Question 1 |
Consider the state-space description of an LTI system with matrices
A=\left[\begin{array}{cc}0 & 1 \\ -1 & -2\end{array}\right], B=\left[\begin{array}{l}0 \\ 1\end{array}\right], C=[3-2], D=1
For the input, \sin (\omega t), \omega \gt 0, the value of \omega for which the steady-state output of the system will be zero, is ___ (Round off to the nearest integer).
A=\left[\begin{array}{cc}0 & 1 \\ -1 & -2\end{array}\right], B=\left[\begin{array}{l}0 \\ 1\end{array}\right], C=[3-2], D=1
For the input, \sin (\omega t), \omega \gt 0, the value of \omega for which the steady-state output of the system will be zero, is ___ (Round off to the nearest integer).
0 | |
1 | |
2 | |
3 |
Question 1 Explanation:
We have, transfer function
\mathrm{T}(\mathrm{s})=\mathrm{C}[\mathrm{SI}-\mathrm{A}]^{-1} \mathrm{~B}+\mathrm{D} \quad (1)
where, [S I-A]=\left[\begin{array}{ll}S & 0 \\ 0 & S\end{array}\right]-\left[\begin{array}{cc}0 & 1 \\ -1 & -2\end{array}\right]
\begin{aligned} & =\left[\begin{array}{cc} \mathrm{s} & -1 \\ 1 & \mathrm{~s}+2 \end{array}\right] \\ \therefore \quad[\mathrm{sl}-\mathrm{A}]^{-1} & =\frac{1}{\mathrm{~s}^{2}+2 \mathrm{~s}+1}\left[\begin{array}{cc} \mathrm{s}+2 & 1 \\ -1 & \mathrm{~s} \end{array}\right] \end{aligned}
From eqn. (1), we get
\begin{aligned} \mathrm{T}(\mathrm{s})&=\left[\begin{array}{ll} 3 & -2 \end{array}\right]\left[\begin{array}{cc} \frac{\mathrm{s}+2}{(\mathrm{~s}+1)^{2}} & \frac{1}{(s+1)^{2}} \\ -\frac{1}{(s+1)^{2}} & \frac{s}{(s+1)^{2}} \end{array}\right]\left[\begin{array}{l} 0 \\ 1 \end{array}\right]+1 \\ &=\left[\begin{array}{ll} 3 & -2 \end{array}\right]\left[\begin{array}{c} \frac{1}{(s+1)^{2}} \\ \frac{s}{(s+1)^{2}} \end{array}\right]+1 \\ &=\frac{3}{(s+1)^{2}}-\frac{2 s}{(s+1)}+1 \\ &=\frac{3-2 s+s^{2}+2 s+1}{s^{2}+2 s+1} \\ &=\frac{s^{2}+4}{s^{2}+2 s+1} \end{aligned}
Put, s=j \omega,
T(j \omega)=\frac{-\omega^{2}+4}{-\omega^{2}+\mathrm{j} 2 \omega+1}
Now, condition for output is zero,
\begin{aligned} -\omega^{2}+4 & =0 \\ \Rightarrow \quad \omega & =2 \mathrm{rad} / \mathrm{sec} . \end{aligned}
\mathrm{T}(\mathrm{s})=\mathrm{C}[\mathrm{SI}-\mathrm{A}]^{-1} \mathrm{~B}+\mathrm{D} \quad (1)
where, [S I-A]=\left[\begin{array}{ll}S & 0 \\ 0 & S\end{array}\right]-\left[\begin{array}{cc}0 & 1 \\ -1 & -2\end{array}\right]
\begin{aligned} & =\left[\begin{array}{cc} \mathrm{s} & -1 \\ 1 & \mathrm{~s}+2 \end{array}\right] \\ \therefore \quad[\mathrm{sl}-\mathrm{A}]^{-1} & =\frac{1}{\mathrm{~s}^{2}+2 \mathrm{~s}+1}\left[\begin{array}{cc} \mathrm{s}+2 & 1 \\ -1 & \mathrm{~s} \end{array}\right] \end{aligned}
From eqn. (1), we get
