Question 1 |
Consider an even polynomial p(s) given by
p(s)=s^4+5s^2+4+K ,
where K is an unknown real parameter. The complete range ofK for which p(s) has all its roots on the imaginary axis is ________.
p(s)=s^4+5s^2+4+K ,
where K is an unknown real parameter. The complete range ofK for which p(s) has all its roots on the imaginary axis is ________.
-4\leq K\leq \frac{9}{4} | |
-3\leq K\leq \frac{9}{2} | |
-6\leq K\leq \frac{5}{4} | |
-5\leq K\leq 0 |
Question 1 Explanation:
p(s)=s^4+5s^2+4+k
\begin{matrix} s^4 & 1& 5 &(4+k) \\ s^3& 0 & 0 & \\ s^2& & & \\ s& & & \\ s^0& & & \end{matrix}
s^4+5s^2+(4+k)=0
4s^3+10s=0
(4s^2+10)=0
s=\pm j\sqrt{5/2}
\begin{matrix} s^4 & 1& 5 &(4+k) \\ s^3& 4 & 10 & \\ s^2& 5/2 & (4+k) & \\ s^1&10-\frac{4(4+k)}{5/2} & & \\ s^0& (4+k) & & \end{matrix}
\begin{aligned} (4+k) & \gt 0\\ k & \gt -4\\ 10 \times \frac{5}{2}& \gt 4(4+K)\\ (4+K)& \lt \frac{25}{4}\\ K& \lt 9/4 \end{aligned}
\Rightarrow All roots be on imaginary axis -4\leq K\leqslant 9/4
\begin{matrix} s^4 & 1& 5 &(4+k) \\ s^3& 0 & 0 & \\ s^2& & & \\ s& & & \\ s^0& & & \end{matrix}
s^4+5s^2+(4+k)=0
4s^3+10s=0
(4s^2+10)=0
s=\pm j\sqrt{5/2}
\begin{matrix} s^4 & 1& 5 &(4+k) \\ s^3& 4 & 10 & \\ s^2& 5/2 & (4+k) & \\ s^1&10-\frac{4(4+k)}{5/2} & & \\ s^0& (4+k) & & \end{matrix}
\begin{aligned} (4+k) & \gt 0\\ k & \gt -4\\ 10 \times \frac{5}{2}& \gt 4(4+K)\\ (4+K)& \lt \frac{25}{4}\\ K& \lt 9/4 \end{aligned}
\Rightarrow All roots be on imaginary axis -4\leq K\leqslant 9/4
Question 2 |
Consider a unity feedback system, as in the figure shown, with an integral compensator \frac{K}{s} and open-loop transfer function
G(s)=\frac{1}{s^2+3s+2}
where k \gt 0. The positive value of K for which there are exactly two poles of the unity feedback system on the j\omega axis is equal to ______ (rounded off to two decimal places).
G(s)=\frac{1}{s^2+3s+2}
where k \gt 0. The positive value of K for which there are exactly two poles of the unity feedback system on the j\omega axis is equal to ______ (rounded off to two decimal places).
2.45 | |
4.28 | |
6.0 | |
6.25 |
Question 2 Explanation:
\frac{Y(s)}{X(s)}=\frac{K}{s^{3}+3 s^{2}+2 s+K}
Two poles of this system lie the system is moro:
System is marginally stable.
\qquad k_{\text{mar}}=3 \times 2=6
Two poles of this system lie the system is moro:
System is marginally stable.
\qquad k_{\text{mar}}=3 \times 2=6
Question 3 |
A unity feedback control system is characterized by the open-loop transfer function
G(s)=\frac{2(s+1)}{s^{3}+ks^{2}+2s+1}
The value of k for which the system oscillates at 2 rad/s is ________.
G(s)=\frac{2(s+1)}{s^{3}+ks^{2}+2s+1}
The value of k for which the system oscillates at 2 rad/s is ________.
0.6 | |
0.69 | |
0.89 | |
0.75 |
Question 3 Explanation:
The given open loop transfer function is,
G(s)=\frac{2(s+1)}{s^{3}+K s^{2}+2 s+1}
The closed loop transfer function is,
T(s)=\frac{G(s)}{1+G(s)}=\frac{2(s+1)}{s^{3}+K s^{2}+4 s+3}
When system oscillates, i.e., when system is marginally stable,
(1)(3)=K(4)
\mathrm{So}, \quad K=0.75
G(s)=\frac{2(s+1)}{s^{3}+K s^{2}+2 s+1}
The closed loop transfer function is,
T(s)=\frac{G(s)}{1+G(s)}=\frac{2(s+1)}{s^{3}+K s^{2}+4 s+3}
When system oscillates, i.e., when system is marginally stable,
(1)(3)=K(4)
\mathrm{So}, \quad K=0.75
Question 4 |
Which one of the following options correctly describes the locations of the roots of the equation s^{4}+s^{2}+1=0 on the complex plane?
Four left half plane (LHP) roots | |
One right half plane (RHP) root, one LHP root and two roots on the imaginary axis | |
Two RHP roots and two LHP roots | |
All four roots are on the imaginary axis |
Question 4 Explanation:
q(s)=s^{4}+s^{2}+1=0
\begin{array}{r|rrr} s^{4} & 1 & 1 & 1\\ s^{3} & 4 & 2 & 0\\ s^{2} & 0.5 & 1 & 0 \\ s^{1} & -6 & 0 & 0 \\ s^{0} & 1 & 0 & 0 \end{array} \frac{d A(s)}{d s}=4 s^{3}+2 s^{1}
There are two sign Changes in the first column of the R-H table and the order of auxiliary equation is 4. So, four poles are symmetric about origin.
\therefore 2 RHP roots and 2 LHP roots.
\begin{array}{r|rrr} s^{4} & 1 & 1 & 1\\ s^{3} & 4 & 2 & 0\\ s^{2} & 0.5 & 1 & 0 \\ s^{1} & -6 & 0 & 0 \\ s^{0} & 1 & 0 & 0 \end{array} \frac{d A(s)}{d s}=4 s^{3}+2 s^{1}
There are two sign Changes in the first column of the R-H table and the order of auxiliary equation is 4. So, four poles are symmetric about origin.
\therefore 2 RHP roots and 2 LHP roots.
Question 5 |
The first two rows in the Routh table for the characteristic equation of a certain closed-loop control system are given as
The range of K for which the system is stable is
The range of K for which the system is stable is
-2.0 \lt K \lt 0.5 | |
0 \lt K \lt 0.5 | |
0\lt K\lt \infty | |
0.5\lt K\lt \infty |
Question 5 Explanation:
\begin{array}{c|cc} s^{3}& 1 & (2 k+3) \\ s^{2}&2 k & 4 \\ s&\frac{4 k^{2}+6 k-4}{2 k} & 0\\ s^{0}&4 \end{array}
For stability, K \gt 0
\begin{aligned} 4 K^{2}+6 K-4& \gt 0 \\ 2 K^{2}+3 K-2& \gt 0 \\ (2 K-1)(K+2)& \gt 0 \\ \Rightarrow \quad K& \gt \frac{1}{2} or, k \lt -2 \\ \text{Hence, }\quad 0.5& \lt K \lt \infty \end{aligned}
For stability, K \gt 0
\begin{aligned} 4 K^{2}+6 K-4& \gt 0 \\ 2 K^{2}+3 K-2& \gt 0 \\ (2 K-1)(K+2)& \gt 0 \\ \Rightarrow \quad K& \gt \frac{1}{2} or, k \lt -2 \\ \text{Hence, }\quad 0.5& \lt K \lt \infty \end{aligned}
There are 5 questions to complete.