Question 1 |
A unity feedback system that uses proportional-integral (\text{PI}
) control is shown in the figure.
The stability of the overall system is controlled by tuning the \text{PI} control parameters K_{P} and K_{I}. The maximum value of K_{I} that can be chosen so as to keep the overall system stable or, in the worst case, marginally stable (rounded off to three decimal places) is ______
The stability of the overall system is controlled by tuning the \text{PI} control parameters K_{P} and K_{I}. The maximum value of K_{I} that can be chosen so as to keep the overall system stable or, in the worst case, marginally stable (rounded off to three decimal places) is ______
1.452 | |
6.325 | |
3.125 | |
7.655 |
Question 1 Explanation:
\begin{aligned} G H &=\left(\frac{s K_{p}+K_{I}}{s}\right)\left(\frac{2}{s^{3}+4 s^{2}+5 s+2}\right) \\ q(s) &=s^{4}+4 s^{3}+5 s^{2}+s\left(2+2 k_{p}\right)+2 k_{I} \\ \text{Necessary}:\qquad K_{p} & \gt -1 ; K_{I} \gt 0\\ &\begin{array}{l|ccc} s^{4} & 1 & 5 & 2 K_{I} \\ s^{3} & 4 & 2+2 K_{p} & \\ s^{2} & \frac{18-2 K_{p}}{4} & 2 K_{I} & \\ s^{1} & \left(9-K_{p}\right)\left(1+K_{p}\right)-8 K_{I} & & \\ s^{0} & 2 K_{I} & & \end{array}\\ \end{aligned}
Sufficient:
\begin{aligned} \frac{18-2 K_{p}}{4}& \gt 0\\ \Rightarrow \quad K_{p}& \lt 9\\ \therefore \quad-1& \lt K_{p} \lt 9\\ \left(18-2 K_{p}\right)\left(2+2 K_{p}\right)-32 K_{I}& \gt 0\\ 32 K_{I}& \lt 36+32 K_{p}-4 K_{p}^{2}\\ \therefore \qquad \qquad 0& \lt K_{I} \lt \frac{36+32 K_{p}-4 K_{p}^{2}}{32} \end{aligned}
\begin{aligned} \text{If }K_{p}&=-1 \Rightarrow k_{I}=0\\ \text{If }K_{p}&=9 \Rightarrow k_{I}=0\\ \end{aligned},
\begin{aligned} \therefore \qquad \qquad \frac{d K_{I}}{d K_{p}}&=0\\ \Rightarrow \qquad \qquad 32-8 K_{p}&=0=0 \Rightarrow K_{p}=4 \end{aligned}
\therefore For K_{p}=4, K_{I} is maximum, which is
K_{I}=\frac{36+32 \times 4-64}{32}=3.125
For K_{p}=4, K_{I} \lt 3.125 for stability
\therefore K_{I \max }=3.125
Sufficient:
\begin{aligned} \frac{18-2 K_{p}}{4}& \gt 0\\ \Rightarrow \quad K_{p}& \lt 9\\ \therefore \quad-1& \lt K_{p} \lt 9\\ \left(18-2 K_{p}\right)\left(2+2 K_{p}\right)-32 K_{I}& \gt 0\\ 32 K_{I}& \lt 36+32 K_{p}-4 K_{p}^{2}\\ \therefore \qquad \qquad 0& \lt K_{I} \lt \frac{36+32 K_{p}-4 K_{p}^{2}}{32} \end{aligned}
\begin{aligned} \text{If }K_{p}&=-1 \Rightarrow k_{I}=0\\ \text{If }K_{p}&=9 \Rightarrow k_{I}=0\\ \end{aligned},
\begin{aligned} \therefore \qquad \qquad \frac{d K_{I}}{d K_{p}}&=0\\ \Rightarrow \qquad \qquad 32-8 K_{p}&=0=0 \Rightarrow K_{p}=4 \end{aligned}
\therefore For K_{p}=4, K_{I} is maximum, which is
K_{I}=\frac{36+32 \times 4-64}{32}=3.125
For K_{p}=4, K_{I} \lt 3.125 for stability
\therefore K_{I \max }=3.125
Question 2 |
Which of the following statements is incorrect?
