CHAPTER # 3DATA AND SIGNALS

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Introduction One of the major functions of physical

layer is to move data in the form of electromagnetic signals across a transmission medium.

Thus, the data must be transformed to electromagnetic signals to be transmitted.

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1. Analog and Digital Data Data can be analog or digital. The term analog refers to information that is

continuous e.g. analog clock hh:mm:ss

The term digital data refers to information that has discrete states.

e.g. digital clock hh:mm

1. Analog data take on continuous values. 2. Digital data take on discrete values.

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Analog and Digital Signals

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2. Periodic and Nonperiodic Signals

Both analog and digital signals can take one of two forms: periodic or nonperiodic

A periodic signal Completes a pattern within a measureable

time frame. Repeats that pattern over subsequent

identical period A nonperiodic signal

Changes without exhibiting a pattern or cycle that repeats over time.

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Periodic and Nonperiodic Signals

In data communications, we commonly use:

Periodic analog signals ( because they need less bandwidth).

and nonperiodic digital signals ( because they can represent variation in data)

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A. Periodic Analog SignalsPeriodic analog signals can be classified as simple or composite.

A simple periodic analog signal, a sine wave, cannot be decomposed into simpler signals.

A composite periodic analog signal is composed of multiple sine waves.

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1) Sine Waves The sine wave is the most fundamental form

of a periodic analog signal.

A sine wave is represented by three parameters: Peak amplitude, Frequency, and Phase.

1. Peak amplitude: it is the absolute value of the highest intensity. It is normally measured in volts.

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2. Frequency: it refers to the number of periods in 1 s. It is formally expressed in Hertz (Hz).

Period is the amount of time, in seconds, a signal needs to complete one cycle (the completion of one full pattern).

Therefore , frequency and period are the inverse of each other.

Note : Frequency is the rate of change with respect to time.

Change in a short span of time means high frequency. Change over a long span of time means low

frequency. If a signal does not change at all, its frequency is zero

Sine Waves

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Units of period and frequency

TABLE 3.1

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Example#1 The power we use at home has a

frequency of 60 Hz. The period of this sine wave can be determined as follows:

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Example# 2Express a period of 100 ms in microseconds. Solution From Table 3.1 we find the equivalents of

1 ms (1 ms is 10−3 s) and 1 s (1 s is 106 μs). We make the following substitutions:.

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Example# 3The period of a signal is 100 ms. What is its frequency in kilohertz? Solution First we change 100 ms to seconds, and

then we calculate the frequency from the period (1 Hz = 10−3 kHz).

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Sine Waves 3. Phase:

It describes the position of the waveform relative to time 0.

It is measured in degree or radian

To look to the phase is in term of shift or offsit:1. A sine wave with a phase 0° is not shifted.2. A sine wave with a phase 90° is shifted to the

left by ¼ cycle.3. A sine wave with a phase 180° is shifted to

the left by ½ cycle.

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Three sine waves with the same amplitude and frequency, but different phases

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The figure below show Two signals with the same phase and frequency, but different

amplitudes

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Two signals with the same amplitude and phase, but different frequencies

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Wavelength Wavelength binds the period or frequency of the

simple sine wave to the propagation speed of the medium.

Wavelength depends on both the frequency and the medium. Wavlength = propgation speed X period =

progation speed/ frequency

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Example# 1In a vacuum, light is propagated with a speed of 3 X 108 m/s. (that speed is lower in air and cable.) . The frequency of red light is 4 X 1014 Wavelength is normally measured in

micrometers.Solution Wavelength= c/f = (3 X 108 ) / (4 X

1014)

= 0.75 X 10-6 m= 0.75 μm

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Time and Frequency Domain

A complete sine wave in the time domain can be represented by one single spike in the frequency domain.

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ExampleThis example Shows three sine waves, each with different amplitude and frequency. All can be represented by three spikes in the frequency domain.

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2) Composite Signals A single-frequency sine wave is not useful in

data communications; we need to send a composite signal, a signal made of many simple sine waves. e.g. if we use single sine wave to convey a

conversation over the telephone. It would just hear a buzz.

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2) Composite Signals According to Fourier analysis, any composite signal

is a combination of simple sine waves with different frequencies, amplitudes, and phases.

A composite signals can be periodic or non periodic. A periodic composite signal can be decomposed into a

series of simple sine waves with discrete frequencies ( with integer values [1,2,3 ,….] ).

A nonperiodic composite signal can be decomposed into a combination of an infinite number of simple sine waves with continuous frequencies. ( with real value)

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A composite periodic signal

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Explanation

The previous figure shows a periodic composite signal with frequency f. This type of signal is not typical of those found in data communications.

