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Systems Analysis and Control

Matthew M. Peet

Illinois Institute of Technology

Lecture 21: Stability Margins and Closing the Loop

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Overview

In this Lecture, you will learn:

Closing the Loop

Effect on Bode Plot

Effect on Stability

Stability Effects

Gain Margin

Phase Margin

Bandwidth

Estimating Closed-Loop Performance using Open-Loop Data

Damping Ratio

Settling Time

Rise Time

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Review

Recall: Frequency Response

Input:

u(t) =Msin(t+)

Output: Magnitude and Phase Shift

y(t) = |G()|Msin(t++ G())

0 2 4 6 8 10 12 14 16 18 202

1.5

1

0.5

0

0.5

1

1.5

2

2.5

Linear Simulation Results

Time (sec)

Amplitude

Frequency Response to sint is given by G()

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Review

Recall: Bode Plot

Definition 1.

The Bode Plot is a pair oflog-log and semi-log plots:

1. Magnitude Plot: 20 log10 |G()| vs. log102. Phase Plot: G() vs. log10

Bite-Size Chucks:

G()=i

Gi()

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Complex Poles and Zeros

We left off with Complex Poles:

G(s) = 1

sn

2+ 2 s

n+ 1

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Complex Poles and Zeros

The amplification at the natural frequency, n, is called resonance.

Figure: Frequency Sweeping with Resonance

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Complex Poles and Zeros

Figure: Frequency Sweeping with Resonance

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Complex Poles and Zeros

Figure: The Tacoma Narrows Bridge

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Closing The Loop

Now we examine the effect of closing the loop on the Frequency Response.

Use simpleUnity Feedback(K= 1).

Closed-Loop Transfer Function:

Gcl() = G()1 +G()

We are most concerned with magnitude:

|Gcl()| = |G()

||1 +G()|

G(s)+

-

y(s)u(s)

Figure: Unity Feedback

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Closing The Loop

On the Bode Plot

20 log

|Gcl()

|= 20 log

|G()

| 20log

|1 +G()

|Which is the combination of The original bode plot The new factor log |1 +G()|

We are most concerned with the effect

of the new term

20 log |1 +G()|

Specifically, as 1 +G() 0

lim1+G()0

20 log |1 +G()| =

An unstable mode!

80

70

60

50

40

30

20

10

0

10

20

Magnit

ude(dB)

101

100

101

102

180

135

90

45

0

Phas

e(deg)

Bode Diagram

Frequency (rad/sec)

Figure: Open Loop: Blue, CL: Green

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Closing The LoopStability Margin

Instability occurs when

1 +G() = 0

For this to happen, we need:

|G()| = 1

G() = 180

Stability Margins measure how far weare from the point(|G| = 1,G= 180).

Definition 2.

The Gain Crossover Frequency, gc is the frequency at which|G(c)| = 1.This is the danger point:

IfG(c) = 180

, we are unstableM. Peet Lecture 21: Control Systems 11 / 34

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Closing The LoopPhase Margin

Definition 3.

The Phase Margin, M is the phase relative to 180 when|G| = 1.

M= |180 G(igc)|

gc is also known as the phase-margin frequency, M

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Closing The LoopGain Margin

Definition 4.

The Phase Crossover Frequency, pc is the frequency (frequencies) at whichG(pc) = 180

.

Definition 5.

The Gain Margin, GM is the gainrelative to 0dB when G= 180.

GM = 20log |G(pc)|GM is the gain (in dB) which will

destabilize the system in closed loop.

pc is also known as the gain-margin frequency, GM

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Closing The LoopStability Margins

Gain and Phase Margin tell how stable the system would be in Closed Loop.

These quantities can be read from the Open-Loop Data.

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Closing The LoopStability Margins

M= 35

GM= 10dBM. Peet Lecture 21: Control Systems 16 / 34

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T i R

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Transient ResponseClosing the Loop

Question: What happens when weClose the Loop?

We want Performance Specs! We only have open-loop data.

M and GMcan help us.

Unity Feedback:

G(s)+

-

y(s)u(s)

We want:

Damping Ratio

Settling Time

Peak Time

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Amplitude

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T i R

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Transient ResponseQuadratic Approximation

Assume the closed loop system is the quadratic

Gcl = 2n

s2 + 2ns+2n

Then the open-loop system must be

Gol = 2n

s2 + 2ns

Assume that our open-loop system is Gol

Use Mto solve for

Set 20 log |Gcl| = 3dB to solve for n

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T i t R

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Transient ResponseDamping Ratio

The Quadratic Approximation gives the Closed-Loop Damping Ratio as

M= tan1 2

22 +

44 + 1

M is from the Open-Loop Data!

