Roots on LHS& RHS
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Q. The forward transfer function of a unity feedback system is
G(s)=K(s2+1)(s+1)(s+2).
The system is stable for
G(s)=K(s2+1)(s+1)(s+2).
The system is stable for
- K < - 1
- K > - 1
- K < - 2
- K > - 1
Q. The close loop transfer function of a system is, T(s)=s3+4s2+8s+16s5+3s4+5s2+4s+3
The number of poles in the right half plane and in left half plane are
The number of poles in the right half plane and in left half plane are
- 3, 2
- 1, 4
- 2, 3
- 4, 1
Q. The number of roots of s3+5s2+7s+3=0 in the left half of the s-plane is
- zero
- one
- two
- three
Q. If s3+3s2+4s+A=0 , then all the roots of this equation are in the left half plane provided that
- A>12
- −3<A<4
- 0<A<12
- 5<A<12
Q. The value of a0 which will ensure that the polynomial s3+3s2+2s+a0 has roots on the left half of the s-plane is
- 11
- 9
- 7
- 5
Q. The range of K for which all the roots of the equation s3+3s2+2s+K=0 are in the left half of the complex s-plane is
- 0 < K < 6
- 0 < K < 16
- 6 < K < 36
- 6 < K < 16
Q. A closed loop system has the characteristic equation given by s3+Ks2+(K+2)s+3=0. For this system to be stable, which one of the following conditions should be satisfied?
- 0 < K < 0.5
- 0.5 < K < 1
- 0 < K < 1
- K > 1
Q. The the feedback system given below, the transfer function G(s)=1(s+1)2. The system CANNOT be stabilized with
- C(s)=1+3s
- C(s)=3+7s
- C(s)=3+9s
- C(s)=1s
Q. The transfer function of a second order real system with a perfectly flat magnitude response of unity has a pole at (2−j3). List all the poles and zeros.
- Poles at (2±j3), no zeros.
- Poles at (±2−j3), one zero at orgin.
- Poles at (2−j3), (−2+j3), zeros at (−2−j3), (2+j3).
- Poles at (2±j3), zeros at (−2±j3).