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Nyquist Plots, Control Design Supplementary Material #10

ECE 415 Fall 2018

Raúl Ordóñez, PhD Professor Dept. Electrical and Computer Engineering University of Dayton Voice: 937-229-3183 E-mail: [email protected]

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

The Nyquist Stability Criterion

• 1932, Harry Nyquist at Bell Labs

• Nyquist criterion: allows to determine closed-loop stability from the open-loop frequency response

• Note! Open-loop frequency response = Bode plots!

• Root locus can also be used to study closed-loop stability

• But, Nyquist allows in addition to define stability margins

- Gain margin (GM)

- Phase margin (PM)

- Delay margin (DM)

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

The Nyquist Stability Criterion (cont.)

• Uses a result from complex variable called the Argument Principle

• Briefly: evaluate open-loop system along a contour covering the entire RHP

• As we evaluate the open-loop system along the contour, we obtain a contour map!

• From the number of RHP open-loop poles and the behavior of the contour map, we can determine closed-loop stability

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

The Nyquist Stability Criterion (cont.)

• Consider the contour covering the entire RHP and the closed-loop system:

G(s) = C(s)P (s)H(s)

• The contour map will encircle the point in the s-plane

C

CG

N = Z � P

times in the cw direction, where

P : # of RHP poles of G(s)

Z : # of RHP poles of T (s)

T (s) = CP (s)

1 + CPH(s)

) Z = N + P

�1 + j0

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Owner
铅笔

© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

The Nyquist Stability Criterion (cont.)

• When we evaluate along the contour :CG(s)

Re[s]

Im[s]

C

R ! 1

G(j!)- Along the Im-axis, we evaluate :

|G(j!)|

\G(j!)

- This is the same as the Bode plot of !G(s)

The Bode plot gives the same information as the Nyquist stability criterion and Nyquist plots!

- For plants with more poles than zeros, this part evaluates to zero

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Nyquist Example 1

P (s) = 1

(s+ 1)2

P = 0 N = 0

Z = N + P = 0 ) Stable for any K > 0

A

A A

B

B B

C

C

C

C(s) = K = 1

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Nyquist Example 2

P = 0 N = 0

A

A

A

B

B

B C

C

C

P (s) = 1

s(s+ 1)2 C(s) = K = 1

Z = N + P = 0 ) Stable for 0 < K < 2

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Nyquist Example 3

C(s) = K = 1P (s) = 2(s� 2)

(s+ 2)(s+ 1)2

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Nyquist Example 3 (cont.)

P = 0

A

A A

B

B

B C

C

C

C(s) = K = 1P (s) = 2(s� 2)

(s+ 2)(s+ 1)2

N = 1

Z = N + P = 1 ) Unstable for K > 0.5 or K < �3 2

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Stability Margins

G(s) = 2.25 (1� s)

(s+ 1)(s+ 2) P = 0 ) Need N = 0 for stability

• Gain Margin (GM):

GM = 20 log

✓ 1

◆ dB

• Phase Margin (PM):

PM = ✓ (in degrees)

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Stability Margins (cont.)

The Gain Margin (GM) tells how much the gain of can be increased without violating closed-loop stability.

G(s)

The Phase Margin (PM) tells how much the phase of can be decreased without violating closed-loop stability.

G(s)

• The PM is strongly linked to damping in the time domain. More soon.

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Stability Margins (cont.)

For systems with no poles in the RHP:

GM = �20 log |G(j!g)| dB

G(s)

where \G(j!g) = �180�

PM = \G(j!c) + 180�

where |G(j!c)| = 1, or 20 log |G(j!c)| = 0dB

• for a stable closed- loop system GM > 0dB

• for a stable closed- loop system PM > 0�

!g

!c

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Stability Margins

G(s) = 2.25 (1� s)

(s+ 1)(s+ 2) crossover frequency

!g

!c

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

More on the Phase Margin

• Consider G(s) = !n s(s+ 2⇣!n)

) T (s) = G(s) 1 +G(s)

= !n

s2 + 2⇣!ns+ !2n

• Can show that

PM = tan�1

0

@ 2⇣qp 1 + 4⇣4 � 2⇣2

1

A

• A useful approximation:

PM ⇡ 100⇣

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

More on the Phase Margin (cont.)

