Question 1 |
The characteristic equation of a system is
s^3 + 3s^2 + (K + 2)s + 3K = 0
In the root locus plot for the given system, as K varies from 0 to \infty, the break-away or break-in point(s) lie within
s^3 + 3s^2 + (K + 2)s + 3K = 0
In the root locus plot for the given system, as K varies from 0 to \infty, the break-away or break-in point(s) lie within
(-1,0) | |
(-2,-1) | |
(-3,-2) | |
(-\infty
,-3) |
Question 1 Explanation:
Q(s)=1+G(s)H(s)=0
s^{3}+3s^{2}+2s+ks+3k=0
-k=\frac{s^{3}+3s^{2}+2s}{s+3}
-\frac{\mathrm{d} k}{\mathrm{d} s}=\frac{(s+3)(3s^{2}+6s+2)-(s^{3}+3s^{2}+2s)}{(s+3)^{2}}=0
3s^{3}+6s^{2}+2s+9s^{2}+18s+6-s^{3}-3s^{2}-2s=0
2s^{3}+12s^{2}+18s+6=0
s=-0.46,-3.87,-1.65
\therefore Break-away point lies between (0, -1), i.e. (-1, 0)
s^{3}+3s^{2}+2s+ks+3k=0
-k=\frac{s^{3}+3s^{2}+2s}{s+3}
-\frac{\mathrm{d} k}{\mathrm{d} s}=\frac{(s+3)(3s^{2}+6s+2)-(s^{3}+3s^{2}+2s)}{(s+3)^{2}}=0
3s^{3}+6s^{2}+2s+9s^{2}+18s+6-s^{3}-3s^{2}-2s=0
2s^{3}+12s^{2}+18s+6=0
s=-0.46,-3.87,-1.65
\therefore Break-away point lies between (0, -1), i.e. (-1, 0)
Question 2 |
A linear time invariant (LTI) system with the transfer function
G(s)=\frac{K(s^{2}+2s+2)}{(s^{2}-3s+2)}
is connected in unity feedback configuration as shown in the figure.
For the closed loop system shown, the root locus for 0 \lt K \lt \infty intersects the imaginary axis for K = 1.5. The closed loop system is stable for
G(s)=\frac{K(s^{2}+2s+2)}{(s^{2}-3s+2)}
is connected in unity feedback configuration as shown in the figure.
For the closed loop system shown, the root locus for 0 \lt K \lt \infty intersects the imaginary axis for K = 1.5. The closed loop system is stable for
K>1.5 | |
1 \lt K \lt 1.5 | |
0 \lt K \lt 1 | |
no positive value of K |
Question 2 Explanation:
Given that,
G(s)=\frac{K\left(s^{2}+2 s+2\right)}{s^{2}-3 s+2}
The root locus plot of the given system is follows:
\therefore System is stable for K \gt 1.5.
G(s)=\frac{K\left(s^{2}+2 s+2\right)}{s^{2}-3 s+2}
The root locus plot of the given system is follows:
\therefore System is stable for K \gt 1.5.
Question 3 |
The forward-path transfer function and the feedback-path transfer function of a single loop negative feedback control system are given as
G(s)=\frac{K(s+2)}{s^{2}+2s+2} and H(s)=1,
respectively. If the variable parameter K is real positive, then the location of the breakaway point on the root locus diagram of the system is __________
G(s)=\frac{K(s+2)}{s^{2}+2s+2} and H(s)=1,
respectively. If the variable parameter K is real positive, then the location of the breakaway point on the root locus diagram of the system is __________
-2 | |
-3.4 | |
-1.5 | |
-6.5 |
Question 3 Explanation:
\begin{aligned} G(s) &=\frac{K(s+2)}{s^{2}+2 s+2} \\ H(s)&=1 \\ 1+G(s) H(s) &=0 \\ 1+\frac{K}{\left(s^{2}\right)+2 s+2}(s+2) &=0\\ K&=-\frac{\left(s^{2}+2 s+2\right)}{(s+2)} \end{aligned}
\frac{d K}{d s}=-\left(\frac{(2 s+2)(s+2)-\left(s^{2}+2 s+2\right)}{(s+2)^{2}}\right)
For break away points, \frac{d K}{d s}=0