We can consider it to be three alarm systems, each with a different frequency.

The analysis of this signal can give us a good understanding of how to decompose signals.

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Decomposition of a composite periodic signal in the time and frequency domains

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The time and frequency domains of a nonperiodic signal

In this case, the composite signal cannot be periodic, because that implies that we are repeating the same word or words with exactly the same tone.This can be the signal created by a microphone or a telephone set when a word or two is pronounced.

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Bandwidth The bandwidth of a composite signal is the

difference between the highest and the lowest frequencies contained in that signal.

e.g. if a composite signal contain frequencies between 1000 and 5000, its bandwidth is 5000-1000 = 4000

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The bandwidth of periodic and non- periodic composite signals

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Example#1 If a periodic signal is decomposed into five sine

waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is its bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V.

Solution Let fh be the highest frequency, fl the lowest

frequency, and B the bandwidth. Then

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Example#2 A periodic signal has a bandwidth of 20 Hz. The

highest frequency is 60 Hz. What is the lowest frequency? Draw the spectrum if the signal contains all frequencies of the same amplitude.

Solution Let fh be the highest frequency, fl the lowest frequency,

and B the bandwidth. Then

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Example#3A nonperiodic composite signal has a bandwidth of 200 kHz, with a middle frequency of 140 kHz and peak amplitude of 20 V. The two extreme frequencies have an amplitude of 0. Draw the frequency domain of the signal.Solution The lowest frequency must be at 40 kHz and the

highest at 240 kHz. Figure 3.15 shows the frequency domain and the bandwidth.

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B- DIGITAL SIGNALS

In addition to being represented by an analog signal, information can also be represented by a digital signal. For example, a 1 can be encoded as a positive voltage and a 0 as zero voltage.

A digital signal can have more than two levels. In this case, we can send more than 1 bit for each level.

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Figure 3.16 Two digital signals: one with two signal levels and the other with four signal levels

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A digital signal has eight levels. How many bits are needed per level? We calculate the number of bits from the formula

Example # 1

Each signal level is represented by 3 bits.

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A digital signal has nine levels. How many bits are needed per level?

We calculate the number of bits by using the formula: Log2 L= number of bits in each levelLog2(9)=3.17bits.

However, this answer is not realistic. The number of bits sent per level needs to be an integer as well as a power of 2. For this example, 4bits can represent one level.

Example #2

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Bit rate and bit interval

Most digital signals are nonperiodic, frequency and period are not appropriate. Another terms instead of frequency is bit rate and instead of period: bit interval(bit duration)Bit rate: number of bits per second bpsBit interval=1/bit rateBit length =propagation speed x Bit interval

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Assume we need to download text documents at the rate of 100 pages per minute. What is the required bit rate of the channel?

SolutionA page is an average of 24 lines with 80 characters in each line. If we assume that one character requires 8bits The bit rate is:

=100x24x80x8/60=25.6Kbps

Example

Fourier analysis can be used to decompose a digital signal

o If the digital signal is periodic (rare in data communications), the decomposed signal has a frequency domain representation with an infinite Bandwidth and discrete frequencies.

o If it is nonperiodic, the decomposed signal still has infinite Bandwidth, but the frequencies are continuous.

Digital Signal as a composite Analog Signal

A digital signal is a composite analog signal with an infinite bandwidth.

Note

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Figure 3.17 The time and frequency domains of periodic and nonperiodic digital signals

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Transmission of Digital Signals

How can we send a digital signal from point A to point B?

We can transmit a digital signal by using one of two

different approach :

1. Baseband transmission

2. Broadband transmission

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Means sending a digital signal over a channel without changing the digital signal to an analog signal

Base band transmission required a low-pass channel (channel with a B-W that starts from zero)

1. Baseband transmission

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Figure 3.19 Bandwidths of two low-pass channels

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Figure 3.20 Baseband transmission using a dedicated medium

Baseband transmission of a digital signal that preserves the shape of the digital signal is possible only if we have a low-pass channel with an infinite or very wide bandwidth.

Note

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Means changing the digital signal to an analog signal for transmission. Modulation use a band-pass channel (a channel with a B-W that doesn't start from Zero). This type of channel is more available than a low-pass channel.

2. Broadband Transmission (modulation)

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If the available channel is a bandpass channel, we cannot send the digital signal

directly to the channel; we need to convert the digital signal to an

analog signal before transmission.

