to 10,000volts by an inductor, usually called an ignition coil, which creates the spark.

By connecting a capacitor in parallel with an inductor, one can form a parallel resonantcircuit as shown in Figure 9.19 . Let us assume the capacitor and the inductor are initiallyat rest, that is, the capacitor has no charge and the inductor has no current. By closingthe switch, a current flow is established inside the inductor and the capacitor charges. Nowwe open the switch again. The sudden open circuit will cause change in the current whichwill result in a change in voltage because of the relationship of Eq. 9.3.1. The change involtage will result in a change in current because of the capacitor relationship discussedin earlier Chapters. Without going into details, which will be covered in the first circuitscourse, we will state that the energy will oscillate between the capacitor and the inductor.This oscillation is called resonance.

When the voltage across the inductor eventually drops to zero (ideally all the storedmagnetic field inside the inductor is transferred to an electric field inside the capacitor), thecapacitor is then discharged through the inductor causing a current flow in the inductor.Eventually all the energy stored in the capacitor is transferred to the inductor and stored ina magnetic field around the inductor. In an ideal situation where there are no losses in thecircuit, the energy exchange between the inductor and the capacitor will go on indefinitely.Of course, in a real circuit there is some resistance, which will prevent this. A pendulum isa good mechanical analog for the resonant circuit. The rate at which this energy exchangetakes place is the resonance frequency and is a function of the inductance and capacitancevalues as shown by the following formula:

f =1

2π√

LC

where f is the resonant frequency of the circuit in Hertz (Hz), L in the inductance in Henryand C is the capacitance in Farad. At the resonant frequency, the parallel LC combinationappears as an open circuit, therefore, causing a large voltage to appear across it. A resonantcircuit in the diode detector is used to select only one carrier frequency (the one that is equalto LC circuits resonant frequency) among hundreds of signals picked up by the antenna.

231

+LC

Figure 9.19 : A parallel LC resonant circuit

A note on the properties of an antenna is appropriate here. A real antenna has circuitproperties of its own. This is one of the reasons for the effort companies spend in designingspecial antennas to receive particular signals. If the antenna is not designed correctly, it canchange the resonant frequency of the LC tuning circuit. The main parameter that affectsthe circuit is the length of the antenna wire. The other significant factor for a single wireantenna is the shape, e.g., helical, loop etc.

Example 9.4. Calculate the resonant frequency of the circuit shown in Figure 9.19 if L = 1µHand C = 1 nF.

Solution:The resonant frequency can be found using Eq. 9.3.2 as

f =1

2π√

LC=

12π√

10−6H × 10−9F≈ 5MHz

‖ X ‖

9.4 A Simple AM Receiver (Crystal Detector)

Shown in Figure 9.21 is the diagram of a simple AM receiver (crystal detector). This receiveris an envelope detector preceded by a tuning circuit. The tuning circuit limits the receivedsignal to a narrow band of frequencies that is determined by a capacitor and inductor. Thedesign of Figure 9.21 is basically the same as the early crystal radios. In the original crystalradio patented in 1904, the semiconductor crystal was galena (lead sulfide) and the diodewas formed by a metal needle contacting the semiconductor crystal, which forms a diode. A

232

crystal radio from the 1920s is shown in Figure 9.20 .

Figure 9.20 : A crystal radio from 1920s.

A simple crystal radio consists of four important components. These are,

• Antenna (to receive the RF signal)

• LC resonant circuit (to tune to a radio station)

• Diode (to half-wave rectify the AM signal)

• RC low-pass filter (to create the AM envelope)

The antenna is usually a long wire for AM radio bands and picks up the electromagneticenergy radiated by broadcasting stations. The LC resonant circuit tunes (selects) the desiredcarrier frequency out of hundreds of frequencies broadcasted by radio stations and rejectsunwanted stations. The resonant frequency can be easily varied by using a variable capacitorin the LC circuit. The diode rectifies the signal and the RC low pass filter removes the highfrequency carrier component. The output of the envelope detector can either be connectedto a headphone or to an audio amplifier for further amplification. The DC value of the AMenvelope is also removed during this final stage.

