Communications I (ELCN 306)

Communications I (ELCN 306)

c© Samy S. Soliman

Electronics and Electrical Communications Engineering DepartmentCairo University, Egypt

Email: [email protected]: http://scholar.cu.edu.eg/samysoliman

Credit Hours System

c© Samy S. Soliman (Cairo University) ELCN 306 Credit Hours System 1 / 79

Outline

1 Introduction to Digital Communications

2 Instantaneous Sampling

3 Pulse Amplitude Modulation

4 Time Division Multiplexing

5 Pulse Code Modulation


Introduction to DigitalCommunications


Introduction

Communication Process

It is a process that involves the transfer of information from one point toanother

Basic Modes of Communication1 Broadcasting

2 Point-to-Point Communication

Activity: Think (Give examples of each mode of communication)Activity: Analyze (What are the differences between the modes ofcommunication)


Introduction

Communication Resources1 Transmitted Power

2 Channel Bandwidth

Activity: Think (Give examples of communication channels and theirclassification)

This gives rise to the need of modulation:

1 Ease of Radiation

2 Simultaneous transmission of several signals

SNR

Signal-to-noise ratio: Ratio of the average signal power to the averagenoise power.


Introduction

Classification of Modulation Process1 Continuous-Wave Modulation

2 Pulse Modulation

3 Digital Modulation

Note

In the following, we will focus on Pulse Modulation (PM), which refersto the Discretization of the signal.It is further classified to

1 Analog PM

2 Digital PM


Introduction


Basics of Communication: Introduction


Why Digital Communication Systems?

Features of Digital Communication Systems

Transmitter sends a waveform from a finite set of possible waveformsduring a limited time

Channel distorts, attenuates and adds noise to the transmitted signal

Receiver decides which waveform was transmitted from the noisyreceived signal

Probability of erroneous decision is an important measure for thesystem performance

Advantages of Digital Communication Systems

The ability to use regenerative repeaters

Different kinds of digital signals are treated identically

Immunity to noise


Designing Digital Communication Systems

Necessary Knowledge/Tools for the Design of DCS

1 Classification of signals

2 Random processes

3 Noise in communication systems

4 Signal transmission through linear systems

5 Bandwidth of signal


Classification of Signals

Signal Classifications

Periodic - Aperiodic

Continuous - Discrete

Analog - Digital

Power - Energy

Deterministic - Random



Energy Signal - Power Signal

Energy Signal: A signal is an energy signal if, and only if, it hasnonzero but finite energy for all time, i.e. 0 < E <∞

E =

∫ ∞−∞|x(t)|2dt = lim

T→∞

∫ T/2

−T/2|x(t)|2dt

Power Signal: A signal is a power signal if, and only if, it has finitebut nonzero power for all time, i.e. 0 < P <∞

P = limT→∞

1

T

∫ T/2

−T/2|x(t)|2dt



Deterministic - Random

Deterministic signal: No uncertainty with respect to the signalvalue at any time.

Random signal: Some degree of uncertainty in signal values before itactually occurs.

1 Thermal noise in electronic circuits due to the random movement ofelectrons.

2 Reflection of radio waves from different layers of ionosphere.

General rule: Periodic and random signals are power signals. Signals thatare both deterministic and non-periodic are energy signals


Instantaneous Sampling



Definition

Sampling is the process in which an analog signal is converted into acorresponding sequence of samples that are usually spaced uniformly intime

Ts = Sampling Period fs = Sampling Frequency/Rate




Using the definition of instantaneous sampling;

gs(t) = g(t)δTs (t)

= g(t)∑

δ(t − nTs)

=∑

g(nTs)δ(t − nTs)

Gs(f ) =∑

g(nTs)e−j2π(nTs)f (1)

Using the propertied of Fourier Transform, it can be shown that:

Gs(f ) = fs∑

G (f − nfs) (2)


