EE451/551:Digital Control

Chapter 4: Stability of Digital Control Systems

Definitions of Stability• Asymptotic Stability (AS): A system is said to be AS if its• Asymptotic Stability (AS): A system is said to be AS if its response to any IC decays to zero asymptotically in the steady state, i.e., lim ( ) 0y k y , ,

• Marginal Stability (MS): A system is said to be MS is its

lim ( ) 0k

y k

Marginal Stability (MS): A system is said to be MS is its response to an IC remains bounded but does not decay to zero, e.g.,

Definitions of Stability• In the absence of unstable pole‐zero cancellation a• In the absence of unstable pole‐zero cancellation, a discrete LTI system is AS if its z‐TF poles lie within the open unit circle and MS if its poles lie within the closed p punit circle with no repeated poles on the unit circle.

Definitions of Stability• Bounded‐Input‐Bounded‐Output Stability (BIBOS): ABounded Input Bounded Output Stability (BIBOS): A system is said to be BIBOS if its response to any bounded input remains bounded, i.e., 

• It can be shown that a discrete LTI system is BIBOS if and 

( ) 0 ( ) 0u yu k b k y k b k y

only if its impulse response sequence is absolutely summable, i.e.,  

• Note in the absence of unstable pole zero cancelation

0( )

ih i

• Note, in the absence of unstable pole‐zero cancelation within a system, the conditions necessary for an LTI system to be AS are equivalent to BIBOS, i.e., all system y q , , ypoles must lie inside the open unit circle 

Internal Stability of Closed‐Loop Systems• The stability of a closed‐loop system’s overall TF is not• The stability of a closed‐loop system s overall TF is not sufficient to ensure stability of all parts of the system, i.e., it is essential that all internal signals in the loop be , g pbounded, e.g., e(k), u(k), and y(k), when bounded external inputs, e.g., r(k) and d(k), are applied to the system

Internal Stability of Closed‐Loop Systems• The z‐TFs associated with the illustrated closed‐loop system are given by:

( ) ( ) ( )ZAS ZASC z G z G z ( ) ( ) ( )1 ( ) ( ) 1 ( ) ( )( ) ( )

( ) ( ) ( ) ( )( )

ZAS ZAS

ZAS ZASC z G z C z G zY z R zU z C z G z D zC z

• The system is Internally Stable (IS) if and only if all of the

( ) ( ) ( ) ( )( )1 ( ) ( ) 1 ( ) ( )

ZAS

ZAS ZAS

U z C z G z D zC zC z G z C z G z

• The system is Internally Stable (IS), if and only if all of the z‐TFs that relate inputs to outputs are BIBOS; implyingthe characteristic poly 1 ( ) ( ) has its zeros withinC z G zthe characteristic poly. 1 ( ) ( ) has its zeros withinthe open unit circle, and the loop gain ( ) ( ) has no

t bl l ll ti i i t id

ZAS

ZAS

C z G zC z G z

unstable pole-zero cancellation, i.e., occurring on or outsidethe unit circle (Theorem 4.5, p. 95)

Questions?

In‐class Exercise• Verify that the closed loop system with T=0 1 is not• Verify that the closed‐loop system with T=0.1 is not internally stable when:

0.5848 0.3549 1s

2

0.5848 0.3549 1( ) , and

0.1828 0.8627 110 0 8149 0 7655

sG s

s s

10 0.8149 0.7655C( )

1 1.334z z

zz z

• Hint: 

1 ( )First find ( ) 1 , thenZAS

G sG z zs

Z

compute ( ) ( ) and verify the conditionsnoted previously (see Theorem 4 5 on p 95)

ZAS

sC z G z

• See p.96 of text for soln.noted previously (see Theorem 4.5 on p. 95)

>> num s=0 5848*[‐0 3549 1];den s=[0 1828 0 8627 1];

Matlab Soln.>> num_s=0.5848 [ 0.3549 1];den_s=[0.1828 0.8627 1];>> G_s=tf(num_s,den_s)Transfer function:‐0.2075 s + 0.5848 Why is this result bad

‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐0.1828 s^2 + 0.8627 s + 1

Why is this result badfor Internal Stability?

