Chi-Cheng Lin, Winona State University

CS412 Introduction to Computer Networking &

Telecommunication

Theoretical Basis of Data Communication

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Topics

Analog/Digital Signals

Time and Frequency Domains

Bandwidth and Channel Capacity

Data Communication Measurements

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Signals

Information must be transformed into electromagnetic signals to be transmitted

Signal formsAnalog or digitalPeriodic or aperiodic

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Analog/Digital Signals

Analog signalContinuous waveformCan have a infinite number of values

in a range Digital signal

DiscreteCan have only a limited number of

valuesE.g., 0 or 1

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Figure 3.1 Comparison of analog and digital signals

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Periodic/Aperiodic Signals

Periodical signalContains continuously repeated

patternPeriod (T): amount of time needed for

the pattern to complete Aperiodical signal

Contains no repetitive signals

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Analog Signals

Simple analog signalSine wave3 characteristics

1. Peak amplitude (A)2. Frequency (f)3. Phase ()

Composite analog signalComposed of multiple sine waves

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Figure 3.2 A sine wave

)2sin()( ftAts

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Figure 3.3 Amplitude

t

s(t): instantaneous amplitude

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Characteristics of Analog Signal

Peak amplitude: highest intensity Frequency (f)

Number of cycles/rate of change per second

Measured in Hertz (Hz), KHz, MHz, GHz, …Period (T): amount of time it takes to

complete one cyclef = 1/T

Phase: position of the waveform relative to time 0

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Figure 3.4 Period and frequency

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Table 3.1 Units of periods and frequenciesTable 3.1 Units of periods and frequencies

Unit Equivalent Unit Equivalent

Seconds (s) 1 s hertz (Hz) 1 Hz

Milliseconds (ms) 10–3 s kilohertz (KHz) 103 Hz

Microseconds (ms) 10–6 s megahertz (MHz) 106 Hz

Nanoseconds (ns) 10–9 s gigahertz (GHz) 109 Hz

Picoseconds (ps) 10–12 s terahertz (THz) 1012 Hz

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Figure 3.5 Relationships between different phases

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Figure 3.6 Sine wave examples

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Figure 3.6 Sine wave examples (continued)

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Figure 3.6 Sine wave examples (continued)

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Characteristics of Analog Signal

Changes in the three characteristics provides the basis for telecommunicationUsed by modems (later …)

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Time Vs. Frequency Domain

The sine waves shown previously are plotted in its time domain.

An analog signal is best represented in the frequency domain.

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Figure 3.7 Time and frequency domains

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Composite Signals

A composite signal can be decomposed into component sine waves - harmonics

The decomposition is performed by Fourier Analysis

Figure 4-13

WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998

Signal with DC Component

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Figure 3.8-3.10 Square wave and the first three harmonics

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Figure 3.11 Frequency spectrum comparison

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Frequency Spectrum and Bandwidth

Frequency spectrumCollection of all component

frequencies it contains Bandwidth

Width of frequency spectrum

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Figure 3.13 Bandwidth

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Example 3Example 3

If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is the bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V.

SolutionSolution

B = fh  fl = 900 100 = 800 HzThe spectrum has only five spikes, at 100, 300, 500, 700, and 900 (see Figure 13.4 )

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Figure 3.14 Example 3

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Example 4Example 4

A 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 integral frequencies of the same amplitude.

SolutionSolution

B = fB = fhh f fll

20 = 60 20 = 60 ffll

ffll = 60 = 60 20 = 40 Hz20 = 40 Hz

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Figure 3.15 Example 4

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Example 5Example 5

A signal has a spectrum with frequencies between 1000 and 2000 Hz (bandwidth of 1000 Hz). A medium can pass frequencies from 3000 to 4000 Hz (a bandwidth of 1000 Hz). Can this signal faithfully pass through this medium?

SolutionSolution

The answer is definitely no. Although the signal can have The answer is definitely no. Although the signal can have the same bandwidth (1000 Hz), the range does not the same bandwidth (1000 Hz), the range does not overlap. The medium can only pass the frequencies overlap. The medium can only pass the frequencies between 3000 and 4000 Hz; the signal is totally lost.between 3000 and 4000 Hz; the signal is totally lost.

