Question 1 |
The root-locus plot of a closed-loop system with unity negative feedback and
transfer function KG(s) in the forward path is shown in the figure. Note that K is
varied from 0 to \infty .
Select the transfer function G(s) that results in the root-locus plot of the closed-loop system as shown in the figure.
Select the transfer function G(s) that results in the root-locus plot of the closed-loop system as shown in the figure.
G(s)=\frac{1}{(s+1)^5} | |
G(s)=\frac{1}{(s^5+1)} | |
G(s)=\frac{s-1}{(s+1)^6} | |
G(s)=\frac{s+1}{(s^6+1)} |
Question 1 Explanation:
Here 5 Root Locus branches are diverging from
same point, this can possible only when if we
have 5 poles in the system at the same point
because Root Locus branch departs from open
loop pole and
Number of Root Locus branches = Number of open loop poles or Number of zero (Whichever is greater).
Here, there are 5 multiple Real poles, and matching with option (A).
Number of Root Locus branches = Number of open loop poles or Number of zero (Whichever is greater).
Here, there are 5 multiple Real poles, and matching with option (A).
Question 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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 5 |
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 5 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}
There are 5 questions to complete.