Signals and Systems

Question 1
The transfer function of a stable discrete-time LTI system is H(z)=\frac{K(z-\alpha )}{z+0.5}, where K and \alpha are real numbers. The value of \alpha (rounded off to one decimal place) with |\alpha| \gt 1, for which the magnitude response of the system is constant over all frequencies, is _____.
A
-1
B
1
C
2
D
-2
GATE EC 2020      Z-Transform
Question 1 Explanation: 
System is all-pass filter.
For digital all-pass filter, condition is
\text{zero}=\frac{1}{\text{pole}^{*}}\; \; ...(i)
By given transfer function,
\text{zero}=\alpha
\text{Pole}=-0.5
using condition(i),
\alpha =\frac{1}{-0.5}=-2
Question 2
X(\omega ) is the Fourier transform of x(t) shown below. The value of \int_{-\infty }^{\infty }|X(\omega )|^2 d\omega (rounded off to two decimal places) is ______.
A
58.61
B
42.45
C
26.12
D
78.92
GATE EC 2020      Fourier Transforms, Frequency Response and Correlation
Question 2 Explanation: 
\begin{aligned} \int_{-\infty }^{\infty }\left | X(\omega ) \right |^{2}d\omega &=2\pi \int_{-\infty }^{\infty }\left | x(t) \right |^{2}dt \\ &=2\pi \int_{-\infty }^{\infty }\left | y(t) \right |^{2}\\ &=2\times 2\pi \int_{-2}^{0}\left | y(t) \right |^{2}dt\\ &=2\times 2\pi\left [ \int_{-2}^{-1}(t+2)^{2}dt+\int_{-1}^{0}(2t+3)^{2}dt \right ]\\&=4\pi \left [\left \{ \frac{(t+2)^{3}}{3} \right \} _{-1}^{-2}+\left \{ \frac{(2t+3)^3}{3\times 2} \right \}_{0}^{-1} \right ]\\&=4\pi \left [ \frac{1-0}{3}+\frac{3^{3}-1}{6} \right ]\\&=4\pi \left [ \frac{1}{3}+\frac{26}{6} \right ]\\&= 4\pi \times \left [ \frac{1}{3}+\frac{26}{6} \right ] \\ &=4\pi \times \left [ \frac{1}{3}+\frac{13}{3} \right ]\\&=4\pi \times \frac{14}{3} \\&=\frac{56\pi }{3}=58.61\end{aligned}
Question 3
A finite duration discrete-time signal x[n] is obtained by sampling the continuous-time signal x(t) = \cos (200 \pi t) at sampling instants t = n/400, n = 0,1,..., 7. The 8-point discrete Fourier transform (DFT) of x[n] is defined as

x[k]=\sum_{n=0}^{7}x[n]e^{-j\frac{\pi kn}{4}}, k=0,1,...,7.

Which one of the following statement is TRUE?
A
All X[k] are non-zero
B
Only X[4] is non-zero
C
Only X[2] and X [6] are non-zero
D
Only X[3] and X [5] are non-zero
GATE EC 2020      DTFS, DTFT and DFT
Question 3 Explanation: 
\begin{aligned}x(t)&=\cos 200\pi t\\ t&=\frac{n}{400}\\ x(n)&=\cos \left ( 200\pi \frac{n}{400} \right )\\&=\cos \left ( \frac{\pi }{2} n\right ); \; \; n=0,1,...,7\\ &=\left \{ \cos 0,\cos \frac{\pi }{2},\cos \pi ,\cos \frac{3\pi }{2} ,\cos 2\pi ,\cos \frac{5\pi }{2},\cos 3\pi ,\cos \frac{7\pi }{2}\right \} \\ &={1,0,-1,0,1,0,-1,0}\xLeftrightarrow{\text{DFT}} X(K) \\ \text{suppose, } y(n)&={1,-1,1,-1}\xLeftrightarrow{\text{DFT}} Y(k)\end{aligned}
[Y(K)]=\left.\begin{matrix} [W_{N}] \end{matrix}\right|_{N=4}[y(n)]=\begin{bmatrix} 1 &1 &1 &1 \\ 1 &-j &-1 &j \\ 1&-1 &1 &-1 \\ 1&j& -1 &-j \end{bmatrix}\begin{bmatrix} 1\\ -1\\ 1\\ -1 \end{bmatrix}=\begin{bmatrix} 0 &0 &4 &0 \end{bmatrix}
Now,as we know,
If for \, \left \{ a,b,c,d \right \}\xLeftrightarrow{\text{DFT}} \left \{ A,B,C,D \right \}
Then for, \left \{ a,0,b,0,c,0,d,0 \right \}\xLeftrightarrow{\text{DFT}}\left \{A,B,C,D,A,B,C,D \right \}
Similarly for y(n)={1,-1,1,-1}\xLeftrightarrow{\text{DFT}}Y(k)=\left \{ 0,0,4,0\right \}
Here, for x(n)=\left \{ 1,0,-1,0,1,0,-1,0 \right \}
X(k)=\left \{ \underset{0}{\uparrow},0,4,0,0,0,4,0 \right \}
Question 4
Which one of the following pole-zero corresponds to the transfer function of an LTI system characterized by the input-output difference equation given below?
y[n]=\sum_{k=0}^{3}(-1)^k x[n-k]
A
A
B
B
C
C
D
D
GATE EC 2020      LTI Systems Continuous and Discrete
Question 4 Explanation: 
\begin{aligned}y(n)&=\sum_{K=0}^{3}(-1)^K x(n-K) \\ &=x(n)-x(n-1)+x(n-2)-x(n-3) \\ y(z)&=X(z)-z^{-1}X(z)+Z^{-2}X(z)-Z^-3X(z) \\ H(z)&=\frac{Y(z)}{X(z)}=1-Z^{-1}+Z^{-2}-Z^{-3}\\ &=\frac{z^{3}-z^{2}+z-1}{z^{3}}=\frac{(z-1)(z^{2}+1)}{Z^{3}}\end{aligned}
Pole Zero plot:
Question 5
The output y[n] of a discrete-time system for an input x[n] is

