Question 1 |

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 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.

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

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 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)

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

linear and time-variant | |

linear and time-invariant | |

non-linear and time-variant | |

non-linear and time-invariant |

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.

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]

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 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.

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

k x(t) , where k is an eigenvalue | |

k e^{j\omega t} x(t), where k is an eigenvalue and e^{j\omega t} is a complex exponential signal | |

x(t) e^{j\omega t}, where e^{j\omega t} is a complex exponential signal | |

k H(\omega) ,where k is an eigenvalue and H(\omega) is a frequency response of the system |

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

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).

c x(ct)*y(ct) | |

x(ct)*y(ct) | |

c x(t)*y(ct) | |

c x(ct)*y(t) |

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)

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

9e^{-\frac{t}{3}}u(t) | |

9e^{-\frac{t}{6}}u(t) | |

9e^{-\frac{t}{3}}u(t)-6e^{-\frac{t}{6}}u(t) | |

54e^{-\frac{t}{6}}u(t)-54e^{-\frac{t}{3}}u(t) |

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}

\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?

cos(t) is an eigen-signal but sin(t) is not | |

cos(t) and sin(t) are both eigen-signals but with different eigenvalues | |

sin(t) is an eigen-signal but cos(t) is not | |

cos(t) and sin(t) are both eigen-signals with identical eigenvalues |

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.

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______.

\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______.

4 | |

5 | |

6 | |

7 |

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

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

This system is

linear and time-invariant | |

non-linear and time-invariant | |

linear and time-varying | |

non-linear and time-varying |

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.

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.