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GATE-2012 ECE Q3 (communication)

Posted By Krishna Sankar On November 1, 2012 @ 10:07 am In GATE | No Comments

Question 3 on Communication from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

## Solution

To answer this question, am using contents from Section 5.1 of  Digital Communication, third edition, John R. Barry, Edward A. Lee, David G. Messerschmitt (buy from Amazon.com [1]buy from Flipkart.com [2]).

Let us first try to understand the minimum bandwidth required to transmit the signal $s(t)$ at symbol rate $\frac{1}{T}$with no inter symbol interference (ISI).

The transmitted signal is,

$s(t) = \sum_{m=-\infty}^{+\infty}a_mg(t-mT)$,

where

$a_m$ is the $m^{\mbox{th}}$ transmitted data symbols,

$\frac{1}{T}$ is the symbol rate and

$g(t)$ is the pulse shape.

The sampled signal is,

$s(kT) = \sum_{m=-\infty}^{+\infty}a_mg(kT-mT)$.

Decomposing the above equation into two parts,

$\begin{array}s(kT)&=&\underbrace{g(0)a_k}_{\mbox{desired}}&+&\underbrace{\sum_{m \ne k}a_mg(kT-mT)}_{\mbox{ISI term}}\end{array}$.

The first term is the desired signal and the second term contributes to inter symbol interference.

To ensure that there is no intersymbol interference, the sampled pulse shape should be

$g(kT) = \delta_k$.

Taking Fourier transform,  this translates to the following criterion

$\begin{array}\frac{1}{T}\sum_{m=-\infty}^{\infty} G$$f-\frac{m}{T}$$&=&1\end{array}$.

Note :  For convenience, take

$\begin{array}B(f)&=&\frac{1}{T}\sum_{m=-\infty}^{\infty} G$$f-\frac{m}{T}$$\end{array}$.

Note :

The proof for this is discussed in bit more detail in Section 9.2 of Digital Communications, 4th edition John G. Proakis (buy from Amazon.com [3], buy from Flipkart.com [4])

Let us assume that pulse shape $g(t)$ has a bandwidth $W$. There are three cases to check

i) When $\begin{array}W&<&\frac{1}{2T}\end{array}$ :

In this case, the plot of $B(f)$ will have non-overlapping replicas of the spectrum $\frac{1}{T}G(f)$ separated by $\frac{1}{T}$ and there is no choice to meet the  $\begin{array}\frac{1}{T}\sum_{m=-\infty}^{\infty} G$$f-\frac{m}{T}$$&=&1\end{array}$ criterion.

ii) When $\begin{array}W&=&\frac{1}{2T}\end{array}$ :

There exists one candidate meeting the criterion,

$G(f)=\{\begin{array}{lrr}T,&&\mbox{|f|<\frac{1}{2T}}\\0,&&\mbox{elsewhere}\end{array}$.

The corresponding pulse shape is

$\begin{array}g(t)&=&\frac{\sin $$\pi t/T$$}{$$\pi t/T$$}=sinc{$$\pi t/T$$}\end{array}$.

Alternately, this can be stated as – for a given single sided bandwidth $W$, the maximum symbol rate which can be achieved for ISI free transmission is $\begin{array}\frac{1}{2T}\end{array}$. To meet this, the pulse shape $g(t)$ has to be a sinc function. Typically usage of the sinc function is not preferred as its tails decay very slowly and a small timing error in the demodulator will result in an infinite series of inter symbol interference components.

iii) When $\begin{array}W&>&\frac{1}{2T}\end{array}$

In this case $B(f)$ consists of overlapping spectrum of $\frac{1}{T}G(f)$ separated by $\frac{1}{T}$ and there exists multiple choices meeting the  $\begin{array}\frac{1}{T}\sum_{m=-\infty}^{\infty} G$$f-\frac{m}{T}$$&=&1\end{array}$ criterion.
A commonly used pulse shaping filter satisfying the criterion while having a faster decay is the raised cosine filters having the following equation,

$g(t) = \left(\frac{sin(\pi t/T)}{\pi t/T}\right)\left(\frac{cos(\alpha \pi t/T)}{1-(2\alpha t/T)^2}\right),\mbox{ } t=-\infty \mbox{ to } +\infty$.

