DTFS, DTFT and DFT

Question 1
For a vector \bar{x}=\left [x[0],x[1],...,x[7] \right ], the 8-point discrete Fourier transform (DFT) is denoted by \bar{X}=DFT(\bar{x})=\left [X[0],X[1],...,X[7] \right ], where
X[k]=\sum_{n=0}^{7}x[n]exp\left ( -j\frac{2 \pi}{8}nk \right )
Here, j=\sqrt{-1}. If \bar{x}==[1,0,0,0,2,0,0,0] and \bar{y}=DFT(DFT(\bar{x})), then the value of y[0] is _________ (rounded off to one decimal place).
A
3.2
B
6.8
C
4
D
8
GATE EC 2022   Signals and Systems
Question 1 Explanation: 
\begin{aligned} X[K]&=\sum_{n=0}^{7}x[n]exp\left ( -j\frac{2\pi}{8}mk \right )\\ \bar{y}&=DFT(DFT(\bar{x})) \; \; where\\ \bar{x}&=[x[0],x[1],...,x[7]]\\ y(0)&=? \;\;x[n]=[1,0,0,0,2,0,0,0]\\ x[n]&\overset{DFT}{\rightarrow}\overset{DFT}{\rightarrow}N\cdot x(-k)=\bar{y}(n)\\ \bar{y}(n)&=N\cdot x(-k)|_{mod.N}=N\cdot x(N.K),N=8\\ \bar{y}(n)&=8[1,0,0,0,2,0,0,0]\\ \bar{y}(n)&=[8,0,0,0,16,0,0,0]\\ y(0)&=8 \end{aligned}
Question 2
The Fourier transform X(j\omega ) of the signal x(t)=\frac{t}{(1+t^2)^2} is _________.
A
\frac{\pi}{2j}\omega e^{-|\omega|}
B
\frac{\pi}{2}\omega e^{-|\omega|}
C
\frac{\pi}{2j} e^{-|\omega|}
D
\frac{\pi}{2} e^{-|\omega|}
GATE EC 2022   Signals and Systems
Question 2 Explanation: 
x(t)=\frac{t}{(1+t^2)^2}
As we know that FT of te^{-|t|} \; \underleftrightarrow{FT} \;\frac{-j4\omega }{(1+\omega ^2)^2}
Duality \frac{-j4\omega }{(1+t ^2)^2} \leftrightarrow 2 \pi(-\omega )e^{-|-\omega |}
\Rightarrow \frac{t}{(1+t^2)^2} \underrightarrow{FT} \frac{-2\pi}{-j4}\omega e^{-|\omega |}
\Rightarrow \;\;\;\rightarrow\frac{\pi}{j2} \omega e^{-|\omega |}
Question 3
Consider the signals x\left [ n \right ]=2^{n-1}u\left [ -n+2 \right ] and y\left [ n \right ]=2^{-n+2}u\left [ n+1 \right ], where u[n] is the unit step sequence. Let X(e^{jw}) and Y(e^{jw}) be the discrete-time Fourier transform of x[n] and y[n], respectively. The value of the integral
\frac{1}{2\pi } \int_{0}^{2\pi }X\left ( e^{j\omega } \right )Y\left ( e^{-j\omega } \right )d\omega
(rounded off to one decimal place) is ______
A
2.4
B
12.5
C
8
D
10.8
GATE EC 2021   Signals and Systems
Question 3 Explanation: 
\begin{aligned} x[n] &=2^{n-1} u[-n+2] \\ y[n] &=2^{-n+2} u[n+1] \\ y[-n] &=2^{n+2} u[-n+1] \\ V &=\frac{1}{2 \pi} \int_{0}^{2 \pi} X\left(e^{j \omega}\right) Y\left(e^{-j \omega}\right) d \omega &\ldots(i)\\ z[n] &=x[n] * y[-n] \\ z[n] & \rightarrow Z\left(e^{j \omega}\right) \\ z[n] &=\frac{1}{2 \pi} \int_{-\pi}^{\pi} Z\left(e^{j \omega}\right) e^{j \omega n} d \omega \\ \text { Put } n=0, \qquad z[0] &=\frac{1}{2 \pi} \int_{0}^{2 \pi} Z\left(e^{j \omega}\right) d \omega &\ldots(ii) \end{aligned}
Compare equations (i) and (ii)
\begin{aligned} z[0] &=V \\ Z\left(e^{j \omega}\right) &=X\left(e^{j \omega}\right) Y\left(e^{-j \omega}\right) \\ \text{Apply IDTFT},\qquad Z[n] &=x[n] * y[-n]=x[n] * p[n]\\ \because \qquad p[n]&=y[-n]=2^{n+2} u[-n+1]\\ z[n] &=\sum_{k=-\infty}^{+\infty} 2^{k-1} u[-k+2] 2^{n-k+2} u[-n+k+1] \\ &=\sum_{k=-\infty}^{2} 2^{k-1} 1 \cdot 2^{n-k+2} u[k+1-n] \\ &=\sum_{k=-\infty}^{2} 2^{k-1+n-k+2} u[k+1-n] \\ z[n] &=\sum_{k=-\infty}^{2} 2^{n+1} u[k-n+1] \\ \text { Put } n=0,\qquad \qquad V &=z[0]=\sum_{k=-\infty}^{2} 2^{1} \cdot u[k+1]=2 \sum_{k=-1}^{2} 1=2[1+1+1+1] \\ V &=8 \end{aligned}
Question 4
Consider two 16-point sequences x[n] and h[n]. Let the linear convolution of x[n] and h[n] be denoted by y[n], while z[n] denotes the 16-point inverse discrete Fourier transform (IDFT) of the product of the 16-point DFTs of x[n] and h[n]. The value(s) of k for which z[k]=y[k] is/are
A
k=0,1,2,,15
B
k=0
C
k=15
D
k=0 and k=15
GATE EC 2021   Signals and Systems
Question 4 Explanation: 
If two' N' point signals x(n) and h(n) are convolving with each other linearly and circularly
then
y(k)=z(k) at k=N-1
where, y(n)= Linear convolution of x(n) and h(n)
z(n)= Circular convolution of x(n) and h(n)
Since, N=16 (Given)
Therefore, \quad y(k)=z(k) at k=N-1=15
Question 5
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   Signals and Systems
Question 5 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 6
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   Signals and Systems
Question 6 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 7
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   Signals and Systems
Question 7 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 8
Let X[k]=k+1, 0\leq k \leq 7 be 8-point DFT of a sequence x[n],
where X[k]= \sum_{n=0}^{N-1} x[n]e^{-\frac{j2nk\pi }{N}}.
The value (correct to two decimal places) of \sum_{n=0}^{3}x[2n] is ______.
A
1.9
B
3
C
4
D
5
GATE EC 2018   Signals and Systems
Question 8 Explanation: 
\begin{aligned} x(k) &=\{1,2,3,4,5,6,7,8\} \\ x(k) &=\sum_{n=0}^{7} x[n] e^{-j \frac{2 \pi}{8} k n} \\ &=\sum_{n=0}^{7} x[n] e^{-i \frac{\pi}{4} k n} \\ \sum_{n=0}^{3} x[2 n] &=\sum_{n=0.2 .4 .6} x[n] \\ &=\frac{1}{2} \sum_{n=0}^{7}\left(x[n]+(-1)^{n} x[n]\right) \\ \sum_{n=0}^{7} x[n] &=x(0)=1 \\ \sum_{n=0}^{7}(-1)^{n} x[n] &=\sum_{n=0}^{7} x[n] e^{-i \frac{\pi}{4} 4 n}=x(4)=5 \\ \sum_{n=0}^{3} x[2 n] &=\frac{1}{2}[1+5]=3 \end{aligned}
Question 9
An LTI system with unit sample response h[n]=5\delta [n]-7\delta [n-1]+7\delta [n-3]-5\delta [n-4] is a
A
Low - pass filter
B
high - pass filter
C
band - pass filter
D
band - stop filter
GATE EC 2017-SET-2   Signals and Systems
Question 9 Explanation: 
\begin{aligned} h[n]=5 \delta[n]&-78[n-1]+7 \delta[n-3]-5 \delta[n-4] & \\ \text { Now, } \quad H\left(e^{j \omega}\right)&=5-7 e^{-j \omega}+7 e^{-3 j \omega}-5 e^{-4 j \omega}\\ \text{Now for }\omega & =0, \\ H\left(e^{j 0}\right)&=5-7+7-5=0\\ \text{and for }\omega & =\pi, &\\ H\left(e^{j \pi}\right) &=5-7(-1)+7(-1)-5(1) \\ &=5+7-7-5=0 \end{aligned}
System is attenuating low and high frequencies whereas passing the mid frequencies. So, its a BPF.
Question 10
Let h[n] be the impulse response of a discrete-time linear time invariant (LTI) filter. The impulse response is given by