Note

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Figure 3.24 Modulation of a digital signal for transmission on a bandpass channel

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An example of broadband transmission using modulation is the sending of computer data through a telephone subscriber line, the line connecting a resident to the central telephone office. These lines are designed to carry voice with a limited bandwidth. The channel is considered a bandpass channel. We convert the digital signal from the computer to an analog signal, and send the analog signal. We can install two converters to change the digital signal to analog and vice versa at the receiving end. The converter, in this case, is called a modem.

Example

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3-4 TRANSMISSION IMPAIRMENTSignals travel through transmission media, which are not perfect. The imperfection causes signal impairment. • This means that the signal at the beginning of the

medium is not the same as the signal at the end of the medium. What is sent is not what is received.

Three causes of impairment are attenuation, distortion, and noise.

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1. Attenuation – a loss of energywhen Signal travels through a medium, it losses some of its energy in overcoming the resistance of the medium. To compensate for this loss, amplifiers are used to amplify the signal.

Decibel: Measure the relative power(attenuation)dB=10 log10 P2/P1

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Suppose a signal travels through a transmission medium and its power is reduced to one-half. This means that P2 is (1/2)P1. In this case, the attenuation (loss of power) can be calculated as

Example #1

A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.

See the book for more Examples

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2. DistortionDistortion : means that signal changes its form or shape.

Each signal component has its own propagation speed through the medium and therefore ,its own delay in arriving final destination

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3. Noises •Thermal noise: is the random motion of electrons in a wire which creates an extra signal not originally sent by the transmitter

•Induced noise: Comes from sources such as motors and appliances.

•Crosstalk noise: Is the effect of one wire on the other.

Impulse Noise: is a spike ( a signal with high energy in a very short time) that comes from power lines, lighting and so on.

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Figure 3.29 Noise

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SNR: ratio between signal power to the noise power

o A high SNR: means the signal is less corrupted by noise

o A low SNR: means the signal is more corrupted by noise.

SNR can be described in db units

SNR db=10 log10 SNR

Signal-to-Noise Ratio

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Figure 3.30 Two cases of SNR: a high SNR and a low SNR

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The power of a signal is 10 mW and the power of the noise is 1 μW; what are the values of SNR and SNRdB ?

SolutionThe values of SNR and SNRdB can be calculated as follows:

Example

SNRdb= 10 log10 10,000 = 40

3.5 Data Rate Limits58

◻ A very important consideration in data communications is how fast we can send data, in bits per second, over a channel. Data rate depends on three factors:⬜ The bandwidth available⬜ The level of the signals we use⬜ The quality of the channel (the level of noise)

Two theoretical formulas were developed to calculate the data rete :

1. By Nyquist for a noiseless channel 2. By Shannon for a noisy channel

Nyquist Theorem (noiseless channel)

59 The relation between bandwidth and data rate in a noiseless channel (throughput)⬜Maximum data rate = 2 x bandwidth x log2 L where L: No of signal levels used to represent data

Example : Consider a noiseless channel with a bandwidth of 3 KHz

transmitting a signal with two signal levels. The maximum bit rate can be calculated as:

◻ Throughput =2*3000 log2 2 = 6000 bps.

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We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need?SolutionWe can use the Nyquist formula as shown:

Example

Since this result is not a power of 2, we need to either increase the number of levels or reduce the bit rate. If we have 128 levels, the bit rate is 280 kbps. If we have 64 levels, the bit rate is 240 kbps.

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Shannon Theorem (Noisy Channel)

The maximum throughput of a noisy channel of bandwidth B with a signal to noisy ratio of S/N is:Maximum throughput = B log2(1+S/N) bps.

Shannon CapacityCapacity = bandwidth x log2(1+SNR)is the capacity of the channel in bps ( max data rate)

Example :Telephone line Bandwidth=3kHz; S/N=30 dB => Max throughput = 3000 * log2(1+1000) =~ 30.000 bps = 28.8 kbps

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We have a channel with a 1-MHz bandwidth. The SNR for this channel is 63. What are the appropriate bit rate and signal level?

SolutionFirst, we use the Shannon formula to find the upper limit.

Example (using both limits)

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The Shannon formula gives us 6 Mbps, the upper limit. For better performance we choose something lower, 4 Mbps, for example. Then we use the Nyquist formula to find the number of signal levels.

Example (continued)

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The Shannon capacity gives us the upper limit; the Nyquist formula tells us how many

signal levels we need.

Note

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3-6 PERFORMANCE

One important issue in networking is the performance of the network—how good is it?

1. Bandwidth

2. Throughput

3. Latency (Delay)

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In networking, we use the term bandwidth in two contexts.❏The first, bandwidth in hertz, refers to the range of frequencies in a composite signal or the range of frequencies that a channel can pass.

❏The second, bandwidth in bits per second, refers to the speed of bit transmission in a channel or link.