\begin{aligned} \mathrm{T}(\mathrm{s})&=\left[\begin{array}{ll} 3 & -2 \end{array}\right]\left[\begin{array}{cc} \frac{\mathrm{s}+2}{(\mathrm{~s}+1)^{2}} & \frac{1}{(s+1)^{2}} \\ -\frac{1}{(s+1)^{2}} & \frac{s}{(s+1)^{2}} \end{array}\right]\left[\begin{array}{l} 0 \\ 1 \end{array}\right]+1 \\ &=\left[\begin{array}{ll} 3 & -2 \end{array}\right]\left[\begin{array}{c} \frac{1}{(s+1)^{2}} \\ \frac{s}{(s+1)^{2}} \end{array}\right]+1 \\ &=\frac{3}{(s+1)^{2}}-\frac{2 s}{(s+1)}+1 \\ &=\frac{3-2 s+s^{2}+2 s+1}{s^{2}+2 s+1} \\ &=\frac{s^{2}+4}{s^{2}+2 s+1} \end{aligned}
Put, s=j \omega,
T(j \omega)=\frac{-\omega^{2}+4}{-\omega^{2}+\mathrm{j} 2 \omega+1}
Now, condition for output is zero,
\begin{aligned} -\omega^{2}+4 & =0 \\ \Rightarrow \quad \omega & =2 \mathrm{rad} / \mathrm{sec} . \end{aligned}
Question 2 |
Consider a lead compensator of the form
K(s)=\frac{1+\frac{s}{a}}{1+\frac{s}{\beta a}}, \quad \beta \gt 1, a \gt 0
The frequency at which this compensator produces maximum phase lead is 4 \mathrm{rad} / \mathrm{s}. At this frequency, the gain amplification provided by the controller, assuming asymptotic Bodemagnitude plot of K(\mathrm{~s}), is 6 \mathrm{~dB}. The values of \alpha, \beta, respectively, are
K(s)=\frac{1+\frac{s}{a}}{1+\frac{s}{\beta a}}, \quad \beta \gt 1, a \gt 0
The frequency at which this compensator produces maximum phase lead is 4 \mathrm{rad} / \mathrm{s}. At this frequency, the gain amplification provided by the controller, assuming asymptotic Bodemagnitude plot of K(\mathrm{~s}), is 6 \mathrm{~dB}. The values of \alpha, \beta, respectively, are
1, 16 | |
2, 4 | |
3, 5 | |
2.66, 2.25 |
Question 2 Explanation:
K(s)=\frac{1+\frac{s}{a}}{1+\frac{s}{\beta}}
Max. phase lead occur at, \omega_{m}=\sqrt{a(a \beta)} =\sqrt{\beta}
Given : \quad \omega_{\mathrm{m}}=4 \mathrm{rad} / \mathrm{sec}
a \sqrt{\beta}=4 \quad ...(1)
Now, \quad K(s)=\frac{1+\frac{s}{a}}{1+\frac{s}{\beta a}}
Put, s=j \omega
\mathrm{K}\left(\mathrm{j} \omega_{\mathrm{n}}\right)=\frac{1+\frac{\mathrm{j} \omega_{m}}{a}}{1+\frac{j \omega_{m}}{\beta a}}=\frac{j w_{m}}{a}
Given : M=6
\begin{aligned} 20 \log \left(\frac{4}{a}\right) & =6 \\ \Rightarrow \quad a & =2 \end{aligned}
From eqn. (1), we get
\begin{aligned} 2 \sqrt{\beta} & =4 \\ \beta & =4 \end{aligned}
Max. phase lead occur at, \omega_{m}=\sqrt{a(a \beta)} =\sqrt{\beta}
Given : \quad \omega_{\mathrm{m}}=4 \mathrm{rad} / \mathrm{sec}
a \sqrt{\beta}=4 \quad ...(1)
Now, \quad K(s)=\frac{1+\frac{s}{a}}{1+\frac{s}{\beta a}}
Put, s=j \omega
\mathrm{K}\left(\mathrm{j} \omega_{\mathrm{n}}\right)=\frac{1+\frac{\mathrm{j} \omega_{m}}{a}}{1+\frac{j \omega_{m}}{\beta a}}=\frac{j w_{m}}{a}
Given : M=6
\begin{aligned} 20 \log \left(\frac{4}{a}\right) & =6 \\ \Rightarrow \quad a & =2 \end{aligned}
From eqn. (1), we get
\begin{aligned} 2 \sqrt{\beta} & =4 \\ \beta & =4 \end{aligned}
Question 3 |
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 3 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 4 |
