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.