233

vam (t)

vRC (t)

+

-

antenna

Figure 9.21 : Crystal Detector with an LC resonance circuit. This circuit was used toreceive AM signals during the early days of radio engineering.

Common consumer receivers are much more complicated than a simple crystal detector.The most common receiver today is the superheterodyne receiver, which was invented in1917 by Edwin Armstrong. The superheterodyne receiver uses several stages of modulationto move the received signal to frequency ranges where more optimal processing can be done,then down to the standard audio range. The modulation to an intermediate frequency (IF)allows the economical use of high precision filters and amplifiers2.

2The concepts of modulation and filtering needed for the super heterodyne receiver will be discussed inthe junior level signals and systems course and the senior level communications course.

234

Problems

P 9.1 Calculate the wavelength of a 100 MHzradio frequency signal (FM Range) infree space.

P 9.2 Find the minimum antenna length re-quired to transmit signals in the 600 kHzto 1000 kHz frequency range.

P 9.3 Calculate the signal power at the carrierfrequency and in the upper and lowersidebands of an AM station with 50%modulation and a total transmitted powerof 2 kW. Assume that a sinusoidal testsignal is broadcasted.

P 9.4 Repeat the previous problem with 80%modulation.

P 9.5 A 1 kHz sinusoidal test signal is broad-casted by an AM radio station operatingat 1 MHz. The test signal has an ampli-tude of 1 V. What is the minimum DCoffset that must be given to the test sig-nal before it is multiplied with the car-rier?

P 9.6 Consider a tuning circuit consisting of a100 µH inductor in parallel with a vari-able capacitor. What should be the ca-pacitance range to be able to tune to theentire AM band (530 kHz to 1650 kHz)?

P 9.7 An AM radio station, whose applicationis limited to broadcasting speech, (e.g.police radios) is operating at a frequencyof 500 kHz. Another nearby AM stationis operating at a higher frequency andit is broadcasting music as well speech.Recall that we can hear music upto 20000Hz. On the other hand, speech is typ-ically limited to a frequency range of5000 Hz. What is the minimum accept-able carrier frequency for the music sta-tion if a guard band of 2 kHz is required

between the two stations?

235

Answers

236

P 1.1. 0.707 ΩP 1.2. a) R1 and R2 are in parallel; b) 15 kΩP 1.3. a) R1 and R2 are in parallel; b) 6 kΩP 1.4. a) R1 and R2, R3 and R5 are in parallel;

b) 5 kΩP 1.5. a) R1 and R2, R3 and R5, R4 and R6

are in parallel; b) 3.33 kΩP 1.6. a) R3, R4, R5 and R6 are all in parallel,

R1 and R2 are in parallel with a conductingwire, hence they are shorted out; b) 2.5 kΩ

P 1.7. 1.14 mA; V1 = 1.14 V , V2 = 2.5 V ,V3 = 5.36 V

P 1.8. 8.81 VP 1.9.

P 1.10.P 1.11.P 1.12.P 1.13. a) 6 V b) -3 VP 1.14. a) ∞ b) 1 kΩ c) 3 kΩP 1.15.P 1.16. R = V 3

03I0V 2

P 1.17. I = 0.81 mA; V1 = 8.1 V ; V2 = 0.9 VP 1.18.P 1.19. 9.2 mAP 1.20. 8.3 kΩP 1.21.P 1.22. on; 2.1 mAP 1.24. on; 27.6 µAP 1.25.P 1.26. on; 7.6 mAP 2.1.P 2.2.P 2.3.P 2.5.P 2.7.P 2.8.P 2.9.

P 2.10.P 2.11.P 2.14.P 2.16.P 2.18.P 2.19.