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 4 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 5 |
Consider a unity-gain negative feedback system consisting of the plant G(s) (given below) and a proportional-integral controller. Let the proportional gain and integral gain be 3 and 1, respectively. For a unit step reference input, the final values of the controller output and the plant output, respectively, are
G(s)=\frac{1}{s-1}
G(s)=\frac{1}{s-1}
\infty, \infty | |
1,0 | |
1,-1 | |
-1,1 |
Question 5 Explanation:
Given plant:
So, \quad OLTF =\frac{(3 s+1)}{s(s-1)}
Closed loop transfer function,
\begin{aligned} \frac{Y(s)}{R(s)} & =\frac{3 s+1}{s^{2}+2 s+1} \\ Y(s) & =\frac{3 s+1}{s\left(s^{2}+2 s+1\right)} \quad\left[\because R(s)=\frac{1}{s}\right] \end{aligned}
Final value of plant,
Y(\infty)=\operatorname{Lims}_{s \rightarrow 0} Y(s)=1
From plant,
\begin{aligned} X(\mathrm{~s}) & =\left[\mathrm{R}(\mathrm{s})-\frac{\mathrm{X}(\mathrm{s})}{\mathrm{s}-1}\right]\left(3+\frac{1}{\mathrm{~s}}\right) \\ X(\mathrm{~s})\left[1+\frac{3 \mathrm{~s}+1}{\mathrm{~s}(\mathrm{~s}-1)}\right] & =\left(\frac{3 \mathrm{~s}+1}{\mathrm{~s}^{2}}\right)\left[\because \mathrm{R}(\mathrm{s})=\frac{1}{\mathrm{~s}}\right] \\ X(\mathrm{~s})\left[\frac{\mathrm{s}^{2}+2 s+1}{\mathrm{~s}(\mathrm{~s}-1)}\right] & =\left(\frac{3 \mathrm{~s}+1}{\mathrm{~s}^{2}}\right) \\ \Rightarrow \quad X(\mathrm{~s}) & =\frac{(3 \mathrm{~s}+1)(\mathrm{s}-1)}{\mathrm{s}\left(\mathrm{s}^{2}+2 \mathrm{~s}+1\right)} \end{aligned}
\therefore Final value of controller,
x(\infty)=\operatorname{LimsX}_{s \rightarrow 0} \mathrm{~s}(\mathrm{~s})=-1
So, \quad OLTF =\frac{(3 s+1)}{s(s-1)}
Closed loop transfer function,
\begin{aligned} \frac{Y(s)}{R(s)} & =\frac{3 s+1}{s^{2}+2 s+1} \\ Y(s) & =\frac{3 s+1}{s\left(s^{2}+2 s+1\right)} \quad\left[\because R(s)=\frac{1}{s}\right] \end{aligned}
Final value of plant,
Y(\infty)=\operatorname{Lims}_{s \rightarrow 0} Y(s)=1
From plant,
\begin{aligned} X(\mathrm{~s}) & =\left[\mathrm{R}(\mathrm{s})-\frac{\mathrm{X}(\mathrm{s})}{\mathrm{s}-1}\right]\left(3+\frac{1}{\mathrm{~s}}\right) \\ X(\mathrm{~s})\left[1+\frac{3 \mathrm{~s}+1}{\mathrm{~s}(\mathrm{~s}-1)}\right] & =\left(\frac{3 \mathrm{~s}+1}{\mathrm{~s}^{2}}\right)\left[\because \mathrm{R}(\mathrm{s})=\frac{1}{\mathrm{~s}}\right] \\ X(\mathrm{~s})\left[\frac{\mathrm{s}^{2}+2 s+1}{\mathrm{~s}(\mathrm{~s}-1)}\right] & =\left(\frac{3 \mathrm{~s}+1}{\mathrm{~s}^{2}}\right) \\ \Rightarrow \quad X(\mathrm{~s}) & =\frac{(3 \mathrm{~s}+1)(\mathrm{s}-1)}{\mathrm{s}\left(\mathrm{s}^{2}+2 \mathrm{~s}+1\right)} \end{aligned}
\therefore Final value of controller,
x(\infty)=\operatorname{LimsX}_{s \rightarrow 0} \mathrm{~s}(\mathrm{~s})=-1
There are 5 questions to complete.