• As a rule of thumb, PM = 30 degrees is the lowest acceptable value

• In some cases, GM and PM are not useful indicators of stability!

- Example: for 1st and 2nd order plants, GM = ∞

• For higher order plants, we may have more than one crossover frequency

• In such cases, one needs to further analyze using all the tools available (root locus, Bode, Nyquist)

• MATLAB commands: nyquist, margin

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Control Design in the Frequency Domain

• Some useful formulas

Resonance peak in dB ( )

Resonance frequency ( )

Bandwidth ( )

Bandwidth ( )

Phase margin ( )

!r = !n p 1� 2⇣2

Ar = �20 log 2⇣ p 1� ⇣2

PM ⇡ 100⇣

⇣  0.7

⇣  0.7

0.3  ⇣  0.8

⇣  0.7

0.3  ⇣  0.8 !BW ⇡ 1.6!c

!BW = (�1.196⇣ + 1.85)!n

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

PD Control

• We know PD control has a stabilizing effect

• How is this reflected in the frequency response?

C(s) = K(⌧ds+ 1)

1

⌧d

• For : phase increase (lead!)

• If we locate so the phase increase occurs close to :

- PM is increased

- “More” stability, better transient

- But, high frequency amplification is undesirable

! > 1

⌧d

1/⌧d !c

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Lead Control

• Approximates PD control: C(s) = K ⌧s+ 1

↵⌧s+ 1 , ↵ < 1

Re[s]

Im[s]

�1 ⌧

� 1 ↵⌧

1

↵⌧

1

⌧ !max

10 log

✓ 1

�max

!max = 1p ↵⌧

�max = sin �1

✓ 1� ↵ 1 + ↵

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Lead Control Procedure

1. Choose from requirements (or response speed requirements)K ess

2. Draw Bode plot of and compute the (old) stability marginsKP (s)

3. Choose new desired PM from overshoot ( ) requirements. Assume⇣

PMnew ⇡ PMold + �max

) �max = (PMnew � PMold) + safety factor

↵ = 1� sin(�max) 1 + sin(�max)

4. Find such that!cnew 20 log |KP (j!cnew)| = �10 log ✓ 1

5. Set ⌧ = 1

!cnew p ↵

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Lead Control Procedure (cont.)

1

↵⌧

1

⌧ !max

10 log

✓ 1

�max

!cold !cnew

PMnew PMold

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Lead Control Example

C(s) = K ⌧s+ 1

↵⌧s+ 1

• Plant: P (s) = 250

s(s+ 2)(s+ 40)(s+ 45)

• Design specs: P.O. < 20% Tr < 0.5s

Ts < 1.2s

Kvel � 10

• MATLAB script: Bode_lead_example.m

Let’s simulate!

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

PI Control

• Helps reduce steady-state error

• But, it can have a destabilizing effect • For : phase decrease (lag!)

• If not carefully used, reduces the PM! (destabilizing effect)

• Need to locate far to the left of so PM is not too affected

• Thus, tends to reduce the closed- loop system bandwidth (opposite of lead control!)

C(s) = K

s

✓ s+

1

⌧i

1

⌧i

! < 1

⌧i

1/⌧i !c

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Lag Control

• Approximates PI control: C(s) = K ⌧s+ 1

↵⌧s+ 1 , ↵ > 1

Re[s]

Im[s]

� 1 ↵⌧

�1 ⌧

1

�20 log(↵)

10

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Lag Control Procedure

1. Choose from requirements (or response speed requirements)K ess

2. Draw Bode plot of and compute the (old) stability marginsKP (s)

3. Choose new desired PM from overshoot ( ) requirements, and add safety factor. Find where this PM can be achieved. Set

!cnew

⌧ = 10

!cnew

4. Find from the relationship↵ 20 log |KP (j!cnew)| = 20 log(↵)

5. Verify new phase margin

24 415_supplemental10.key - November 20, 2018

© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Lag Control Procedure (cont.)