y[n]=\begin{matrix} max\\ -\infty \leq k\leq n \end{matrix}\; |x[k]|.

The unit impulse response of the system is
A
0 for all n
B
1 for all n
C
unit step signal u[n].
D
unit impulse signal \delta[n].
GATE EC 2020      LTI Systems Continuous and Discrete
Question 6
Let h[n] be a length-7 discrete-time finite impulse response filter, given by

h[0]=4, h[1]=3, h[2]=2, h[3]=1,
h[-1]=-3, h[-2]=-2, h[-3]=-1,

and h[n] is zero for |n|\geq 4. A length-3 finite impulse response approximation g[n] of g[n] has to be obtained such that

E(h,g)=\int_{-\pi}^{\pi}|H(e^{j\omega })-G(e^{j\omega })|^2d\omega

is minimized, where H(e^{j\omega }) and G(e^{j\omega }) are the discrete-time Fourier transforms of h[n] and g[n], respectively. For the filter that minimizes E(h,g), the value of 10g[-1]+g[1], rounded off to 2 decimal places, is ________
A
-18.05
B
-32.45
C
-27.00
D
-20.25
GATE EC 2019      Digital Filters
Question 6 Explanation: 
From Parseval's theorem,
\sum_{n=-\infty}^{\infty}|x[n]|^{2}=\frac{1}{2 \pi} \int_{-\pi}^{\pi}\left|X\left(e^{j \omega}\right)\right|^{2} d \omega
So,
\int_{-\pi}^{\pi}\left|H\left(e^{j \omega}\right)-G\left(e^{j \omega}\right)\right|^{2} d \omega=2 \pi \sum_{n=-3}^{3}|h(n)-g(n)|^{2}
The solution of g(n) that minimizes E(h, g) also
minimizes \sum_{n=-3}^{3}|h(n)-g(n)|^{2}
\sum_{n=-3}^{3}|h(n)-g(n)|^{2}
=|4-g(0)|^{2}+|3-g(1)|^{2}+|-3-g(-1)|^{2}+10
The solution of g(n) that minimizes the above
equation is
g(n)=(-3,4,3\}
So, 10 g(-1)+g(1)=10(-3)+3=-27
Question 7
It is desired to find a three-tap causal filter which gives zero signal as an output to an input of the form

x[n]=c_1 \; exp\left ( -\frac{j\pi n}{2} \right )+c_2 \; exp\left ( \frac{j\pi n}{2} \right )

where c_1 \; and \; c_2 are arbitrary real numbers. The desired three-tap filter is given by

h[0]=1,\;\; h[1]=a, \;\;h[2]=b
and
h[n]=0\;\; for \; n \lt 0 \; or \; n\gt 2

What are the values of the filter taps a and b if the output is y[n]=0 for all n, when x[n] is as given above?

