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 1/34 Process Control Nyquist Plot

Outline 1. Cauchy’s principle 2. Nyquist plots for analysis and design 3. Example 2/34 Process Control Nyquist Plot

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 3/34 Process Control Nyquist Plot

1. Cauchy’s principle 4/34 Process Control Nyquist Plot

Cauchy’s Principle for Function F(s) ◮ Draw a closed contour C 1 in s plane Im( s ) C 1 Re( s ) ◮ such that no zeros/poles of F(s) lies on C 1 ◮ Let Z zeros, P poles of F(s) lie within C 1 ◮ Evaluate F(s) at all points on C 1 clockwise ◮ If s i is complex, F(s i ) is also complex ◮ A plot of Im(F(s)) vs. Re(F(s)) is C 2 . 5/34 Process Control Nyquist Plot

MCQ: The C 2 curve C 2 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 6/34 Process Control Nyquist Plot

Cauchy’s Principle for Function F(s) Im( s ) Im( F ( s ) ) C 1 C 2 Re( s ) Re( F ( s ) ) ◮ Because C 1 is closed, C 2 also is closed. ◮ Cauchy’s Principle: C 2 will encircle origin of F plane N times in clockwise direction ◮ N = Z − P ◮ Z and P are number of zeros and poles of F(s) within C 1 7/34 Process Control Nyquist Plot

Counting encirclements of (0,0) Im( F ( s ) ) N =? Re( F ( s ) ) C 2 N = 2 8/34 Process Control Nyquist Plot

Counting encirclements of (0,0) - ctd Im( F ( s ) ) N =? Re( F ( s ) ) C 2 N = 0 9/34 Process Control Nyquist Plot

Counting encirclements of (0,0) - ctd Im( F ( s ) ) N =? C 2 Re( F ( s ) ) N = 0 10/34 Process Control Nyquist Plot

2. Nyquist plots for analysis and design 11/34 Process Control Nyquist Plot

Let us apply to control system design 12/34 Process Control Nyquist Plot

MCQ: Closed loop Stability y G = b r K + 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 13/34 Process Control Nyquist Plot

MCQ: Closed loop Stability y G = b r K + 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 14/34 Process Control Nyquist Plot

MCQ: Closed loop Stability y G = b r K + 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 15/34 Process Control Nyquist Plot

Recall: Design of proportional controller K y G = b r K + a − Kb(s) KG(s) a(s) y(s) = 1 + KG(s)r(s) = r(s) 1 + Kb(s) a(s) ◮ Zeros of 1 + Kb(s) a(s) = poles of closed loop system. ◮ Want them in left half plane for stability. 16/34 Process Control Nyquist Plot

Encirclement Criterion for Stability ◮ Let C 1 cover all of RHP Im(s) C 1 Re(s) ◮ For closed loop stability, no. of zeros of 1 + Kb(s) a(s), inside RHP (C 1 ), should be zero 17/34 Process Control Nyquist Plot

Encirclement Criterion for Stability Im(s) C 1 Re(s) ◮ For stability, no. of zeros of 1 + Kb(s) a(s), inside RHP (C 1 ) should be zero ◮ Let 1 + Kb(s) a(s) have Z zeros and P poles, inside C 1 ◮ For stability, Z = 0 18/34 Process Control Nyquist Plot

Encirclement Criterion for Stability - Ctd ◮ Let F(s) = 1 + Kb(s) a(s) have Z zeros and P poles, inside C 1 ◮ C 1 covers the entire right half plane ◮ For stability, Z = 0 ◮ Evaluate F(s) = 1 + Kb(s) a(s) along C 1 ◮ plot it and call it C 2 in the F plane ◮ For stability, N = Z − P = − P ◮ P is the number of poles of F(s) inside C 1 ◮ P is the number of unstable poles of F(s) ◮ C 2 should encircle − P times for stability 19/34 Process Control Nyquist Plot

Encirclement Criterion for Stability - Ctd ◮ C 2 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) a(s) + Kb(s) ◮ P = No. of unstable poles of 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 20/34 Process Control Nyquist Plot

How to Calculate K Using Nyquist Plot? ◮ Evaluate 1+Kb(s) a(s) along the unit circle (C 1 ), plot C 2 Im( s ) Im( F ( s ) ) C 1 C 2 Re( s ) Re( F ( s ) ) ◮ C 2 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 21/34 Process Control Nyquist Plot