Lead compensator is used to reduce the settling time. | |
Lag compensator is used to reduce the steady state error. | |
Lead compensator may increase the order of a system. | |
Lag compensator always stabilizes an unstable system. |
Question 2 Explanation:
In_ case of high type systems Lag compensator
fails to give stability.
Question 3 |
Which of the following can be pole-zero configuration of a phase-lag controller (lag compensator)?
A | |
B | |
C | |
D |
Question 3 Explanation:
Phase lag controller transfer function is
G_{C}(s)=\frac{s+Z}{s+P} ;|Z|\gt|P|
G_{C}(s)=\frac{s+Z}{s+P} ;|Z|\gt|P|
Question 4 |
The transfer function of a first-order controller is given as
G_{c}(s)=\frac{K(s+a)}{s+b}
where K, a and b are positive real numbers. The condition for this controller to act as a phase lead compensator is
G_{c}(s)=\frac{K(s+a)}{s+b}
where K, a and b are positive real numbers. The condition for this controller to act as a phase lead compensator is
a \lt b | |
a \gt b | |
K \lt ab | |
K \gt ab |
Question 4 Explanation:
For phase lead compensator
\begin{aligned} G_{c}(s)&=\frac{(1+s \tau)}{(1+\alpha s \tau)} \quad ; \quad \alpha \lt 1\\ \text{Here,}\quad \tau&=\frac{1}{a}\\ \text{and}\quad \alpha \tau&=\frac{1}{b}\\ \text{or,}\quad \alpha&=\frac{a}{b} \lt 1\\ \text{or,}\quad a& \lt b \end{aligned}
\begin{aligned} G_{c}(s)&=\frac{(1+s \tau)}{(1+\alpha s \tau)} \quad ; \quad \alpha \lt 1\\ \text{Here,}\quad \tau&=\frac{1}{a}\\ \text{and}\quad \alpha \tau&=\frac{1}{b}\\ \text{or,}\quad \alpha&=\frac{a}{b} \lt 1\\ \text{or,}\quad a& \lt b \end{aligned}
Question 5 |
A lead compensator network includes a parallel combination of R and C in the feed-forward path. If the transfer function of the compensator is G(s)=\frac{s+2}{s+4} , the value of RC is ________.
1 | |
1.5 | |
0.5 | |
2 |
Question 5 Explanation:
G_{C}(s)=\frac{s+2}{s+4} \quad\dots(i)
For lead compensator,
Transfer function =\frac{1+s \tau}{1+\alpha s \tau} \quad\dots(ii)
Where,
\tau= Lead time constant =R_{1} C
and \alpha=\frac{R_{2}}{R_{1}+R_{2}}
Comparing equation (i) and (ii), we get
\tau=\frac{1}{2} \quad and \quad \alpha \tau=\frac{1}{4}
or, \alpha=\frac{1}{2}
\therefore RC time constant =0.5
For lead compensator,
Transfer function =\frac{1+s \tau}{1+\alpha s \tau} \quad\dots(ii)
Where,
\tau= Lead time constant =R_{1} C
and \alpha=\frac{R_{2}}{R_{1}+R_{2}}
Comparing equation (i) and (ii), we get
\tau=\frac{1}{2} \quad and \quad \alpha \tau=\frac{1}{4}
or, \alpha=\frac{1}{2}
\therefore RC time constant =0.5
Question 6 |
For the following feedback system G(s)=\frac{1}{(s+1)(s+2)}. The 2% settling time of the step response is required to be less than 2 seconds.
Which one of the following compensators C(s) achieves this ?
