Question 1 |
The magnitude and phase plots of an LTI system are shown in the figure. The transfer function of the system is
2.51 \mathrm{e}^{-0.032 \mathrm{~s}} | |
\frac{\mathrm{e}^{-2.514 \mathrm{~s}}}{\mathrm{~s}+1} | |
1.04 \mathrm{e}^{-2.514 \mathrm{~s}} | |
2.51 e^{-1.047\mathrm{~s}} |
Question 1 Explanation:
Transfer function of transportation lag system,
\mathrm{T}(\mathrm{s})=\mathrm{Ke}^{-\mathrm{ST} T_{\mathrm{d}}}
From magnitude plot,
\begin{aligned} & 20 \log \mathrm{K}=8 \\ & \Rightarrow \quad \mathrm{K}=2.51 \end{aligned}
From angle plot,
at \omega=1 \mathrm{rad} / \mathrm{sec} ., \theta=-60^{\circ}
We have,
\begin{aligned} \theta & =-\omega \mathrm{T}_{\mathrm{d}} \times \frac{180^{\circ}}{\pi} \\ -60^{\circ} & =-1 \times \mathrm{T}_{\mathrm{d}} \times \frac{180^{\circ}}{\pi} \\ \Rightarrow \quad \mathrm{T}_{\mathrm{d}} & =\frac{\pi}{3}=1.047 \\ \therefore \quad \mathrm{T}(\mathrm{s}) & =2.51 \mathrm{e}^{-1.047\mathrm{~s}} \end{aligned}
\mathrm{T}(\mathrm{s})=\mathrm{Ke}^{-\mathrm{ST} T_{\mathrm{d}}}
From magnitude plot,
\begin{aligned} & 20 \log \mathrm{K}=8 \\ & \Rightarrow \quad \mathrm{K}=2.51 \end{aligned}
From angle plot,
at \omega=1 \mathrm{rad} / \mathrm{sec} ., \theta=-60^{\circ}
We have,
\begin{aligned} \theta & =-\omega \mathrm{T}_{\mathrm{d}} \times \frac{180^{\circ}}{\pi} \\ -60^{\circ} & =-1 \times \mathrm{T}_{\mathrm{d}} \times \frac{180^{\circ}}{\pi} \\ \Rightarrow \quad \mathrm{T}_{\mathrm{d}} & =\frac{\pi}{3}=1.047 \\ \therefore \quad \mathrm{T}(\mathrm{s}) & =2.51 \mathrm{e}^{-1.047\mathrm{~s}} \end{aligned}
Question 2 |
A continuous-time system that is initially at rest is described by
\frac{d y(t)}{d t}+3 y(t)=2 x(t)
where x(t) is the input voltage and y(t) is the output voltage. The impulse response of the system is
\frac{d y(t)}{d t}+3 y(t)=2 x(t)
where x(t) is the input voltage and y(t) is the output voltage. The impulse response of the system is
3 e^{-2 t} | |
\frac{1}{3} e^{-2 t} u(t) | |
2 \mathrm{e}^{-3 \mathrm{t}} \mathrm{u}(\mathrm{t}) | |
2 \mathrm{e}^{-3 \mathrm{t}} |
Question 2 Explanation:
Given :
\frac{\mathrm{dy}(\mathrm{t})}{\mathrm{dt}}+3 \mathrm{y}(\mathrm{t})=x(\mathrm{t})
Taking Laplace transform,
\begin{aligned} Y(\mathrm{~s})[\mathrm{s}+3] & =2 X(\mathrm{~s}) \\ \frac{Y(s)}{X(\mathrm{~s})} & =\frac{2}{s+3} \end{aligned}
We have impulse response
=\mathrm{L}^{-1} \text { (Transfer function) }
So, taking inverse Laplace transform,
y(t)=2 e^{-3 t} u(t)
\frac{\mathrm{dy}(\mathrm{t})}{\mathrm{dt}}+3 \mathrm{y}(\mathrm{t})=x(\mathrm{t})
Taking Laplace transform,
\begin{aligned} Y(\mathrm{~s})[\mathrm{s}+3] & =2 X(\mathrm{~s}) \\ \frac{Y(s)}{X(\mathrm{~s})} & =\frac{2}{s+3} \end{aligned}
We have impulse response
=\mathrm{L}^{-1} \text { (Transfer function) }
So, taking inverse Laplace transform,
y(t)=2 e^{-3 t} u(t)
Question 3 |
For the block diagrm shown in the figure, the transfer function \frac{Y(s)}{R(s)} is
\frac{2 s+3}{s+1} | |
\frac{3 s+2}{s-1} | |
\frac{s+1}{3 \mathrm{~s}+2} | |
\frac{3 s+2}{s+1} |
Question 3 Explanation:
Signal flow graph:
Forward paths,
\begin{aligned} & \mathrm{P}_{1}=3, \quad \Delta_{1}=1 \\ & \mathrm{P}_{2}=\frac{2}{\mathrm{~S}}, \Delta_{2}=1 \end{aligned}
Loops: L_1=\frac{1}{S}
Using Masson's graph formula,
\begin{aligned} \frac{Y(s)}{R(s)} & =\frac{P_{1} \Delta_{1}+P_{1} \Delta_{2}}{1-L_{1}} \\ & =\frac{3+\frac{2}{S}}{1-\frac{1}{S}} \\ & =\frac{3 S+2}{S-1} \end{aligned}
Forward paths,
\begin{aligned} & \mathrm{P}_{1}=3, \quad \Delta_{1}=1 \\ & \mathrm{P}_{2}=\frac{2}{\mathrm{~S}}, \Delta_{2}=1 \end{aligned}
Loops: L_1=\frac{1}{S}
Using Masson's graph formula,
\begin{aligned} \frac{Y(s)}{R(s)} & =\frac{P_{1} \Delta_{1}+P_{1} \Delta_{2}}{1-L_{1}} \\ & =\frac{3+\frac{2}{S}}{1-\frac{1}{S}} \\ & =\frac{3 S+2}{S-1} \end{aligned}
Question 4 |
For the closed-loop system shown, the transfer function \dfrac{E\left ( s \right )}{R\left ( s \right )} is
\frac{G}{1+GH} | |
\frac{GH}{1+GH} | |
\frac{1}{1+GH} | |
\frac{1}{1+G} |
Question 4 Explanation:
\begin{aligned} \frac{E(s)}{R(s)} &=\frac{R(s)-H \times C(s)}{R(s)}=1-H \times \frac{C(s)}{R(s)} \\ &=1-\frac{H G}{1+G H}=\frac{1+G H-G H}{1+G H} \\ &=\frac{1}{1+G H} \end{aligned}
Question 5 |
Which of the options is an equivalent representation of the signal flow graph shown here?
A | |
B | |
C | |
D |
Question 5 Explanation:
Simplifying given signal flow graph
There are 5 questions to complete.