1. Cauchys Principle for Function F ( z ) 1. Cauchys Principle for - - PowerPoint PPT Presentation

1. Cauchy’s Principle for Function F (z)

1. Cauchy’s Principle for Function F (z)

• Draw a closed contour C1 in Z plane: no ze-

ros/poles of F (z) should lie on C1.

1. Cauchy’s Principle for Function F (z)

• Draw a closed contour C1 in Z plane: no ze-

ros/poles of F (z) should lie on C1.

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

1. Cauchy’s Principle for Function F (z)

• Draw a closed contour C1 in Z plane: no ze-

ros/poles of F (z) should lie on C1.

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

• Let z zeros and p poles of F (z) lie within the

closed contour.

1. Cauchy’s Principle for Function F (z)

• Draw a closed contour C1 in Z plane: no ze-

ros/poles of F (z) should lie on C1.

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

• Let z zeros and p poles of F (z) lie within the

closed contour.

• Evaluate F (z) at all points on the curve C1 in

the clockwise direction.

1. Cauchy’s Principle for Function F (z)

• Draw a closed contour C1 in Z plane: no ze-

ros/poles of F (z) should lie on C1.

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

• Let z zeros and p poles of F (z) lie within the

closed contour.

• Evaluate F (z) at all points on the curve C1 in

the clockwise direction.

Digital Control

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Kannan M. Moudgalya, Autumn 2007

2. Cauchy’s Principle for Function F (z)

2. Cauchy’s Principle for Function F (z)

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

2. Cauchy’s Principle for Function F (z)

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

• Plot the evaluated values in another plane, Im[F (z)]
• vs. Re[F (z)].

2. Cauchy’s Principle for Function F (z)

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

• Plot the evaluated values in another plane, Im[F (z)]
• vs. Re[F (z)].
• The new curve also will be a closed contour, call it

C2.

2. Cauchy’s Principle for Function F (z)

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

• Plot the evaluated values in another plane, Im[F (z)]
• vs. Re[F (z)].
• The new curve also will be a closed contour, call it

C2.

• Cauchy’s Principle: C2 will encircle origin of F

plane N times in clockwise direction.

2. Cauchy’s Principle for Function F (z)

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

• Plot the evaluated values in another plane, Im[F (z)]
• vs. Re[F (z)].
• The new curve also will be a closed contour, call it

C2.

• Cauchy’s Principle: C2 will encircle origin of F

plane N times in clockwise direction. N = z−p

Digital Control

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Kannan M. Moudgalya, Autumn 2007

3. Example: Cauchy’s Principle

3. Example: Cauchy’s Principle

x Im(z) Re(z) C1

F (z) = z − 0.5

3. Example: Cauchy’s Principle

x Im(z) Re(z) C1

F (z) = z − 0.5 z = ejω

3. Example: Cauchy’s Principle

x Im(z) Re(z) C1

F (z) = z − 0.5 z = ejω F

• ejω

= ejω − 0.5

3. Example: Cauchy’s Principle

x Im(z) Re(z) C1

F (z) = z − 0.5 z = ejω F

• ejω

= ejω − 0.5

• ejω is a circle of ra-

dius 1, centred at 0.

3. Example: Cauchy’s Principle

x Im(z) Re(z) C1

F (z) = z − 0.5 z = ejω F

• ejω

= ejω − 0.5

• ejω is a circle of ra-

dius 1, centred at 0.

• Subtract 0.5 from ev-

ery point.

3. Example: Cauchy’s Principle

x Im(z) Re(z) C1

F (z) = z − 0.5 z = ejω F

• ejω

= ejω − 0.5

• ejω is a circle of ra-

dius 1, centred at 0.

• Subtract 0.5 from ev-

ery point.

• It is a circle shifted to

the left by 0.5.

3. Example: Cauchy’s Principle

x Im(z) Re(z) C1

F (z) = z − 0.5 z = ejω F

• ejω

= ejω − 0.5

• ejω is a circle of ra-

dius 1, centred at 0.

• Subtract 0.5 from ev-

ery point.

• It is a circle shifted to

the left by 0.5.

Im(F(z)) Re(F(z)) C2

3. Example: Cauchy’s Principle

x Im(z) Re(z) C1

F (z) = z − 0.5 z = ejω F

• ejω

= ejω − 0.5

• ejω is a circle of ra-

dius 1, centred at 0.

• Subtract 0.5 from ev-

ery point.

• It is a circle shifted to

the left by 0.5.

Im(F(z)) Re(F(z)) C2

Direction: same as C1,

3. Example: Cauchy’s Principle

x Im(z) Re(z) C1

F (z) = z − 0.5 z = ejω F

• ejω

= ejω − 0.5

• ejω is a circle of ra-

dius 1, centred at 0.

• Subtract 0.5 from ev-

ery point.

• It is a circle shifted to

the left by 0.5.