>> T=0.1;GZAS_z=c2d(G_s,T,'zoh')Transfer function:‐0.076 z + 0.1014

>> C_z=tf(num_z,den_z,T)Transfer function:‐10 z^2 + 15.8 z ‐ 6.238

‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐z^2 ‐ 1.58 z + 0.6238

‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐z^2 ‐ 2.334 z + 1.334

>> zpk(GZAS_z)Zero/pole/gain:‐0.075997 (z‐1.334)

>> zpk(C_z)Zero/pole/gain:‐10 (z‐0.8149) (z‐0.7655)

‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐(z‐0.8149) (z‐0.7655)

‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐(z‐1.334) (z‐1)

Matlab Soln.: YoR(z)

>> YoR_z=minreal(C_z*GZAS_z/CE_z,1e‐5)Transfer function:0.76‐‐‐‐‐‐‐‐z ‐ 0 24z  0.24

>> step(YoR_z)

The closed‐loop system appears BIBO Stable…pp

Matlab Soln.: UoR(z)

U R i l(C /CE 1 5)>> UoR_z=minreal(C_z/CE_z,1e‐5)Transfer function:‐10 z^2 + 15.8 z ‐ 6.238‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐z^2 ‐ 1.574 z + 0.3201

>> zpk(UoR_z)Zero/pole/gain:Zero/pole/gain:‐10 (z‐0.8149) (z‐0.7655)‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐(z‐1.334) (z‐0.24)

>> step(UoR_z)But the output stability requires an infinite control effort!infinite control effort!

Routh‐Hurwitz Criterion• The R‐H criterion establishes conditions for left‐half plane (LHP) polynomial roots and cannot be directly applied to the stability of discrete‐time systems

• The bilinear transform maps the inside of the unit circle to the LHP, and permits application of the R‐H criterion and other s‐domain techniques

Bilinear Transform• Under the bilinear transform ( 2), a discrete systemT

1 0

Under the bilinear transform ( 2), a discrete systemwith the z-domain characteristic polynomial:

( ) n n

T

1 01 0 1

1( )

maps into the w domain as shown:

n nn n wz w

F z a z a z a z

1

maps into the w-domain as shown:

1 1( )n nw wF w a a

1 0

01 wa

1( )1 1n nF w a a

w w 0 1

aw

Bilinear Transform• Note, under the bilinear transform: 

Mapping of s to z

Mapping of z to w

Note: is the imaginary part of ww

Routh‐Hurwitz Criterion

Routh‐Hurwitz Criterion

Example: Analyze Stability Using RH

Example: Analyze Stability Using RH

This implies that stable range of gain for the system is 0 20.3K

Matlab Verification>> num_s=[1];den_s=conv([1 0],[1 1]);>> G s tf(num s den s);>> G_s=tf(num_s,den_s);>> T=0.1;GZAS_z=c2d(G_s,T,'zoh')Transfer function:0.004837 z + 0.004679

Note: Marginal stability occurs at K 20.3,due to a closed-loop pole at 1z

‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐z^2 ‐ 1.905 z + 0.9048

>> K=20 34;Hcl z=K*GZAS z/(1+K*GZAS z);>> K=20.34;Hcl_z=K GZAS_z/(1+K GZAS_z);>> pole(Hcl_z)ans =0.9032 + 0.4292i0.9032 ‐ 0.4292i1.0000          0.9048          

>> zero(Hcl z)>> zero(Hcl_z)ans =‐0.96721.00000 90480.9048

>> step(Hcl_z)

Questions?

