Question 1 |
Let Y(s) be the unit-step response of a causal system having a transfer function
G(s)=\frac{3-s}{(s+1)(s+3)}
that is, Y(s)=\frac{G(s)}{s}. The forced response of the system is
G(s)=\frac{3-s}{(s+1)(s+3)}
that is, Y(s)=\frac{G(s)}{s}. The forced response of the system is
u(t)-2e^{-t}u(t)+e^{-3t}u(t) | |
2u(t)-2e^{-t}u(t)+e^{-3t}u(t) | |
2u(t) | |
u(t) |
Question 1 Explanation:
Given, \quad G(s)=\frac{3-s}{(s+1)(s+3)}
\therefore \quad Y(s)=\frac{G(s)}{s}=\frac{3-s}{s(s+1)(s+3)}
Using partial fractions, we get,
\begin{aligned} Y(s)&=\frac{A}{s}+\frac{B}{(s+1)}+\frac{C}{(s+3)} \\ A\left(s^{2}+4 s+3\right)&+B\left(s^{2}+3 s\right)+C\left(s^{2}+s\right)=3-s \\ A+B+C&=0\\ 4 A+3 B+C&=-1 \\ \text{and }3 A&=3 \\ \text{Therefore, }&\text{we get,}\\ A=1, B&=-2 \text { and } C=1\\ \text{So, }\quad Y(s)&=\frac{1}{s}-\frac{2}{(s+1)}+\frac{1}{(s+3)} \\ \text{and}\quad \mathrm{y}(t)&=u(t)-2 e^{-t} u(t)+e^{-3 t} u(t) \\ \end{aligned}
Forced response,
y_{t}(t)=u(t) \Rightarrow \text { option }(D)
\therefore \quad Y(s)=\frac{G(s)}{s}=\frac{3-s}{s(s+1)(s+3)}
Using partial fractions, we get,
\begin{aligned} Y(s)&=\frac{A}{s}+\frac{B}{(s+1)}+\frac{C}{(s+3)} \\ A\left(s^{2}+4 s+3\right)&+B\left(s^{2}+3 s\right)+C\left(s^{2}+s\right)=3-s \\ A+B+C&=0\\ 4 A+3 B+C&=-1 \\ \text{and }3 A&=3 \\ \text{Therefore, }&\text{we get,}\\ A=1, B&=-2 \text { and } C=1\\ \text{So, }\quad Y(s)&=\frac{1}{s}-\frac{2}{(s+1)}+\frac{1}{(s+3)} \\ \text{and}\quad \mathrm{y}(t)&=u(t)-2 e^{-t} u(t)+e^{-3 t} u(t) \\ \end{aligned}
Forced response,
y_{t}(t)=u(t) \Rightarrow \text { option }(D)
Question 2 |
The transfer function of a causal LTI system is H(s) = 1/s. If the input to the system is x(t) =[sin(t) / \pi t] u(t) , where u(t) is a unit step function, the system output y(t) as t \rightarrow \infty is______.
0 | |
0.5 | |
1 | |
2 |
Question 2 Explanation:
\begin{aligned} H(s) &=\frac{1}{s} \Rightarrow \text { integrator } \\ x(t) &=\frac{\sin t}{\pi t} u(t) \\ \text { So. } \quad y(t) &=\int_{0}^{t} \frac{\sin \tau}{\pi \tau} d t \\ y(t)_{t \rightarrow \infty} &=\frac{1}{\pi} \int_{0}^{\infty} \frac{\sin \tau}{\tau} d t \\ \int_{0}^{\infty} \frac{\sin \tau}{\tau} d t &=\frac{\pi}{2} \\ \text { So. } \quad y(t)_{t \rightarrow \infty} &=\frac{1}{\pi} \times \frac{\pi}{2}=\frac{1}{2}=0.5 \end{aligned}
Question 3 |
Consider the following statements for continuous-time linear time invariant (LTI) systems.
I. There is no bounded input bounded output (BIBO) stable system with a pole in the righthalf of the complex plane.
II. There is non causal and BIBO stable system with a pole in the right half of the complex plane.
Which one among the following is correct?
I. There is no bounded input bounded output (BIBO) stable system with a pole in the righthalf of the complex plane.
II. There is non causal and BIBO stable system with a pole in the right half of the complex plane.
Which one among the following is correct?
