COMMUNICATIONTHEORY 1

ECM 221

CHAPTER 4:

DIGITAL COMMUNICATION SYSTEMS

Introduction of digital modulation

• If the information signal is digital and the amplitude (V) of the carrier is varied proportional to the information signal ASK is produced.

• If the frequency (f) of the carrier is varied proportional to the information signal FSK is produced.

• If the phase () of the carrier is varied proportional to the information signal PSK is produced.

• If both the amplitude (V) and the phase () of the carrier is varied proportional to the information signal QAM is produced.

• ASK, FSK, PSK and QAM are all forms of digital modulation.

)2sin()( tfVt

Application of digital modulation

• Low-speed voice band data communications modems, such as those found in most personal computers.

• High-speed data transmission systems, such as broadband digital subscriber lines (DSL)

• Digital microwave and satellite communication systems• Cellular telephone Personal Communication Systems (PCS).

INTRODUCTION

Digital transmission • The transmittal of digital signals between two or more points in a

communications system. • Can be binary or any other form of discrete-level digital pulses.• The original source information may be in digital form or it could be

analog signals that have been converted to digital pulses prior to transmission and converted back to analog signals in the receiver.

• Physical facility: a pair of wires, coaxial cable or an optical fiber cable is required to interconnect the various points within the system.

• Digital pulses cannot be propagated through a wireless transmission system such as Earth’s atmosphere or free space (vacuum).

• Today, digital transmission systems are used to carry not only digitally encoded voice and video signals but also digital source information directly between computers and computer networks.

• Digital transmission systems use both metallic and optical fiber cables for their transmission medium

ADVANTAGES OF DIGITAL TRANSMISSION

• Noise immunity• Inherently less susceptible to interference because it is not necessary to

evaluate the precise amplitude, frequency or phase to ascertain its logic condition.

• Better suited for processing and combining using a technique called multiplexing.

• Digital signal processing (DSP) is the processing of analog signals using digital methods and includes bandlimiting the signal with filters, amplitude equalization and phase shifting.

• Much simpler to store and the transmission rate can be easily changed to adapt different environments and to interface with different types of equipment.

• More resistant to additive noise• They use signal regeneration rather than signal amplification.• Can be transported longer distances than analog signals..• Simpler to measure and evaluate • Easier to compare the error performance of one digital system to another

digital system• Transmission errors can be detected and corrected more easily and more

accurate.

DISADVANTAGES OF DIGITAL SIGNAL

• The transmission of digitally encoded analog signals requires significantly more bandwidth than simply transmitting the original analog signal.

• Bandwidth is one of the most important aspects of any communications system because it is costly and limited.

• Analog signals must be converted to digital pulses transmission and converted back.

• Requires precise time synchronization between the clocks in the transmission and receivers

• Incompatible with older analog transmission systems

Information capacity, bits, bit rate, baud and M-ary encoding

• Information capacity refer to Hartley’s Law (Already covered in Chapter 1).

• Binary digit or bits: the most basic digital symbol used to represent information.

• Bit rate is simply the number of bits transmitted during one second and is expressed in bits per second (bps).

• Shannon law: (refresh)

N

S1log3.32

orN

S1log

10

2

BI

BI

M-ary encoding

• M-ary is a term derived from the word binary.• M represents a digit that corresponds to the number of conditions, levels

and combinations possible for a given number of binary variables.• The number of bits necessary to produce a given number of conditions is

expressed mathematically as:N = log2 M

where N = number of bits necessary M = number of conditions, levels, or combinations possible with N bits.

• Simplified: 2N = M

Baud and minimum bandwidth

• Baud is a term that is often misunderstood and commonly confused with bit rate (bps).

• Bit rate refers to the rate of the change of a digital information signal, which usually binary.

• Baud is also a rate of change, however baud refers to the rate of change of a signal on the transmission medium after encoding and modulation have occurred.

• Baud = 1/ts where baud = symbol rate (baud per second)

ts= time of one signaling element (seconds)• Nyquist bandwidth:

– Binary digital signals can be propagated thru an ideal noiseless transmission medium at a rate equal to two times the bandwidth of the medium.

Baud and minimum bandwidth

• Nyquist bandwidth:– Binary digital signals can be propagated thru an ideal noiseless transmission

medium at a rate equal to two times the bandwidth of the medium.– The minimum theoretical bandwidth necessary to propagate a signal is called

Nyquist bandwidth or sometimes the minimum Nyquist frequency.– Thus:

• fb= 2 B where fb = bit rate in bps, B = ideal Nyquist bandwidth.