In‐class ExerciseA l th t bilit f th t b l i• Analyze the stability of the system below using RH criterion when T=1s:

Matlab VerificationG_w=d2c(GZAS_z,'tustin')Transfer function:

num_s=[1];den_s=conv([1 0],[1 1]);( ) Transfer function:

‐0.03788 s^2 ‐ 0.3864 s + 0.9242‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐

s^2 + 0.9242 s

G_s=tf(num_s,den_s);T=1;GZAS_z=c2d(G_s,T,'zoh')Transfer function:0.3679 z + 0.2642

zpk(G_w)Zero/pole/gain:‐0 037883 (s+12 2) (s‐2)

0.3679 z   0.2642‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐z^2 ‐ 1.368 z + 0.3679

k(GZAS ) ‐0.037883 (s+12.2) (s‐2)‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐

s (s+0.9242)

zpk(GZAS_z)Zero/pole/gain:0.36788 (z+0.7183)‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ Note: Matlab uses "s" rather than "w" for the bilinear transform variable(z‐1) (z‐0.3679)

2

Form the Characteristic Polynomial as shown: 1 ( ) 1 0 03788 0 9242 0 3864 0 9242 0KG K K K 21 ( ) 1 0.03788 0.9242 0.3864 0.9242 0Form the RH Array as shown:

KG w K w K w K

2 1 0.03788 26.4w K K

Note: This implies stability for 0 2.39,with marginal stability occurring at 2.39

KK

1

0

0.03788 6.0.9242 0.3864 2.39

0.9242 0

ww K Kw K K

Matlab Verification>> zero(1+K*G_w) %  Find the poles of the Characteristic Polynominal in the w‐planeans=[‐0.0005 + 1.5585i,‐0.0005 ‐ 1.5585i]

>> omega=2*atan(1.5585/2) % Calculate the inverse of the relation:   

omega =1 32391.3239

>> T_oscilation=2*pi/omega % Calculate the period of oscillation at pt. of MS

T_oscilation =4.7461

Matlab Verification>> Hcl_z=feedback(2.392*GZAS_z,1,‐1) % Calculate the closed‐loop TF at gain of MS Transfer function:Transfer function:0.88 z + 0.6321

‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐z^2 ‐ 0.4879 z + 0.9999>> step(Hcl_z,0:T:50*T);axis([0 50*T, 0 2])

Jury Stability Test

Note, for an nth order system thearray has 2n‐3 rows (always odd)

0 2 2 , k n n kc c c c

Jury Stability Test

Jury Stability Test

Jury Stability Test

Jury Stability Test

those of the prior RH-based in-class exercise.

M tl b V ifi tiMatlab Verification:>> Hcl_z=feedback(2.392*GZAS_z,1,‐1)Transfer function:

>> abs(pole(Hcl_z))ans =

0.88 z + 0.6321‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐z^2 ‐ 0.4879 z + 0.9999

[1.0000,1.0000]

>> angle(pole(Hcl z)) % equal to wT>> angle(pole(Hcl_z)) % equal to wTans =[1.3243,‐1.3243]

Jury Stability Test

Jury Stability Test

Matlab Verification:Matlab Verification:>> roots([1 ‐1.8 1.05 ‐.20]) ans =[ 0.8000,0.5000,0.5000]

Questions?

• The Nyquist Criterion allows determination of the Nyquist Criterion

number of unstable closed‐loop poles, denoted Z (assumed unknown), based on the number of unstable 

l l d d ( d k ) dopen loop poles, denoted P (assumed known), and an analysis of the loop gain frequency response plotted in a polar form called the Nyquist plotpolar form called the Nyquist plot   

• Theorem 4.7 ‐ Nyquist Criterion:Let be the number of clockwise encirclements of the pointN

Let be the number of clockwise encirclements of the point

1,0 for the Nyquist plot when traversing a stability contour,called the N q ist path here the loop gain has open loop

N

P

called the Nyquist path, where the loop gain has open-looppoles inside the Ny

Pquist path. Then the system has Z closed-

loop poles lying inside the Nyquist path (thus unstable) given by:loop poles lying inside the Nyquist path (thus unstable) given by: Z P N

• Corollary: An open‐loop stable system is closed‐loop Nyquist Criterion

stable if and only if it does not encircle the (‐1,0) point, i.e., N=0.