Both I and II are true | |
Both I and II are not true | |
Only I is true | |
Only II is true |
Question 3 Explanation:
A 8180 stable system can have poles in right
half of complex plane, if it is a non-causal
system. So, statement-I is wrong.
A causal and BIBO stable system should have all poles in the left half of complex plane. So, statement-II is correct.
A causal and BIBO stable system should have all poles in the left half of complex plane. So, statement-II is correct.
Question 4 |
A signal 2cos(\frac{2\pi t}{3})-cos(\pi t) is the input to an LTI system with the transfer function
H(s)=e^{s}+e^{-s}.
If C_{k} denotes the k^{th} coefficient in the exponential Fourier series of the output signal, then C_{3} is equal to
H(s)=e^{s}+e^{-s}.
If C_{k} denotes the k^{th} coefficient in the exponential Fourier series of the output signal, then C_{3} is equal to
0 | |
1 | |
2 | |
3 |
Question 4 Explanation:
Given. H(s)=e^{s}+e^{-s}
H\left(e^{i \omega}\right)=e^{j \omega}+e^{-j \omega}=2 \cos \omega
\begin{aligned} \text{if}\quad x_{1}(t)&=2 \cos \left(\frac{2 \pi}{3} t\right) \\ \omega_{b} &=\frac{2 \pi}{3} \\ H\left(\dot{\mu}_{0}\right) &=2 \cos \left(\frac{2 \pi}{3}\right)=2\left(-\frac{1}{2}\right)=-1 \\ y_{1}(t) &=2 \cos \left(\frac{2 \pi}{3} t+180^{\circ}\right) \\ \text{if}\quad x_{2}(t)&=\cos \pi t \\ \omega_{b} &=\pi \\ H\left(e^{i \theta_{0}}\right) &=2 \cos (\pi)=-2 \\ y_{2}(t) &=2 \cos \left(\pi t+180^{\circ}\right) \\ y(t)=2 \cos &\left(\frac{2 \pi}{3} t+\pi\right)-2 \cos (\pi t+\pi) \\ \omega_{1}&=\frac{2 \pi}{3}, \omega_{2}=\pi \\ \omega_{1} &=3 \quad T_{2}=2 \\ \therefore \quad T_{0} &=6 \\ \Rightarrow \quad & 0_{0}=\frac{2 \pi}{T_{0}}=\frac{\pi}{3} \end{aligned}
\begin{aligned} &\mathcal{V}(t)=2 \cos \left(2 \omega_{0} t+\pi\right)-2 \cos \left(3 \omega_{0} t+\pi\right)\\ &\mathcal{N}(t)=e^{j(2 \operatorname{cod} t \pi)}+e^{-i\left(2 \omega_{0} \theta^{\prime}+\pi\right)}-e^{j\left(3 \omega_{0} t+\pi\right)}-e^{-i\left(3 \omega_{0} t+\pi\right)}\\ &y(t)=-e^{j\left(2 \omega_{0} t\right)}-e^{-j\left(2 \omega_{0} t\right)}+e^{j\left(3 \omega_{0} t\right)}+e^{-j\left(3 \omega_{0} t\right)}\\ &\therefore \quad C_{3}=1 \end{aligned}
H\left(e^{i \omega}\right)=e^{j \omega}+e^{-j \omega}=2 \cos \omega
\begin{aligned} \text{if}\quad x_{1}(t)&=2 \cos \left(\frac{2 \pi}{3} t\right) \\ \omega_{b} &=\frac{2 \pi}{3} \\ H\left(\dot{\mu}_{0}\right) &=2 \cos \left(\frac{2 \pi}{3}\right)=2\left(-\frac{1}{2}\right)=-1 \\ y_{1}(t) &=2 \cos \left(\frac{2 \pi}{3} t+180^{\circ}\right) \\ \text{if}\quad x_{2}(t)&=\cos \pi t \\ \omega_{b} &=\pi \\ H\left(e^{i \theta_{0}}\right) &=2 \cos (\pi)=-2 \\ y_{2}(t) &=2 \cos \left(\pi t+180^{\circ}\right) \\ y(t)=2 \cos &\left(\frac{2 \pi}{3} t+\pi\right)-2 \cos (\pi t+\pi) \\ \omega_{1}&=\frac{2 \pi}{3}, \omega_{2}=\pi \\ \omega_{1} &=3 \quad T_{2}=2 \\ \therefore \quad T_{0} &=6 \\ \Rightarrow \quad & 0_{0}=\frac{2 \pi}{T_{0}}=\frac{\pi}{3} \end{aligned}