1. Bandwidth

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Is a measure of how fast we can actually send data through a network.

The bandwidth is potential measurement of a link; the throughput is an actual measurement of how fast we can send data.

Throughput less than Bandwidth

2. Throughput

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A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network?

SolutionWe can calculate the throughput as

Example

The throughput is almost one-fifth of the bandwidth in this case.

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Latency defines how long it takes for an entire message to completely arrive at the destination from the time the first bit is sent out from the source

3. Latency ( Delay)

Latency (Delay) = propagation time + transmission time+queuing time + processing time

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• It is the time required for a bit to travel from the source to the destination

• Propagation speed depend on the medium and on the frequency of the signal

Example : light propagate by 3x108m/s in vacuum. It is lower in air ; it is much lower in cable.

1) Propagation time

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• It is the time required for transmission of a message .

• It depends on the size of the message and the bandwidth of the channel.

2. Transmission time

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What are the propagation time and the transmission time for a 2.5-kbyte message (an e-mail) if the bandwidth of the network is 1 Gbps? Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 108 m/s.

SolutionWe can calculate the propagation and transmission time as shown on the next slide:

Example #1 :

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Note that in this case, because the message is short and the bandwidth is high, the dominant factor is the propagation time, not the transmission time. The transmission time can be ignored.

Example #1 (continued)

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What are the propagation time and the transmission time for a 5-Mbyte message (an image) if the bandwidth of the network is 1 Mbps? Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 108 m/s.

SolutionWe can calculate the propagation and transmission times as shown on the next slide.

Example # 2

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Note that in this case, because the message is very long and the bandwidth is not very high, the dominant factor is the transmission time, not the propagation time. The propagation time can be ignored.

Example #2 (continued)

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• It is the time needed for each end device to hold the message before it can be processed.

• It changes with the load imposed on the

network, if there is heavy traffic on the network , the queuing time increases.

3. Queuing time

McGraw-Hill © The McGraw-Hill Companies, Inc., 2000

DIGITAL AND ANALOG TRANSMISSION

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Digital modulation and multiplexing

• Baseband transmission: the signal occupies frequencies from zero up to a maximum that depends on the signaling rate.• It is common for wires.

• Passband transmission: the signal occupies a band of frequencies around the frequency of the carrier signal. • It is common for wireless and optical channels.

• Multiplexing: Channels are often shared by multiple signals.

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Line codes: (a) Bits, (b) NRZ, (c) NRZI, (d) Manchester, (e) Bipolar or AMI.

1. Baseband transmission

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Clock Recovery

• The receiver must know when one symbol ends and the next symbol begins to correctly decode the bits.

• One strategy is to send a separate clock signal to the receiver.

1. NRZ: using a positive voltage to represent 1 and a negative voltage to represent 0

2. Manchester: mix the clock signal with the data signal by XORing them together.

3. NRZI: coding a 1 as a transition and a 0 as no transition, or vice versa.

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4. 4B/5B: Every 4 bits is mapped into a5-bit pattern with a fixed translation table. The five bit patterns are chosen so that there will never be a run of more than three consecutive 0s.

5. Balanced Signals: Signals that have as much positive voltage as negative voltage even over short periods of time are called balanced signals.

6. Bipolar encoding: represent a logical 1, (say +1 V or −1 V) with 0 V representing a logical zero. To send a 1, the transmitter alternates between the +1 V and −1 V levels.

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2- Passband Transmission

• We can take a baseband signal that occupies 0 to B Hz and shift it up to occupy a passband of S to S +B Hz.

• Digital modulation is accomplished with passband transmission by regulating or modulating a carrier signal that sits in the passband.

1. ASK (Amplitude Shift Keying), two different amplitudes are used to represent 0 and 1.

2. FSK (Frequency Shift Keying), two or more different tones are used.

3. PSK (Phase Shift Keying): the carrier wave is systematically shifted 0 or 180 degrees at each symbol period.

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Passband Transmission

(a) A binary signal. (b) Amplitude shift keying. (c) Frequency shift keying. (d) Phase shift keying.

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Multiplexing

1- Frequency Division Multiplexing

FDM: divides the spectrum into frequency bands, with each user having exclusive possession of some band.

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2- Time Division Multiplexing• TDM : The users take turns (in a round-robin fashion), each one

periodically getting the entire bandwidth for a little burst of time.

• Bits from each input stream are taken in a fixed time slot and output to the aggregate stream.

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3 - Code Division Multiplexing• CDM : is a form of spread spectrum communication.