3 (\frac{1}{s+5}) | |
5 (\frac{0.03}{s}+1) | |
2 (s+4) | |
4 (\frac{s+8}{s+3}) |
Question 6 Explanation:
Given open loop transfer function is
\begin{aligned} G(s)&=\frac{1}{(s+1)(s+2)} \\ \therefore T(s)&=\frac{G(s)}{1+G(s)}=\frac{1}{(s+1)(s+2)+1} \\ T(s)&=\frac{1}{s^{2}+3 s+3} &\ldots(i) \end{aligned}
Comparing equation (i) with standard transfer function
\xi \omega_{n}=\frac{3}{2}=1.5
\therefore 2% settling time
\tau_{s}=\frac{4}{\xi \omega_{n}}=\frac{4}{1.5}=2.67>2 \mathrm{sec}
Thus, in order to make settling time \left(\tau_{s}\right) less than 2 sec, PD controller should be used. Hence, option (C) is the correct choice.
Where, C(s)=2(s+4)
\therefore New transfer function =T^{\prime}(G)
=\frac{C(s) G(s)}{1+C(s) G(s)}=\frac{2(s+4)}{s^{2}+3 s+2+2 s+8}
or T^{\prime}(s)=\frac{2(s+4)}{s^{2}+5 s+10}
\therefore \quad \tau_{s}=\frac{4}{\xi \omega_{n}}=\frac{4}{5 / 2}=\frac{4}{2.5}=1.6
\begin{aligned} G(s)&=\frac{1}{(s+1)(s+2)} \\ \therefore T(s)&=\frac{G(s)}{1+G(s)}=\frac{1}{(s+1)(s+2)+1} \\ T(s)&=\frac{1}{s^{2}+3 s+3} &\ldots(i) \end{aligned}
Comparing equation (i) with standard transfer function
\xi \omega_{n}=\frac{3}{2}=1.5
\therefore 2% settling time
\tau_{s}=\frac{4}{\xi \omega_{n}}=\frac{4}{1.5}=2.67>2 \mathrm{sec}
Thus, in order to make settling time \left(\tau_{s}\right) less than 2 sec, PD controller should be used. Hence, option (C) is the correct choice.
Where, C(s)=2(s+4)
\therefore New transfer function =T^{\prime}(G)
=\frac{C(s) G(s)}{1+C(s) G(s)}=\frac{2(s+4)}{s^{2}+3 s+2+2 s+8}
or T^{\prime}(s)=\frac{2(s+4)}{s^{2}+5 s+10}
\therefore \quad \tau_{s}=\frac{4}{\xi \omega_{n}}=\frac{4}{5 / 2}=\frac{4}{2.5}=1.6
Question 7 |
The transfer function of a compensator is given as
G_{c}(s)=\frac{s+a}{s+b}
The phase of the above lead compensator is maximum at
G_{c}(s)=\frac{s+a}{s+b}
The phase of the above lead compensator is maximum at
\sqrt{2} rad/s | |
\sqrt{3}rad/s | |
\sqrt{6} rad/s | |
1/\sqrt{3} rad/s |
Question 7 Explanation:
For phase to be maximum,
\begin{aligned} \frac{\partial P}{\partial \omega}&=0 \\ \frac{1}{1+\frac{\omega^{2}}{a^{2}}}- \frac{1/b}{1+\frac{\omega^{2}}{b^{2}}}&=0 \\ \frac{1}{1+\omega^{2}}- \frac{1/b}{1+\frac{\omega^{2}}{4}}&=0\\ \frac{1}{1+\omega^{2}}-\frac{2}{\omega^{2}+4} &=0 \\ \omega^{2}+4-2-2 \omega^{2} &=0 \\ \omega^{2} &=2 \\ \omega &=\sqrt{2} \mathrm{rad} / \mathrm{sec} \end{aligned}
\begin{aligned} \frac{\partial P}{\partial \omega}&=0 \\ \frac{1}{1+\frac{\omega^{2}}{a^{2}}}- \frac{1/b}{1+\frac{\omega^{2}}{b^{2}}}&=0 \\ \frac{1}{1+\omega^{2}}- \frac{1/b}{1+\frac{\omega^{2}}{4}}&=0\\ \frac{1}{1+\omega^{2}}-\frac{2}{\omega^{2}+4} &=0 \\ \omega^{2}+4-2-2 \omega^{2} &=0 \\ \omega^{2} &=2 \\ \omega &=\sqrt{2} \mathrm{rad} / \mathrm{sec} \end{aligned}
Question 8 |
The transfer function of a compensator is given as
G_{c}(s)=\frac{s+a}{s+b}
G_{c}(S) is a lead compensator if
G_{c}(s)=\frac{s+a}{s+b}
G_{c}(S) is a lead compensator if
a =1, b = 2 | |
a = 3, b = 2 | |
a = -3, b = -1
| |
a = 3, b = 1 |
Question 8 Explanation:
G_{c}(s)=\left(\frac{s+a}{s+b}\right)
Phase P=\tan ^{-1}(\omega / a)-\tan ^{-1} \omega / b
for lead comparators phase must be + ve.
