systeem en regeltechniek deel 4: stabiliteit van regelsystemen · deel 4 blok 8. stabiliteit van...

1

Systeem en RegeltechniekSysteem en RegeltechniekFMT / FMT / MechatronicaMechatronica

Gert van Schothorst

Philips Centre for Technical Training (CTT)Philips Centre for Industrial Technology (CFT)Hogeschool van Utrecht - PTGroep

Deel 4:Deel 4: StabiliteitStabiliteit van van regelsystemenregelsystemenBlok 8:Blok 8: Stabiliteit Stabiliteit van van regelsystemen regelsystemen -- TheorieTheorie

2Philips Centre for Technical Training (CTT) / Philips Centre for Industrial Technology (CFT) / HvU PTGroep – Gert van Schothorst

Cursus Systeem en RegeltechniekCursus Systeem en RegeltechniekOverzichtOverzicht

Deel 1 Blok 1. InleidingWo. 14-04 Blok 2. Basisprincipes modelvorming massa-veersystemen

Blok 3. De regelaar als veer-demper combinatie

Deel 2 Blok 4. Frequentie-domein beschrijvingWo. 21-04 Blok 5. Basisconcepten in de regeltheorie

Deel 3 Blok 6. Verdere inleiding in de regeltheorieWo. 28-04 Blok 7. De PD regelaar als veer-demper combinatie

Deel 4 Blok 8. Stabiliteit van regelsystemenWo. 12-05 Blok 9. De PID regelaar in het frequentie domein

Deel 5 Blok 10. Bandbreedte en verstoringsonderdrukkingWo. 19-05 Blok 11. Toepassing: Tunen PID regelaar mechatronisch systeem

Deel 6 Blok 12. Diverse onderwerpenWo. 26-05 Blok 13. Terugblik

Toepassing: PID regelaarontwerp

Extra regeltechniek

2


Why becomes a control system unstable?Why becomes a control system unstable?

?Stability of control systems


Stable Instable

Stability of control systems

IntroductionIntroduction

3


IntroductionIntroduction

K+

-xexs

KK

xxs +

=1

Kx/xs

00

10.5

-1∞



IntroductionIntroductionStability of control systems

dssFsx

sH == 1)()(

)(

H(s)Fs x+

-xs e

C(s)

)( ksC p==sesF

)()(

d

x

Fservo

Transfer functions:

kp /ds

kp /dsxxs +

=1 d/kp·s

1

+ 1=

τ s1+ 1

= When stable?

4


Stability in the time domainStability in the time domain

First order system(damper with P controller):

usy 11+= τ

Assume: u(t)=0, y(0)=y0

Question: y(t) ?


d

x

Fservo



Initial condition:01

0⋅−⋅= τeky

ky =0

0=+⋅ tt keke λλλτ

0)1( =+ tkeλτλ

0≠k 0≠teλ

τλ 1−=


uys =+ )1(τ1

1+=

suy

τ

Laplace domain:

uyy =+�τ

Time domain:

u=0 y(0)=y0

Initial condition:

tekty λ=)(tkety λλ=)(

.Solution:

τt

eyty−⋅= 0)(

Substitute in differential equation:

5



First order system(damper with P controller):

usy 11+= τ

Assume: u(t)=0, y(0)=y0

Question: y(t) ?


d

x

Fservo

Solution: y(t)=y0e-t/�

stable if: � >0!!!


Intermezzo: poles and zerosIntermezzo: poles and zeros

Transfer function examples:


11)( ssH +=τFirst order system:

cdsmssH ++

= 21)(Second order system:

)()( skksC vp+=PD controller:

22 2)(

oossMsH ωβω ++

=2s

22 2 nnss ωβω ++2oω 2

nωFourth order system w. zeros:

Standard form: )((s - z1 ) (s - z2 ) … (s - zm )

KsH =(s - p1 ) (s - p2 ) … (s - pn )

6


Intermezzo: poles and zerosIntermezzo: poles and zeros

Zeros are roots of the numerator polynomial

Poles are roots of the denominator polynomial


)((s - z1 ) (s - z2 ) … (s - zm )

KsH =(s - p1 ) (s - p2 ) … (s - pn )

zerospoles

n: number of poles = order of the system

(values of s for which numerator becomes zero)