(a) Chip sequences for four stations. (b) Signals the sequences represent

McGraw-Hill © The McGraw-Hill Companies, Inc., 2000

CHAPTER # 7TRANSMISSION MEDIA

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Transmission medium and physical layer

7.88

• A transmission media defined as anything that carry

information between a source to a destination

• Located below the physical layer and are directly controlled

by the physical layer

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Classes of Transmission Media

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7-1 GUIDED MEDIA

Guided media, which are those that provide a conduit from one device to another, include twisted-pair cable, coaxial cable, and fiber-optic cable.

Twisted –pair cables and coaxial cable:use metallic (copper) conductors that transport signals in the form of electric currentOptical fiber : transport signals in the form of the light

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1. Twisted-pair cable

• One of the wire used to carry signal and the other as a ground. The receiver uses the difference between the two.• If the two wires are parallel, the effect of interference

noise and crosstalk is big • Twisting the pair of wire balance the effect of unwanted

signal and reduce it.

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Applications of Twisted pair

Used in 1. Telephone lines to provide voice and data channels

2. The DSL lines that are used by the telephone companies to provide high-data-rate connections

3. Local area networks, such as 10-base-Tand 100base-T

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2. Coaxial cable

• Coax cable carries signals of higher frequency ranges than those in Twisted pair cable because the two media are constructed quite differently.

• The outer conductor serves both as a shield against noise and as second conductor, which complete the circuit

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Applications of coaxial cable

1.Analog telephone network where a single cable could carry 10,000 voice signals. Later it was used in Digital telephone networks where cable can carry 600Mbps 2.Cable TV network: hybrid network use coaxial cable only at the network boundaries , near the consumer.

3.Traditional Ethernet LANs.

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3. Fiber Optic Cable

• Is made of glass or plastic and transmit signals in the form of light.

• Light travels in a straight line as long as it is moving through a single uniform substance. If a ray of light traveling through one substance enters another substance of different density , the ray change direction as shown:

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Optical fiber

Fiber Optical : uses reflection to guide light through a channel. A glass or plastic core is surrounded by a cladding of less dense glass or plastic

Back to the book for advantages and disadvantages

7.97

Used in :

1.Cable TV network: hybrid network use a combination of optical fiber and coax cable. Optical provides the backbone while coaxial cable provide the connation to the user.

2.Local area networks such as ( fast Ethernet)

3.Backbone networks because its wide bandwidth

Applications for Fiber Optic cable

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Propagation modes using fiber optics

• Multimode Fiber: any light ray incident on the boundary above the critical angle will be reflected internally, many different rays will be bouncing around at different angles.

• Single-mode Fiber: light can propagate only in a straight line, without bouncing.

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Fiber cable composition

(a) Side view of a single fiber.(b) End view of a sheath with three fibers.

Core: 50 microns for multi-mode, 8-10 microns for single modeCladding: glass with a lower refraction index, to keep the light

in the core

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7- 2 UNGUIDED MEDIA - wireless

• Unguided media transport electromagnetic waves without using a physical conductor.

• Signals are normally broadcast through free space and thus are available to anyone who has a device capable of receiving them

• Unguided signals can travel from the source to destination in several ways:

1. Ground propagation2. Sky propagation3. Line – of – sight propagation

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7.102

The Electromagnetic Spectrum

The electromagnetic spectrum and its uses for communication.

7.103

1. Radio Transmission

(a) In the VLF, LF, and MF bands, radio waves follow the curvature of the earth.

(b) In the HF band, they bounce off the ionosphere.

7.104

• Frequency ranges: 3 KHz to 1 GHz• Omnidirectional • Susceptible to interference by other antennas using same

frequency or band• Ideal for long-distance broadcasting• May penetrate walls• Apps: AM and FM radio, TV, maritime radio, cordless

phones, paging

1. Radio Transmission

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• Frequencies between 1 and 300 GHz• Unidirectional.• Narrow focus requires sending and receiving antennas to be

aligned.

• Issues:

• Line-of-sight (curvature of the Earth; obstacles)• Cannot penetrate walls

2. Microwaves

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Similar to terrestrial microwave except the signal travels from a ground station on earth to a satellite and back to another ground station.

Satellite receives on one frequency, amplifies or repeats signal and transmits on another frequency

Satellite is relay station Applications

Television Long distance telephone Private business networks

2. Satellite Microwaves

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Communication Satellites The principal satellite bands.

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• Frequencies between 300 GHz and 400 THz.• Short-range communication in a closed area.• High frequencies cannot penetrate walls.• Requires line-of-sight propagation.

• Advantage: prevents interference between systems in adjacent rooms.

• Disadvantage: cannot use for long-range communication or outside a building due to sun’s rays.

3. Infrared

The End 7.109