# 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=1, x=2, x=1, h=1.
Let y[n] be the linear convolution of x[n] and h[n]. Given that y = 3 and y = 4, the value of the expression (10y+y 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=1 ; y=2+a ; y=1+2 a+b \\ &y=a+2 b ; y=b \\ &\begin{aligned} \text { Given. } \quad y&=2+a=3 \Rightarrow a=1 \\ \text { Given, } \quad y&=1+2 a+b=4 \Rightarrow b=1 \\ \text { So, } \quad y&=a+2 b=3 \\ \text { and } \quad y&=b=1 \end{aligned}\\ &\text { So. } 10 y+y=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 is _____.
 A 16 B 6 C 10 D 12
GATE EC 2014-SET-2   Signals and Systems
Question 8 Explanation:
$\begin{array}{l} x[n]=\left\{\begin{array}{ll} n & \text { for } 0 \leq n \leq 10 \\ 0 & \text { otherwise } \end{array}\right. \\ y[n]=\sum_{K=\infty}^{\infty} x(k) x(n-k) \\ y=\sum_{k=0}^{10} x(k) x(4-k) \\ =x(0) x(4)+x(1) x(3)+x(2) x(2)+x(3) x(1)+0 \\ =0+(1 \times 3)+(2 \times 2)+(3 \times 1) \\ =10 \end{array}$
 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 = 1 and y = 1/2, then g 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 &=\sum_{k=-\infty}^{\infty} h(-k] g[k] \\ y &=h g \\ 1 &=1 g \\ g &=1 \\ y &=\sum_{k=-\infty}^{\infty} h(1-k] g[k] \\ y &=h g(0]+h 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 \\ g &=0 \end{aligned}
There are 10 questions to complete. 