Question 1 |
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 1 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
Question 2 |
The block diagram of a system is shown in the figure
If the desired transfer function of the system is
\frac{C(s)}{R(s)}=\frac{s}{s^{2}+s+1}
then G(s) is
If the desired transfer function of the system is
\frac{C(s)}{R(s)}=\frac{s}{s^{2}+s+1}
then G(s) is
1 | |
s | |
1/s | |
\frac{-s}{s^{3}+s^{2}-s-2} |
Question 2 Explanation:
For all given values of G(s) in options, G(s) does not satisfy the desired condition. But option (B) is official answer given by IIT.
Question 3 |
The closed-loop transfer function of a system is T(s)=\frac{4}{(s^{2}+0.4s+4)}. The steady
state error due to unit step input is _____.
0 | |
0.15 | |
0.75 | |
1 |
Question 3 Explanation:
Closed loop transfer function,
T(s)=\frac{4}{(s^2+0.4s+4)}
or, \frac{G(s)}{1+G(s)}
=\left ( \frac{4}{s^2+0.4s+4} \right )\;\;\;\;[H(s)=1]
or, (s^2+0.4s+4)=4+4G(s)
or, G(s)=\frac{4}{s(s+0.4)}
=Open llop transfer function ( for unity feedback system)
Given, input =u(t)=r(t)
\therefore \;\;\; R(s)=1/s
Steady state error
e_{ss}=\lim_{s \to 0}\left [ \frac{sR(s)}{1+G(s)H(s)} \right ]
\;\;=\lim_{s \to 0}\left [ \frac{sR(s)}{1+G(s)} \right ]\;\;\; [For H(s)=1]
\;\;=\lim_{s \to 0}\left [ \frac{s \times \frac{1}{s}}{1+\frac{4}{s(s+0.4)}} \right ]
\;\;=\lim_{s \to 0}\left [ \frac{s(s+0.4)}{s^2+0.4s+4} \right ]=0
\therefore Steady state error for step input, e_{ss}=0
T(s)=\frac{4}{(s^2+0.4s+4)}
or, \frac{G(s)}{1+G(s)}
=\left ( \frac{4}{s^2+0.4s+4} \right )\;\;\;\;[H(s)=1]
or, (s^2+0.4s+4)=4+4G(s)
or, G(s)=\frac{4}{s(s+0.4)}
=Open llop transfer function ( for unity feedback system)
Given, input =u(t)=r(t)
\therefore \;\;\; R(s)=1/s
Steady state error
e_{ss}=\lim_{s \to 0}\left [ \frac{sR(s)}{1+G(s)H(s)} \right ]
\;\;=\lim_{s \to 0}\left [ \frac{sR(s)}{1+G(s)} \right ]\;\;\; [For H(s)=1]
\;\;=\lim_{s \to 0}\left [ \frac{s \times \frac{1}{s}}{1+\frac{4}{s(s+0.4)}} \right ]
\;\;=\lim_{s \to 0}\left [ \frac{s(s+0.4)}{s^2+0.4s+4} \right ]=0
\therefore Steady state error for step input, e_{ss}=0
Question 4 |
The open-loop transfer function of a dc motor is given as \frac{\omega (s)}{V_a(s)}=\frac{10}{1+10s}. When
connected in feedback as shown below, the approximate value of K_a that will
reduce the time constant of the closed loop system by one hundred times as
compared to that of the open-loop system is
1 | |
5 | |
10 | |
100 |
Question 4 Explanation:
\frac{\omega (s)}{V_a(s)}=\frac{10}{1+10s}=\frac{10}{1+\tau s}
\tau =10
Closed loop systems,
\frac{\omega (s)}{R(s)}=\frac{\frac{10K_a}{1+10s}}{1+\frac{10K_a}{1+10s}}
=\frac{10K_a}{1+10K_a+10s}
=\frac{\frac{10K_a}{1+10K_a}}{1+\frac{10}{1+10K_a}s}
According to quetion,
\frac{10}{1+10K_a}=\frac{1}{100}\times 10
1+10K_a=100
K_a=9.9
\tau =10
Closed loop systems,
\frac{\omega (s)}{R(s)}=\frac{\frac{10K_a}{1+10s}}{1+\frac{10K_a}{1+10s}}
=\frac{10K_a}{1+10K_a+10s}
=\frac{\frac{10K_a}{1+10K_a}}{1+\frac{10}{1+10K_a}s}
According to quetion,
\frac{10}{1+10K_a}=\frac{1}{100}\times 10
1+10K_a=100
K_a=9.9
Question 5 |
As shown in the figure, a negative feedback system has an amplifier of gain 100
with \pm10% tolerance in the forward path, and an attenuator of value 9/100 in
the feedback path. The overall system gain is approximately :
10\pm 1% | |
10\pm 2% | |
10\pm 5% | |
10\pm 10% |
Question 5 Explanation:
G=100\pm 10%
\frac{\Delta G}{G}=10% or 0.1
H=\frac{9}{100}
Overall gain,
T=\frac{G}{1+GH}
T=\frac{100}{1+100 \times \frac{9}{100}}=10
Sensitivity w.r.t. 'G'=\frac{1}{1+GH} \times 10%
\;\;\;=\frac{1}{1+10 \times 1}\times 10%
\;\;\;=\frac{10}{11}\cong 1%
\frac{\Delta G}{G}=10% or 0.1
H=\frac{9}{100}
Overall gain,
T=\frac{G}{1+GH}
T=\frac{100}{1+100 \times \frac{9}{100}}=10
Sensitivity w.r.t. 'G'=\frac{1}{1+GH} \times 10%
\;\;\;=\frac{1}{1+10 \times 1}\times 10%
\;\;\;=\frac{10}{11}\cong 1%
There are 5 questions to complete.