• The actual bandwidth necessary to propagate a given bit rate depends on several factors:– Type of encoding -- system noise– Modulation used -- desired error performance– The types of filter used

Baud and minimum bandwidth

• If multilevel signaling is used, the Nyquist formulation for channel capacity is:

fb = 2 B log 2 M

where: fb = channel capacity (bps)

B = minimum Nyquist bandwidth (hertz)M = number of discrete signal or voltage levels.

Amplitude Shift Keying (ASK)

• The simplest digital modulation technique is amplitude- shift keying (ASK), where a binary information signal directly modulates the amplitude of an analog carrier.

• In ASK, a carrier wave is switched ON and OFF by the input data or binary signals. During a ‘mark’ (binary ‘1’), a carrier wave is transmitted and during a ‘space’ (Binary ‘0’), the carrier is suppressed.

• Hence, it also known as ON- OFF keying (OOK). • Mathematically, amplitude- shift keying is

Where:– (ask)(t) = amplitude- shift keying wave

– νm(t) = digital information (modulating) signals (volts)– A/2 = unmodulated carrier amplitude (volts)– ωc = analog carrier radian frequency

)cos(

2)(1)()( tA

tvtv cmask

Amplitude Shift Keying (ASK)

Frequency Shift Keying (FSK)

• Frequency-shift keying (FSK) is another relatively simple, low-performance type of digital modulation.

• FSK is a form of constant-amplitude angle modulation similar to standard frequency modulation (FM) except the modulating signal is a binary signal that varies between two discrete voltage levels rather than a continuously changing analog waveform.

• The general expression for FSK is: νfsk(t) = νccos { 2π [ fc + νm (t)∆f ] } • Where νfsk(t) = binary FSK waveform

vc = peak analog carrier amplitude (volts)

fc = analog carrier center frequency (volts)

∆f = peak change (shift) in the analog carrier frequency (hertz) νm(t) = binary input (modulating) signal (volts)

Frequency Shift Keying (FSK)

• The modulating signal is a normalized binary waveform where a logic 1= +1V and logic 0 = -1V.

• For logic 1 or mark input :νfsk(t) = Vccos [ 2 ( fc + ∆f )]

• For logic 0 or space input :νfsk(t) = Vccos [ 2 ( fc - ∆f )]

• The mark frequency is the higher frequency ( fc + ∆f ).• The space frequency is the lower frequency ( fc - ∆f ).

Frequency Shift Keying (FSK)

Phase Shift Keying (PSK)

• The simplest form of PSK is binary phase-shift keying (BPSK), where N=1 and M=2.

• Therefore, BPSK 2 phase (21=2) are possible for the carrier.• One phase represents a logic 1, and the other phase represents a logic 0.• As the input digital signal changes state (ie, from a 1 to a 0 or from 0 to 1),

the phase will switch normally 0 to 180.

Phase Shift Keying (PSK)

Phase Shift Keying (PSK)

error

PULSE MODULATION

A process of sampling analog information signals and then converting those samples into discrete pulses and transporting the pulses from a source to a destination over a physical transmission medium.

The predominant methods of pulse modulation1) Pulse width modulation (PWM)• Called pulse duration modulation (PDM) or pulses length modulation (PLM)• The technique of varying the width of the constant-amplitude pulse proportional to

the amplitude of the modulating signal.2) Pulse position modulation (PPM)

The higher the amplitude of the sample, the farther to the right the pulse is positioned within a prescribed time slot.

3) Pulse amplitude modulation (PAM)• The amplitude of a constant width• Constant position pulse is varied according to the amplitude of the sample of the

analog signal.• Waveform resemble the original analog signal more than the waveforms for PWM

or PPM4) Pulse code modulation (PCM)• Sampled and converted to a serial n-bit binary code for transmission.• Each code has the same number of bits• Same length of the time for transmission

PULSE CODE MODULATION (PCM)

• PCM is only digitally encoded modulation technique that commonly use for digital transmission.

• With PCM, the pulses are of fixed length and fixed amplitude.

• It is a binary system where a pulse or lack of a pulse within a prescribed time slot represents either a logic 1 or a logic 0

Figure 10.1: Pulse Modulation: (a) analog signal (b) sample pulse (c) PWM (d) PPM (e) PAM (f) PCM

Figure 10.2: Simplified block diagram of a single-channel, simplex PCM transmission system

From The Block Diagram 1 The Simple-and-hold circuit; • It periodically samples the analog signal and converts those

samples to a multilevel PAM signal

2 Analog-to-digital converter (ADC);• Convert the PAM samples parallel PCM codes which are converted

to serial binary data in the parallel-to-serial converter. After that, the outputted onto the transmission line as serial digital pulse.