for this \frac{\omega}{a} \gt \frac{\omega}{b} \Rightarrow a \lt b
So a=1, b=2
Phase P=\tan ^{-1}(\omega / a)-\tan ^{-1} \omega / b
for lead comparators phase must be + ve.
for this \frac{\omega}{a} \gt \frac{\omega}{b} \Rightarrow a \lt b
So a=1, b=2
Question 9 |
A unity negative feedback closed loop system has a plant with the transfer function G(s)=\frac{1}{s^{2}+2s+2} and a controller G_{c}(s) in the
feed forward path. For a unit set input, the transfer function of the controller that
gives minimum steady state error is
G_{c}(s)=\frac{s+1}{s+2} | |
G_{c}(s)=\frac{(s+1)(s+4)}{(s+2)(s+3)} | |
G_{c}(s)=\frac{s+2}{s+1} | |
G_{c}(s)=1+\frac{2}{s}+3s |
Question 9 Explanation:
Steady state error,
\begin{aligned} e_{s s} &=\lim _{s \rightarrow 0} \frac{s R(s)}{1+G(s) G_{c}(s)} \\ r(t) &=u(t) \\ R(s) &=\frac{1}{s} \\ e_{ss} &=\lim _{s \rightarrow 0} \frac{s \cdot \frac{1}{s}}{1+G(s) G_{c}(s)} \\ e_{ss}&=\lim _{s \rightarrow 0} \frac{1}{1+G(s) G_{c}(s)} \\ \text { Taking, } \quad G_{c}(s) &=\frac{s+1}{s+2}, e_{s s}=\frac{2}{3} \\ \text { Taking, } \quad G_{c}(s) &=\frac{s+2}{s+1}, e_{s s}=\frac{1}{3} \\ \text { Taking, } \quad G_{c}(s) &=\frac{(s+1)(s+4)}{(s+2)(s+3)}, e_{s s}=\frac{3}{5} \\ \text { Taking, } \quad & G_{c}(s)=1+\frac{2}{s}+3 s, e_{s s}=0 \end{aligned}
\begin{aligned} e_{s s} &=\lim _{s \rightarrow 0} \frac{s R(s)}{1+G(s) G_{c}(s)} \\ r(t) &=u(t) \\ R(s) &=\frac{1}{s} \\ e_{ss} &=\lim _{s \rightarrow 0} \frac{s \cdot \frac{1}{s}}{1+G(s) G_{c}(s)} \\ e_{ss}&=\lim _{s \rightarrow 0} \frac{1}{1+G(s) G_{c}(s)} \\ \text { Taking, } \quad G_{c}(s) &=\frac{s+1}{s+2}, e_{s s}=\frac{2}{3} \\ \text { Taking, } \quad G_{c}(s) &=\frac{s+2}{s+1}, e_{s s}=\frac{1}{3} \\ \text { Taking, } \quad G_{c}(s) &=\frac{(s+1)(s+4)}{(s+2)(s+3)}, e_{s s}=\frac{3}{5} \\ \text { Taking, } \quad & G_{c}(s)=1+\frac{2}{s}+3 s, e_{s s}=0 \end{aligned}
Question 10 |
The magnitude plot of a rational transfer function G(s) with real coefficients is
shown below. Which of the following compensators has such a magnitude plot ?
Lead compensator | |
Lag compensator | |
PID compensator | |
Lead-lag compensator |
There are 10 questions to complete.