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Digital Signals 0s and 1s Bit interval and bit rate

Bit interval: time required to send 1 bit

Bit rate: #bit intervals in one second

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Example 6Example 6

A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval)

SolutionSolution

The bit interval is the inverse of the bit rate.

Bit interval = 1/ 2000 s = 0.000500 s = 0.000500 x 106 s = 500 s

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Digital Signal - Decomposition

A digital signal can be decomposed into an infinite number of simple sine waves (harmonics), each with a different amplitude, frequency, and phaseA digital signal is a composite signal A digital signal is a composite signal with an infinite bandwidth.with an infinite bandwidth.

Significant spectrum Components required to reconstruct the

digital signal

Figure 4-20

WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998

Harmonics of a Digital Signal

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Bandwidth-Limited Signals (a) A binary signal and its root-

mean-square Fourier amplitudes.

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Bandwidth-Limited Signals (2)

(b) – (e) Successive approximations to the original signal.

Figure 4-21

WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998

Exact and Significant Spectrums

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Channel Capacity Channel capacity

Max. bit rate a transmission medium can transfer

Nyquist theorem C = 2H log2V

where C: channel capacity (bit per second)H: bandwidth (Hz)V: signal levels (2 for binary)

C is proportional to H Significant bandwidth puts a limit on channel capacity

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Figure 3.18 Digital versus analog

To transmit 6bps, we need a bandwidth = 3 - 0 = 3Hz

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Channel Capacity

Nyquist theorem is for noiseless (error-free) channels.

Shannon CapacityC = H log2(1 + S/N)

where C: (noisy) channel capacity (bps)H: bandwidth (Hz)S/N: signal-to-noise ratio

dB = 10 log10 S/N

In practice, we have to apply both for determining the channel capacity.

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Example 7Example 7

Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. The maximum bit rate can be calculated as

BitBit Rate = 2 Rate = 2 3000 3000 log log22 2 = 6000 bps 2 = 6000 bps

Example 8Example 8

Consider the same noiseless channel, transmitting a signal with four signal levels (for each level, we send two bits). The maximum bit rate can be calculated as:

Bit Rate = 2 x 3000 x logBit Rate = 2 x 3000 x log22 4 = 12,000 bps 4 = 12,000 bps

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Example 9Example 9

Consider an extremely noisy channel in which the value of the signal-to-noise ratio is almost zero. In other words, the noise is so strong that the signal is faint. For this channel the capacity is calculated as

C = B logC = B log22 (1 + S (1 + S//N) = B logN) = B log22 (1 + 0) (1 + 0)

= B log= B log22 (1) = B (1) = B 0 = 0 0 = 0

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Example 10Example 10

We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is usually 35dB, i.e., 3162. For this channel the capacity is calculated as

C = B logC = B log22 (1 + S (1 + S//N) = 3000 logN) = 3000 log22 (1 + 3162) (1 + 3162)

= 3000 log= 3000 log22 (3163) (3163)

C = 3000 C = 3000 11.62 = 34,860 bps 11.62 = 34,860 bps

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Example 11Example 11

We have a channel with a 1 MHz bandwidth. The S/N for this channel is 63; what is the appropriate bit rate and signal level?

SolutionSolution

C = B logC = B log22 (1 + S (1 + S//N) = 10N) = 1066 log log22 (1 + 63) = 10 (1 + 63) = 1066 log log22 (64) = 6 Mbps (64) = 6 Mbps

Then we use the Nyquist formula to find the number of signal levels.

4 Mbps = 2 4 Mbps = 2 1 MHz 1 MHz log log22 LL L = 4 L = 4

First, we use the Shannon formula to find our upper First, we use the Shannon formula to find our upper limit.limit.

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Data Communication Measurements

ThroughputHow fast data can pass through an entity

Propagation speedDepends on medium and signal frequency

Propagation time (propagation delay)Time required for one bit to travel from

one point to another Wavelength

Propagation speed = wavelength X frequency

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Figure 3.25 Throughput

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Figure 3.26 Propagation time

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Figure 3.27 Wavelength