\begin{array}{l} (2 s+2)(s+2)-\left(s^{2}+2 s+2\right)=0 \\ 2 s^{2}+6 s+4-s^{2}-2 s-2=0 \\ s^{2}+4 s+2=0 \end{array}
\begin{aligned} s&=\frac{-4 \pm \sqrt{16-8}}{2} \\ &=\frac{-4 \pm 2 \sqrt{2}}{2}=-2 \pm \sqrt{2} \\ s&=-0.58 ; s=-3.41 \\ \text{But,}\quad s&=-3.41 \text{ lies on root locus }\\ \text{Hence}\quad s&=-3.41 \end{aligned}
\frac{d K}{d s}=-\left(\frac{(2 s+2)(s+2)-\left(s^{2}+2 s+2\right)}{(s+2)^{2}}\right)
For break away points, \frac{d K}{d s}=0
\begin{array}{l} (2 s+2)(s+2)-\left(s^{2}+2 s+2\right)=0 \\ 2 s^{2}+6 s+4-s^{2}-2 s-2=0 \\ s^{2}+4 s+2=0 \end{array}
\begin{aligned} s&=\frac{-4 \pm \sqrt{16-8}}{2} \\ &=\frac{-4 \pm 2 \sqrt{2}}{2}=-2 \pm \sqrt{2} \\ s&=-0.58 ; s=-3.41 \\ \text{But,}\quad s&=-3.41 \text{ lies on root locus }\\ \text{Hence}\quad s&=-3.41 \end{aligned}
Question 4 |
The open-loop transfer function of a unity-feedback control system is
G(s)=\frac{K}{s^{2}+5s+5}
The value of K at the breakaway point of the feedback control system's root-locus plot is ________
G(s)=\frac{K}{s^{2}+5s+5}
The value of K at the breakaway point of the feedback control system's root-locus plot is ________
0.65 | |
1.25 | |
2.25 | |
3.15 |
Question 4 Explanation:
Characteristic equation 1+G(s)H(s)=0
1+\frac{K}{s^2+5s+5}=0
K=-s^2-5s-5
For break away point \frac{dK}{ds}=0
\frac{dK}{ds}=-2s+5=0\;\Rightarrow \; s=-2.5
According to magnitude condition,
\begin{aligned} |G(s)H(s)|_{s=-2.5} &=1 \\ |G(s)H(s)|_{s=-2.5} &=\frac{|K|}{|(2.5)^2 +5 \times -2.5+5|}=1 \\ |K|&=|(6.25+5-12.5)|=1.25 \\ K&=\pm 1.25 \end{aligned}
1+\frac{K}{s^2+5s+5}=0
K=-s^2-5s-5
For break away point \frac{dK}{ds}=0
\frac{dK}{ds}=-2s+5=0\;\Rightarrow \; s=-2.5
According to magnitude condition,
\begin{aligned} |G(s)H(s)|_{s=-2.5} &=1 \\ |G(s)H(s)|_{s=-2.5} &=\frac{|K|}{|(2.5)^2 +5 \times -2.5+5|}=1 \\ |K|&=|(6.25+5-12.5)|=1.25 \\ K&=\pm 1.25 \end{aligned}
Question 5 |
For the system shown in the figure, s = -2.75 lies on the root locus if K is _______.
0 | |
0.3 | |
0.8 | |
1.2 |
Question 5 Explanation:
The open loop transfer function of the system:
G(s) H(s)=\frac{K(s+3) \times 10}{(s+2)}
In order to lie a point s=-2.75, the angle condition must be satisfy and therefore the magnitude condition is
\begin{aligned} &\mid G(s) H(s) \|_{s=-275}=1 \\ &=\left.\frac{10 K \sqrt{s^{2}+3^{2}}}{\sqrt{s^{2}+2^{2}}}\right|_{s=-2.75}=1 \\ &=\frac{10 K \sqrt{(3-2.75)^{2}}}{\sqrt{(2-2.75)^{2}}}=1 \\ &=\frac{0.25 \times 10 K}{0.75} or K=0.3 \end{aligned}
G(s) H(s)=\frac{K(s+3) \times 10}{(s+2)}
In order to lie a point s=-2.75, the angle condition must be satisfy and therefore the magnitude condition is
\begin{aligned} &\mid G(s) H(s) \|_{s=-275}=1 \\ &=\left.\frac{10 K \sqrt{s^{2}+3^{2}}}{\sqrt{s^{2}+2^{2}}}\right|_{s=-2.75}=1 \\ &=\frac{10 K \sqrt{(3-2.75)^{2}}}{\sqrt{(2-2.75)^{2}}}=1 \\ &=\frac{0.25 \times 10 K}{0.75} or K=0.3 \end{aligned}
Question 6 |
The open-loop transfer function of a plant in a unity feedback configuration is given as G(s)=\frac{K(s+4)}{(s+8)(s^{2}-9)} . The value of the gain K(> 0) for which -1+ j2 lies on the root locus is_______.