Im(F(z)) Re(F(z)) C2

Direction: same as C1, with one encirclement

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Kannan M. Moudgalya, Autumn 2007

4. Example 2: Cauchy’s Principle

4. Example 2: Cauchy’s Principle

F (z) = 1 z

4. Example 2: Cauchy’s Principle

F (z) = 1 z z = ejω

4. Example 2: Cauchy’s Principle

F (z) = 1 z z = ejω F

• ejω

= e−jω

4. Example 2: Cauchy’s Principle

F (z) = 1 z z = ejω F

• ejω

= e−jω

• It is a unit circle with counter clockwise direction.

4. Example 2: Cauchy’s Principle

F (z) = 1 z z = ejω F

• ejω

= e−jω

• It is a unit circle with counter clockwise direction.
• One pole inside C1, encirclement = −1.
• N = z − p = 0 − 1 = −1

Digital Control

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Kannan M. Moudgalya, Autumn 2007

5. Counting of Encirclements of (0,0)

5. Counting of Encirclements of (0,0)

C2 N = 2 N = 0 Re(F(z)) Im(F(z)) Im(F(z)) N = 0 Re(F(z)) Im(F(z)) Re(F(z)) C2 C2

Digital Control

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Kannan M. Moudgalya, Autumn 2007

6. Design of Proportional Controller

6. Design of Proportional Controller

+

− y K r G = b a

6. Design of Proportional Controller

+

− y K r G = b a

y(n) = KG(z) 1 + KG(z)r(n)

6. Design of Proportional Controller

+

− y K r G = b a

y(n) = KG(z) 1 + KG(z)r(n) = K b(z) a(z) 1 + K b(z) a(z) r(n)

6. Design of Proportional Controller

+

− y K r G = b a

y(n) = KG(z) 1 + KG(z)r(n) = K b(z) a(z) 1 + K b(z) a(z) r(n)

• Zeros of 1+K b(z)

a(z) = poles of closed loop system.

6. Design of Proportional Controller

+

− y K r G = b a

y(n) = KG(z) 1 + KG(z)r(n) = K b(z) a(z) 1 + K b(z) a(z) r(n)

• Zeros of 1+K b(z)

a(z) = poles of closed loop system.

• Want them inside unit circle for stability.

Digital Control

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Kannan M. Moudgalya, Autumn 2007

7. Encirclement Criterion for Stability

7. Encirclement Criterion for Stability

x Im(z) Re(z) C1

7. Encirclement Criterion for Stability

x Im(z) Re(z) C1

Want zeros of 1 + K b(z) a(z) inside unit circle C1 for stability.

7. Encirclement Criterion for Stability

x Im(z) Re(z) C1

Want zeros of 1 + K b(z) a(z) inside unit circle C1 for stability.

• Let 1 + K b(z)

a(z) have

7. Encirclement Criterion for Stability

x Im(z) Re(z) C1

Want zeros of 1 + K b(z) a(z) inside unit circle C1 for stability.

• Let 1 + K b(z)

a(z) have – a total n zeros and n poles.

7. Encirclement Criterion for Stability

x Im(z) Re(z) C1

Want zeros of 1 + K b(z) a(z) inside unit circle C1 for stability.

• Let 1 + K b(z)

a(z) have – a total n zeros and n poles. – Z zeros outside unit circle,

7. Encirclement Criterion for Stability

x Im(z) Re(z) C1

Want zeros of 1 + K b(z) a(z) inside unit circle C1 for stability.

• Let 1 + K b(z)

a(z) have – a total n zeros and n poles. – Z zeros outside unit circle, n − Z inside.

7. Encirclement Criterion for Stability

x Im(z) Re(z) C1

Want zeros of 1 + K b(z) a(z) inside unit circle C1 for stability.

• Let 1 + K b(z)

a(z) have – a total n zeros and n poles. – Z zeros outside unit circle, n − Z inside. – P poles outside unit circle,

7. Encirclement Criterion for Stability

x Im(z) Re(z) C1

Want zeros of 1 + K b(z) a(z) inside unit circle C1 for stability.

• Let 1 + K b(z)

a(z) have – a total n zeros and n poles. – Z zeros outside unit circle, n − Z inside. – P poles outside unit circle, n − P inside.

7. Encirclement Criterion for Stability

x Im(z) Re(z) C1

Want zeros of 1 + K b(z) a(z) inside unit circle C1 for stability.

• Let 1 + K b(z)

a(z) have – a total n zeros and n poles. – Z zeros outside unit circle, n − Z inside. – P poles outside unit circle, n − P inside. – P is number of unstable poles

7. Encirclement Criterion for Stability

x Im(z) Re(z) C1

Want zeros of 1 + K b(z) a(z) inside unit circle C1 for stability.

• Let 1 + K b(z)

a(z) have – a total n zeros and n poles. – Z zeros outside unit circle, n − Z inside. – P poles outside unit circle, n − P inside. – P is number of unstable poles

• For stability, Z = 0

7. Encirclement Criterion for Stability

x Im(z) Re(z) C1

Want zeros of 1 + K b(z) a(z) inside unit circle C1 for stability.