A
a=1, b=1
B
a=0, b=-1
C
a=-1, b=1
D
a=0, b=1
GATE EC 2019      DTFS, DTFT and DFT
Question 7 Explanation: 
\begin{aligned} x(n) &=c_{1} e^{-j \frac{\pi}{2} n}+c_{2} e^{j \frac{\pi}{2} n} \\ \omega_{0} &=\frac{\pi}{2} \mathrm{rad} / \mathrm{s} \\ H\left(e^{i \omega}\right) &=1+a e^{-j \omega}+b e^{-j 2 \omega}\\ \text{To get }y(n)=0& \\ H\left(e^{j \omega_{o}}\right)&=\left.H\left(e^{j \omega}\right)\right|_{\omega=\frac{\pi}{2}}=0\\ 1+a e^{-i \frac{\pi}{2}}+b e^{-i 2 \frac{\pi}{2}}&=0 \\ 1-j a-b&=0 \end{aligned}
From the given options, a=0 and b=1
Question 8
Consider a six-point decimation-in-time Fast Fourier Transform (FFT) algorithm, for which the signal-flow graph corresponding to X[1] is shown in the figure. Let W_6=exp\left ( -\frac{j2 \pi}{6} \right ). In the figure, what should be the values of the coefficients a_1,a_2,a_3 in terms of W_6 so that X[1] is obtained correctly?

A
a_1=-1,a_2=W_6,a_3=W_6^2
B
a_1=1,a_2=W_6^2,a_3=W_6
C
a_1=1,a_2=W_6,a_3=W_6^2
D
a_1=-1,a_2=W_6^2,a_3=W_6
GATE EC 2019      DTFS, DTFT and DFT
Question 8 Explanation: 
\begin{array}{l} X(k)=\sum_{n=0}^{N-1} x(n) e^{-j \frac{2 \pi}{N} k n} \\ X(1)=\sum_{n=0}^{5} x(n) W_{6}^{n}\\ =x(0)+x(1) W_{6}+x(2) M_{6}^{2}+x(3) n_{6}^{3}+x(4) M_{6}^{4}+x(5) n_{6}^{5} \ldots(i) \\ \end{array}
From the given flow graph,
X(k)=[x(0)-x(3)] a_{1}+[x(1)-x(4)] a_{2}+[x(2)-x(5)] a_{3}
By comparing equations (i) and (ii), we get.
a_{1}=1, a_{2}=W_{6}, a_{3}=W_{6}^{2}
Question 9
Consider the signal f(t)=1+2cos(\pi t)+3sin\left ( \frac{2 \pi}{3}t \right )+4cos\left (\frac{\pi}{2}t+\frac{\pi}{4} \right ), where t is in seconds. Its fundamental time period, in seconds, is ____________
A
8
B
12
C
16
D
20
GATE EC 2019      Basics of Signals and Systems
Question 9 Explanation: 
\begin{aligned} f(t)=1+2 \cos (\pi t)&+3 \sin \left(\frac{2 \pi}{3} t\right)+4 \cos \left(\frac{\pi}{2} t+\frac{\pi}{4}\right) \\ \omega_{1}&=\pi \\ \omega_{2}&=\frac{2 \pi}{3} \\ \omega_{3}&=\frac{\pi}{2} \\ \omega_{0}&=G C D\left(\pi, \frac{2 \pi}{3}, \frac{\pi}{2}\right)=\frac{\pi}{6} \end{aligned}
Fundamental period,
N=\frac{2 \pi}{\omega_{0}}=\frac{2 \pi}{(\pi / 6)}=12
Question 10
Let Y(s) be the unit-step response of a causal system having a transfer function
G(s)=\frac{3-s}{(s+1)(s+3)}

that is, Y(s)=\frac{G(s)}{s}. The forced response of the system is
A
u(t)-2e^{-t}u(t)+e^{-3t}u(t)
B
2u(t)-2e^{-t}u(t)+e^{-3t}u(t)
C
2u(t)
D
u(t)
GATE EC 2019      Laplace Transform
Question 10 Explanation: 
Given, \quad G(s)=\frac{3-s}{(s+1)(s+3)}
\therefore \quad Y(s)=\frac{G(s)}{s}=\frac{3-s}{s(s+1)(s+3)}
Using partial fractions, we get,
\begin{aligned} Y(s)&=\frac{A}{s}+\frac{B}{(s+1)}+\frac{C}{(s+3)} \\ A\left(s^{2}+4 s+3\right)&+B\left(s^{2}+3 s\right)+C\left(s^{2}+s\right)=3-s \\ A+B+C&=0\\ 4 A+3 B+C&=-1 \\ \text{and }3 A&=3 \\ \text{Therefore, }&\text{we get,}\\ A=1, B&=-2 \text { and } C=1\\ \text{So, }\quad Y(s)&=\frac{1}{s}-\frac{2}{(s+1)}+\frac{1}{(s+3)} \\ \text{and}\quad \mathrm{y}(t)&=u(t)-2 e^{-t} u(t)+e^{-3 t} u(t) \\ \end{aligned}
Forced response,
y_{t}(t)=u(t) \Rightarrow \text { option }(D)


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