# LTI Systems Continuous and Discrete

 Question 1
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   Signals and Systems
Question 1 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 2
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   Signals and Systems
Question 2 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 3
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   Signals and Systems
Question 3 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 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   Signals and Systems
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   Signals and Systems
 Question 6
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 6 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 7
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 7 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 8
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 8 Explanation:
If the input to the system is eigen signal output is also the same eigen signal.
 Question 9
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 9 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 10
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 10 Explanation:
$x(t)*\delta (-t-t_0)=x(-t)*\delta (t+t_0)=x(-t-t_0)$
There are 10 questions to complete.