!cold!cnew

PMnew PMold1

�20 log(↵)

10

⌧ 10

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Frequency Design Example: Quanser DC Motor

P (s) = 781588.9154

s(s+ 14420)(s+ 33.05)• From MATLAB:

• Or, can also simplify: P̂ (s) = 54.2193

s(s+ 33.05)

• Design a lead compensator to achieve: PO < 5%

Tp < 0.5s

Ts < 0.6s

• Also, take the arm oscillation into account! Try to reduce the oscillation

• Two approaches

‣ Lead design: dc_motor_pos_gear_Bode_lead.m

‣ Lag design: dc_motor_pos_gear_Bode_lag.m Let’s simulate!

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Sensitivity Revisited

• Example: DC motor position control

- What happens if there is an output disturbance?

- E.g., press load

• How can we know how the disturbance will affect the performance?

- Consider frequency domain lead and RL lead design

Let’s simulate!

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© Raúl Ordóñez, 2012 Dept. ECE, University of Dayton

Sensitivity Revisited (cont.)

TRY = Y (s)

R(s) =

CP (s)

1 + CPH(s)

TDiY = Y (s)

Di(s) =

P (s)

1 + CPH(s)

TDoY = Y (s)

Do(s) =

1

1 + CPH(s) =

E(s)

R(s) = STCP (s) = S(s)

TDmY = Y (s)

Dm(s) = � CP (s)

1 + CPH(s) = �TRY

28 415_supplemental10.key - November 20, 2018

© Raúl Ordóñez, 2012 Dept. ECE, University of Dayton

DC Motor Root Locus Lead Design Sensitivity

S(s) = Y (s)

Do(s) =

1

1 + CP (s) TRY =

Y (s)

R(s) =

CP (s)

1 + CP (s)

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© Raúl Ordóñez, 2012 Dept. ECE, University of Dayton

DC Motor Freq. Domain Lead Design Sensitivity

S(s) = Y (s)

Do(s) =

1

1 + CP (s) TRY =

Y (s)

R(s) =

CP (s)

1 + CP (s)

Text

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415_supplemental10.key - November 20, 2018

ppts/supplemental8.pdf

ECE 415 Control Systems Supplementary Material 8 The Root Locus Control Design Method

Raúl Ordóñez, PhD Professor Dept. Electrical and Computer Engineering University of Dayton Voice: 937-229-3183 E-mail: [email protected]

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

Control Design Using Root Locus

• We have seen many cases where a simple proportional (P) controller is not enough

- Reason: it can only move the poles along the plant’s RL, but cannot change the RL itself!

- Thus, we need to use dynamic compensation instead:

‣ Lead (PD)

‣ Lag (PI)

‣ Lead-lag (PID)

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

Lead Compensation

• Approximates Proportional-Derivative (PD) control! (as p gets larger)

• Intuitively

- Has a stabilizing effect

- Lowers rise time

- Increases damping, decreases overshoot

C(s) = K s+ z

s+ p , z < p

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

Lead Compensation: Example

P (s) = 1

s(s+ 1)

CP (s) = K

• Want for small rise time!n = 2

• Result: small damping, so large overshoot

4 415_supplemental8.key - November 6, 2018

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

Lead Compensation: Example

P (s) = 1

s(s+ 1) • PD controller “pulls” the root locus to the left: more stable!