• Let 1 + K b(z)

a(z) have – a total n zeros and n poles. – Z zeros outside unit circle, n − Z inside. – P poles outside unit circle, n − P inside. – P is number of unstable poles

• For stability, Z = 0

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Kannan M. Moudgalya, Autumn 2007

8. Encirclement Criterion for Stability - Continued

8. Encirclement Criterion for Stability - Continued

• Evaluate 1 + K b(z)

a(z) along C1 and plot it: Called C2

8. Encirclement Criterion for Stability - Continued

• Evaluate 1 + K b(z)

a(z) along C1 and plot it: Called C2

• C2 encircles origin N = (n − Z) − (n − P )

8. Encirclement Criterion for Stability - Continued

• Evaluate 1 + K b(z)

a(z) along C1 and plot it: Called C2

• C2 encircles origin N = (n − Z) − (n − P ) = P − Z

times

8. Encirclement Criterion for Stability - Continued

• Evaluate 1 + K b(z)

a(z) along C1 and plot it: Called C2

• C2 encircles origin N = (n − Z) − (n − P ) = P − Z

times

• Want Z = 0 for stability.

8. Encirclement Criterion for Stability - Continued

• Evaluate 1 + K b(z)

a(z) along C1 and plot it: Called C2

• C2 encircles origin N = (n − Z) − (n − P ) = P − Z

times

• Want Z = 0 for stability. i.e., N = P for stability

8. Encirclement Criterion for Stability - Continued

• Evaluate 1 + K b(z)

a(z) along C1 and plot it: Called C2

• C2 encircles origin N = (n − Z) − (n − P ) = P − Z

times

• Want Z = 0 for stability. i.e., N = P for stability

P = No. of unstable poles of 1 + K b(z) a(z)

8. Encirclement Criterion for Stability - Continued

• Evaluate 1 + K b(z)

a(z) along C1 and plot it: Called C2

• C2 encircles origin N = (n − Z) − (n − P ) = P − Z

times

• Want Z = 0 for stability. i.e., N = P for stability

P = No. of unstable poles of 1 + K b(z) a(z) = No. of unstable poles of a(z) + Kb(z) a(z)

8. Encirclement Criterion for Stability - Continued

• Evaluate 1 + K b(z)

a(z) along C1 and plot it: Called C2

• C2 encircles origin N = (n − Z) − (n − P ) = P − Z

times

• Want Z = 0 for stability. i.e., N = P for stability

P = No. of unstable poles of 1 + K b(z) a(z) = No. of unstable poles of a(z) + Kb(z) a(z) = No. of unstable poles of b(z) a(z)

8. Encirclement Criterion for Stability - Continued

• Evaluate 1 + K b(z)

a(z) along C1 and plot it: Called C2

• C2 encircles origin N = (n − Z) − (n − P ) = P − Z

times

• Want Z = 0 for stability. i.e., N = P for stability

P = No. of unstable poles of 1 + K b(z) a(z) = No. of unstable poles of a(z) + Kb(z) a(z) = No. of unstable poles of b(z) a(z) = no. of open loop unstable poles

8. Encirclement Criterion for Stability - Continued

• Evaluate 1 + K b(z)

a(z) along C1 and plot it: Called C2

• C2 encircles origin N = (n − Z) − (n − P ) = P − Z

times

• Want Z = 0 for stability. i.e., N = P for stability

P = No. of unstable poles of 1 + K b(z) a(z) = No. of unstable poles of a(z) + Kb(z) a(z) = No. of unstable poles of b(z) a(z) = no. of open loop unstable poles N should be equal to the number of open loop unstable poles

Digital Control

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Kannan M. Moudgalya, Autumn 2007

9. Procedure to Calculate K Using Nyquist Plot

9. Procedure to Calculate K Using Nyquist Plot

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

9. Procedure to Calculate K Using Nyquist Plot

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

9. Procedure to Calculate K Using Nyquist Plot

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

• Evaluate 1+K b(z)

a(z) along the unit circle (C1) and plot C2

9. Procedure to Calculate K Using Nyquist Plot

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

• Evaluate 1+K b(z)

a(z) along the unit circle (C1) and plot C2

• C2 should encircle origin P times = open loop unstable poles

9. Procedure to Calculate K Using Nyquist Plot

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

• Evaluate 1+K b(z)

a(z) along the unit circle (C1) and plot C2

• C2 should encircle origin P times = open loop unstable poles
• K has to be known to do this

9. Procedure to Calculate K Using Nyquist Plot

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

• Evaluate 1+K b(z)

a(z) along the unit circle (C1) and plot C2

• C2 should encircle origin P times = open loop unstable poles
• K has to be known to do this
• Want to convert this into a design approach to calculate K

