Lecture 27 Nyquist Plot Process Control Prof. Kannan M. Moudgalya - - PowerPoint PPT Presentation

Lecture 27 Nyquist Plot

Process Control

• Prof. Kannan M. Moudgalya

IIT Bombay Thursday, 3 October 2013

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Outline

• 1. Cauchy’s principle
• 2. Nyquist plots for analysis and design
• 3. Example

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Motivation for Nyquist plot

◮ Bode analysis some times gave

unpredictable results

◮ When K was decreased, system became

unstable

◮ Nyquist solved this problem ◮ Using Cauchy principle ◮ Nyquist plot analysis

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• 1. Cauchy’s principle

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Cauchy’s Principle for Function F(s)

◮ Draw a closed contour C1 in s plane

Re(s) C1 Im(s)

◮ such that no zeros/poles of F(s) lies on C1 ◮ Let Z zeros, P poles of F(s) lie within C1 ◮ Evaluate F(s) at all points on C1 clockwise ◮ If si is complex, F(si) is also complex ◮ A plot of Im(F(s)) vs. Re(F(s)) is C2.

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MCQ: The C2 curve

C2 curve

• 1. cannot cut itself
• 2. can cut itself any number of times
• 3. will be a closed contour
• 4. is an arbitrary one, cannot say much

Answer: 3

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Cauchy’s Principle for Function F(s)

Im(F (s)) Re(s) C1 C2 Re(F (s)) Im(s)

◮ Because C1 is closed, C2 also is closed. ◮ Cauchy’s Principle: C2 will encircle origin

• f F plane N times in clockwise direction

◮ N = Z − P ◮ Z and P are number of zeros and poles of

F(s) within C1

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Counting encirclements of (0,0)

C2 N =? Im(F (s)) Re(F (s))

N = 2

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Counting encirclements of (0,0) - ctd

C2 N =? Re(F (s)) Im(F (s))

N = 0

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Counting encirclements of (0,0) - ctd

C2 N =? Re(F (s)) Im(F (s))

N = 0

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• 2. Nyquist plots for analysis and design

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Let us apply to control system design

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MCQ: Closed loop Stability

+

− K y r G = b a

The closed loop system is stable when

• 1. The poles of KG/(1 + KG) are in left half

plane

• 2. The zeros of KG/(1 + KG) are in left half

plane

• 3. The poles of KG/(1 + KG) are in right

half plane

• 4. The zeros of KG/(1 + KG) are in left half

plane Answer: 1

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MCQ: Closed loop Stability

+

− K y r G = b a

The closed loop system is stable when

• 1. The poles of 1 + KG are in left half plane
• 2. The zeros of 1 + KG are in left half plane
• 3. The poles of 1 + KG are in right half plane
• 4. The zeros of 1 + KG are in right half plane

Answer: 2

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MCQ: Closed loop Stability

+

− K y r G = b a

The closed loop system is stable when

• 1. The poles of a + Kb are in left half plane
• 2. The zeros of a + Kb are in left half plane
• 3. The zeros of b are in right half plane
• 4. The zeros of a are in left half plane

Answer: 2

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Recall: Design of proportional controller K

+

− K y r G = b a

y(s) = KG(s) 1 + KG(s)r(s) = Kb(s) a(s) 1 + Kb(s) a(s) r(s)

◮ Zeros of 1 + Kb(s)

a(s) = poles of closed loop system.

◮ Want them in left half plane for stability.

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Encirclement Criterion for Stability

◮ Let C1 cover all of RHP

C1 Re(s) Im(s)

◮ For closed loop stability, no. of zeros of

1 + Kb(s) a(s), inside RHP (C1), should be zero

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Encirclement Criterion for Stability

C1 Re(s) Im(s)

◮ For stability, no. of zeros of 1 + Kb(s)

a(s), inside RHP (C1) should be zero

◮ Let 1 + Kb(s)

a(s) have Z zeros and P poles, inside C1

◮ For stability, Z = 0

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Encirclement Criterion for Stability - Ctd

◮ Let F(s) = 1 + Kb(s)

a(s) have Z zeros and P poles, inside C1

◮ C1 covers the entire right half plane ◮ For stability, Z = 0 ◮ Evaluate F(s) = 1 + Kb(s)

a(s) along C1

◮ plot it and call it C2 in the F plane ◮ For stability, N = Z − P = −P ◮ P is the number of poles of F(s) inside C1 ◮ P is the number of unstable poles of F(s) ◮ C2 should encircle −P times for stability

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Encirclement Criterion for Stability - Ctd

◮ C2 should encircle −P times for stability ◮ P = No. of unstable poles of F(s) ◮ P = No. of unstable poles of 1 + Kb(s)

a(s)

◮ P = No. of unstable poles of

a(s) + Kb(s) a(s)

◮ Any connection with open loop system? ◮ P = No. of unstable poles of b(s)

a(s)

◮ P = No. of unstable roots of a(s) = 0 ◮ P = no. of open loop unstable poles ◮ For stability, N = −P, P being the

number of open loop unstable poles

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How to Calculate K Using Nyquist Plot?

