LTI Systems Continuous and Discrete

Question 1
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   Signals and Systems
Question 1 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 2
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   Signals and Systems
Question 3
Consider the parallel combination of two LTI systems shown in the figure.

The impulse responses of the systems are
h_{1}(t)=2\delta (t+2)-3\delta (t+1)
h_{2}(t)=\delta(t-2)
If the input x(t) is a unit step signal, then the energy of y(t) is ____________.
A
1
B
3
C
7
D
9
GATE EC 2017-SET-2   Signals and Systems
Question 3 Explanation: 
Given that,
\begin{aligned} h_1(t)&=2\delta (t+2)-3\delta (t+1)\\ h_2(t)&=\delta (t-2) \end{aligned}
Overall impulse response is,
\begin{aligned} h(t)&=h_1(t)+h_2(t)\\ &=2\delta (t+2)-3\delta (t+1) +\delta (t-2) \end{aligned}
If input, x(t)=u(t), then the output will be,
\begin{aligned} y(t)&=x(t)*h(t)\\ &=u(t)*[2\delta (t+2)-3\delta (t+1) \delta (t-2)]\\ &=2u(t+2)-3u(t+1)+u(t-2) \end{aligned}
By plotting y(t), we get,

The energy of y(t) can be given as,
E_y=\int_{-\infty }^{\infty }y^2(t)dt
By plotting y^2(t), we get,

E_y=\text{Area under the plot of }y^2(t)=(4 \times 1)+(3 \times 1)=7J
Question 4
Two discrete-time signals x[n] and h[n] are both non-zero for n = 0, 1, 2 and are zero otherwise. It is given that
x[0]=1, x[1]=2, x[2]=1, h[0]=1.
Let y[n] be the linear convolution of x[n] and h[n]. Given that y[1] = 3 and y[2] = 4, the value of the expression (10y[3]+y[4] is _________.
A
20
B
18
C
31
D
37
GATE EC 2017-SET-1   Signals and Systems
Question 4 Explanation: 
\begin{array}{l} x[n]=\{1,2,1\} \\ h(n]=\{ 1, a, b\} \\ y[n]=\{A, 3,4, B, C\} \end{array}


\begin{aligned} &y[0]=1 ; y[1]=2+a ; y[2]=1+2 a+b \\ &y[3]=a+2 b ; y[4]=b \\ &\begin{aligned} \text { Given. } \quad y[1]&=2+a=3 \Rightarrow a=1 \\ \text { Given, } \quad y[2]&=1+2 a+b=4 \Rightarrow b=1 \\ \text { So, } \quad y[3]&=a+2 b=3 \\ \text { and } \quad y[4]&=b=1 \end{aligned}\\ &\text { So. } 10 y[3]+y[4]=10 \times 3+1=31 \end{aligned}
Question 5
Which one of the following is an eigen function of the class of all continuous-time, linear, timeinvariant systems (u(t) denotes the unit-step function)?
A
e^{j\omega_{0}t }u(t)
B
cos(\omega _{0}t)
C
e^{j\omega_{0}t }
D
sin(\omega _{0}t)
GATE EC 2016-SET-1   Signals and Systems
Question 5 Explanation: 
If the input to the system is eigen signal output is also the same eigen signal.
Question 6
The impulse response of an LTI system can be obtained by
A
differentiating the unit ramp response
B
differentiating the unit step response
C
integrating the unit ramp response
D
integrating the unit step response
GATE EC 2015-SET-3   Signals and Systems
Question 6 Explanation: 
\begin{aligned} IR&= \frac{d}{dt}(SR)\\ &\text{Step response,} \\ y(t)&=\int_{-\infty }^{t}h(\tau )\cdot d\tau \\ h(t)&= \frac{d}{dt}y(t) \end{aligned}
Question 7
The result of the convolution x(-t)*\delta (-t-t_{o}) is
A
x(t+t_{0})
B
x(t-t_{0})
C
x(-t+t_{o})
D
x(-t-t_{o})
GATE EC 2015-SET-1   Signals and Systems
Question 7 Explanation: 
x(t)*\delta (-t-t_0)=x(-t)*\delta (t+t_0)=x(-t-t_0)
Question 8
Consider a discrete-time signal
x[n]=\left\{\begin{matrix} n & for \; 0\leq n\leq 10\\ 0&otherwise \end{matrix}\right.
If y[n] is the convolution of x[n] with itself, the value of y[4] is _____.
A
16
B
6
C
10
D
12
GATE EC 2014-SET-2   Signals and Systems
Question 8 Explanation: 
\begin{array}{l} x[n]=\left\{\begin{array}{ll} n & \text { for } 0 \leq n \leq 10 \\ 0 & \text { otherwise } \end{array}\right. \\ y[n]=\sum_{K=\infty}^{\infty} x(k) x(n-k) \\ y[4]=\sum_{k=0}^{10} x(k) x(4-k) \\ =x(0) x(4)+x(1) x(3)+x(2) x(2)+x(3) x(1)+0 \\ =0+(1 \times 3)+(2 \times 2)+(3 \times 1) \\ =10 \end{array}
Question 9
Two systems with impulse responses h_{1}\left ( t \right ) and h_{2}\left ( t \right ) are connected in cascade. Then the overall impulse response of the cascaded system is given by
A
product of h_{1}\left ( t \right ) and h_{2}\left ( t \right )
B
sum of h_{1}\left ( t \right ) and h_{2}\left ( t \right )
C
convolution of h_{1}\left ( t \right ) and h_{2}\left ( t \right )
D
subtraction of h_{2}\left ( t \right ) from h_{1}\left ( t \right )
GATE EC 2013   Signals and Systems
Question 9 Explanation: 
The overall impulse response h(t) of the cascade system is given by:
h(t)=h_{1}(t) * h_{2}(t)
Question 10
Let y[n] denote the convolution of h[n] and g[n], where h[n]=(1/2)^{n}u[n] and g[n] is a causal sequence. If y[0] = 1 and y[1] = 1/2, then g[1] equals
A
0
B
1/2
C
1
D
3/2
GATE EC 2012   Signals and Systems
Question 10 Explanation: 
\begin{aligned} h[n]&=\left(\frac{1}{2}\right)^{n} u[n] \\ g[n]&=? ? \end{aligned}


y[n]=\sum_{k=-\infty}^{\infty} h[n-k] g[k]


\begin{aligned} y[0] &=\sum_{k=-\infty}^{\infty} h(-k] g[k] \\ y[0] &=h[0] g[0] \\ 1 &=1 g[0] \\ g[0] &=1 \\ y[1] &=\sum_{k=-\infty}^{\infty} h(1-k] g[k] \\ y[1] &=h[1] g(0]+h[0] g(1] \end{aligned}
h[1-k] will be zero for k \gt 1 and g[k] will be z
for k \lt 0 as it is causal sequence.
\begin{aligned} \frac{1}{2} &=\frac{1}{2} \times 1+1 g[1] \\ g[1] &=0 \end{aligned}
There are 10 questions to complete.
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