9. Procedure to Calculate K Using Nyquist Plot

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

• Evaluate 1+K b(z)

a(z) along the unit circle (C1) and plot C2

• C2 should encircle origin P times = open loop unstable poles
• K has to be known to do this
• Want to convert this into a design approach to calculate K
• Evaluate 1 + K b(z)

a(z) − 1 = K b(z) a(z) along C1, plot, call it C3

9. Procedure to Calculate K Using Nyquist Plot

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

• Evaluate 1+K b(z)

a(z) along the unit circle (C1) and plot C2

• C2 should encircle origin P times = open loop unstable poles
• K has to be known to do this
• Want to convert this into a design approach to calculate K
• Evaluate 1 + K b(z)

a(z) − 1 = K b(z) a(z) along C1, plot, call it C3

Digital Control

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Kannan M. Moudgalya, Autumn 2007

10. Procedure to Calculate K Using Nyquist Plot

10. Procedure to Calculate K Using Nyquist Plot

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

10. Procedure to Calculate K Using Nyquist Plot

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

10. Procedure to Calculate K Using Nyquist Plot

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

• For stability, plot of Kb(z)/a(z), C3, should encircle the

point (−1, 0), P times

10. Procedure to Calculate K Using Nyquist Plot

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

• For stability, plot of Kb(z)/a(z), C3, should encircle the

point (−1, 0), P times

• Still need to know K

10. Procedure to Calculate K Using Nyquist Plot

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

• For stability, plot of Kb(z)/a(z), C3, should encircle the

point (−1, 0), P times

• Still need to know K
• Evaluate b(z)

a(z) along C1 and plot.

10. Procedure to Calculate K Using Nyquist Plot

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

• For stability, plot of Kb(z)/a(z), C3, should encircle the

point (−1, 0), P times

• Still need to know K
• Evaluate b(z)

a(z) along C1 and plot. Call it C4.

10. Procedure to Calculate K Using Nyquist Plot

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

• For stability, plot of Kb(z)/a(z), C3, should encircle the

point (−1, 0), P times

• Still need to know K
• Evaluate b(z)

a(z) along C1 and plot. Call it C4.

• For stability, C4 should encircle point (−1/K, 0), P times

10. Procedure to Calculate K Using Nyquist Plot

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

• For stability, plot of Kb(z)/a(z), C3, should encircle the

point (−1, 0), P times

• Still need to know K
• Evaluate b(z)

a(z) along C1 and plot. Call it C4.

• For stability, C4 should encircle point (−1/K, 0), P times
• C4 is the Nyquist plot

Digital Control

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Kannan M. Moudgalya, Autumn 2007

11. Example of Nyquist Plot to Design Controller

11. Example of Nyquist Plot to Design Controller

G(z) = b(z) a(z) = 1 z(z − 1)

11. Example of Nyquist Plot to Design Controller

G(z) = b(z) a(z) = 1 z(z − 1)

C1 Im(z) Re(z) A B D E

11. Example of Nyquist Plot to Design Controller

G(z) = b(z) a(z) = 1 z(z − 1)

C1 Im(z) Re(z) A B D E

• C1 should not go through pole/zero

11. Example of Nyquist Plot to Design Controller

G(z) = b(z) a(z) = 1 z(z − 1)

C1 Im(z) Re(z) A B D E

• C1 should not go through pole/zero
• Indent it with a semicircle of radius → 0

11. Example of Nyquist Plot to Design Controller

G(z) = b(z) a(z) = 1 z(z − 1)

C1 Im(z) Re(z) A B D E

• C1 should not go through pole/zero
• Indent it with a semicircle of radius → 0
• Number of unstable poles, P = 0

11. Example of Nyquist Plot to Design Controller

G(z) = b(z) a(z) = 1 z(z − 1)

C1 Im(z) Re(z) A B D E

• C1 should not go through pole/zero
• Indent it with a semicircle of radius → 0
• Number of unstable poles, P = 0
• Evaluate b(z)

a(z) along main C1

11. Example of Nyquist Plot to Design Controller

G(z) = b(z) a(z) = 1 z(z − 1)

C1 Im(z) Re(z) A B D E

• C1 should not go through pole/zero
• Indent it with a semicircle of radius → 0
• Number of unstable poles, P = 0
• Evaluate b(z)

a(z) along main C1

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Kannan M. Moudgalya, Autumn 2007

12. Example of Nyquist Plot to Design Controller

12. Example of Nyquist Plot to Design Controller

G(z) = b(z) a(z) = 1 z(z − 1)

12. Example of Nyquist Plot to Design Controller

G(z) = b(z) a(z) = 1 z(z − 1) G

• ejω

= 1 ejω(ejω − 1)

