Question 1 |
The open loop transfer function of a unity gain negative feedback system is given by G(s)=\frac{k}{s^2+4s-5}.
The range of k for which the system is stable, is
The range of k for which the system is stable, is
k \gt 3 | |
k \lt 3 | |
k \gt 5 | |
k \lt 5 |
Question 1 Explanation:
Characteristic equation:
\begin{aligned} 1+G(s)H(s)&=0\\ 1+\frac{k}{s^2+4s-5}&=0\\ s^2+4s+k-5&=0 \end{aligned}
R-H criteria:
\left.\begin{matrix} s^2\\ s^1\\ s^0 \end{matrix}\right| \begin{matrix} 1 & k-5\\ 4 & 0\\ k-5 & \end{matrix}
Hence, for stable system,
k-5 \gt 0 \;\; \Rightarrow \; k \gt 5
\begin{aligned} 1+G(s)H(s)&=0\\ 1+\frac{k}{s^2+4s-5}&=0\\ s^2+4s+k-5&=0 \end{aligned}
R-H criteria:
\left.\begin{matrix} s^2\\ s^1\\ s^0 \end{matrix}\right| \begin{matrix} 1 & k-5\\ 4 & 0\\ k-5 & \end{matrix}
Hence, for stable system,
k-5 \gt 0 \;\; \Rightarrow \; k \gt 5
Question 2 |
The root locus of the feedback control system having the characteristic equation
s^{2}+ 6Ks +2s +5 =0 \; where \; K \gt 0, enters into the real axis at
s^{2}+ 6Ks +2s +5 =0 \; where \; K \gt 0, enters into the real axis at
s=-1 | |
s=-\sqrt{5} | |
s=-5 | |
s=\sqrt{5} |
Question 2 Explanation:
G(s)H(s)=\frac{6Ks}{s^2+2s+5}
The point at which root locus enters real axis (break away point) is given by
\frac{dK}{ds}=0
K=-\frac{(s^2+2s+5)}{6s}
\frac{dK}{ds}=0\;\Rightarrow \;s=\pm \sqrt{5}
\therefore \;\;s=-\sqrt{5}
The point at which root locus enters real axis (break away point) is given by
\frac{dK}{ds}=0
K=-\frac{(s^2+2s+5)}{6s}
\frac{dK}{ds}=0\;\Rightarrow \;s=\pm \sqrt{5}
\therefore \;\;s=-\sqrt{5}
Question 3 |
The gain at the breakaway point of the root locus of a unity feedback system with open loop transfer function G(s)=\frac{Ks}{(s-1)(s-4)} is
1 | |
2 | |
5 | |
9 |
Question 3 Explanation:
OLTS\Rightarrow G(s)=\frac{Ks}{(s-1)(s-4)}
Now, characteristic equation,
1+G(s)H(s)=0
\frac{Ks}{(s-1)(s-4)}+1=0
\Rightarrow \;\; Ks+(s^2-5s+4)=0
\;\;K=-\frac{(s^2-5s+4)}{s}=\left [ s-5+\frac{4}{s} \right ]
For break away point : \frac{dK}{ds}=0
\frac{dK}{ds}=-\left [ 1-0-\frac{4}{s^2} \right ]=0
We get, s=\pm 2
Therefore valid break away point is s=2
Now gain at s=2 is
K=(Product of distances from all the poles to break away point)/(Product of distances from all the zero to break away point)
Gain, K=\frac{1 \times 2}{2}=1
Now, characteristic equation,
1+G(s)H(s)=0
\frac{Ks}{(s-1)(s-4)}+1=0
\Rightarrow \;\; Ks+(s^2-5s+4)=0
\;\;K=-\frac{(s^2-5s+4)}{s}=\left [ s-5+\frac{4}{s} \right ]
For break away point : \frac{dK}{ds}=0
\frac{dK}{ds}=-\left [ 1-0-\frac{4}{s^2} \right ]=0
We get, s=\pm 2
Therefore valid break away point is s=2
Now gain at s=2 is
K=(Product of distances from all the poles to break away point)/(Product of distances from all the zero to break away point)
Gain, K=\frac{1 \times 2}{2}=1
Question 4 |
An open loop transfer function G(s) of a system is
G(s)=\frac{K}{s(s+1)(s+2)}
For a unity feedback system, the breakaway point of the root loci on the real axis occurs at,
G(s)=\frac{K}{s(s+1)(s+2)}
For a unity feedback system, the breakaway point of the root loci on the real axis occurs at,
-0.42 | |
-1.58 | |
-0.42 and -1.58 | |
none of the above |
Question 4 Explanation:
G_1(s)=\frac{k}{s(s+1)(s+2)}
\Rightarrow \;\; 1+\frac{k}{s(s+1)(s+2)}=0
\Rightarrow \;\; s(s^2+3s+2)+k=0
\Rightarrow \;\; s^3+3s^2+2s+k=0
\Rightarrow \;\; k=-s^3-3s^2-2s\;\;\;...(i)
\Rightarrow \;\;\frac{dk}{ds}=0
\;\;\;\;\frac{dk}{ds}=-3s^2-6s-2=0
\;\;\;\;s=-0.422,-1.577
Only s=-0.422 lie on root locus. Therefore brakaway point is s=-0.42.
\Rightarrow \;\; 1+\frac{k}{s(s+1)(s+2)}=0
\Rightarrow \;\; s(s^2+3s+2)+k=0
\Rightarrow \;\; s^3+3s^2+2s+k=0
\Rightarrow \;\; k=-s^3-3s^2-2s\;\;\;...(i)
\Rightarrow \;\;\frac{dk}{ds}=0
\;\;\;\;\frac{dk}{ds}=-3s^2-6s-2=0
\;\;\;\;s=-0.422,-1.577
Only s=-0.422 lie on root locus. Therefore brakaway point is s=-0.42.
Question 5 |
The open loop poles of a third order unity feedback system are at 0, -1, -2. Let the frequency corresponding to the point where the root locus of the system transits to unstable region be K. Now suppose we introduce a zero in the open loop transfer function at -3, while keeping all the earlier open loop poles intact. Which one of the following is TRUE about the point where the root locus of the modified system transits to unstable region?
It corresponds to a frequency greater than K | |
It corresponds to a frequency less than K | |
It corresponds to a frequency K | |
Root locus of modified system never transits to unstable region |
Question 5 Explanation:
\frac{K(s+3)}{s(s+1)(s+2)}
P=3, \; Z=1
P-Z=3-1=2
asymptotes angle =90^{\circ}, 180^{\circ}
Centroid =\frac{(-3)-(-3)}{2}=0
as root locus never cross asymptotes so, will remain in left sideof x-axis.
P=3, \; Z=1
P-Z=3-1=2
asymptotes angle =90^{\circ}, 180^{\circ}
Centroid =\frac{(-3)-(-3)}{2}=0
as root locus never cross asymptotes so, will remain in left sideof x-axis.
There are 5 questions to complete.