20.18 | |
25.54 | |
30.56 | |
33.35 |
Question 6 Explanation:
\begin{aligned} G(s) &=\frac{K(s+4)}{(s+8)\left(s^{2}-9\right)} \\ &=\frac{K(s+4)}{(s+8)(s+3)(s-3)} \end{aligned}
For the point (-1+2 j) to lie on the root locus the angle condition must be satisfy
i.e. \angle G(s) H(s)=\pm(2 Q+1) 180^{\circ}
Taking LHS:
\begin{array}{l} \left.\angle G(s) H(s)\right|_{s=-1+2 j} \\ =\frac{\angle K+\angle(s+4)}{\angle(s+8)+\angle(s+3)+\angle(s-3)} \\ =\frac{\angle K+\angle(-1+2 j+4)}{\angle(-1+2 j+8)+\angle(-1+2 j+3)+\angle(-1+2 j-3)} \\ =\frac{0+\tan ^{-1}\left(\frac{2}{3}\right)}{\tan ^{-1}\left(\frac{2}{7}\right)+\tan ^{-1}(1)+180^{\circ}-\tan ^{-1}\left(\frac{1}{2}\right)} \\ =33.69^{\circ}-15.945^{\circ}-45^{\circ}-180^{\circ}+26.565^{\circ} \\ \simeq 180^{\circ} \end{array}
As angle condition is satisfied the value of system gain K can be obtained by using magnitude
condition
\begin{array}{l} i.e. |G(s) H(s)|_{s=-1+2 j}=1 \\ =\frac{K \sqrt{(-1+4)^{2}+2^{2}}}{\sqrt{(-1+8)^{2}+2^{2}} \cdot \sqrt{(-1+3)^{2}+2^{2}} \cdot \sqrt{(-1-3)^{2}+2^{2}}} \\ =1 \\ =\frac{K \sqrt{9+4}}{\sqrt{49+4} \sqrt{4+4} \sqrt{16+4}}=1 \\ =\frac{K \sqrt{13}}{\sqrt{53} \sqrt{8} \sqrt{20}}=1 \\ K=\frac{\sqrt{53 \times 8 \times 20}}{\sqrt{13}}=25.54 \end{array}
For the point (-1+2 j) to lie on the root locus the angle condition must be satisfy
i.e. \angle G(s) H(s)=\pm(2 Q+1) 180^{\circ}
Taking LHS:
\begin{array}{l} \left.\angle G(s) H(s)\right|_{s=-1+2 j} \\ =\frac{\angle K+\angle(s+4)}{\angle(s+8)+\angle(s+3)+\angle(s-3)} \\ =\frac{\angle K+\angle(-1+2 j+4)}{\angle(-1+2 j+8)+\angle(-1+2 j+3)+\angle(-1+2 j-3)} \\ =\frac{0+\tan ^{-1}\left(\frac{2}{3}\right)}{\tan ^{-1}\left(\frac{2}{7}\right)+\tan ^{-1}(1)+180^{\circ}-\tan ^{-1}\left(\frac{1}{2}\right)} \\ =33.69^{\circ}-15.945^{\circ}-45^{\circ}-180^{\circ}+26.565^{\circ} \\ \simeq 180^{\circ} \end{array}
As angle condition is satisfied the value of system gain K can be obtained by using magnitude
condition
\begin{array}{l} i.e. |G(s) H(s)|_{s=-1+2 j}=1 \\ =\frac{K \sqrt{(-1+4)^{2}+2^{2}}}{\sqrt{(-1+8)^{2}+2^{2}} \cdot \sqrt{(-1+3)^{2}+2^{2}} \cdot \sqrt{(-1-3)^{2}+2^{2}}} \\ =1 \\ =\frac{K \sqrt{9+4}}{\sqrt{49+4} \sqrt{4+4} \sqrt{16+4}}=1 \\ =\frac{K \sqrt{13}}{\sqrt{53} \sqrt{8} \sqrt{20}}=1 \\ K=\frac{\sqrt{53 \times 8 \times 20}}{\sqrt{13}}=25.54 \end{array}
Question 7 |
A unity negative feedback system has the open-loop transfer function G(s)=\frac{K}{s(s+1)(s+3)}. The value of the gain K (\gt0) at which the root locus crosses the imaginary axis is ________.
