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MM5 Stability Analysis

Readings: • Section 4.4 (stability, p.212-223);• Section 4.3 (steady-state tracking & system type, p.200-210)• Section 3.5 (effects of zeros & add. Poles, p.131-138)• Extra reading materials (p.40-60)

What have we talked in MM4?

• Poles vs time reponses• Feedback charactersitics•Matlab: pzmap(), sgrid

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MM4 : Poles vs Performance

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Pole locations Time response

MM4: First-order System

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4

t

t

e)ss

L(y(t)

e)s

L(y(t)

ssG

1

1

1)1(

1 :response Step

11 :response Impulse

, :constant time,1:pole

0 assume,1

1)(

Time constant – why?

63%

Time response is determined by the time constantSystem pole is the negative of inverse time constant

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MM4: Second-Order System

0 if , :polescomplex

10 if ,1 :polescomplex

1 if , :poles )(identical real

1 if,1 :poles )(different real

1:poles

0,0 assume,2

)(

2,1

22,1

2,1

22,1

22,1

22

2

n

nn

n

nn

nn

n

nn

n

jp

jp

ξp

p

p

ssG

21,

3.0%,355.0%,16

7.0%,5

6.46.4

8.1

ndd

p

p

ns

nr

t

M

t

t

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MM4: Summary of Pole vs Performance

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MM4: Plot of Pole Locations

s1=tf(1,[1 2 1]); s2=tf(1,[1 1.6 1]);s3=tf(1,[1 1.0 1]);s4=tf(1,[1 0 1]);pzmap(s1,s2,s3,s4)sgrid

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Goals for this lecture (MM5)

Stability analysis Definition of BIBO Pole locations Routh criteron

Steady-state errors Final Theorem DC-Gain Stead-state errors

Effects of zeros and additional poles

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MM4: Summary of Pole vs Performance

0 if , :polescomplex

10 if ,1 :polescomplex

1 if , :poles )(identical real

1 if,1 :poles )(different real

1:poles

0,0 assume,2

)(

2,1

22,1

2,1

22,1

22,1

22

2

n

nn

n

nn

nn

n

nn

n

jp

jp

ξp

p

p

ssG

, :constant time,1:pole

0 assume,1

1)(

s

sG

How about if these are not satisfied?

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System Stability Definitions BIBO stability Internal stability ...

Determination methods:

Impulse response function/sequence Roots of characteristic equation (poles) Routh’s stability criterion Gain and phase margins Nyquist stability criterion

BIBO Stability A system is said to have bounded input-bounded output

(BIBO) stability if every bounded input results in a bounded output (regardless of what goes on inside the system)

The continuous (LTI) system with impuse response h(t) is BIBO stable if and only if h(t) is absolutely integrallable

All system poles locate in the left half s-plane

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)())((

|)(|

sHthL

dh

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Characteristic equation

All poles (roots of the chracteristic equation) of the continuous system are strictly in the LHP of the s-plane - asymptotic internal stability

(Matlab: roots(den))

BIBO Stability – Characteristic Equation

02 :eauqtion sticcharacteri,2

)(

0 :eauqtion sticcharacteri,)(

2222

2

0

0

0

nn

nn

n ss

sG

sasa

sbsG

n

i

iin

i

ii

m

i

ii

BIBO Stability – Execise (I)

Are these systems BIBO stable?

Intuitive explanation Theoretical analysis

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1

2

3

4

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Motivation: Testing stability without calculating poles Criterion: For a stable system, there is no changes in sign and

no zeros in the first column of the Routh array.

BIBO Stability – Routh Criterion (I)

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BIBO Stability – Routh Criterion (II)

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BIBO Stability – Examples

See page 46-49 on the extra readings

BIBO Stability – Execise (II) How about the stability of your project systems?

