Question 1 |
Which of the following statement(s) is/are true?
If an LTI system is causal, it is stable | |
A discrete time LTI system is causal if and only if its response to a step input u[n] is 0 for \mathrm{n} \lt 0 | |
If a discrete time LTI system has an impulse response h[n] of finite duration the system is stable | |
If the impulse response 0 \lt |h[n]| \lt 1 for all n, then the L T I system is stable. |
Question 1 Explanation:
For causal system, impuse response
h(n)=0 ; \quad n \lt 0
Therefore, for step input also
h(n)=0 ; n \lt 0
h(n)=0 ; \quad n \lt 0
Therefore, for step input also
h(n)=0 ; n \lt 0
Question 2 |
Consider the system as shown below
where y(t)=x(e^t) . The system is
where y(t)=x(e^t) . The system is
linear and causal. | |
linear and non-causal. | |
non-linear and causal | |
non-linear and non-causal |
Question 2 Explanation:
We know, a linear system follows the law of
superposition.
It is a combination of two laws:
(i) Law of additivity:
Both results are same, hence, it follows law of additivity.
(ii) Law of Homogeneity:
Here also both results are same, hence it follows law of Homogeneity.
Therefore, System is linear.
We know, a causal system is independent of future values of input at each & every instant of time them system will be causal.
Given : y(t)=x(e^t)
Put t=0
y(0)=x(e^0)=x(1)
Because its depends on future value.
Therefore, system is non-causal.
It is a combination of two laws:
(i) Law of additivity:
Both results are same, hence, it follows law of additivity.
(ii) Law of Homogeneity:
Here also both results are same, hence it follows law of Homogeneity.
Therefore, System is linear.
We know, a causal system is independent of future values of input at each & every instant of time them system will be causal.
Given : y(t)=x(e^t)
Put t=0
y(0)=x(e^0)=x(1)
Because its depends on future value.
Therefore, system is non-causal.
Question 3 |
Let a causal LTI system be governed by the following differential equation y(t)+\frac{1}{4}\frac{dy}{dt}=2x(t) , where x(t) and x(t) are the input and output respectively.
Its impulse response is
2e^{-\frac{1}{4}t}u(t) | |
2e^{-{4}t}u(t) | |
8e^{-\frac{1}{4}t}u(t) | |
8e^{-{4}t}u(t) |
Question 3 Explanation:
Given:
y(t)+\frac{1}{4}\frac{dy}{dt}=2x(t)
Taking Laplace transform,
Y(s)+\frac{1}{4}(sY(s))=2X(s)
Now, H(s)=\frac{Y(s)}{X(s)}=\frac{2}{\frac{s}{4}+1}=\frac{8}{s+4}
Taking inverse Laplace, transform,
h(t)=8e^{-4t}u(t)
y(t)+\frac{1}{4}\frac{dy}{dt}=2x(t)
Taking Laplace transform,
Y(s)+\frac{1}{4}(sY(s))=2X(s)
Now, H(s)=\frac{Y(s)}{X(s)}=\frac{2}{\frac{s}{4}+1}=\frac{8}{s+4}
Taking inverse Laplace, transform,
h(t)=8e^{-4t}u(t)
Question 4 |
If the input x(t) and output y(t) of a system are related as y\left ( t \right )=\text{max}\left ( 0,x\left ( t \right ) \right ), then the system is
linear and time-variant | |
linear and time-invariant | |
non-linear and time-variant | |
non-linear and time-invariant |
Question 4 Explanation:
\begin{aligned} y(t) &=\max (0, x(t)) \\ &=\left\{\begin{array}{cl} 0, & x(t) \lt 0 \\ x(t), & x(t)\gt 0 \end{array}\right. \end{aligned}
Linearity check:
at input x_{1}(t)=-2, output y_{1}(t)=0
at input x_{2}(t)=1, output y_{2}(t)=1
\therefore system is non-linear because it violates law of additivity.
Check for time-invariance :
Delayed \mathrm{O} / \mathrm{P}:
y\left(t-t_{0}\right)=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right) \gt 0 \\ 0 & x\left(t-t_{0}\right)\lt 0 \end{array}\right.
\mathrm{O} / \mathrm{P} of system when input is x\left(t-t_{0}\right)=f(t)
y_{1}(t)=\left\{\begin{array}{cl} f(t), & f(t) \gt 0 \\ 0, & f(t)\lt 0 \end{array}=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right)\gt 0 \\ 0, & x\left(t-t_{0}\right)\lt 0 \end{array}\right.\right.
Therefore, system is time-invariant.
Linearity check:
at input x_{1}(t)=-2, output y_{1}(t)=0
at input x_{2}(t)=1, output y_{2}(t)=1
\therefore system is non-linear because it violates law of additivity.
Check for time-invariance :
Delayed \mathrm{O} / \mathrm{P}:
y\left(t-t_{0}\right)=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right) \gt 0 \\ 0 & x\left(t-t_{0}\right)\lt 0 \end{array}\right.
\mathrm{O} / \mathrm{P} of system when input is x\left(t-t_{0}\right)=f(t)
y_{1}(t)=\left\{\begin{array}{cl} f(t), & f(t) \gt 0 \\ 0, & f(t)\lt 0 \end{array}=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right)\gt 0 \\ 0, & x\left(t-t_{0}\right)\lt 0 \end{array}\right.\right.
Therefore, system is time-invariant.
Question 5 |
Which of the following options is true for a linear time-invariant discrete time system that
obeys the difference equation:
y[n]-ay[n-1]=b_0x[n]-b_1x[n-1]
y[n]-ay[n-1]=b_0x[n]-b_1x[n-1]
y[n] is unaffected by the values of x[n - k]; k \gt 2 | |
The system is necessarily causal. | |
The system impulse response is non-zero at infinitely many instants. | |
When x[n] = 0, n \lt 0, the function y[n]; n \gt 0 is solely determined by the function x[n]. |
Question 5 Explanation:
\begin{aligned}
y(n)-ay(n-1)&=b_{0}x(n)-b_{1}x(n-2) \\ &\text{By applying ZT,} \\ Y(z)-az^{-1}Y(z)&=b_{0}X(z)-b_{1}z^{-1}X(z)\\ \Rightarrow \, \, H(z)&=\frac{Y(z)}{X(z)}=\frac{b_{0}-b_{1}z^{-1}}{1-az^{-1}}
\end{aligned}
By taking right-sided inverse ZT,
h(n)=b_{0}a^{n}u(n)-b_{1}a^{n-1}u(n-1)
By taking left-sided inverse ZT,
h(n)=-b_{0}a^{n}u(-n-1)+b_{1}a^{n-1}u(-n)
Thus system is not necessarily causal.
The impulse response is non-zero at infinitely many instants.
By taking right-sided inverse ZT,
h(n)=b_{0}a^{n}u(n)-b_{1}a^{n-1}u(n-1)
By taking left-sided inverse ZT,
h(n)=-b_{0}a^{n}u(-n-1)+b_{1}a^{n-1}u(-n)
Thus system is not necessarily causal.
The impulse response is non-zero at infinitely many instants.
There are 5 questions to complete.