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GATE-2012 ECE Q3 (communication)

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

Q3. In a baseband communications link, frequencies upto 3500Hz are used for signalling. Using a raised cosine pulse with 75% excess bandwidth and for no inter-symbol interference, the maximum possible signaling rate in symbols per second is,

(A) 1750

(B) 2625

(C) 4000

(D) 5250

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, buy from Flipkart.com).

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, buy from Flipkart.com)

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

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

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

GATE-2012 ECE Q26 (electronic devices)

Question 26 on Electronic Devices from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q26. The source of a silicon ( ${\begin{array}n_i&=&10^{10}&\mbox{per cm^3\end{array}$ ), n-channel MOS transistor has an area of ${\mbox{1 sq \mu m}$ and a depth of ${\mbox{1 \mu m}$ . If the dopant density in the source is ${\begin{array}10^{19}\mbox{/cm^3\end{array}$ , the number of holes in the source region with above volume is approximately

(A) ${\mbox{10^7}$

(B) ${\mbox{100}$

(C) ${\mbox{10}$

(D) ${\mbox{0}$

Solution

To answer this question, had to dig up the copy of Solid State Electronic Devices, Ben G Streetman, Sanjay Kumar Banarjee (buy from Amazon.com, buy from Flipkart.com) and am referring to the content from Section 3.3 Carrier Concentrations.

Electrons in solids obey Fermi-Dirac statistics. The function $f(E)$ Fermi-Dirac distribution function, gives the probability that an available energy state at $E$ will be occupied by an electron at absolute temperature $T$ ,

$f(E) = \frac{1}{1+e^{(E-E_F)/kT}}$ .

The quantity $E_f$ is the Fermi level and $\begin{array}k&=&\mbox{8.63x10^{-5} eV/K} & = & \mbox{1.38x10^{-23} J/K}\end{array}$ is the Boltzmann’s constant.

The concentration of electrons in the conduction band is,

$n_0 = \int_{E_c}^{\infty}f(E)N(E)dE$ , where $N(E)dE$ is the density of states $cm^{-3}$ in the energy range $dE$ .

The integral is equivalently stated as,

$alt=$ , where $alt=$ is the effective density of states at conduction band edge $alt=$ .

The Fermi function at $alt=$ is approximately,

$\begin{array}f(E_c)&=&\frac{1}{1+e^{(E_c-E_F)/kT}}&\simeq e^{-(E_c-E_F)/kT}&\end{array}$ .

So in this condition the concentration of electrons in the conduction band is,

$alt=$ .

Similarly, the concentration of holes in the valence band is,

$alt=$ , where $alt=$ is the effective density of states at valence band edge $alt=$ .

The term,

$\begin{array}1-f(E_v)&=&\frac{1}{1-1+e^{(E_v-E_F)/kT}}&\simeq e^{-(E_F-E_v)/kT}&\end{array}$ .

So in this condition the concentration of holes in the valence band is,

$alt=$ .

The product of $alt=$ and $alt=$ at equilibrium is a constant for a particular material and temperature even if doping is varied.

$alt=$ , where

$alt=$ is the gap between the conduction band and valence band.

Similarly, for an intrinsic material, the product of $alt=$ and $alt=$ is,

$alt=$

For an intrinsic material, the electron and hole concentration are equal i.e. $alt=$ .

The constant product of electron and hole concentration can be written as

$alt=$

With the above understanding, let us try to find the solution to the problem. In our question,

$alt=$ and $alt=$ .

The hole concentration is,
$alt=$ .

The volume of the source region is

$alt=$ .

The number of holes is,

$alt=$

Based on the above, the right choice is (D) 0.

References

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

[2] Solid State Electronic Devices, Ben G Streetman, Sanjay Kumar Banarjee (buy from Amazon.com, buy from Flipkart.com)

[3] Valence band http://en.wikipedia.org/wiki/Valence_band

[4] Conduction band http://en.wikipedia.org/wiki/Conduction_band

GATE-2012 ECE Q36 (math)

Question 36 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q36. A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is

(A) 1/3

(B) 1/2

(D) 3/4

Solution

Let us start by finding the sample space. The possible sample space with odd number of tosses till a head appears for the first time is,

$\begin{array}S=\{H,&TTH,& TTTTH,& TTTTTTH,& ...&\}\end{array}$

Given that the coin is fair,

$p(H)=p(T)=\frac{1}{2}$ .

The total probability is,

$\begin{array}{lll}p(S)&=&\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+...\\&=&\frac{1}{2}$1+\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...$\\&=&\frac{1}{2}$1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...$\end{array}$

Using Taylor series,

$\begin{array}\frac{1}{1-x}&=&\sum_{n=0}^{\infty}x^n,&\mbox{for } |x| < 1\end{array}$ .