Instantaneous Sampling: Nyquist Rate and Reconstruction

From (2), assuming fs = 2B, where B is the bandwidth of the originalsignal,

Gs(f ) = 2B∑

G (f − n(2B))

Recall that a band-limited signal is time-unlimited.In order to reconstruct the signal, a LPF is used such that

Greconstructed(f ) =1

2BGs(f ), −B < f < B (3)

= G (f )

Applying the inverse Fourier Transform, one can obtain

greconstructed(t) =∑

g( n

2B

)sinc(2Bt − n) (4)


Instantaneous Sampling: Interpolation


Instantaneous Sampling: Sampling Theorem

Notes

The expression in (4) represents an Interpolation Formula forreconstructing the original signal from the sequence of sampled values

The sinc(2Bt) function plays the rule of Interpolation Function,where each sample is multiplied by a delayed version of it, resultingwaveforms added together to obtain greconstructed(t) = g(t)

Theorem (Sampling Theorem)

A band-limited signal of finite energy, which has no frequency componentshigher than B Hz, is completely described by the values of the signal atinstants of time separated by 1

2B seconds.The signal may be completely recovered from the knowledge of its samples.

Nyquist rate = 2B Nyquist interval = 12B


Instantaneous Sampling: Aliasing

Definition of Aliasing

Aliasing is the phenomenon of a high-frequency component in thespectrum of the signal, seemingly taking on the identity of a lowerfrequency in the spectrum of its sampled version

Aliasing occurs if fs < 2B


Instantaneous Sampling: Combating Aliasing

To combat the effects of aliasing;

1 An anti-aliasing LPF is used prior to sampling to attenuate thenon-essential high-frequency components of the signal

2 The filtered signal is sampled at a rate slightly higher than theNyquist rate


Instantaneous Sampling: Combating Aliasing

The use of fs > 2B has the benefit of;1 Easing the design of the reconstruction filter used to recover the

original signal from its sampled version, such that the filter has1 A passband that extends from 0 to B2 A transition band that extends from B to fs − B


Instantaneous Sampling: Reconstruction LPF


Instantaneous Sampling: Summary

Theorem (Sampling Theorem)

A band-limited signal of finite energy, which has no frequency componentshigher than B Hz, is completely described by the values of the signal atinstants of time separated by 1

2B seconds.The signal may be completely recovered from the knowledge of its samples.

Nyquist rate = 2B Nyquist interval = 12B

Definition of Aliasing

Aliasing is the phenomenon of a high-frequency component in thespectrum of the signal, seemingly taking on the identity of a lowerfrequency in the spectrum of its sampled version

Aliasing occurs if fs < 2B


Instantaneous Sampling: Example 1

x(t) is a band-limited signal with maximum frequency ωm = 1000π. Animpulse train is used to sample the signal.Which of the following sampling periods guarantees proper reconstructionusing a LPF

1 Ts = 0.5× 10−3

2 Ts = 2× 10−3

3 Ts = 10−4

Solution

For proper reconstruction using a LPF,

ωs ≥ 2ωm

2π

Ts≥ 2000π

Ts ≤ 10−3 ⇒ (1) and (3)


Instantaneous Sampling: Example 2

Find the Nyquist Rate for

1 x(t) = 1 + cos(2000πt) + sin(4000πt)

2 y(t) = sin(4000πt)πt

3 z(t) = y(t)2

Solution

1 B = 2000 Hz ⇒ fs = 2B = 4000 Hz ⇒ Ts = 0.25× 10−3 sec

2 sinc function in time domain ⇒ rectangular function in frequencydomain with ωm = 4000π rad/sec ⇒ B = 2000 Hz ⇒fs = 2B = 4000 Hz

3 sinc2 function in time domain ⇒ triangular function in frequencydomain with ωm = 8000π rad/sec ⇒ B = 4000 Hz ⇒fs = 2B = 8000 Hz


Pulse Modulation


Pulse Amplitude Modulation (PAM)