• But: PD controller will amplify high frequency noise (more later...)CPD(s) = K(s+ 2)

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

Lead Compensation: Example

P (s) = 1

s(s+ 1) • Can retain the good effects if we add a pole far to the left

• Results are almost identical with the lead controller CL(s) = K

s+ 2

s+ 20

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

Lag Compensation

• Approximates Integral (I) or Proportional-Integral (PI) control

• Intuitively:

- May help reduce steady-state error

- Often, will use a lead controller for better transient performance - then a lag controller for better steady-state performance

- Note: pure I or PI control can have a destabilizing effect - so lag may be preferable

C(s) = s+ z

s+ p , z > p

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

Lag Compensation: Example

P (s) = 1

s(s+ 1) Clead(s) = 36

s+ 2

s+ 20

• Since the plant is Type 1, can track a step with zero steady-state error

• How about tracking a ramp?

Kvel = lim s!0

sClead(s)P (s) = 36 2

20 = 3.6 ) ess =

1

Kvel = 0.28

• To improve ramp tracking performance, use a lag compensator:

Clag(s) = s+ 0.1

s+ 0.01 ) Kvel = lim

s!0 sClead(s)Clag(s)P (s) = 36

) ess = 1

KV el = 0.028

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

Lag Compensation: Example

P (s) = 1

s(s+ 1) Clead(s) = 36

s+ 2

s+ 20 Clag(s) =

s+ 0.1

s+ 0.01

Slowly decaying pole!

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

Lag Compensation: Example

P (s) = 1

s(s+ 1) Clead(s) = 36

s+ 2

s+ 20 Clag(s) =

s+ 0.1

s+ 0.01

• Slowly decaying transient

• What can we do?

- Redesign lag compensator, move pole-zero to the left

- Or, redesign lead compensator, increase gain

10

© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

Lag Compensation: Example

P (s) = 1

s(s+ 1) Clag(s) =

s+ 0.1

s+ 0.01 Clead(s) = 70

s+ 2

s+ 20

K = 36

K = 70

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Lead Design: Manual Approach

• Given: plant, desired dominant closed-loop location. Example:

sd = �5 + j5P (s) = s+ 7

s(s+ 10)

✓zi : angles from zeros of P (s) to sd

✓pi : angles from poles of P (s) to sd

12 415_supplemental8.key - November 6, 2018

© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Lead Design: Manual Approach (cont.)

• Given: plant, desired dominant closed-loop location. Example:

sd = �5 + j5P (s) = s+ 7

s(s+ 10)

✓zi : angles from zeros of P (s) to sd

✓pi : angles from poles of P (s) to sd

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Lead Design: Manual Approach (cont.)

• Given: plant, desired dominant closed-loop location. Example:

sd = �5 + j5P (s) = s+ 7

s(s+ 10)

Pick the lead controller zero “arbitrarily”

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Lead Design: Manual Approach (cont.)

• Given: plant, desired dominant closed-loop location. Example:

sd = �5 + j5P (s) = s+ 7

s(s+ 10)

Now figure out where the pole should be!

15

© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Lead Design: Manual Approach (cont.)

• Given: plant, desired dominant closed-loop location. Example:

sd = �5 + j5P (s) = s+ 7

s(s+ 10)

• Root locus angle condition: X

angle of zeros� X

angle of poles = 180�

16 415_supplemental8.key - November 6, 2018

© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Lead Design: Manual Approach (cont.)

• Given: plant, desired dominant closed-loop location. Example:

sd = �5 + j5P (s) = s+ 7

s(s+ 10)

• Root locus angle condition:

(✓z1 + ✓z)� (✓p1 + ✓p2 + ✓p) = 180�

17

© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Lead Design: Manual Approach (cont.)

• Given: plant, desired dominant closed-loop location. Example:

sd = �5 + j5P (s) = s+ 7

s(s+ 10)

• Root locus angle condition:

✓p = (✓z1 + ✓z)� (✓p1 + ✓p2)� 180�

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Lead Design: Manual Approach (cont.)