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BIBO Stability – Objectives of Control

Performance: • time domain specifications• Frequency specifications• Dynamic transient responses • Steady-state responses• Continuous control systems• Digital control systems

Control design Objectives:• Closed-loop stability• Good command response• Disturbance attenuation• Robustness

Control design Objectives:• Closed-loop stability• Good command response• Disturbance attenuation• Robustness

BIBO Stability – Stabilizing Control

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See page 49-50 on the extra readings

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Goals for this lecture (MM5) Stability analysis Definition of BIBO Pole locations Routh criteron

Steady-state errors Final Theorem DC-Gain Stead-state errors

Effects of zeros and additional poles

MM4: Example : First-order System

An design problem: control sys1 so as to have the same performance as sys2

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Design tasks: • ”Speed-up” the response• Eliminate the steady-state error

S1=tf(0.95,[10 1]);S2=tf(1,[1 1]); Step(s1,s2)

Steady-State Error Objective:

to know whether or not the response of a system can approach to the reference signal as time increases

Assumption: The considered system is stable

Analysis method: Transfer function + final-value Theorem

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Steady-State Error – Final-Value Theorem

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S1=tf(0.95,[10 1]);S2=tf(1,[1 1]); Step(s1,s2) )0(1))(1(lim

1)(),())(1(lim

))()()((lim))()((lim)(

0

0

00

GsGs

sRsRsGs

sRsGsRssYsRse

s

s

ss

DC-Gain

Steady-State Error – System Types

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Position-error constant Velocity constant Acceleration constant System types (type 0, type I, type II)

)(lim

)(lim

)(lim

2

0

0

0

sGsK

ssGK

sGK

osa

osv

osp

Revisit of example: First-order System (II)

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9/0.9510/9

What’s the tpye of original system? Derive the transfer function of the closed-loop system What’s the time constant and DC-gain of the CL system? What’s the feedforward gain so that there is no steady-state

error?

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Goals for this lecture (MM5) Stability analysis Definition of BIBO Pole locations Routh criteron

Steady-state errors Final Theorem DC-Gain Stead-state errors

Effects of zeros and additional poles

System Zeros C

hapt

er 6

The dynamic behavior of a transfer function model can be characterized by the numerical value of its poles and zeros

{zi} are the “zeros” and {pi} are the “poles”

in order to have a physically realizable system

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1 2

1 2(6-7)m m

n n

b s z s z s zG s

a s p s p s p

n m

1:poles

0,0 assume,2

)(

22,1

22

2

nn

n

nn

n

p

ssG

, :constant time,1:pole

0 assume,1

1)(

s

sG

How about the effects of zero(s) to system performance?

Effect of Zero in the Left Half s-PlaneC

hapt

er 6

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s0=tf(1,[1 1 1]); s1=tf([1 1],[1 1 1]); s2=tf([0.25 1],[1 1 1]); s3=tf([0.1 1],[1 1 1]);s4=tf([0.05 1],[1 1 1]);s5=tf([4 1],[1 1 1]);step(s0,s1,s2,s3,s4,s5), grid

An additional zero in the left half-plane will increase the overshootIf the zero is within a factor of 4 of the real part of the complex poles

Effect of Zero in the Right Half s-PlaneC

hapt

er 6

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s0=tf(1,[1 1 1]); s1=tf([-1 1],[1 1 1]); s2=tf([-0.25 1],[1 1 1]); s3=tf([-0.1 1],[1 1 1]);s4=tf([-0.05 1],[1 1 1]);s5=tf([-4 1],[1 1 1]);step(s0,s1,s2,s3,s4,s5), grid

An additional zero in the right half-plane will depress the overshootand may cause the step response to start out in the wrong direction

Nonminimum-phase

Cha

pter

6

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Effect of Zeros (I)

)+(1)+(1

)+K(1=G(s)21

3

sss

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s0=tf(1,[4 5 1]); s1=tf([16 1],[4 5 1]); s2=tf([8 1],[4 5 1]); s3=tf([4 1],[4 5 1]);s4=tf([2 1],[4 5 1]);s5=tf([1 1],[4 5 1]);s6=tf([0.5 1],[4 5 1]);s7=tf([-1 1],[4 5 1]);s8=tf([-4 1],[4 5 1]);step(s0,s1,s2,s3,s4,s5,s6,s7,s8); grid; figure; pzmap(s0,s1,s2,s3,s4,s5,s6,s7,s8)

Effect of Zeros (II)

Execise Five Determine system’s stability on slide p.13 Determine your project system’s stability, see

slide p.17 Steady-state error analysis, see slide p.25

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