12. Example of Nyquist Plot to Design Controller

G(z) = b(z) a(z) = 1 z(z − 1) G

• ejω

= 1 ejω(ejω − 1) = 1 ej 3

2ω

ej 1

2ω − e−j 1 2ω

12. Example of Nyquist Plot to Design Controller

G(z) = b(z) a(z) = 1 z(z − 1) G

• ejω

= 1 ejω(ejω − 1) = 1 ej 3

2ω

ej 1

2ω − e−j 1 2ω = −je−j 3 2ω

2 sin ω

2

12. Example of Nyquist Plot to Design Controller

G(z) = b(z) a(z) = 1 z(z − 1) G

• ejω

= 1 ejω(ejω − 1) = 1 ej 3

2ω

ej 1

2ω − e−j 1 2ω = −je−j 3 2ω

2 sin ω

2

= −j

• cos 3

2ω − j sin 3 2ω

• 2 sin ω

2

12. Example of Nyquist Plot to Design Controller

G(z) = b(z) a(z) = 1 z(z − 1) G

• ejω

= 1 ejω(ejω − 1) = 1 ej 3

2ω

ej 1

2ω − e−j 1 2ω = −je−j 3 2ω

2 sin ω

2

= −j

• cos 3

2ω − j sin 3 2ω

• 2 sin ω

2

= − sin 3

2ω

2 sin 1

2ω − j cos 3 2ω

2 sin 1

2ω

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Kannan M. Moudgalya, Autumn 2007

13. Example of Nyquist Plot - Continued

13. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

13. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

13. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

13. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

G

• ejω

= −sin 3

2ω

2 sin ω

2

− j cos 3

2ω

2 sin 1

2ω

13. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

G

• ejω

= −sin 3

2ω

2 sin ω

2

− j cos 3

2ω

2 sin 1

2ω

• At point A, ω = 180o, G = 0.5

13. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

G

• ejω

= −sin 3

2ω

2 sin ω

2

− j cos 3

2ω

2 sin 1

2ω

• At point A, ω = 180o, G = 0.5
• At point B, ω = 120o

13. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

G

• ejω

= −sin 3

2ω

2 sin ω

2

− j cos 3

2ω

2 sin 1

2ω

• At point A, ω = 180o, G = 0.5
• At point B, ω = 120o

G = − sin 3

2120

2 sin 1

2120 − j cos 3 2120

2 sin 1

2120

13. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

G

• ejω

= −sin 3

2ω

2 sin ω

2

− j cos 3

2ω

2 sin 1

2ω

• At point A, ω = 180o, G = 0.5
• At point B, ω = 120o

G = − sin 3

2120

2 sin 1

2120 − j cos 3 2120

2 sin 1

2120

= − sin 180 2 sin 60

13. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

G

• ejω

= −sin 3

2ω

2 sin ω

2

− j cos 3

2ω

2 sin 1

2ω

• At point A, ω = 180o, G = 0.5
• At point B, ω = 120o

G = − sin 3

2120

2 sin 1

2120 − j cos 3 2120

2 sin 1

2120

= − sin 180 2 sin 60 − j cos 180 2 sin 60

13. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

G

• ejω

= −sin 3

2ω

2 sin ω

2

− j cos 3

2ω

2 sin 1

2ω

• At point A, ω = 180o, G = 0.5
• At point B, ω = 120o

G = − sin 3

2120

2 sin 1

2120 − j cos 3 2120

2 sin 1

2120

= − sin 180 2 sin 60 − j cos 180 2 sin 60 = j0.5774

13. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

G

• ejω

= −sin 3

2ω

2 sin ω

2

− j cos 3

2ω

2 sin 1

2ω

• At point A, ω = 180o, G = 0.5
• At point B, ω = 120o

G = − sin 3

2120

2 sin 1

2120 − j cos 3 2120

2 sin 1

2120

= − sin 180 2 sin 60 − j cos 180 2 sin 60 = j0.5774

Digital Control

13

Kannan M. Moudgalya, Autumn 2007

14. Example of Nyquist Plot - Continued

14. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

14. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

14. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

14. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

G

• ejω

= −sin 3

2ω

2 sin ω

2

− j cos 3

2ω

2 sin 1

2ω

14. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

G

• ejω

= −sin 3

2ω

2 sin ω

2

− j cos 3

2ω

2 sin 1

2ω

• At point D, ω = 60o, G = −

1 2×1

2 −

j0

14. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

G

• ejω

= −sin 3

2ω

2 sin ω

2

− j cos 3

2ω

2 sin 1

2ω

• At point D, ω = 60o, G = −

1 2×1

2 −

j0 = −1

14. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

G

• ejω

= −sin 3

2ω

2 sin ω

2

− j cos 3

2ω

2 sin 1

2ω

• At point D, ω = 60o, G = −

1 2×1

2 −

j0 = −1

• At point E, ω → 0,

14. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

G

• ejω

= −sin 3

2ω

2 sin ω

2

− j cos 3

2ω

2 sin 1

2ω

• At point D, ω = 60o, G = −

1 2×1

2 −

j0 = −1

• At point E, ω → 0, G = −0/0 −

j∞.

14. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

G

• ejω

= −sin 3

2ω

2 sin ω

2

− j cos 3

2ω

2 sin 1

2ω

• At point D, ω = 60o, G = −

1 2×1

2 −

j0 = −1

• At point E, ω → 0, G = −0/0 −

j∞.