Substituting,

$\begin{array}{lll}p(S)&=&\frac{1}{2}*\frac{1}{$1-\frac{1}{4}$}\\&=&\frac{2}{3}\end{array}$

Matlab/Octave example

Felt it would be nice to get similar results using a simple simulation model

% Script to find the probability that odd number of tosses
% are required to get a head for the first time

clear all; close all;
N = 5*10^5;

% HEAD = 0, TAIL = 1;
% definfing the pattern corresponding to odd tosses 
% till the first head - limiting to max of 21 tosses
% converted to integer (for easy comparison)
pattern_v = [ ...
   sum(2.^0         .*  [0 ]);                      ... % H
   sum(2.^(2:-1:0)  .*  [1 1 0 ]);                  ... % TTH
   sum(2.^(4:-1:0)  .*  [1 1 1 1 0 ]);              ... % TTTTH 
   sum(2.^(6:-1:0)  .*  [1 1 1 1 1 1 0 ]);          ... % TTTTTTH 
   sum(2.^(8:-1:0)  .*  [1 1 1 1 1 1 1 1 0 ]);      ... % TTTTTTTTH ...
   sum(2.^(10:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 0 ]);  ...  
   sum(2.^(12:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 1 1 0 ]); ...
   sum(2.^(14:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 ]); ... 
   sum(2.^(16:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 ]); ... 
   sum(2.^(18:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 ]); ...
   sum(2.^(20:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 ]); ...
  ];

% finding the probability for each event in the  sample space
event_v = [1:2:21];
for ii=1:length(event_v)
   kk           = event_v(ii);
   x            = rand(N,kk)>0.5;
   xVal         = sum((ones(N,1)*[2.^([kk-1:-1:0])]).*x,2);
   matchCnt(ii) = size(find(xVal==pattern_v(ii)),1);
end
totalProb = sum(matchCnt./N)

Figure : Probability of odd number of tosses till first head

Based on the above, the right choice is (C) 2/3.

References

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

[2] Wiki entry on Taylor series http://en.wikipedia.org/wiki/Taylor_series

GATE-2012 ECE Q46 (math)

Question 46 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q46. The maximum value of ${f(x) = x^3 - 9x^2 + 24x + 5}$ in the interval [1, 6] is

(A) 21

(B) 25

(D) 46

Solution

Let us start by finding the critical points of the function $f(x)$ .

The first derivative is,

$f'(x) = 3x^2 - 18x + 24$ .

Solving by setting $f'(x) =0$ .

$\begin{array}{ll}3x^2 - 18x + 24 & = 0 \\ (x-4)(x-2)&=0\end{array}$ .

Taking second derivative, $f''(x) = 6x - 18$ .

At $\begin{array}{lll}x=4, &f''(x) &= +6 & (+ve)\end{array}$ and this corresponds to a local minimum.

At $\begin{array}{lll}x=2, &f''(x) &= -6 & (-ve)\end{array}$ and this corresponds to a local maximum.

The value of the function at these critical points are,

$\begin{array}f(2)&=&2^3 - 9*2^2 + 24*2 + 5&=&25 &\mbox{(local maximum)}\\f(4)&=&4^3 - 9*4^2 + 24*4 + 5&=&21&\mbox{(local minimum)}\end{array}$ .

Hold on, we are not done yet… 🙂

From the wiki entry on Extereme Value Theorem,

In calculus, the extreme value theorem states that if a real-valued function f is continuous in the closed and bounded interval [a,b], then f must attain its maximum and minimum value, each at least once. That is, there exist numbers c and d in [a,b] such that:

$\begin{array}f(c) &\ge & f(x)& \ge& f(d), & \mbox{for all x} & \in & \[a,b\]\end{array}$ .

Further, from wiki entry on finding maxima and minima,

If a function is continuous on a closed interval, then by the extreme value theorem global maxima and minima exist. Furthermore, a global maximum (or minimum) either must be a local maximum (or minimum) in the interior of the domain, or must lie on the boundary of the domain. So a method of finding a global maximum (or minimum) is to look at all the local maxima (or minima) in the interior, and also look at the maxima (or minima) of the points on the boundary; and take the biggest (or smallest) one.

For our example, the function $f(x) = x^3 - 9x^2 + 24x + 5$ is real valued and is continuous in the closed and bounded interval [1,6].

Finding out the value of the function at the boundaries,

$\begin{array}f(1)&=&1^3 - 9*1^2 + 24*1 + 5 &= & 21\\f(6)&= & 6^3 - 9*6^2 + 24*6 + 5 & =& 41\end{array}$ .

So the global maximum of the function $f(x) = x^3 - 9x^2 + 24x + 5$ in the interval [1, 6] is $\max{$25, 41$} = 41$ and it occurs at the boundary i.e. x = 6.