7 | |
10 | |
12 | |
15 |
Question 7 Explanation:
G(s)=\frac{K}{s(s+1)(s+3)}
The characteristic equation
\begin{array}{l} =1+G(s) H(s)=0 \\ =s(s+1)(s+3)+K=0 \\ =s^{3}+4 s^{2}+3 s+K=0 \end{array}
Using Routh's tabular form
\begin{array}{c|cc} s^{3} & 1 & 3 \\ s^{2} & 4 & K \\ s^{1} & \frac{12-K}{4} & 0 \\ s^{0} & K & \end{array}
In order to cross the imaginary axis, system
should be marginally stable
\begin{aligned} \frac{12-K}{4}&=0\\ \text{or }\quad K&=12 \end{aligned}
The characteristic equation
\begin{array}{l} =1+G(s) H(s)=0 \\ =s(s+1)(s+3)+K=0 \\ =s^{3}+4 s^{2}+3 s+K=0 \end{array}
Using Routh's tabular form
\begin{array}{c|cc} s^{3} & 1 & 3 \\ s^{2} & 4 & K \\ s^{1} & \frac{12-K}{4} & 0 \\ s^{0} & K & \end{array}
In order to cross the imaginary axis, system
should be marginally stable
\begin{aligned} \frac{12-K}{4}&=0\\ \text{or }\quad K&=12 \end{aligned}
Question 8 |
In the root locus plot shown in the figure, the pole/zero marks and the arrows have been removed. Which one of the following transfer functions has this root locus ?
\frac{s+1}{(s+2)(s+4)(s+7)} | |
\frac{s+4}{(s+1)(s+2)(s+7)} | |
\frac{s+7}{(s+1)(s+2)(s+4)} | |
\frac{(s+1)(s+2)}{(s+4)(s+7)} |
Question 8 Explanation:
Since the root locus always emerges from the break-away points i.e. when two poles are at a part of same root locus lie. So \sigma =-1 and -2 will surely be poles.
Also we know that locus emerges from poles and terminates at zero. So, s=-4 is a zero and s=-7 is a pole
\therefore T(s)=\frac{(s+4)}{(s+1)(s+2)(s+7)}
Question 9 |
The root locus plot for a system is given below. The open loop transfer function
corresponding to this plot is given by
G(S)H(S)=k\frac{s(s+1)}{(s+2)(s+3)} | |
G(S)H(S)=k\frac{(s+1)}{(s+2)(s+3)^{2}} | |
G(S)H(S)=k\frac{1}{s(s-1)(s+2)(s+3)} | |
G(S)H(S)=k\frac{(s+1)}{(s+2)(s+3)} |
Question 9 Explanation:
From plot we can observe that one pole terminates
at one zero at position -1 and three poles
terminates to \infty. It means there are four poles and
1 zero. Pole at -3 goes on both sides. It means
there are two poles at -3.
Question 10 |
The feedback configuration and the pole-zero locations of G(s)=\frac{s^{2}-2s+2}{s^{2}+2s+2} are shown below. The root locus for negative values of k , i.e. for -\infty \lt k \lt 0, has breakaway/break-in points and angle of departure at pole P (with respect to the positive real axis) equal to
\pm \sqrt{2} \; and \; 0^{\circ} | |
\pm \sqrt{2} \; and \; 45^{\circ} | |
\pm \sqrt{3} \; and \; 0^{\circ} | |
\pm \sqrt{3} \; and \; 45^{\circ} |
Question 10 Explanation:
\begin{aligned} 1+G(s) H(s) &=0 \\ 1+\frac{K\left(s^{2}-2 s+2\right)}{s^{2}+2 s+2} &=0 \\ k &=-\frac{s^{2}+2 s+2}{s^{2}-2 s+2} \\ \text{put }\quad \frac{\partial K}{\partial s}&=0\text{ we have}\\ \left(s^{2}-2 s+2\right)(s+1) &-\left(s^{2}+2 s+2\right) \\ (s-1) =0 \\ 2 s^{2}-4 s^{2}+4 &=0 \\ 2 s^{2} &=+4 \\ s &=\pm \sqrt{2} \\ \text{Angle of departure is}\\ \phi_{D}=\phi\\ \text { where } &=\Sigma \phi_{2}-\Sigma \phi_{p} \\ &=225^{\circ} \\ \phi_{D} &=\pm 225^{\circ} \end{aligned} \\ \text{No option is matching.}
There are 10 questions to complete.