• Use L’Hospital’s rule

14. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

G

• ejω

= −sin 3

2ω

2 sin ω

2

− j cos 3

2ω

2 sin 1

2ω

• At point D, ω = 60o, G = −

1 2×1

2 −

j0 = −1

• At point E, ω → 0, G = −0/0 −

j∞.

• Use L’Hospital’s rule

G = −

3 2 cos 3 2ω

21

2 cos 1 2ω − j∞

14. Example of Nyquist Plot - Continued

C1 Im(z) Re(z) A B D E

E Im Re A B D

• Transfer function is given by

G

• ejω

= −sin 3

2ω

2 sin ω

2

− j cos 3

2ω

2 sin 1

2ω

• At point D, ω = 60o, G = −

1 2×1

2 −

j0 = −1

• At point E, ω → 0, G = −0/0 −

j∞.

• Use L’Hospital’s rule

G = −

3 2 cos 3 2ω

21

2 cos 1 2ω − j∞

= −3 2 − j∞

Digital Control

14

Kannan M. Moudgalya, Autumn 2007

15. Example of Nyquist Plot - Small Semicircle

15. Example of Nyquist Plot - Small Semicircle

C1 Im(z) Re(z) A B D E

15. Example of Nyquist Plot - Small Semicircle

C1 Im(z) Re(z) A B D E

E Im Re A B D

15. Example of Nyquist Plot - Small Semicircle

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

15. Example of Nyquist Plot - Small Semicircle

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• Indentation around 1

15. Example of Nyquist Plot - Small Semicircle

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• Indentation around 1
• Semicircle with centre at (1,0), radius ε →

15. Example of Nyquist Plot - Small Semicircle

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• Indentation around 1
• Semicircle with centre at (1,0), radius ε →
• Because z = 1 + εejφ, φ goes from 90o

15. Example of Nyquist Plot - Small Semicircle

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• Indentation around 1
• Semicircle with centre at (1,0), radius ε →
• Because z = 1 + εejφ, φ goes from 90o

to 0o

15. Example of Nyquist Plot - Small Semicircle

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• Indentation around 1
• Semicircle with centre at (1,0), radius ε →
• Because z = 1 + εejφ, φ goes from 90o

to 0o to −90o

15. Example of Nyquist Plot - Small Semicircle

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• Indentation around 1
• Semicircle with centre at (1,0), radius ε →
• Because z = 1 + εejφ, φ goes from 90o

to 0o to −90o G(z) = G

• 1 + εejφ

15. Example of Nyquist Plot - Small Semicircle

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• Indentation around 1
• Semicircle with centre at (1,0), radius ε →
• Because z = 1 + εejφ, φ goes from 90o

to 0o to −90o G(z) = G

• 1 + εejφ

= 1 (1 + εejφ) εejφ

15. Example of Nyquist Plot - Small Semicircle

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• Indentation around 1
• Semicircle with centre at (1,0), radius ε →
• Because z = 1 + εejφ, φ goes from 90o

to 0o to −90o G(z) = G

• 1 + εejφ

= 1 (1 + εejφ) εejφ = ∞e−jφ 1 as ε → 0

15. Example of Nyquist Plot - Small Semicircle

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• Indentation around 1
• Semicircle with centre at (1,0), radius ε →
• Because z = 1 + εejφ, φ goes from 90o

to 0o to −90o G(z) = G

• 1 + εejφ

= 1 (1 + εejφ) εejφ = ∞e−jφ 1 as ε → 0

• G goes from −90o

15. Example of Nyquist Plot - Small Semicircle

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• Indentation around 1
• Semicircle with centre at (1,0), radius ε →
• Because z = 1 + εejφ, φ goes from 90o

to 0o to −90o G(z) = G

• 1 + εejφ

= 1 (1 + εejφ) εejφ = ∞e−jφ 1 as ε → 0

• G goes from −90o through 0o

15. Example of Nyquist Plot - Small Semicircle

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• Indentation around 1
• Semicircle with centre at (1,0), radius ε →
• Because z = 1 + εejφ, φ goes from 90o

to 0o to −90o G(z) = G

• 1 + εejφ

= 1 (1 + εejφ) εejφ = ∞e−jφ 1 as ε → 0

• G goes from −90o through 0o to 90o

15. Example of Nyquist Plot - Small Semicircle

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• Indentation around 1
• Semicircle with centre at (1,0), radius ε →
• Because z = 1 + εejφ, φ goes from 90o

to 0o to −90o G(z) = G

• 1 + εejφ

= 1 (1 + εejφ) εejφ = ∞e−jφ 1 as ε → 0

• G goes from −90o through 0o to 90o
• Nyquist plot is complete!