3 Repeaters;• are placed at prescribed distance to regenerate the digital pulse.• In receiver a serial-to-parallel converter converts the serial pulses

to to parallel PCM code.4 Digital-to-analog converter (DAC);• To converts the parallel PCM codes to multilevel PAM signals5 The hold circuit is basically a low-pass filter that converts the PAM

signals back to the original analog form.

PCM Sampling

• Function of sampling circuit in PCM transmitter is to periodically sample the continually changing analog input voltage and convert those samples to a series of constant-amplitude pulse that can more easily be converted to binary PCM code.

• • ü Natural sampling• When tops o the sample pulses retain the natural shape during the

sample interval, making it difficult for ADC to convert the sample to PCM code.

• With natural sampling, the frequency spectrum of the sampled output is different from that of an ideal sample.

• • ü Flat-top sampling• The most common method use for sampling voice signals in PCM,

which is accomplished in a sample-and-hold circuit.

Figure 10.3: Natural sampling (a) input analog signal (b) sample pulse (c) sampled output

Figure 10.4: Flat-top sampling(a) input analog signal (b) sample pulse (c) sampled output

(a)

(b)

(c)

Example 10.1• For the sample and hold circuit shown in figure 10.5a, determine the largest

value capacitor that can be used. Use an output impedance for Z1 of 10 , an on resistance for Q1 of 10 , an acquisition time of 10 s, a maximum peak to peak input of 10 V, a maximum output current from Z1 of 10 mA, and an accuracy of 1%.

• Solution• Current through a capacitor• • • • Rearranged and solve for C• • • • Where;• C = maximum capacitance • i = maximum output current form Z1, 10 mA• dv = maximum change in voltage across C1, which equals 10 V• dt = change time, which equals to aperture time, 10 s

dt

dvCi

dv

dtiC

• Therefore, •

• The charge time constant for C1 when Q1 is on is • • • Where • • = one charge time constant (s)• R= output impedance of Z1 plus the on resistance of Q1 ()• C= capacitance value of C1 (F)• • Rearranging and solving for C gives us• • • • The charge time of capacitor C1 is also dependent on the accuracy desired

from the device. The percent accuracy and its required RC time constant are summarized as follows:

nFV

smAC 10

10

)10)(10(max

RC

RC

max

Accuracy (%) Charge time

10 2.3

1 4.6

0.1 6.9

0.01 9.2

 For an accuracy of 1%    

 

nFs

C 7.108)20(6.4

10

Figure 10.6

• Figure 10.6a- – The output spectrum includes two original inputs – audio and the

fundamental frequency of the sampling pulse– Shows the sum and difference frequencies (fs fa) all the harmonics of fs

and fa and their associated cross products.– It is made up of a series of harmonically related sine waves. – None of the side frequencies form one harmonic will spill into the

sidebands of another harmonic and aliasing does not occur. • Figure 10.6b • Shows the results when an analog input frequency greater than fs/2

modulates fs. • Aliasing or foldover distortion has occurred. It cannot be removed through

any technique.

• Example 10.2

• For a PCM system with a maximum audio input frequency of 4 kHz, determine the minimum sample rate and the alias frequency produced if a 5 kHz audio signal were allowed to enter the sample-and-hold circuit.

• Solution• Using Nyquist’s sampling

• Therefore

• If a 5 kHz audio frequency entered the sampled-and-hold circuit, the alias frequency of 3 kHz has been introduced into the original audio spectrum.

as ff 2

kHzf s 8

10.2

• With PCM, an analog input signal is sampled, then converted to a serial binary code and transmitted to receiver where it is converted back to the original analog signal. PCM uses n-bit codes. Table 10.1 shows an n-bit PCM code where n = 3. The most significant bit is used to represent the sign of the sample

• Logic 1 = positive; logic 0 = negative. • The two remaining bits represent the magnitude. There are four codes

possible for positive numbers and four codes for negative numbers. The total is eight possible code L = 2n = 23 = 8

QUANTIZATION AND THE FOLDED BINARY CODE

• Quantization; is the process of converting an infinite number of possibilities to a finite number of conditions.

• Converting an analog signal to a PCM code with a limited number of combinations of requires quantization.

• It also a process of rounding off the amplitude of flat-top samples to a manageable number of level.

• For example, a sine wave with a peak amplitude of 5V varies between +5V and -5V.

Table 10.2:

– -The total voltage range is subdivided into a smaller number of sub-ranges.