Definition

PAM is the simplest and most basic form of analog pulsemodulation

The amplitudes of regularly spaced pulses are varied in proportionto the corresponding sample values of a continuous message signal

The message is instantaneously sampled every Ts seconds

The duration of each sample is lengthened to a constant value forT seconds (Sample and Hold)

Train of Pulses

The pulses can be rectangular or any appropriate shape

s(t) =∑n

m(nTs)h(t − nTs)


Flat-Top PAM


PAM: Sampling

The instantaneously sampled version of m(t) is given as

ms(t) =∑n

m(nTs)δ(t − nTs)

After some mathematical manipulation, the PAM signal can be expressedas

s(t) = ms(t) ∗ h(t)

PAM Signal

s(t) = ms(t) ∗ h(t)

S(f ) = Ms(f ).H(f )


PAM: Flat-Top Rectangular Pulse

For rectangular pulses

h(t) =

{1, 0 < t < T

0, elsewhere

H(f ) = Tsinc(fT )e−jπfT

Recall thatMs(f ) = fs

∑k

M(f − kfs)


PAM: Time Domain


PAM: Demodulation

Message Recovery: LPF

The first step in the recovery of the original signal is through using aLow-Pass Reconstruction FilterThe spectrum of the filter output is

LPF{Ms(f ).H(f )} = M(f )LPF{sinc(fT )e−jπfT}

This is an amplitude distorted and T/2 delayed version of the message

Message Recovery: Equalizer

An equalizer is connected in cascade with the LPF and it is used to correctthe amplitude distortionThe magnitude response of the equalizer is given as

1

Tsinc(fT )=

πf

sin(πfT )


PAM: Duty Cycle of Pulses

Duty Cycle = TTs

The presence of T causes the presence of amplitude distortion

As T ⇑, distortion ⇑For duty cycle T

Ts≤ 0.1, the amplitude distortion is less than 0.5%

and the use of equalization can be ignored.


Another PAM


Other Pulse Modulation Schemes


Time Division Multiplexing


Multiplexing

Purpose: To enable the joint utilization of the communication resourcesby a plurality of independent message sources, without creating mutualinterference among them.Resources: Physical channel, Time, Bandwidth, etc.



LPFLow-pass anti-aliasing filters.CommutatorTakes a narrow sample of each of the N input messages at a rate fs ≥ 2WSequentially interleaves these N samples inside a sampling interval Ts ⇒This expands the bandwidth by a factor of N



Pulse ModulatorTransforms the multiplexed signal into a form suitable for transmissionover the channelPulse DemodulatorMakes decisions to transform the received pulses into their correspondingsamples.



DecommutatorDemultiplex the samples to their corresponding destinationOperates in synchronism with the commutator at the transmitter. Suchsynchronization is essential for proper operationLPFLow-Pass reconstruction filter


Pulse Code Modulation


Pulse Code Modulation: Basic Elements


PCM: Introduction

Definition

A message signal is represented by a sequence of coded pulses

Accomplished by representing the signal in discrete form in both timeand amplitude


PCM Transmitter: Sampling

Definition

It is the process of transforming a message signal m(t) into an analogdiscrete signal m(nTs) with a sampling frequency fs which is higher thantwice the highest frequency component W of the message signal

Ensure perfect reconstruction at the Receiver

Narrow rectangular pulses ⇒ instantaneous sampling

Proceeded by an anti-aliasing filter

Reduces the continuously varying message signal to a limited numberof discrete values per second


PCM Transmitter: Quantization

Definition

It is the process of transforming the sample amplitude m(nTs) into adiscrete amplitude ν(nTs) taken from a finite set of possible amplitudes


Uniform Mid-Rise Quantization

Quantizer Characteristic: Mid-Rise Staircase

The origin lies in the middle of a rise


Uniform Mid-Tread Quantization

Quantizer Characteristic: Mid-Tread Staircase

The origin lies in the middle of a tread


Quantization Error

Definition

It is the difference between the input signal, m, and the output signal ν

q = m − ν

Notes:

Maximum error: qmax = ±1

2step size

Step size: ∆ =max−min

LAs the step width ⇓, the quantization error ⇓It is better to use binary weighted number of levels, i.e. L = 2R

bits/sample


Signal-to-Noise Ratio (SNR)

The signal-to-noise ratio (SNR) is one of the performance measures usedto describe communication systems.Quantization error is usually more significant than pulse detection errors.