• Given: plant, desired dominant closed-loop location. Example:

sd = �5 + j5P (s) = s+ 7

s(s+ 10)

• Now, compute the pole location:

p = 5

tan ✓p + 5

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Lead Design: Manual Approach (cont.)

• Given: plant, desired dominant closed-loop location. Example:

sd = �5 + j5P (s) = s+ 7

s(s+ 10)

• Finally, compute gain from magnitude condition:

K = 1��� s+zs+pP (s)

��� s=sd

20 415_supplemental8.key - November 6, 2018

© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Lead Design: Manual Approach (cont.)

• Given: plant, desired dominant closed-loop location. Example:

sd = �5 + j5P (s) = s+ 7

s(s+ 10)

• Lead compensator:

C(s) = K s+ z

s+ p

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Cookbook Design Recipes: Lead Control

• Given the plant , find lead controller for a desired location of dominant closed-loop poles

P (s) C(s) = K s+ z

s+ p sd = re

±j�

• Let G(s) = C(s)P (s)

1) From the required determine dc-gain ess G(0)

) C(0) = G(0) P (0)

= a0

) a0 = K z

p

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Cookbook Design Recipes: Lead Control

2) Choose z = a0 a1

p = 1

b1 K =

a0p

z

where M = |P (sd)|

� = \P (sd) 6= 0�, 180�

a1 = sin(�) + a0M sin(� � �)

rM sin(�)

b1 = � ✓ sin(� + �) + a0M sin(�)

r sin(�)

23

© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Design Example: Lead Control

• Plant: P (s) = 250

s(s+ 2)(s+ 40)(s+ 45)

• Design specs: P.O. < 20% Tr < 0.5s

Ts < 1.2s

Kvel � 10

• How to do this?

‣ Convert the design specs to desired regions in the s-plane for closed-loop poles!

‣ Then, come up with a compensator that makes the root locus go through the desired location!

24 415_supplemental8.key - November 6, 2018

© Raúl Ordóñez, 2011 Dept. ECE, University of Dayton

Transient Performance Summary (Reminder)

T (s) = !2n

s2 + 2⇣!ns+ !2n

� = ⇣!n

!d = !n p

1� ⇣2

Criterion Computation

Peak time

Percent overshoot

Rise time

Settling time

Tp = ⇡

!d

Ts ⇡ 4

⇣!n

PO = e � ⇡⇣p

1�⇣2

Tr ⇡ 2.16⇣ + 0.6

!n

25

© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Design Example: Lead Control (cont.)

) ⇣ �

s ln2(1/PO)

⇡2 + ln2(1/PO) P.O. < 20%

) ✓  62.9�

Ts < 1.2s ) Ts = 4

⇣!n < 1.2

) ⇣!n > 4

1.2 = 3.3

= 0.46

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© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Design Example: Lead Control (cont.)

) ⇣ �

s ln2(1/PO)

⇡2 + ln2(1/PO) P.O. < 20%

) ✓  62.9�

Ts < 1.2s ) Ts = 4

⇣!n < 1.2

) ⇣!n > 4

1.2 = 3.3

= 0.46

27

© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Design Example: Lead Control (cont.)

Tr < 0.5 ) Tr ⇡ 2.16⇣ + 0.6

!n < 0.5

) !n > 2.16⇣ + 0.6

0.5

Fix ⇣ = 0.5

28 415_supplemental8.key - November 6, 2018

© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Design Example: Lead Control (cont.)

Tr < 0.5 ) Tr ⇡ 2.16⇣ + 0.6

!n < 0.5

Fix

Kvel � 10 ) Kvel = lim s!0

sC(s)P (s) � 10

Minimum controller’s dc-gain!

) C(0) � 10 lims!0 sP (s)

= 144

⇣ = 0.5 ) ✓ = cos�1(⇣) = 60�

) !n > 2.16⇣ + 0.6

0.5 = 3.4

29

© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Design Example: Lead Control (cont.)