◮ Evaluate

1+Kb(s) a(s) along the unit circle (C1), plot C2

Im(F (s)) Re(s) C1 C2 Re(F (s)) Im(s)

◮ C2 should encircle origin −P times, P =

• no. of open loop unstable poles

◮ But we do not yet know the value of K ◮ Want a design approach to calculate K

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How to Calculate K Using Nyquist Plot?

◮ Plot 1 + Kb(s)

a(s) − 1 = Kb(s) a(s) along C1: C3

Im(F (s)) Re(s) C1 C2 Re(F (s)) Im(s) C3 C2 Im(F (s)) Re(F (z))

◮ For stability, plot of Kb(s)/a(s), called C3,

should encircle −P times the pt. (−1, 0)

◮ Still need to know K ◮ Evaluate b(s)/a(s) along C1 and plot: C4 ◮ C4 should encircle (−1/K, 0), −P times ◮ C4 is the Nyquist plot

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• 3. Example

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Example 1

Find the range of proportional controller K that will make the closed loop system stable for the plant G(s) = 30 (s + 1)(s + 2)(s + 3)

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Contour C1 for Example 1

R1 → ∞

Re(s) Im(s) C12 C13 C11

◮ Split C1 contour

into three parts

◮ Call them C11,

C12 and C13

◮ Evaluate G(s)

along each

◮ Plot them on

G(s) plane

◮ Call the

corresponding plots C41, C42 and C43

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We get this Nyquist Plot

R2

Im(G(s)) Re(G(s)) C41 C43 (−0.5, 0) at ω = √ 11 (5, 0) At ω = 1, (0, −3)

R1 R3 R4

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Contour C1 for Example 1

R1 → ∞

Re(s) Im(s) C12 C13 C11

◮ Split C1 contour

into three parts

◮ Call them C11,

C12 and C13

◮ Evaluate G(s)

along each

◮ Plot them on

G(s) plane

◮ Call the

corresponding plots C41, C42 and C43

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Evaluate G(s) for s = jω

◮ G(jω) =

30 (jω + 1)(jω + 2)(jω + 3)

◮ = 30(−jω + 1)(−jω + 2)(−jω + 3)

(ω2 + 1)(ω2 + 4)(ω2 + 9)

◮ = 30

jω3 − 6ω2 − jω11 + 6 (ω2 + 1)(ω2 + 4)(ω2 + 9)

◮ = 30[−6ω2 + 6] + j(ω3 − 11ω)

(ω2 + 1)(ω2 + 4)(ω2 + 9)

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Evaluation of G(s) on C11

◮ G(s) = 30

−6ω2 + 6 (ω2 + 1)(ω2 + 4)(ω2 + 9)

◮ +j30

(ω3 − 11ω) (ω2 + 1)(ω2 + 4)(ω2 + 9)

◮ ω = 0: G(jω) = 30

6 4 × 9 + j30 × 0 = 5

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Evaluation of G(s) on C11

◮ G(s) = 30

−6ω2 + 6 (ω2 + 1)(ω2 + 4)(ω2 + 9)

◮ +j30

(ω3 − 11ω) (ω2 + 1)(ω2 + 4)(ω2 + 9)

◮ 1 > ω > 0: Re G(jω) > 0, Im G(jω) < 0 ◮ ω = 1: Re G(jω) = 0

◮ Im G(jω) = 30

−10 2 × 5 × 10 = −3

◮ G = (0, −3)

◮ At ω =

√ 11, Im G(jω) = 0

◮ Re G(jω) =

−66 + 6 12 × 15 × 20 = −0.5

◮ G(jω) = (−0.5, 0) 30/34 Process Control Nyquist Plot

We get this Nyquist Plot

R2

Im(G(s)) Re(G(s)) C41 C43 (−0.5, 0) at ω = √ 11 (5, 0) At ω = 1, (0, −3)

R1 R3 R4

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Why does it end at 0?

Recall C1:

R1 → ∞

Re(s) Im(s) C12 C13 C11

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What we learnt today

◮ Cauchy condition ◮ What is a Nyquist plot ◮ Nyquist stability condition ◮ An example

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Thank you

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