Linear Time Invariant Systems

Question 1
Consider the system as shown below

where y(t)=x(e^t) . The system is
A
linear and causal.
B
linear and non-causal.
C
non-linear and causal
D
non-linear and non-causal
GATE EE 2022   Signals and Systems
Question 1 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.
Question 2
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
A
2e^{-\frac{1}{4}t}u(t)
B
2e^{-{4}t}u(t)
C
8e^{-\frac{1}{4}t}u(t)
D
8e^{-{4}t}u(t)
GATE EE 2022   Signals and Systems
Question 2 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)
Question 3
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
A
linear and time-variant
B
linear and time-invariant
C
non-linear and time-variant
D
non-linear and time-invariant
GATE EE 2021   Signals and Systems
Question 3 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.
Question 4
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]
A
y[n] is unaffected by the values of x[n - k]; k \gt 2
B
The system is necessarily causal.
C
The system impulse response is non-zero at infinitely many instants.
D
When x[n] = 0, n \lt 0, the function y[n]; n \gt 0 is solely determined by the function x[n].
GATE EE 2020   Signals and Systems
Question 4 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.
Question 5
A continuous-time input signal x(t) is an eigenfunction of an LTI system, if the output is
A
k x(t) , where k is an eigenvalue
B
k e^{j\omega t} x(t), where k is an eigenvalue and e^{j\omega t} is a complex exponential signal
C
x(t) e^{j\omega t}, where e^{j\omega t} is a complex exponential signal
D
k H(\omega) ,where k is an eigenvalue and H(\omega) is a frequency response of the system
GATE EE 2018   Signals and Systems
Question 5 Explanation: 
Eigen function is a type of input for which output is constant times of input.
i.e.

Where,
x(t)= System input = eigen function
H(s)= transfer function of system
y(t)= system output
Here,
y(t)=H(s)|_{s=a}\; e^{at}=k \cdot x(t)
where,
k= eigen-value =H(s)|_{s=a}
x(t)= eigen-function input
Question 6
Let z(t)=x(t) * y(t) , where "*" denotes convolution. Let c be a positive real-valued constant. Choose the correct expression for z(ct).
A
c x(ct)*y(ct)
B
x(ct)*y(ct)
C
c x(t)*y(ct)
D
c x(ct)*y(t)
GATE EE 2017-SET-1   Signals and Systems
Question 6 Explanation: 
Time scaling property of convolution.
If, x(t)*y(t)=z(t)
Then, x(ct)*y(ct)=\frac{1}{c} z(ct)
z(ct)=c \times x(ct) * y(ct)
Question 7
Consider a causal LTI system characterized by differential equation \frac{dy(t)}{dt}+\frac{1}{6}y(t)=3x(t). The response of the system to the input x(t)=3e^{-\frac{t}{3}}u(t). where u(t) denotes the unit step function, is
A
9e^{-\frac{t}{3}}u(t)
B
9e^{-\frac{t}{6}}u(t)
C
9e^{-\frac{t}{3}}u(t)-6e^{-\frac{t}{6}}u(t)
D
54e^{-\frac{t}{6}}u(t)-54e^{-\frac{t}{3}}u(t)
GATE EE 2016-SET-2   Signals and Systems
Question 7 Explanation: 
The differential equation
\begin{aligned} \frac{dy(t)}{dt} &+\frac{1}{6}y(t)=3x(t) \\ \text{So, }sY(s)&+\frac{1}{6}Y(s) =3X(s) \\ Y(s) &=\frac{3X(s)}{\left ( s+\frac{1}{6} \right )} \\ X(s) &=\frac{9}{\left ( s+\frac{1}{3} \right )} \\ \text{So, } Y(s)&=\frac{9}{\left ( s+\frac{1}{3} \right )\left ( s+\frac{1}{6} \right )} \\ &=\frac{54}{\left ( s+\frac{1}{6} \right )} -\frac{54}{\left ( s+\frac{1}{3} \right )}\\ \text{So, }y(t) &= (54e^{-1/6t}-54e^{-1/3t})u(t) \end{aligned}
Question 8
The output of a continuous-time, linear time-invariant system is denoted by T{x(t)} where x(t) is the input signal. A signal z(t) is called eigen-signal of the system T, when T\{z(t)\} = \gamma z(t), where \gamma is a complex number, in general, and is called an eigenvalue of T. Suppose the impulse response of the system T is real and even. Which of the following statements is TRUE?
A
cos(t) is an eigen-signal but sin(t) is not
B
cos(t) and sin(t) are both eigen-signals but with different eigenvalues
C
sin(t) is an eigen-signal but cos(t) is not
D
cos(t) and sin(t) are both eigen-signals with identical eigenvalues
GATE EE 2016-SET-1   Signals and Systems
Question 8 Explanation: 
Given that impulse response is real and even, Thus H(j\omega ) will also be real and even.

Since, H(j\omega ) is real and even thus,
H(j\omega )=H(-j\omega )
Now, \cos (t) is input i.e. \frac{e^{jt}+e^{-jt}}{2} is input
Output will be
\frac{H(j1)e^{jt}+H(-j1)e^{-jt}}{2}=H(j1)\left ( \frac{e^{jt}+e^{-jt}}{2} \right )= H(j1) \cos (t)
If, \sin (t) is input i.e. \frac{e^{jt}+e^{-jt}}{2} is input
Output will be
\frac{H(j1)e^{jt}+H(-j1)e^{-jt}}{2}=H(j1)\left ( \frac{e^{jt}-e^{-jt}}{2j} \right )= H(j1) \sin (t)
So, \sin (t) and \cos (t) are eigen signal with same eigen values.
Question 9
Consider the following state-space representation of a linear time-invariant system.
\dot{x}(t)=\begin{bmatrix} 1 & 0\\ 0&2 \end{bmatrix}x(t), y(t)=c^{T}x(t), c=\begin{bmatrix} 1\\ 1 \end{bmatrix} and x(0)=\begin{bmatrix} 1\\ 1 \end{bmatrix}
The value of y(t) for t= log_{e}2 is______.
A
4
B
5
C
6
D
7
GATE EE 2016-SET-1   Signals and Systems
Question 10
Consider a continuous-time system with input x(t) and output y(t) given by

y(t) = x(t) cos(t)

This system is
A
linear and time-invariant
B
non-linear and time-invariant
C
linear and time-varying
D
non-linear and time-varying
GATE EE 2016-SET-1   Signals and Systems
Question 10 Explanation: 
\begin{aligned} y(t)&=x(t)\cos (t)\\ &\text{To check linearity,}\\ y_1(t)&=x_1(t)\cos (t)\\ &[y_1(t) \text{ is output for }x_1(t)]\\ y_2(t)&=x_2(t) \cos (t)\\ &[y_2(t) \text{ is output for }x_2(t)]\\ \text{so, the}& \text{ output for }(x_1(t)+ x_2(t)) \text{ will be}\\ y(t)&=[x_1(t)+ x_2(t)]\cos (t)\\ &=y_1(t)+y_2(t) \end{aligned}
So, the system is linear, to check time invariance.
The delayed output,
y(t-t_0)=x(t-t_0)\cos (t-t_0)
The output for delayed input,
y(t, t_0)=x(t-t_0)\cos (t)
Since, y(t-t_0)\neq y(t,t_0)
System is time varying.
There are 10 questions to complete.