Electronics and Communication Engineering - Automatic Control Systems - Discussion
Discussion Forum : Automatic Control Systems - Section 4 (Q.No. 27)
27.
If Nyquist plot has a encirclement of (-1 + jo) point, the system is stable if
Discussion:
5 comments Page 1 of 1.
Snehil said:
9 years ago
Z = N + P.
If no of encirclements is 1, then the pole on RHS should also be 1 so that Z = 0.
Answer should be (b).
If no of encirclements is 1, then the pole on RHS should also be 1 so that Z = 0.
Answer should be (b).
Abhimanyu Sharma said:
9 years ago
The answer should be option B. Because given then there is one encirclement. So for stable system Z=0 means one pole should lie on RHS. As N=P-Z where P is number of poles lie RHS of s Plane.
Nandini said:
8 years ago
In Nyquis plot N=Z-P, N=no.of encirclements about origin, P=poles, Z=zeros, i.e.,if P=0 in order to get the system stable, therefore there will be no poles in the RHS plane.
N = 0 (no encirclement), so Z = P = 0 and Z = P.
If N = 0, P must be zero therefore system is stable.
N = 0 (no encirclement), so Z = P = 0 and Z = P.
If N = 0, P must be zero therefore system is stable.
Barman said:
8 years ago
But when you take g(s) h (s) you have to count a number of encirclement about the point (-1, 0) not about the origin.
Midhun Joy said:
2 years ago
I think option B is the right one.
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