The frequency response is,

$G(f)=\{\begin{array}{llr}T,&&\mbox{0<|f|\le\frac{1-\alpha}{2T}}\\\frac{T}{2}\{1+\cos$\frac{\pi T}{\alpha}$$|f|-\frac{1-\alpha}{2T}$$$\},&&\mbox{\frac{1-\alpha}{2T}<|f|\le \frac{1+\alpha}{2T}}\\0,&&\mbox{|f|>\frac{1+\alpha}{2T}}\end{array}$.

With a raised cosine pulse shape, the bandwidth $W$ is larger than the minimum required pulse shape and is related as,

$\begin{array}W&=&\frac{1+\alpha}{2T}\end{array}$.

Then term $\alpha$ is called the excess bandwidth factor. For example,

• $\alpha=3/4$ translates to 75% excess bandwidth
• $\alpha=1$ translates to 100% excess bandwidth
• $\alpha=0$, the raised cosine pulse shape reduces to sinc pulse shape.

In our question,

$\alpha=3/4$and $\begin{array}W&=&\frac{1+\alpha}{2T}&=&3500\end{array}$ and the goal is to find the maximum possible symbol rate $\frac{1}{T}$.

Substituting for $W$$\alpha$ and solving for $\frac{1}{T}$,

$\begin{array}{lll}\frac{1+\frac{3}{4}}{2T}&=&3500\\\frac{1}{T}&=&3500\frac{2}{(7/4)}\\&=&4000&\mbox{symbols per second}\end{array}$.

Matlab example

% script for plotting the time and frequency response of raised cosine pulse shape
% filter with
% a) alpha = 0 (sinc pulse)
% b) alpha = 3/4
%
clear all; close all;

fs = 10;
% defining the sinc filter
sincNum = sin(pi*[-fs:1/fs:fs]); % numerator of the sinc function
sincDen = (pi*[-fs:1/fs:fs]); % denominator of the sinc function
sincDenZero = find(abs(sincDen) < 10^-10);
sincOp = sincNum./sincDen;
sincOp(sincDenZero) = 1; % sin(pix/(pix) =1 for x =0

alpha = 0;
cosNum = cos(alpha*pi*[-fs:1/fs:fs]);
cosDen = (1-(2*alpha*[-fs:1/fs:fs]).^2);
cosDenZero = find(abs(cosDen);
cosOp(cosDenZero) = pi/4;
gt_alpha0 = sincOp.*cosOp;
GF_alpha0 = (1/fs)*fft(gt_alpha0,1024);

alpha = 0.75;
cosNum = cos(alpha*pi*[-fs:1/fs:fs]);
cosDen = (1-(2*alpha*[-fs:1/fs:fs]).^2);
cosDenZero = find(abs(cosDen);
cosOp(cosDenZero) = pi/4;
gt_alpha0 = sincOp.*cosOp;
GF_alpha0 = (1/fs)*fft(gt_alpha0,1024);

alpha = 0.75;
cosNum = cos(alpha*pi*[-fs:1/fs:fs]);
cosDen = (1-(2*alpha*[-fs:1/fs:fs]).^2);
cosDenZero = find(abs(cosDen)
cosOp(cosDenZero) = pi/4;
gt_alpha_p75 = sincOp.*cosOp;
GF_alpha_p75 = (1/fs)*fft(gt_alpha_p75,1024);

Figure : Time domain plot

Figure : Frequency domain plot

Based on the above, the right choice is (C) 4000 symbols per second.

## References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf [5]

[2] Digital Communication, Third edition, John R. Barry, Edward A. Lee, David G. Messerschmitt (buy from Amazon.com [1]buy from Flipkart.com [2])

[3] Digital Communications, Fourth edition John G. Proakis (buy from Amazon.com [3]buy from Flipkart.com [4])

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