SNR

It is the ratio of the useful signal power to the noise power.

Assuming a uniform quantizer with ±mp peak levels, the averagequantization noise level can be evaluated as

Nq = q̃2 =∆2

12=

m2p

3L2

Quantizer’s Output SNR

SNR =m̃2

Nq=

3L2

m2p

P


Non-Uniform Quantization

Motivation

The SNR is a function of the signal average power, it can be differentfrom one user to another. It is needed to have SNR levels close toeach other.

The solution is to use smaller quantization steps for smaller signalamplitudes.

Achieved through compressing the signal (µ-Law or A-Law), thenapplying a uniform quantizer. This is equivalent to non-uniformquantization.

At the reconstruction end, and inverse process is applied usingexpander.

The combined system is called Compander.


Non-Uniform Quantization

µ-Law Quantizer

y =ln(1 + µm̂)

ln(1 + µ)

A-Law Quantizer

y =

Am̂

1 + ln(A), 0 ≤ m̂ ≤ 1/A

1 + ln(Am̂)

1 + ln(A), 1/A ≤ m̂ ≤ 1

SNR ' 3L2

[ln(1 + µ)]2


Non-Uniform µ-Law Quantization


Non-Uniform A-Law Quantization


Encoding (Digital Baseband Modulation)

1 Encoding is used to make the transmitted signal more robust to noise,interference and other channel impairments.

2 It translates the discrete set of sample values to a more appropriateform.

3 Binary codes give the maximum advantage over the effects of noise ina transmission medium, because a binary symbol withstands arelatively high level of noise and it is easy to generate.


Encoding: Line Codes

Line codes are used for the electrical representation of binary data stream.

1 Unipolar NRZ signaling

2 Polar NRZ signaling

3 Unipolar RZ signaling

4 Bipolar BRZ signaling (Alternate Mark Inversion)

5 Split-Phase signaling (Manchester Code)


Line Codes

Activity: Identify each of the following line codes


Line Codes

Line codes usually differ in:

1 Spectral characteristics (power spectral density and bandwidthefficiency): BW should be as small as possible + no DC component.

2 Power Efficiency: for a given BW and a specified detection errorprobability, the transmitted power should be as small as possible.

3 Error detection capability (Interference and noise immunity):should be possible to detect and preferably correct errors.

4 Bit synchronization capability: should be possible to extract timingor clock information from the line code.

5 Implementation cost and complexity


Bit Rate - Transmission Bandwidth - Output SNR

A baseband signal with maximum power P Watts and bandwidth B Hz,sampled at the Nyquist rate, 2B Hz, and quantized into L = 2R PCMlevels, using a uniform quantizer with ±mp peak levels, to betransmitted over a channel of efficiency η bits/sec/Hz

Bit Rate

Rb = 2BR

Transmission Bandwidth

BT =Rb

η

Output SNR

SNR =3P

m2p

22R SNR|dB = 10 log(SNR) = 10 log

(3P

m2p

)+ 6R dB


Regeneration

1 This is one of the most important features of PCM

2 It provides the ability to control the effects of distortion and noise

3 It is done by a chain of Regenerative Repeaters located atsufficiently close spacings along the transmission route

Notes

While it is not possible to compensate for the quantization process due tothe lost values, the regeneration process can overcome the effects of:

Attenuation

Distortion

Random noise

This is done through:

Detection

Regeneration


Block Diagram of a Regenerative Repeater

The input, at A, is a distorted PCM wave.The output, at B, is the regenerated PCM wave.