– -For a three bit sign-magnitude code with eight possible combinations – (four +ve and four –ve)

– -The leftmost bit is the sign bit– -The two rightmost bits represent magnitude– -It is called as a folded binary code – the code on the bottom half of the

table are a mirror image of the codes on the top half.– -The magnitude difference between adjacent step is called the

quantization interval or quantum

– -Resolution V is defined as the voltage of the minimum step size, which is equal to the voltage of the least significant bit f the PCM code. The equation can be written as

– -The smaller the magnitude of a quantum, the better (smaller) the resolution and the more accurately the quantized signal will resemble the original analog sample.

1max

L

VV

010

101

111

101

011

001

0 1 0 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1

Analog input signal

Sampling pulse

Sampled waveform

Quantized signal

PCM pulses

Example on how to illustrate an analog waveform signal can be coded into 3 bits code using single mode for transmission using PCM technique.

Solution:

For a 3 bits code of single mode transmission system using PCM technique.

Figure 10.8:1) Each sample voltage is rounded of to the closest available level and then converted to its corresponding PCM code.

2) The PAM signal in the transmitter is essentially the same PAM signal produced in the receiver.

3) Any round-off errors (quantization error Qe or quantization noise Qn) in the transmitter signal are reproduced when the code is converted back to analog in the receiver.

Figure 10.9 shows the same analog input signal shown in figure 10.8 except the signal is being sampled at a much higher rate.

The folded PCM code is =

The quality of the PAM signal can be improved by

a) Using a PCM code with more bits, reducing the magnitude of a quantum and improving the resolution.

b) Sampling the analog signal at a faster rate

resolution

voltagesample

Figure 10.10: The quantized signal is a staircase function

Example 10.3:

• For a PCM coding scheme shown in Figure 10.8, determine the quantized voltage, quantization error (Qe) and PCM code for the analog sample voltage of 1.07 V.

• Solution:• To determine the quantized level, divide the sample

voltage by resolution and then round the answer off to the nearest quantization level

• The quantization error is the difference between the original sample voltage and the quantized level,

07.0107.1 eQ

107.11

07.1

PCM transmission bit rate (R) and Transmission bandwidth (TB)

• R – the rate of information transmission (bit/s). It is depends on the sampling frequency and the number of bit per sample used to encode the signal. Transmission bandwidth (TB) is equal to the transmission bit rate, but the unit is hertz

sec/bitsfnR s

HzfnTB s

Dynamic Range (DR)

• The ratio of the largest possible magnitude to the smallest possible magnitude (other than 0V) that can be decoded by the digital to analog converter in the receiver.

• Where;• DR = dynamic range (unit less ratio)• Vmin = the quantum value (resolution)• Vmax = the maximum voltage magnitude that can be discerned by

the DACs in the receiver• So

• In dB,

min

max

V

VDR

resolution

VDR max

min

maxlog20V

VDR

• Example 10.4• For a PCM system with the following parameters, determine (a) minimum

sample rate, (b) minimum number of bits used in the PCM code, (c) resolution and (d) quantization error.

• Maximum analog input frequency = 4 kHz• Maximum decoded voltage at the receiver = 2.55 V• Minimum dynamic range = 46 dB

• Solution• The minimum sample rate is

• The absolute of dynamic range

• The minimum number of bit is

kHzkHzff as 8)4(22

min

maxlog2046V

VdB

5.199DR

63.72log

)15.199log(

n

• To choose the number of bit must be greater than minimum value which is equal to 8. it is also require a bit for the sign bit. Therefore the total number of bit is nine. The total number of the PCM code is 29 = 512

• The actual dynamic range

•

• Note: the actual dynamic range is not include the sign bit• The resolution is determined by dividing the maximum positive or maximum

negative voltage by the number of positive or negative nonzero PCM codes

• The maximum quantization error is

dB

DR ndB

13.48

)12(log20)(

VV

L

VV

n01.0

12

55.2

121 8

maxmax

VVresolution

Qe 005.02

01.0

2

• PCM Line Speed• Line speed is a data rate at which serial PCM bits are

clocked out of the PCM encoder into a transmission line.• Where,

• Line speed = the transmission rate in bits per second• Samples/second = sample rate (fs)• Bit/sample = number of bits in the compressed PCM

code.

• Example 10.8 • For a single channel PCM system wit a sample rate fs =

6000 samples per second and a seven bit compressed of PCM code, determine the line speed.

• Solution•

sample

bits

ond

samplesspeedline

sec

bps

sample

bits

ond

samplesspeedline

00042

7

sec

6000