A Handy approximation is

=M100

Only valid out to =.7. Given M, we find closed-loop .

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T i t R

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Transient ResponseBandwidth and Natural Frequency

We find closed-loop from Phase

margin. We can find closed-loopNatural

Frequency n from the closed-loopBandwidth.

Definition 6.The Bandwidth, BW is the frequencyat gain20 log |G(BW) = |20log |G(0)|3dB.

Closely related to crossover frequency.

The Bandwidth measures the range of frequencies in the output.

For 2nd order, Bandwidth is related to natural frequency by

BW =n

(1 22) +

44 42 + 2

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Transient Response

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Transient ResponseFinding Closed-Loop Bandwidth from Open-Loop Data

Question: How to find closed-loop bandwidth?Finding the closed-loop bandwidth from open-loop data is tricky.

Have to find the frequency when the Bode plot intersects this curve.

Heuristic: Check the frequency at6dB and see if phase is=180.

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Finding Closed Loop Bandwidth from Open Loop Data

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Finding Closed-Loop Bandwidth from Open-Loop DataExample

80

70

60

50

40

30

20

10

0

10

20

Magnitude(dB)

101

100

101

102

180

135

90

45

0

Phase(deg)

Bode Diagram

Frequency (rad/sec)

At phase 135,5dB, we get closed loop BW=1.

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Transient Response

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Transient ResponseBandwidth and Settling Time

We can use the expression Ts = 4n

to get

BW = 4

Ts

(1 22) +

44 42 + 2

Given closed-loop and BW, we can find Ts.

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Transient Response

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Transient ResponseBandwidth and Peak Time

We can use the expression Tp =

n

12to get

BW =

Tp

1 2

(1 22) +

44 42 + 2

Given closed-loop and BW, we can find Tp.M. Peet Lecture 21: Control Systems 25 / 34

Transient Response

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Transient ResponseBandwidth and Rise Time

Using an expression for Tr, we get a relationship between BW Tr and .

Given closed-loop and BW, we can find Tr.

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Transient Response

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Transient ResponseExample

Question: Using Frequency Response Data, find Tr, Ts, Tp after unityfeedback.

First Step: Find the phase Margin.

Frequency at 0dB is gc=2 G(2) = 145

M= 180

145

= 35

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Transient Response

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Transient ResponseExample

Step 2: Closed-Loop Damping Ratio

=M100

=.35

Step 3: Closed-Loop Bandwidth

Intersect at=(|G| = 6dB,G= 170)

Frequency at intersection is

BW=3.7

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Transient Response

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Transient ResponseExample

Step 4: Settling Time and Peak Time

BW = 3.7, =.35

BWTs = 20 implies Ts = 5.4s

BWTp = 4.9 implies Tp = 1.32

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Transient Response

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Transient ResponseExample

Step 6: Experimental Validation.

Use the plant

G(s) = 50

s(s+ 3)(s+ 6)

We find Tp = 1.6s

predicted 1.32

Tr =.7s predicted.535

Ts = 4s Predicted 5.4

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Figure: Step Response

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Steady-State Error

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y

Finally, we want steady-state error. Steady-State Step response is

lims0G(s) = lim0G()

Steady-state response is theLow-Frequency Gain,|G(0)|.

Close The Loop to get steady-state error

ess = 1

1 +

|G(0)

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Steady-State Error

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yExample

A Lag Compensator

lim0

20log G() = 20dB

So lim0|G()

|= 10.

Steady-State Error:

ess= 1

1 +G(0)=

1

11=.091

20

10

0

10

20

30

40

Ma

gnitude(dB)

100

101

102

103

104

105

90

45

0

Phase(deg)

Bode Diagram

Frequency (rad/sec)

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Summary

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y

What have we learned today? Closing the Loop

Effect on Bode Plot

Effect on Stability

Stability Effects

Gain Margin Phase Margin

Bandwidth

Estimating Closed-Loop Performance using Open-Loop Data

Damping Ratio Settling Time

Rise Time

Next Lecture: Compensation in the Frequency Domain

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