• Lets’s make some choices:

) ⇣!n = 4 > 3.3⇣ = 0.5, !n = 8 > 3.4

30

© Raúl Ordóñez, 2013 Dept. ECE, University of Dayton

RL Design Example: Lead Control (cont.)

• Lets’s make some choices:

) ⇣!n = 4 > 3.3⇣ = 0.5, !n = 8 > 3.4

• This results in desired closed-loop pole locations:

sd = �a+ jb

a = ⇣!n = 4

b = a tan(✓) = 6.9

31 415_supplemental8.key - November 6, 2018

ppts/supplemental9.pdf

ECE 415 Control Systems Supplementary Material 9 Frequency Response of Linear Systems

Raúl Ordóñez, PhD Professor Dept. Electrical and Computer Engineering University of Dayton Voice: 937-229-3183 E-mail: [email protected]

1

© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Frequency Response

• So far, studied behavior in terms of poles and zeros

• Poles and zeros: direct relationship with time response

• Now, we are interested in the response of a linear system to a sinusoidal input with frequency ω: Frequency Response!

• Why?

- Frequency response can be easily found experimentally (if stable)

- Then, find transfer function from frequency response: system ID!

- Also, shows bandwidth and response to noise, etc

- Can do control design in the frequency domain!

2

© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Frequency Response (cont.)

• System: ← Assume stable, with poles at �p1, . . . ,�pn

• Input: u(t) = A sin(!t) L{u(t)} = U(s) = A !

s2 + !2

• With zero initial conditions:

Y (s) = G(s)U(s)

= c1

s+ p1 + · · ·+ cn

s+ pn +

↵0 s+ j!

+ ↵⇤0

s� j!

Due to plant Due to sin input

G(s) = Y (s)

U(s)

3

© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Frequency Response (cont.)

Y (s) = c1

s+ p1 + · · ·+ cn

s+ pn +

↵0 s+ j!

+ ↵⇤0

s� j!

y(t) =c1e �p1t + · · ·+ cne�pnt

+ 2|↵0| sin(!t+ �) ← Decay to zero!

• Can show that

2|↵0| = A|G(j!)| = A p

Re2[G(j!)] + Im2[G(j!)]

� = \G(j!) = tan�1 ✓ Im[G(j!)]

Re[G(j!)]

4 415_supplemental9.key - November 20, 2018

Owner
铅笔

© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Frequency Response (cont.)

y(t) =c1e �p1t + · · ·+ cne�pnt

+ 2|↵0| sin(!t+ �)

• After the transient (in “steady-state”):

y(t) = A|G(j!)| sin(!t+ \G(j!))

• For a linear system, a sin input “goes through,” with the same frequency, but magnitude and phase changed by the system!

Feeding a sinusoidal to a linear system is the same as evaluating G(s)|s=j!

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Bode Plots

• Plots of magnitude vs frequency and phase shift vs frequency (in scale)

• Magnitude is in decibels (dB): from communications

|G(j!)|dB = 20 log10(|G(j!)|)

|G(j!)|dB = 10 log10 ✓ Po Pi

◆ ← Output power ← Input power

• In terms of voltages:

|G(j!)|dB = 20 log10 ✓ Vo Vi

◆ � P = V 2R

log10

• In general:

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Bode Plots (cont.)

G(s) = K (s+ z1)(s+ z2) . . . (s+ zm)

(s+ p1)(s+ p2) . . . (s+ pn)

G(j!) = |G(j!)|\G(j!)

• Example: G(s) = 1

s2 + 2s+ 4

|G(j!)||G(s)| = |G(� + j!)|

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Bode Plots (cont.)

G(j!) = |G(j!)|\G(j!)

