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)?
Based on the given information, which of the four systems is/are definitely NOT LTI (linear and time-invariant)?
S_1 | |
S_2 | |
S_3 | |
S_4 |
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).
0 | |
0.4 | |
0.6 | |
0.8 |
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 ______
0 | |
0.5 | |
1 | |
1.5 |
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]
y[n]=\sum_{k=0}^{3}(-1)^k x[n-k]
A | |
B | |
C | |
D |
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:
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
y[n]=\begin{matrix} max\\ -\infty \leq k\leq n \end{matrix}\; |x[k]|.
The unit impulse response of the system is
0 for all n | |
1 for all n | |
unit step signal u[n]. | |
unit impulse signal \delta[n]. |
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 ____________.
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 ____________.
1 | |
3 | |
7 | |
9 |
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
\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 _________.
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 _________.
20 | |
18 | |
31 | |
37 |
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}
\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)?
e^{j\omega_{0}t }u(t) | |
cos(\omega _{0}t) | |
e^{j\omega_{0}t } | |
sin(\omega _{0}t) |
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
differentiating the unit ramp response | |
differentiating the unit step response | |
integrating the unit ramp response | |
integrating the unit step response |
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
x(t+t_{0}) | |
x(t-t_{0}) | |
x(-t+t_{o}) | |
x(-t-t_{o}) |
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.