• Procedure for counting encirclements  of (‐1,0):

1. Starting on the minus real axis at an infinite radius, move toward the (‐1,0) point

2. Count all lines of stability contour crossed, counting h l h d ( d d)each line with an upward (or downward) pointing 

arrow as a positive (or negative) encirclement

3 Th t b f li t d i th b f3. The net number of lines counted is the number of encirclements

Example Nyquist Plot – Continuous System

( )pG s

0N

X X0P 0P

0Z N PNyquist path for s or w domains Nyquist plot (same for z and w domains)

0Z N P

More Detailed Analysis

(note, true for all gai0 n K!)

Example Nyquist Plot – Discrete System

( )ZASG z

0N

( )ZASG z

0N

X X0P 0P

0Z N PNyquist path for s or w domains Nyquist plot (same for z and w domains)

0Z N P

More Detailed Analysis

Note: In the following discussion, ( ) ( )ZASG z G z

More Detailed Analysis

Matlab Soln.: Nyquist of GZAS(z)

Note, as the gain K is i d thi i tincreased, this point will cross the ‐1 pt. causing N=2 and the

>> nyquist(G_z)causing N=2, and the system to become

unstable,F K 1 N 0

,since Z=N+P=2

For K=1, N=0Z=0

This will occur for K=1/0.423=2.36, as

b fapprox. before 

Questions?

In‐class Exercise• Analyze the stability of the system below usingAnalyze the stability of the system below using the Nyquist criterion in the w‐domain, i.e., using  the bilinear transform when T=1s:the bilinear transform, when T 1s:

As shown previously:G w=d2c(GZAS z 'tustin') zpk(G w)G_w=d2c(GZAS_z, tustin )Transfer function:‐0.03788 s^2 ‐ 0.3864 s + 0.9242‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐

zpk(G_w)Zero/pole/gain:‐0.037883 (s+12.2) (s‐2)‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐

s^2 + 0.9242 s s (s+0.9242)

0P

Matlab Verification

Note, as the gain K is i d hi i

>> nyquist(G w)

increased, this point will cross the ‐1 pt. causing N=2 and the>> nyquist(G_w)

>> axis([‐2 2 ‐2 2])causing N=2, and the system to become

unstable,For K=1 N=0 ,since Z=N+P=2

For K=1, N=0Z=0

This will occur for K=1/0.405=2.69, as

b fapprox. before 

• Gain Margin (GM) is defined as the factor 1/a by which the loop Gain and Phase Margin

g ( ) / y pgain must change to force the system to become marginally stable; similarly, Phase Margin (PM) is defined as the angle mthrough which the Nyquist plot must be rotated such that itthrough which the Nyquist plot must be rotated such that it intersects the ‐1 point:  Nyquist Plot of Loop Gain

( ) ( )ZASL z KG z( ) ( )ZAS

where  GZAS_zTransfer function:0.3679 z + 0.2642‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐z^2 ‐ 1.368 z + 0.3679

and K=1 is assumedand K=1 is assumedto make the plot

• The role of the Gain Margin (GM) and Phase MarginPerturbation Model View of GM and PM

The role of the Gain Margin (GM) and Phase Margin (PM) may become clearer when viewed from the model perturbation perspective shown below:

1The system perturbation ( ) mjG z GM PMa

e

represents the gain or phase variations needed to push a stablesystem into instability, and can be used to determine the relativestability of a given control system design

Using Matlab to Compute GM and PM>> [GM,PM]=margin(GZAS z)[ , ] g ( _ )GM =2.3927

PMPM =30.3747

>> margin(GZAS_z)

Note: Matlab can alsodisplay the GM andPM directly on aPM directly on aNyquist plot usinghe plot options

Questions?