Digital Control

15

Kannan M. Moudgalya, Autumn 2007

16. Example - Controller Design Using Nyquist Plot

16. Example - Controller Design Using Nyquist Plot

C1 Im(z) Re(z) A B D E

16. Example - Controller Design Using Nyquist Plot

C1 Im(z) Re(z) A B D E

E Im Re A B D

16. Example - Controller Design Using Nyquist Plot

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

16. Example - Controller Design Using Nyquist Plot

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• P = no. of unstable poles of G; P = 0

16. Example - Controller Design Using Nyquist Plot

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• P = no. of unstable poles of G; P = 0
• C4 should encircle (−1/K, 0) point,

16. Example - Controller Design Using Nyquist Plot

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• P = no. of unstable poles of G; P = 0
• C4 should encircle (−1/K, 0) point, P

times for stability.

16. Example - Controller Design Using Nyquist Plot

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• P = no. of unstable poles of G; P = 0
• C4 should encircle (−1/K, 0) point, P

times for stability.

• Should not encircle (−1/K, 0), as P =

16. Example - Controller Design Using Nyquist Plot

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• P = no. of unstable poles of G; P = 0
• C4 should encircle (−1/K, 0) point, P

times for stability.

• Should not encircle (−1/K, 0), as P =
• If − 1

K is to the left of (−1, 0), stable

16. Example - Controller Design Using Nyquist Plot

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• P = no. of unstable poles of G; P = 0
• C4 should encircle (−1/K, 0) point, P

times for stability.

• Should not encircle (−1/K, 0), as P =
• If − 1

K is to the left of (−1, 0), stable

• −1 > −1/K > −∞

16. Example - Controller Design Using Nyquist Plot

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• P = no. of unstable poles of G; P = 0
• C4 should encircle (−1/K, 0) point, P

times for stability.

• Should not encircle (−1/K, 0), as P =
• If − 1

K is to the left of (−1, 0), stable

• −1 > −1/K > −∞
• −1 > −1/K and −1/K > −∞

16. Example - Controller Design Using Nyquist Plot

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• P = no. of unstable poles of G; P = 0
• C4 should encircle (−1/K, 0) point, P

times for stability.

• Should not encircle (−1/K, 0), as P =
• If − 1

K is to the left of (−1, 0), stable

• −1 > −1/K > −∞
• −1 > −1/K and −1/K > −∞
• 1 < 1/K and 1/K < ∞

16. Example - Controller Design Using Nyquist Plot

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• P = no. of unstable poles of G; P = 0
• C4 should encircle (−1/K, 0) point, P

times for stability.

• Should not encircle (−1/K, 0), as P =
• If − 1

K is to the left of (−1, 0), stable

• −1 > −1/K > −∞
• −1 > −1/K and −1/K > −∞
• 1 < 1/K and 1/K < ∞
• K < 1 and K > 0

16. Example - Controller Design Using Nyquist Plot

C1 Im(z) Re(z) A B D E

E Im Re A B D

G(z) = 1 z(z − 1)

• P = no. of unstable poles of G; P = 0
• C4 should encircle (−1/K, 0) point, P

times for stability.

• Should not encircle (−1/K, 0), as P =
• If − 1

K is to the left of (−1, 0), stable

• −1 > −1/K > −∞
• −1 > −1/K and −1/K > −∞
• 1 < 1/K and 1/K < ∞
• K < 1 and K > 0
• 1 > K > 0

Digital Control

16

Kannan M. Moudgalya, Autumn 2007

17. Scilab Code: nyquist ex1.sce

17. Scilab Code: nyquist ex1.sce

1

/ / U p d a t e d ( 2 4 − 7 − 0 7 )

2

/ / 7 . 2

3 4 H = t r f u (1 ,[1 −1 0] , −1); 5 w = −5:0.4:5; 6 f r e = w/(2∗%pi ) ; 7 nyquist (H, f r e ) ; 8 xset ( ’ window ’ ,1) 9 nyquist (H) 10 xset ( ’ window ’ ,2) 11 w = −1.3:0.3:3.3; 12 f r e = w/(2∗%pi ) ; 13 f r e = −0.557:0.02:0.557; 14 f r e = 0 . 0 1 : 0 . 0 1 : 1 ;

Digital Control

17

Kannan M. Moudgalya, Autumn 2007

15 nyquist (H, f r e ) ;

Digital Control

18

Kannan M. Moudgalya, Autumn 2007

18. Scilab Code: trfu.sci

18. Scilab Code: trfu.sci

1

/ / U s e r d e f i n e d f u n c t i

• n

2

/ / F o r m s a t r a n s f e r f u n c t i

• n

3

/ / S c i l a b : C o e f f i c i e n t s a r e g i v e n i n i n c r e a s i n g

4

/ / M a t l a b : C o e f f i c i e n t s a r e g i v e n i n d e c r e a s i n g

5

/ / H e n c e c o e f f i c i e n t s a r e f l i p p e d h e r e

6 7

/ / I n p u t a r g u m e n t s : ( 1 ) N u m e r a t

• r

c o e f f i c i e n t s (

8

/ / ( 2 ) D e n

• m

i n a t

• r

c o e f f i c i e n t s

9

/ / ( 3 ) V a r i a b l e t o s p e c i f y d o m a i n

10 11

/ / U p d a t e d ( 2 7 − 7 − 0 7 )