\begin{aligned} &\mathcal{V}(t)=2 \cos \left(2 \omega_{0} t+\pi\right)-2 \cos \left(3 \omega_{0} t+\pi\right)\\ &\mathcal{N}(t)=e^{j(2 \operatorname{cod} t \pi)}+e^{-i\left(2 \omega_{0} \theta^{\prime}+\pi\right)}-e^{j\left(3 \omega_{0} t+\pi\right)}-e^{-i\left(3 \omega_{0} t+\pi\right)}\\ &y(t)=-e^{j\left(2 \omega_{0} t\right)}-e^{-j\left(2 \omega_{0} t\right)}+e^{j\left(3 \omega_{0} t\right)}+e^{-j\left(3 \omega_{0} t\right)}\\ &\therefore \quad C_{3}=1 \end{aligned}
Question 5 |
A first-order low-pass filter of time constant T is excited with different input signals (with zero initial conditions up to t = 0). Match the excitation signals X, Y, Z with the corresponding time responses for t\geq 0:
X\rightarrowR, Y\rightarrowQ, Z\rightarrowP | |
X\rightarrowQ, Y\rightarrowP, Z\rightarrowR | |
X\rightarrowR, Y\rightarrowP, Z\rightarrowQ | |
X\rightarrowP, Y\rightarrowR, Z\rightarrowQ |
Question 5 Explanation:
For 1st order system
G(s)=\frac{1}{s T} ; H(s)=1
Impulse response \quad A(s)=1
\begin{aligned} n(s)&=\left(\frac{G(s)}{1+G(s) H(s)} R(s)\right) \\ &=\left(\frac{1}{1+s T}\right)=\frac{1}{T} e^{-t / T} \text { for } t \geq 0 \\ \text { Unit step response } &\quad A(s)=\frac{1}{s}\\ \eta(s) &=\frac{1}{s(1+s T)}=\frac{(1+s T)-(s T)}{s(1+s T)} \\ &=\frac{1}{s}-\frac{T}{(1+s T}=\frac{1}{s}-\frac{T}{T\left(s+\frac{1}{T}\right)} \end{aligned}
\begin{aligned} y(t)&=1-e^{-t / T} \text { for } t \geq 0 \\ \text { Ramp response } &\quad R(s)=\frac{1}{s^{2}}\\ \eta(s)&=\frac{1}{s^{2}(1+s T)}=\frac{1}{s^{2}}-\frac{T}{s}+\frac{T}{s+\frac{1}{T}} \\ y(t)&=t-T\left(1-e^{-t / T}\right) \quad\text{ for }t \geq 0 \end{aligned}
G(s)=\frac{1}{s T} ; H(s)=1
Impulse response \quad A(s)=1
\begin{aligned} n(s)&=\left(\frac{G(s)}{1+G(s) H(s)} R(s)\right) \\ &=\left(\frac{1}{1+s T}\right)=\frac{1}{T} e^{-t / T} \text { for } t \geq 0 \\ \text { Unit step response } &\quad A(s)=\frac{1}{s}\\ \eta(s) &=\frac{1}{s(1+s T)}=\frac{(1+s T)-(s T)}{s(1+s T)} \\ &=\frac{1}{s}-\frac{T}{(1+s T}=\frac{1}{s}-\frac{T}{T\left(s+\frac{1}{T}\right)} \end{aligned}
\begin{aligned} y(t)&=1-e^{-t / T} \text { for } t \geq 0 \\ \text { Ramp response } &\quad R(s)=\frac{1}{s^{2}}\\ \eta(s)&=\frac{1}{s^{2}(1+s T)}=\frac{1}{s^{2}}-\frac{T}{s}+\frac{T}{s+\frac{1}{T}} \\ y(t)&=t-T\left(1-e^{-t / T}\right) \quad\text{ for }t \geq 0 \end{aligned}
There are 5 questions to complete.