Signals and Systems

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      DTFS, DTFT and DFT
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 outputs of four systems (S_1,S_2,S_3,S_4) corresponding to the input signal \sin (t), for all time t, are shown in the figure.
Based on the given information, which of the four systems is/are definitely NOT LTI (linear and time-invariant)?

A
S_1
B
S_2
C
S_3
D
S_4
GATE EC 2022      LTI Systems Continuous and Discrete
Question 2 Explanation: 


Since, LTI systems does not change the frequency of sinusoidal input. So S3 and S4 are definitely not LTI as input and output sinusoidal frequencies are different.
Question 3
Let x_1(t)=e^{-t}u(t) and x_2(t)=u(t)-u(t-2) , where u(\cdot) denotes the unit step function. If y(t) denotes the convolution of x_1(t) and x_2(t), then \lim_{t\rightarrow \infty }y(t)= _________ (rounded off to one decimal place).
A
0
B
0.4
C
0.6
D
0.8
GATE EC 2022      LTI Systems Continuous and Discrete
Question 3 Explanation: 
\begin{aligned} x_1(t)=e^{-1}U(t)&\underleftrightarrow{L.T.}\frac{1}{(s+1)}\\ x_2(t)=U(t)-U(t-2)&\underleftrightarrow{L.T.}\frac{1}{s}-\frac{e^{-2s}}{s}=\frac{1-e^{-2s}}{s}\\ y(t)=x_1(t)\otimes x_2(t)&\underleftrightarrow{L.T.} \frac{(1-e^{-2s})}{s(s+1)} \\ y(t)&\underleftrightarrow{L.T.}\frac{1-e^{-2s}}{s(s+1)}\\ \lim_{t \to \infty }y(t)&=\lim_{s \to 0 }\frac{s(1-e^{-2s})}{s(s+1)}=0 \end{aligned}
Question 4
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      DTFS, DTFT and DFT
Question 4 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 5
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      DTFS, DTFT and DFT
Question 5 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 6
For a unit step input u[n], a discrete-time \text{LTI} system produces an output signal \left ( 2\delta \left [ n+1 \right ] +\delta \left [ n \right ]+\delta \left [ n-1 \right ]\right ). Let y[n] be the output of the system for an input \left ( \left ( \dfrac{1}{2} \right )^n u\left [ n \right ]\right ). The value of y[0] is ______
A
0
B
0.5
C
1
D
1.5
GATE EC 2021      LTI Systems Continuous and Discrete
Question 6 Explanation: 


The impulse response h(n)=s(n)-s(n-1)
\begin{aligned} h(n)=2 \delta(n+1)+\delta(n)&+ \delta(n-1)-2 \delta(n)-\delta(n-1)-\delta(n-1-1) \\ &=2 \delta(n+1)+\delta(n)+\delta(n-1)-2 \delta(n)-\delta(n-1)-\delta(n-2) \\ h(n) &=2 \delta(n+1)-\delta(n)-\delta(n-2) \\ \text{For input }\qquad x_{1}(n)&=(1 / 2)^{n} u(n) \text{ then output } y_{1}(n)=x_{1}(n) * h(n)\\ y_{1}(n)&=\left(\frac{1}{2}\right)^{n} u(n) *[2 \delta(n+1)-\delta(n)-\delta(n-2)]\\ y_{1}(n)&=\left(\frac{1}{2}\right)^{n} u(n) * 2 \delta(n+1)-\left(\frac{1}{2}\right)^{n} u(n) * \delta(n)-\left(\frac{1}{2}\right)^{n} u(n) * \delta(n-2) \\ y_{1}(n)&=2\left(\frac{1}{2}\right)^{n+1} u(n+1)-\left(\frac{1}{2}\right)^{n} u(n)-\left(\frac{1}{2}\right)^{n-2} u(n-2) \\ \left.y_{1}(n)\right|_{n=0} &=2\left(\frac{1}{2}\right)^{1} u(1)-\left(\frac{1}{2}\right)^{0} u(0)-\left(\frac{1}{2}\right)^{-2} u(-2) \\ &=1-1 \\ y_{1}(0) &=0 \end{aligned}
Question 7
The exponential Fourier series representation of a continuous-time periodic signal x(t) is defined as
x\left ( t \right )=\sum_{k=-\infty }^{\infty }a_{k}e^{jk\omega _{0}t}
where \omega _0 is the fundamental angular frequency of x(t) and the coefficients of the series are a_{k}. The following information is given about x(t) and a_{k}.

I. x(t) is real and even, having a fundamental period of 6
II. The average value of x(t) is 2
III. a_{k}=\left\{\begin{matrix} k, & 1\leq k\leq 3\\ 0,& k> 3 \end{matrix}\right.

The average power of the signal x(t) (rounded off to one decimal place) is ____________
A
14
B
45
C
63
D
32
GATE EC 2021      Fourier Transforms, Frequency Response and Correlation
Question 7 Explanation: 
1. x(t) is real and even so a_{k} is also real and even a_{k}=a_{-k}
2. Average of x(t) is 2 i.e., a_{0}=2.
\begin{array}{llll} 3. \;x(t) \rightarrow a_{k}=k &1 \leq k \leq 3 \quad& a_{1}=1 \quad& a_{-1}=1 \\ \quad \quad \quad \quad \;\;\;\;\;\;\;\;0 & k>3 & a_{2}=2 & a_{-2}=2 \\ && a_{3}=3 & a_{-3}=3 \end{array}
4. \quad T_{0}=6
Parsval's Power Theorem
\begin{aligned} \frac{1}{T} \int_{0}^{T}\left|x(t)^{2}\right| d t &=\sum_{n=-\infty}^{+\infty}\left|a_{k}\right|^{2} \\ P(t) &=\sum_{n=-\infty}^{+\infty}\left|a_{k}\right|^{2}=\left|a_{-3}\right|^{2}+\left|a_{-2}\right|^{2}+\left|a_{-1}\right|^{2}+\left|a_{0}\right|^{2}+\left|a_{1}\right|^{2}+\left|a_{2}\right|^{2}+\left|a_{3}\right|^{2} \\ &=2\left|a_{1}\right|^{2}+2\left|a_{2}\right|^{2}+2\left|a_{3}\right|^{2}+\left|a_{0}\right|^{2} \\ &=2 \times 1^{2}+2 \times(2)^{2}+2(3)^{2}+(2)^{2} \\ &=2+8+18+4 \\ P_{x(t)} &=32 \end{aligned}
Question 8
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      DTFS, DTFT and DFT
Question 8 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 9
Consider a real-valued base-band signal x(t), band limited to \text{10 kHz}. The Nyquist rate for the signal y\left ( t \right )=x\left ( t \right )x\left ( 1+\dfrac{t}{2} \right ) is
A
\text{15 kHz}
B
\text{30 kHz}
C
\text{60 kHz}
D
\text{20 kHz}
GATE EC 2021      Sampling
Question 9 Explanation: 






\mathrm{NR}=2 \times f_{\mathrm{max}}=2 \times 15=30 \mathrm{kHz}
Question 10
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 10 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


There are 10 questions to complete.