12

/ / S y s t e m i s c

• n

t i n u

• u

s = > a i s n o t p a s s e d

13

/ / S y s t e m i s d i s c r e t e = > a = − 1

14

/ / S y s t e m i s d i s c r e t i z e d ( s a m p l e d s y s t e m ) = > a =

Digital Control

19

Kannan M. Moudgalya, Autumn 2007

15

/ / U s e s s y s l i n

16 17 function trafunc = t r f u (num, den , a ) 18

i f argn (2) == 2

19

d = ’ c ’ ;

20

e l s e i f a == −1

21

d = ’d ’ ;

22

else

23

d = a

24

end ;

25 num = clean (num ) ; 26 den = clean ( den ) ; 27 num1 = poly (num( length (num): −1:1) , ’ x ’ , ’ c o e f f ’ ) ; 28 den1 = poly ( den ( length ( den ): −1:1) , ’ x ’ , ’ c o e f f ’ ) ; 29 trafunc = syslin (d , num1 , den1 ) ; 30 endfunction ;

Digital Control

20

Kannan M. Moudgalya, Autumn 2007

19. Stability Margins

19. Stability Margins

(−1, 0) B O PM

1 GM

Im(z) Re(z) A

19. Stability Margins

(−1, 0) B O PM

1 GM

Im(z) Re(z) A

• Nyquist plot (C3) drawn.

19. Stability Margins

(−1, 0) B O PM

1 GM

Im(z) Re(z) A

• Nyquist plot (C3) drawn. If passes through (−1, 0), unstable.

19. Stability Margins

(−1, 0) B O PM

1 GM

Im(z) Re(z) A

• Nyquist plot (C3) drawn. If passes through (−1, 0), unstable.
• Gain margin = 1/OA.

19. Stability Margins

(−1, 0) B O PM

1 GM

Im(z) Re(z) A

• Nyquist plot (C3) drawn. If passes through (−1, 0), unstable.
• Gain margin = 1/OA. If plant transfer function is multiplied

by 1/OA,

19. Stability Margins

(−1, 0) B O PM

1 GM

Im(z) Re(z) A

• Nyquist plot (C3) drawn. If passes through (−1, 0), unstable.
• Gain margin = 1/OA. If plant transfer function is multiplied

by 1/OA, Nyquist plot will go through (−1, 0) point

19. Stability Margins

(−1, 0) B O PM

1 GM

Im(z) Re(z) A

• Nyquist plot (C3) drawn. If passes through (−1, 0), unstable.
• Gain margin = 1/OA. If plant transfer function is multiplied

by 1/OA, Nyquist plot will go through (−1, 0) point and become unstable.

19. Stability Margins

(−1, 0) B O PM

1 GM

Im(z) Re(z) A

• Nyquist plot (C3) drawn. If passes through (−1, 0), unstable.
• Gain margin = 1/OA. If plant transfer function is multiplied

by 1/OA, Nyquist plot will go through (−1, 0) point and become unstable. Can handle unmodelled gains.

• Phase margin = ∠AOB.

19. Stability Margins

(−1, 0) B O PM

1 GM

Im(z) Re(z) A

• Nyquist plot (C3) drawn. If passes through (−1, 0), unstable.
• Gain margin = 1/OA. If plant transfer function is multiplied

by 1/OA, Nyquist plot will go through (−1, 0) point and become unstable. Can handle unmodelled gains.

• Phase margin = ∠AOB. If rotated clockwise,

19. Stability Margins

(−1, 0) B O PM

1 GM

Im(z) Re(z) A

• Nyquist plot (C3) drawn. If passes through (−1, 0), unstable.
• Gain margin = 1/OA. If plant transfer function is multiplied

by 1/OA, Nyquist plot will go through (−1, 0) point and become unstable. Can handle unmodelled gains.

• Phase margin = ∠AOB. If rotated clockwise, will go through

(−1, 0)

19. Stability Margins

(−1, 0) B O PM

1 GM

Im(z) Re(z) A

• Nyquist plot (C3) drawn. If passes through (−1, 0), unstable.
• Gain margin = 1/OA. If plant transfer function is multiplied

by 1/OA, Nyquist plot will go through (−1, 0) point and become unstable. Can handle unmodelled gains.

• Phase margin = ∠AOB. If rotated clockwise, will go through

(−1, 0) and become unstable.

• Can handle unmodelled delay.

19. Stability Margins

(−1, 0) B O PM

1 GM

Im(z) Re(z) A

• Nyquist plot (C3) drawn. If passes through (−1, 0), unstable.
• Gain margin = 1/OA. If plant transfer function is multiplied

by 1/OA, Nyquist plot will go through (−1, 0) point and become unstable. Can handle unmodelled gains.

• Phase margin = ∠AOB. If rotated clockwise, will go through

(−1, 0) and become unstable.

• Can handle unmodelled delay. e−jωD has phase = −ωD.

Digital Control

21

Kannan M. Moudgalya, Autumn 2007