How to Calculate K Using Nyquist Plot? ◮ Plot 1 + Kb(s) a(s) − 1 = Kb(s) a(s) along C 1 : C 3 Im( F ( s ) ) Im( s ) Im( F ( s ) ) C 1 C 2 C 3 C 2 Re( F ( z ) ) Re( s ) Re( F ( s ) ) ◮ For stability, plot of Kb(s) / a(s), called C 3 , should encircle − P times the pt. ( − 1 , 0) ◮ Still need to know K ◮ Evaluate b(s) / a(s) along C 1 and plot: C 4 ◮ C 4 should encircle ( − 1 / K , 0), − P times ◮ C 4 is the Nyquist plot 22/34 Process Control Nyquist Plot

3. Example 23/34 Process Control Nyquist Plot

Example 1 Find the range of proportional controller K that will make the closed loop system stable for the plant 30 G(s) = (s + 1)(s + 2)(s + 3) 24/34 Process Control Nyquist Plot

Contour C 1 for Example 1 ◮ Split C 1 contour Im(s) into three parts C 12 ◮ Call them C 11 , C 11 C 12 and C 13 ◮ Evaluate G(s) R 1 → ∞ along each Re(s) ◮ Plot them on C 13 G(s) plane ◮ Call the corresponding plots C 41 , C 42 and C 43 25/34 Process Control Nyquist Plot

We get this Nyquist Plot Im(G(s)) C 43 R 4 R 1 (5 , 0) R 3 R 2 ( − 0 . 5 , 0) Re(G(s)) √ at ω = 11 C 41 At ω = 1 , (0 , − 3) 26/34 Process Control Nyquist Plot

Contour C 1 for Example 1 ◮ Split C 1 contour Im(s) into three parts C 12 ◮ Call them C 11 , C 11 C 12 and C 13 ◮ Evaluate G(s) R 1 → ∞ along each Re(s) ◮ Plot them on C 13 G(s) plane ◮ Call the corresponding plots C 41 , C 42 and C 43 27/34 Process Control Nyquist Plot

Evaluate G(s) for s = j ω 30 ◮ G(j ω ) = (j ω + 1)(j ω + 2)(j ω + 3) ◮ = 30( − j ω + 1)( − j ω + 2)( − j ω + 3) ( ω 2 + 1)( ω 2 + 4)( ω 2 + 9) j ω 3 − 6 ω 2 − j ω 11 + 6 ◮ = 30 ( ω 2 + 1)( ω 2 + 4)( ω 2 + 9) ◮ = 30[ − 6 ω 2 + 6] + j( ω 3 − 11 ω ) ( ω 2 + 1)( ω 2 + 4)( ω 2 + 9) 28/34 Process Control Nyquist Plot

Evaluation of G(s) on C 11 − 6 ω 2 + 6 ◮ G(s) = 30 ( ω 2 + 1)( ω 2 + 4)( ω 2 + 9) ( ω 3 − 11 ω ) ◮ +j30 ( ω 2 + 1)( ω 2 + 4)( ω 2 + 9) 6 ◮ ω = 0: G(j ω ) = 30 4 × 9 + j30 × 0 = 5 29/34 Process Control Nyquist Plot

Evaluation of G(s) on C 11 − 6 ω 2 + 6 ◮ G(s) = 30 ( ω 2 + 1)( ω 2 + 4)( ω 2 + 9) ( ω 3 − 11 ω ) ◮ +j30 ( ω 2 + 1)( ω 2 + 4)( ω 2 + 9) ◮ 1 > ω > 0: Re G(j ω ) > 0, Im G(j ω ) < 0 ◮ ω = 1: Re G(j ω ) = 0 − 10 ◮ Im G(j ω ) = 30 2 × 5 × 10 = − 3 ◮ G = (0 , − 3) √ ◮ At ω = 11, Im G(j ω ) = 0 − 66 + 6 ◮ Re G(j ω ) = 12 × 15 × 20 = − 0 . 5 ◮ G(j ω ) = ( − 0 . 5 , 0) 30/34 Process Control Nyquist Plot

We get this Nyquist Plot Im(G(s)) C 43 R 4 R 1 (5 , 0) R 3 R 2 ( − 0 . 5 , 0) Re(G(s)) √ at ω = 11 C 41 At ω = 1 , (0 , − 3) 31/34 Process Control Nyquist Plot

Why does it end at 0? Im(s) C 12 C 11 R 1 → ∞ Re(s) C 13 Recall C 1 : 32/34 Process Control Nyquist Plot

What we learnt today ◮ Cauchy condition ◮ What is a Nyquist plot ◮ Nyquist stability condition ◮ An example 33/34 Process Control Nyquist Plot

Thank you 34/34 Process Control Nyquist Plot