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From the below given Nyquist plot, calculate the number of open-loop poles on the right-hand side of the s-plane for the closed-loop system to be stable.
- a)1
- b)2
- c)0
- d)-1
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
From the below given Nyquist plot, calculate the number of open-loop p...
Principle arguments
- It states that if there are “P” poles and “Z” zeroes for a closed, random selected path then the corresponding G(s)H(s) plane encircles the origin with P – Z times.
- Encirclements in s – plane and GH – plane are shown below.
- In GH plane Anti clockwise encirclements are taken as positive and clockwise encirclements are taken as negative.
It is applied to the total RH plane by selecting a closed path with r = ∞
Nyquist stability completely deals with the right half of s – plane.
N(0, 0) = P – Z
N(0, 0): Number of encirclements around critical point (- 1, 0)
N(0, 0) = P – Z
N(0, 0): Number of encirclements around critical point (- 1, 0)
P: Open loop poles
Z: Open-loop zeroes.
Note:
1) To get the Closed-loop stability we require 1 + GH plane but available is GH plane, hence the origin is shifted to “-1” to get the closed-loop stability.
2) To become the system stable there should not be any closed-loop pole in the right of s – plane.
3) The closed-loop pole is the same as that of the zeroes of Characteristic Equation which must be zero in the right. i.e, Z = 0
N = P is the criteria.
Calculation:
From the given Nyquist plot there is one encirclement about ( -1, 0 ) in the Anti-clockwise direction.
So, N = 1
Now to satisfy the stability criteria N should be equal to P.
N = P = 1
So the number of open-loop poles in the Right-hand side of the system is 1.
Most Upvoted Answer
From the below given Nyquist plot, calculate the number of open-loop p...
Principle arguments
- It states that if there are “P” poles and “Z” zeroes for a closed, random selected path then the corresponding G(s)H(s) plane encircles the origin with P – Z times.
- Encirclements in s – plane and GH – plane are shown below.
- In GH plane Anti clockwise encirclements are taken as positive and clockwise encirclements are taken as negative.
It is applied to the total RH plane by selecting a closed path with r = ∞
Nyquist stability completely deals with the right half of s – plane.
N(0, 0) = P – Z
N(0, 0): Number of encirclements around critical point (- 1, 0)
N(0, 0) = P – Z
N(0, 0): Number of encirclements around critical point (- 1, 0)
P: Open loop poles
Z: Open-loop zeroes.
Note:
1) To get the Closed-loop stability we require 1 + GH plane but available is GH plane, hence the origin is shifted to “-1” to get the closed-loop stability.
2) To become the system stable there should not be any closed-loop pole in the right of s – plane.
3) The closed-loop pole is the same as that of the zeroes of Characteristic Equation which must be zero in the right. i.e, Z = 0
N = P is the criteria.
Calculation:
From the given Nyquist plot there is one encirclement about ( -1, 0 ) in the Anti-clockwise direction.
So, N = 1
Now to satisfy the stability criteria N should be equal to P.
N = P = 1
So the number of open-loop poles in the Right-hand side of the system is 1.
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