Matlab example

clear all; close all;
x = [1:.1:6];
fx = x.^3 - 9*x.^2 + 24*x + 5;
plot(x,fx);
xlabel('x'); ylabel('fx');
grid on; axis([1 6 20 45]);
title('fx  =  x^3 - 9x^2 + 24x + 5');

Figure : Plot of the function $f(x) = x^3 - 9x^2 + 24x + 5$ in the interval [1,6]

From the figure, it is relatively easy to observe that the maximum is hit at x=6 (and not at the local maximum point of x=2).

Based on the above, the right choice is (C) 41.

References

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

[2] Wiki entry on Extreme Value Theorem

http://en.wikipedia.org/wiki/Extreme_value_theorem

[3] Wiki entry on Maxima and Minima

http://en.wikipedia.org/wiki/Maxima_and_minima

GATE-2012 ECE Q24 (math)

Question 24 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q24. Two independent random variables X and Y are uniformly distributed in the interval [-1, 1]. The probability that max[X,Y] is less than 1/2 is

(A) 3/4

(B) 9/16

(D) 2/3

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GATE-2012 ECE Q25 (math)

Question 25 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q25. If ${x=\sqrt{-1}}$ , then the value of ${x^x}$ is,

(a) $e^{-\pi/2}$

(b) $e^{\pi/2}$

(d) 1

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GATE-2012 ECE Q6 (digital)

Question 6 on digital circuit from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q6. Consider the given circuit

In this circuit, the race around

(A) does not occur

(B) occurs when CLK=0

(D) occurs when CLK=1 and A=B=0

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Solved objective questions (GATE)

Using the services of a new author ‘RV’, we are starting a new series of articles in the blog. Typically in India, many of the competitive examinations pertaining to Engineering (GATE, IES) and rectuitment by private and public sector companies (ISRO, BSNL, BEL, BHEL) uses examination with objective questions for the first level screening. We are hoping that this Objective Question Answer series, mostly discussing topics pertaining to Electronics and Communication Engineering, will help those preparing for those examination. Kindly do give your feedback via comment and/or email. Thanks, Krishna

Tag: GATE

GATE-2012 ECE Q3 (communication)

Q3. In a baseband communications link, frequencies upto 3500Hz are used for signalling. Using a raised cosine pulse with 75% excess bandwidth and for no inter-symbol interference, the maximum possible signaling rate in symbols per second is,

(A) 1750

(B) 2625

(C) 4000

(D) 5250

Solution

References

GATE-2012 ECE Q26 (electronic devices)

(A) ${\mbox{10^7}$

(B) ${\mbox{100}$

(C) ${\mbox{10}$

(D) ${\mbox{0}$

Solution

References

GATE-2012 ECE Q36 (math)

Q36. A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is

Solution

References

GATE-2012 ECE Q46 (math)

Q46. The maximum value of ${f(x) = x^3 - 9x^2 + 24x + 5}$ in the interval [1, 6] is

Solution

Based on the above, the right choice is (C) 41.

References

GATE-2012 ECE Q24 (math)

Q24. Two independent random variables X and Y are uniformly distributed in the interval [-1, 1]. The probability that max[X,Y] is less than 1/2 is

GATE-2012 ECE Q25 (math)

Q25. If ${x=\sqrt{-1}}$ , then the value of ${x^x}$ is,

GATE-2012 ECE Q6 (digital)

Q6. Consider the given circuit

Solved objective questions (GATE)

Continue reading “Solved objective questions (GATE)”

Q3. In a baseband communications link, frequencies upto 3500Hz are used for signalling. Using a raised cosine pulse with 75% excess bandwidth and for no inter-symbol interference, the maximum possible signaling rate in symbols per second is,

(A) 1750

(B) 2625

(C) 4000

(D) 5250

Solution

References

Q26. The source of a silicon (), n-channel MOS transistor has an area of and a depth of . If the dopant density in the source is , the number of holes in the source region with above volume is approximately

(A)

(B)

(C)

(D)

Solution

References

Q36. A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is

Solution

References

Q46. The maximum value of in the interval [1, 6] is

Solution

Based on the above, the right choice is (C) 41.

References

Q24. Two independent random variables X and Y are uniformly distributed in the interval [-1, 1]. The probability that max[X,Y] is less than 1/2 is

Q25. If , then the value of is,

Q6. Consider the given circuit

(A) ${\mbox{10^7}$

(B) ${\mbox{100}$

(C) ${\mbox{10}$

(D) ${\mbox{0}$

Q46. The maximum value of ${f(x) = x^3 - 9x^2 + 24x + 5}$ in the interval [1, 6] is

Q25. If ${x=\sqrt{-1}}$ , then the value of ${x^x}$ is,