Question 1 |
The characteristic equation of a linear time-invariant (LTI) system is given by
\Delta (s)=s^4+3s^3+3s^2+s+k=0
The system is BIBO stable if
\Delta (s)=s^4+3s^3+3s^2+s+k=0
The system is BIBO stable if
0 \lt k \lt \frac{12}{9} | |
k \gt 3 | |
0 \lt k \lt \frac{8}{9} | |
k \gt 6 |
Question 1 Explanation:
Routh array is
For BIBO stability,
\frac{\left ( \frac{8}{3}-3k \right )}{\left ( \frac{8}{3} \right )} \gt 0
\Rightarrow k \lt \frac{8}{9}
and k \gt 0
\therefore \;\; 0 \lt k \lt \frac{8}{9}
For BIBO stability,
\frac{\left ( \frac{8}{3}-3k \right )}{\left ( \frac{8}{3} \right )} \gt 0
\Rightarrow k \lt \frac{8}{9}
and k \gt 0
\therefore \;\; 0 \lt k \lt \frac{8}{9}
Question 2 |
The number of roots of the polynomial,
s^{7}+s^{6}+7s^{5}+14s^{4}+31s^{3}+73s^{2}+25s+200,
in the open left half of the complex plane is
s^{7}+s^{6}+7s^{5}+14s^{4}+31s^{3}+73s^{2}+25s+200,
in the open left half of the complex plane is
3 | |
4 | |
5 | |
6 |
Question 2 Explanation:
Characteristic equation, s^7+s^6+7s^5+14s^4+31s^3+73s^2+25s+200=0
Auxillary equation, A(s)=8s^4+48s^2+200
\frac{d}{ds}A(s)=32s^3+96s
Total number of poles =7
Two sign change above auxillary equation=2 poles in RHS.
Two sign changes below auxillary equation implies out of 4 symmetric roots about origin, two poles are in LHS and two poles are in RHS.
Therefore 3 poles in LHS and 4 poles in RHS.
Auxillary equation, A(s)=8s^4+48s^2+200
\frac{d}{ds}A(s)=32s^3+96s
Total number of poles =7
Two sign change above auxillary equation=2 poles in RHS.
Two sign changes below auxillary equation implies out of 4 symmetric roots about origin, two poles are in LHS and two poles are in RHS.
Therefore 3 poles in LHS and 4 poles in RHS.
Question 3 |
The range of K for which all the roots of the equation s^{3}+3s^{2}+2s+K=0 are in the left half of the complex s-plane is
0 \lt K \lt 6 | |
0 \lt K \lt 16 | |
6 \lt K \lt 36 | |
6 \lt K \lt 16 |
Question 3 Explanation:
From the given equation, s^3+3s^2+2s+K=0
Using Routh's criterion, we get
K \lt 6 and K \gt 0 or 0 \lt K \lt 6
Using Routh's criterion, we get
K \lt 6 and K \gt 0 or 0 \lt K \lt 6
Question 4 |
A closed loop system has the characteristic equation given by s^{3}+Ks^{2}+(K+2)s+3=0. For this system to be stable, which one of the following conditions should be satisfied?
0 \lt K \lt 0.5 | |
0\lt K \lt 1 | |
0 \lt K \lt 1 | |
K \gt 1 |
Question 4 Explanation:
Characteristic equation is,
s^3+Ks^2+(K+2)s+3=0
For this system to be stable, using Routh's criterion, we can write,
3 \lt K(K+2)
or, K^2+2K-3 \gt 0
or, (K+3)(K-1) \gt 0
Here, the valid answer will be out of all the options given.
i.e K \gt 1.
s^3+Ks^2+(K+2)s+3=0
For this system to be stable, using Routh's criterion, we can write,
3 \lt K(K+2)
or, K^2+2K-3 \gt 0
or, (K+3)(K-1) \gt 0
Here, the valid answer will be out of all the options given.
i.e K \gt 1.
Question 5 |
The open loop transfer function of a unity feedback control system is given by
G(s)=\frac{K(s+1)}{s(1+Ts)(1+2s)}, K\gt 0,T \gt 0
The closed loop system will be stable if,
G(s)=\frac{K(s+1)}{s(1+Ts)(1+2s)}, K\gt 0,T \gt 0
The closed loop system will be stable if,
0\lt T \lt\frac{4(K+1)}{K-1} | |
0 \lt K \lt \frac{4(T+2)}{T-2} | |
0 \lt K \lt \frac{T+2}{T-2} | |
0 \lt T \lt \frac{8(K+1)}{K-1} |
Question 5 Explanation:
Open loop transfer function:
G(s)=\frac{K(s+1)}{s(1+Ts)(1+2s)}; \; K \gt 0 \; and \; t \gt 0
For closed loop system stability, characteristic equation is,
1+G(s)H(s)=0
1+\frac{K(s+1)}{s(1+Ts)(1+2s)}\cdot 1=0
s(1+Ts)(1+2s)+K(s+1)=0
2Ts^3+(2+T)s^2+(1+K)s+K=0
Using Routh's criteria,
For stability, K \gt 0
and (2+T)(1+K)-2TK \gt 0
K(2+T-2T)+(2+T) \gt 0
or
-(T-2)K+2(2+T) \gt 0
-K \gt -\frac{(2+T)}{(T-2)} or K \lt \frac{T+2}{(T-2)}
Hence for stability,
0 \lt K \lt \frac{T+2}{T-2}
G(s)=\frac{K(s+1)}{s(1+Ts)(1+2s)}; \; K \gt 0 \; and \; t \gt 0
For closed loop system stability, characteristic equation is,
1+G(s)H(s)=0
1+\frac{K(s+1)}{s(1+Ts)(1+2s)}\cdot 1=0
s(1+Ts)(1+2s)+K(s+1)=0
2Ts^3+(2+T)s^2+(1+K)s+K=0
Using Routh's criteria,
For stability, K \gt 0
and (2+T)(1+K)-2TK \gt 0
K(2+T-2T)+(2+T) \gt 0
or
-(T-2)K+2(2+T) \gt 0
-K \gt -\frac{(2+T)}{(T-2)} or K \lt \frac{T+2}{(T-2)}
Hence for stability,
0 \lt K \lt \frac{T+2}{T-2}
There are 5 questions to complete.