P 2.21.P 2.23.P 2.24.P 3.1. Vp = 10 V ; Vp−p = 20 V ; VDC = 0 V

for all four waveforms.P 3.2. a)f = 100 Hz, T = 0.01 s; b)f =

100 Hz, T = 0.01 s; c) f = 200 Hz, T =0.05 s; d) f = 100/2π Hz, T = 2π/100 s

P 3.3. a) 0 V ; b) 20 V ; c) −20 V ; d) −20 VP 3.4. a) 0; b)π/4 rad; c)π/4±π rad; d)−π/4±

π radP 3.9. 100 Hz and 5 VP 4.3. R = 144 Ω.P 4.5.

(a) VRMS = VDC = 0 V, VRMS = 10/√

2 V

(b) VRMS = VDC = 0 V, VRMS = 10/√

2 V

(c) VRMS = VDC = 0 V, VRMS = 10/√

2 V

(d) VRMS = VDC = 0 V, VRMS = 10/√

2 V

(e) VRMS = VDC = 2 V, VRMS =√

54 V

(f) VRMS = VDC = 2 V, VRMS = 2 V

P 4.11.

(a) Vdc = 12√

2/π V

(b) VRMS = 6√

2 V

P 4.12.

(a) VRMS = 120V

(b) IRMS = 1.41 A

(c) Power Factor = 1/√

2

(d) Apparent Power = 120.√

2 VA

(e) Real Power = 120 W.

P 4.13. Power Factor = 0.8.P 4.14.

(a) Power Factor = 0.707

237

(b) Apparent Power = 0.4 VA

(c) Real Power = 0.282 W.

P 4.15.

(a) vL(t) = (−10V ) sin(1200t + 6)

(b) VRMS = 10/√

2 V,IRMS = 70.7 mA

(c) Power Factor = 0

(d) Apparent Power = 0.5 VA

(e) Real Power = 0 W.

P 4.16.

(a) R = 240 Ω

(b) P = 30 (3/8 + 1/(4π)) W.

P 6.1. a) 1.2 V ; b) 1.2/√

2 ' 0.849 VP 6.2. 0.5 WP 6.3. 2.5 WP 6.4. a) 200 W; b) 40 dBP 6.5. 4000P 6.6. a) 10 kHz; b) 10 kHz; c) 50 kHz and

100 kHz; d) 0.52%P 6.7. a) 12/

√2 V ; b) 0.042 V

P 6.8. 7.3 WP 6.9. 1 mAP 7.1. a) inverting, b) - 10 , c)1/

√2 V

P 7.2. 1 kΩP 7.3. a) non-inverting, b) 11 , c)∞P 7.4. a) - 10, b) noneP 7.5. a) - 10, b) noneP 7.6. a) i1 = 0, i2 = 0, if = −vout/Rf , in =

vs/Rn; b)vout = −(Rf/Rn)vs, c) Rn

P 7.7. a) 11, b) none, c) Since the current flow-ing into the non-inverting input of the op-amp is negligibly small, the input resistanceis infinitely large regardless of the value ofthe resistor connected.

P 7.8. vout = −4.7v1 − 1.4v2

P 7.9. V1 = 3.83V ; V2 = 12.68V .P 7.10. 0.75W

P 7.11. vout = −22vs + 23VDC

P 7.12. vout = Rf

Rn(v2 − v1)

P 7.13. 6.16WP 7.14. vout =

√−vs

P 7.15. R1R1+Rf

[R2C

dvoutdt + vout

]= vs

P 8.1. 0.5 WP 8.2. 1/

√2

P 8.3. 1 WP 8.4. 1.066 WP 8.5. a)7.071 V, 3.578 V, 1.897 V; b) 33.2 WP 9.1. 3 mP 9.2. 250 mP 9.3. 1.78 kW and 0.11 kWP 9.4. 1.52 kW and 0.24 kWP 9.5. 1 VP 9.6. 93 pF to 0.9 nFP 9.7. 527 kHz

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