20 log |G(j!)| = 20 log ����K

(s+ z1) . . . (s+ zm)

(s+ p1) . . . (s+ pn)

����

= 20 log |K(s+ z1) . . . (s+ zm)|� 20 log |(s+ p1) . . . (s+ pn)|

= 20 log |K|+ 20 log |s+ z1|+ · · ·+ 20 log |s+ zm| � 20 log |s+ p1|� · · ·� 20 log |s+ pn|

In dB, we just need to add each individual term’s log magnitude!

8 415_supplemental9.key - November 20, 2018

© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Bode Plots (cont.)

G(j!) = |G(j!)|\G(j!)

We add all the angles of the zeros and subtract the angles of the poles

\G(j!) = \ ✓ K

(s+ z1) . . . (s+ zm)

(s+ p1) . . . (s+ pn)

= \(s+ z1) + · · ·+ \(s+ zm)� \(s+ p1)� · · ·� \(s+ pn)

• Let’s look at the behavior of all terms individually

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Bode Plots (cont.)

G(s) = K

K = 100

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Bode Plots (cont.)

\G(j!) = �90�

|G(j!)|dB = �20 log(!)

G(j!) = 1

j!

G(s) = 1

s

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Bode Plots (cont.)

G(s) = s

G(j!) = j!

\G(j!) = 90�

|G(j!)|dB = 20 log(!)

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Bode Plots (cont.)

G(s) = 1

⌧s+ 1

G(j!) = 1

j⌧! + 1

|G(j!)|dB = �20 log |j⌧! + 1|

\G(j!) = � tan�1(⌧!)

1

10

0.1

⌧ = 0.1

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Bode Plots (cont.)

G(s) = 1

⌧s+ 1

G(j!) = 1

j⌧! + 1

|G(j!)|dB = �20 log |j⌧! + 1|

\G(j!) = � tan�1(⌧!)

1

10

0.1

�3dB

�20dB

�45�

⌧ = 0.1

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Bode Plots (cont.)

G(s) = ⌧s+ 1

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Bode Plots (cont.)

G(s) = 1

1 !2n

s2 + 2⇣!n s+ 1

⇣ = 1

⇣ = 1

⇣ = 0.1

⇣ = 0.1

�40dB

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Bode Plots (cont.)

G(s) = 1

1 !2n

s2 + 2⇣!n s+ 1

⇣ = 0.3

!n!r

Ar

• A resonance peak exists if

⇣ < 1p 2 ⇡ 0.707

• Resonance frequency:

!r = !n p 1� 2⇣2

• Peak magnitude (dB):

Ar = �20 log 2⇣ p 1� ⇣2

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Example: PD Controller Frequency Response

= Kp

✓ Kd Kp

s+ 1

= Kp(⌧s+ 1), ⌧ = Kd Kp

C(s) = Kp +Kds

Example: Kp = 10,Kd = 0.1

The PD controller acts as a high-pass filter!

It amplifies high frequencies, such as in noise.

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Example: Filtering Sound

Chirp System Output

Let’s simulate!

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Bandwidth

• Consider a typical closed-loop system:

TRY (s) = CP (s)

1 + CPH(s)

• The magnitude Bode plot of typically looks likeTRY (s)

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Bandwidth (cont.)

• For low frequencies:

|TRY (j!)| ⇡ 1

• For high frequencies:

|TRY (j!)| ⇡ 0

�3

!BW

The bandwidth is the maximum frequency at which the output will track a sinusoidal input satisfactorily.

!BW Y (s) R(s)

• That is, the output will not be attenuated more than times the input magnitude.

�3dB = 1p 2 = 0.707

) |Y (s)| ⇡ |R(s)|

) |Y (s)| ⇡ 0

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© Raúl Ordóñez, 2018 Dept. ECE, University of Dayton

Bandwidth (cont.)

• Consider tracking a train of steps

• How fast can the steps occur while still obtaining good closed-loop tracking?

‣ The bandwidth gives the answer!

! < !BW ) can track ! > !BW ) cannot track

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415_supplemental9.key - November 20, 2018