Question 1 |

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 1 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 2 |

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 2 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 3 |

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 3 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 4 |

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 4 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 5 |

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 5 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 6 |

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 7 |

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 7 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.

Question 8 |

The following discrete-time equations result from the numerical integration of the differential equations of an un-damped simple harmonic oscillator with state variables x and y. The integration time step is h.

\frac{x_{k+1}-x_{k}}{h}=y_{k}

\frac{y_{k+1}-y_{k}}{h}=-x_{k}

For this discrete-time system, which one of the following statements is TRUE?

\frac{x_{k+1}-x_{k}}{h}=y_{k}

\frac{y_{k+1}-y_{k}}{h}=-x_{k}

For this discrete-time system, which one of the following statements is TRUE?

The system is not stable for h\gt 0 | |

The system is stable for h \gt \frac{1}{\pi } | |

The system is stable for 0 \lt h \lt \frac{1}{2\pi } | |

The system is stable for \frac{1}{2\pi } \lt h \lt \frac{1}{\pi } |

Question 9 |

For linear time invariant systems, that are Bounded Input Bounded Output stable, which one of the following statements is TRUE?

The impulse response will be integrable, but may not be absolutely integrable. | |

The unit impulse response will have finite support. | |

The unit step response will be absolutely integrable | |

The unit step response will be bounded. |

Question 10 |

The impulse response g(t) of a system, G , is as shown in Figure (a). What is the maximum value attained by the impulse response of two cascaded blocks of G as shown in Figure (b)?

\frac{2}{3} | |

\frac{3}{4} | |

\frac{4}{5} | |

1 |

Question 10 Explanation:

\begin{aligned} g(t)&=u(t)-u(t-1)\\ G(s)&=\frac{1}{s}-\frac{e^{-s}}{s}\\ G(s) \times G(s)&=g(t)* g(t) \end{aligned}

Maximum value =1

Maximum value =1

There are 10 questions to complete.