Regenerative Repeaters

1 Except for delay, the regenerated signal is exactly the same as thesignal originally transmitted (under ideal circumstances)

2 Regenerative repeaters perform the following functions:

EqualizationTo compensate for the effects of amplitude and phase distortionTimingTo sample the equalized pulses at instants of maximum signal-to-noiseratio (SNR)Decision MakingBy comparing each sample to a threshold so that a ”clean” pulse, thatrepresents ”0” or ”1”, is retransmitted.


Decoding

Definition

It is the process of generating a pulse whose amplitude is the linear sum ofall the pulses in the codeword, with each pulse being weighted by its placevalue in the code.

Note:

This process is done after regenerating the received pulses one lasttime

Decoding results a Quantized PAM signal


Filtering

Definition

It is the process of passing the decoder’s output through a LPF with cutofffrequency equal to the message bandwidth.

Note: For an error-free transmission path, the recovered signal is similarto the original signal with the exception of distortion introduced by thequantization process.


Example 1

Question

A signal m(t) band-limited to 3 KHz is sampled at a rate 33.3% higherthan the Nyquist rate. The maximum acceptable error in the sampleamplitude (the maximum quantization error) is 0.5% of the peakamplitude mp. The quantized samples are binary coded.Find the minimum bit rate to transmit the encoded binary signal.If 24 such signals are time-division-multiplexed, determine the bit rate ofthe multiplexed signal.


Example 2

Question

The ASCII code has 128 characters (symbols) which are binary coded. If acertain computer generates 100,000 characters per second, determine thefollowing:

1 The number of bits required per character

2 The number of bits per second required to transmit the computeroutput

3 For error-detection, an additional bit, called parity bit, is added to thecode of each character. Modify the answers of the previous 2 partsaccordingly.


Example 3

Question

A CD records audio signals using PCM. Assume the audio signalbandwidth is 15 KHz.

1 What is the Nyquist rate?

2 If the Nyquist samples are quantized into L = 65, 536 levels and thenbinary coded. What is the number of bits required per sample?

3 Determine the number of bits per second required to encode theaudio signal

4 If the transmission channel supports a transmission rate of 2 bps/Hz.What is the minimum bandwidth required to transmit the encodedsignal?

5 If practical CDs use 44, 100 samples per second. Determine thetransmission bit rate and the minimum bandwidth required totransmit the encoded signal


Example 4

Question

Five telemetry signals, each of bandwidth 1 KHz, are to be transmittedsimultaneously by binary PCM. The maximum tolerable error in sampleamplitudes is 0.2% of the peak signal amplitude. The signal must besampled at least 20% above the Nyquist rate. Framing andsynchronization require additional 0.5% extra bits.Determine the minimum possible data rate that must be transmitted andthe minimum bandwidth required for transmission (assuming channelefficiency of 2 bps/Hz).


Example 5

Question

A TDM system is used to multiplex six signal. Two of them havebandwidths of 40 Hz, two of them have a bandwidths of 100 Hz and thelast two have bandwidths of 500 Hz. The higher frequency signals aresampled at a rate of 1600 samples/second. This sampling rate is dividedby 2R1 and 2R2 to obtain the sampling rate of the first and second pairs oflower frequency signals, respectively.

1 Find the maximum value of R1 and R2

2 Using these R1 and R2, design a multiplexing system that multiplexesthe first two signals into a new sequence, and then multiplexes thissequence and the next two signals into a new sequence, and finallymultiplexes the new sequence with the remaining two signals.


References

B. P. Lathi and Zhi Ding (2010)Modern Digital and Analog Communication Systems, 4th Edition.Oxford University Press.

Simon Haykin and Michael Moher (2010)Communication Systems, 5th Edition.John Wiley.


Thank YouQuestions ?

[email protected]

http://scholar.cu.edu.eg/samysoliman


Communications I (ELCN 306)

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