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CHAPTER 4: DIGITAL TRANSMISSION SYSTEM

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BLOCK DIAGRAM

A/D

CONVERTER

LINE

CODING

PULSE

TXN

Analog

signalOutput

PCM

signal

Bit

stream

Pulse Analog ModulationPAM, PWM, PPM

Pulse Digital ModulationPCM

- Sampling- Quantizing

- Encoding - Binary code

- Grey code

DM

NRZ

RZ

Manchester

Etc.

ASK

FSK

PSK

MultiplexingFDM

TDM

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In any digital transmission system, it will involve the analog todigital conversion since most of baseband signals are in analogform.

Therefore, digital modulation is required in order to convert theanalog signal to a digital form.

The advantages of digital system compared to the analogsystem: Easy to regenerate using the regenerative repeaters.

Less noise and distortion (high SNR).

Easy to process and to store QP, DSP, RAM, ROM, etc.

Small size, low cost.

Low power.

INTRODUCTION

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However, there are some disadvantages of digital system:

Require wider bandwidth.

Digital signal detection needs synchronization

Require a coder to convert the analog signal to a digital signal.

Might need partial changes to the available analog channels.

In order to transmit the analog signal in digital form, we need to

convert the signal into pulses.

Thus, a periodic pulse train is required as a carrier signal.The carrier pulses will be changed by the instantaneous value of

the analog modulating signal yield apulse modulation.

INTRODUCTION

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Pulse modulation includes many different methods of converting

information into pulse form for transferring them from a source to a

destination.

It can be categorized into two:

Pulse analog modulation

PAM Pulse Amplitude Modulation

PWM Pulse Width Modulation

PPM Pulse Position Modulation

Pulse digital modulation

PCM Pulse Code Modulation

DM Delta Modulation

INTRODUCTION

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Advantages of pulse modulation:

Noise immunity

Inexpensive digital circuitry

Can be time-division multiplexed with other pulse modulated signals.

Digital pulse stream can be stored

Error detected and correction is easily implemented

Disadvantages of pulse modulation:

Special encoding and decoding technique may be necessary to increasetransmission rates, thus making the pulse stream more difficult to recover.

INTRODUCTION

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SAMPLING THEOREM

Any analog signals can be digitized through sampling and quantization.

The sampling rate must be sufficiently large so that the analog signal

can be reconstructed from the samples with sufficient accuracy.

The sampling theorem is the basis for determining the proper sampling

rate for a given signal, has a deep significance in signal processing and

communication theory.

Generally, it is a process of changing the analog signal to a discrete

form without losing any information.It is the first part that has to be done in the process of analog to digital

conversion.

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SAMPLING THEOREM

Sampling theorem is very important in signal analysis, processing andtransmission.

It allows us to replace a continuous-time signal by a discrete sequence

of numbers.The continuous-time signal is sampled, and sample values are used tomodify certain parameters of a periodic pulse train.

We may vary the amplitudes, widths or positions of the pulses inproportion to the sample values of the signal.

Accordingly we have three types ofpulse analog modulation; Pulse Amplitude Modulation (PAM)

Pulse Width Modulation (PWM)

Pulse Position Modulation (PPM)

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SAMPLING THEOREM

9

Sampling process can be done by multiplying the information

signal, vm(t) with the sampling signal (pulse signal), s(t) as given

by:

The sampling signal, s(t) is shown:

Where,

)()()( tstvtv ms y!

X Ts

A

amplitudepulse

intervalsampling

timesampling

!

!

!

A

Ts

X

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Sampling frequency, fs is given by:

It must be sufficiently large so that no information will loss oraliasing

occurs during the transmission.

Therefore, information signal must be sampled at least twice of the

maximum frequency of the information signal.

This rate is known as Nyquist rate and it is given by:

SAMPLING THEOREM

s

s

Tf

1!

frequencymax.

frequencysamplingwhere,2

max

max

!

!u

f

fffss

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If the information signal is sampled at Nyquist rate (fsu 2fmax),thus the

spectrum can be shown below:

Iffs < 2fmax:

SPECTRUM OF THE SAMPLED SIGNAL

fm

fs-f

mfs

2fs

fs+f

m2f

s-f

m2f

s+f

m

V

fm

fs-f

mfs

2fs

fs+f

m

2fs-f

m

2fs+f

m

Valiasing

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1. Two signals have their time for one full cycle of 0.01s and 0.02srespectively. Calculate the sampling frequency, fs required foreach signal.

2. Determine the Nyquist rate for the following signal:a)

b)

Answer:

1. 200Hz 2. a) 5000Hz b) 200Hz

EXAMPLES

ttv

tttv

m

m

T

TT

200sin200)(

5000cos3000cos5.2)(

!

!