h[0]=1/3; \; h[1]=1/3; \; h[2]=1/3; \; and \; h[n]=0 for n \lt 0 \; and \; n \gt 2.

Let H(\omega) be the discrete-time Fourier system transform (DTFT) of h[n], where \omega is the normalized angular frequency in radians. Given that H(\omega_{o})=0 and 0 \lt \omega _{o} \lt \pi, the value of \omega_{0} (in radians) is equal to __________.
A
1.25
B
2.10
C
3.50
D
4.75
GATE EC 2017-SET-1   Signals and Systems
Question 10 Explanation: 
\begin{aligned} \text{since }\quad h[n]&=\frac{1}{3} \delta[n]+\frac{1}{3} \delta[n-1]+\frac{1}{3} \delta[n-2] \\ \text{So. }\quad H\left(e^{j \omega}\right)&=\frac{1}{3} e^{-j \omega}[1+2 \cos \omega] \\ H\left(e^{j \omega_{0}}\right)&=0, \quad\text{ for }\left(1+2 \cos \omega_{0}\right)=0 \\ \text{or} \quad \cos \omega_{0}&=-\frac{1}{2} \\ \text{or}\quad \omega_{0}&=\frac{2 \pi}{3}=2.10 \mathrm{radians} \end{aligned}
There are 10 questions to complete.