(values of s for which denominator becomes zero)


Stability in the frequency domainStability in the frequency domain

)(1

11)(

1psKssH −=+=τ

Poles determine stability!!!

zerospoles


First order system(damper with P controller) d

x

Fservo

Transfer function instandard form:

Pole: � s+1=0 � s=-1/�

p1=-1/� stable if p1<0

Zero: -

7


Zero: -

Poles: mc

md

mdp 42

121

2

2

2,1 −±−=

Stable if d>0, or: Re(p)<0


zerospoles))((

)(21 psps

KsH

−−=


cdsmssH ++

= 2)(1

Second order system(mass-spring-damper):


c

dM

Fdisturb

x



Zero: z1 = - kp/kv

Pole: -

Always stable


zerospoles

)(K

sC =


PD controller



dtdkk vp+

controllere Fservo)()( skksC vp+=

(s - z1 )1

8



zerospoles


PD controller in series with mass-spring-damper:

Transfer function in standard form:

)()(

skksC vp+

=cdsms

sH ++2)(

K (s - z1 )))(( 21 psps −−)(sC =sH )(

Zero: z1 = - kp/kv

Pole:mc

md

mdp 42

121

2

2

2,1 −±−=


Re

Im


PolePole--zero map in szero map in s--planeplane

9


0ω=mc

412

−md

mc ( )

21 d/m

kp/kv

Re

Im


PolePole--zero map in szero map in s--planeplane


x(t)=exp(st)

LHP

Im(s)

RHP

Re(s)�

��

�

� � ��


Stability of poles in sStability of poles in s--planeplane

10


21

Mskp

xs e+-

xFservo

controller process


Stability of closed loop systemsStability of closed loop systems

M

x

kp

Mechanical equivalent:

Closed loop system(simple mass with P controller):


p

p

s kMsk

xx

+= 2

Poles: Ms2+kp=0, or

Mk

jsMk

s pp ±=�−=2



Transfer function:

Marginally stable!!!

21

Mskp

xs e+-

xFservo

controller process

11




Closed loop system(simple mass with PD controller):

21

Mskp+kvs

xs e+-

xFservo

controller process

Mechanical equivalent:

M

x

kp

kv


Exercise PolesExercise Poles

For the controlled mechanical system of the previous slide, take M=1 Kg, kp=2 N/m and kv=2 Ns/m.

1. Calculate the poles of the open-loop system.

2. Calculate the poles of the closed-loop system.

3. Is the system stable?


12


)(2 sHMs

kskex pv =

+=Open-loop:

Closed-loop:)(1

)(sH

sHxx

s +=

pv

pv

pv

Pv

kskMsksk

Msksk

Msksk

HH

+++

=++

+

=+ 2

2

2

11sx

x =

Closed-loop poles: 1+H(s)=0, or;

02 =++ pv kskMs



21

Mskp+kvsxs e+

-

x

Fservo

controller process

Stable if kv > 0


-100101

-50

0

50amplitude [dB]

102

101 102

frequency [Hz]phase [deg]

0

-200

-100

phase (appr. -175 deg)

amplitude (appr 0.7)0.5

0

1

1.5

-1.5

-1

-0.5

-1.5 -1 0 0.5 1 1.5-0.5

Im

Re

H(s)

Bode plot:- log(|H|) vs. log(f)- ∠H vs. log(f)

Nyquist plot:- Re(H) vs. Im(H)

in complex plane

Intermezzo: Intermezzo: NyquistNyquist plotplotStability of control systems

13


1.5

0.50 1.51-0.5-1-1.5

0.5

0

-0.5

-1

-1.5

1

Re

Im

� �


Intermezzo:Intermezzo: NyquistNyquist plotplot


Make Bode and Nyquist diagrams for the following systems:

1. A mass (double integrator)2. A double integrator in series with Kp/Kv controller3. A first order system


Exercise Exercise Nyquist Nyquist plotsplots

14



Towards the Towards the Nyquist Nyquist stability criterionstability criterion

Stability:• Poles of closed loop system in LHP• Poles of closed loop are solutions of 1 + H(s) = 0• Nyquist plot of open loop H(s) can be used to evaluate

stability...

Yet another exercise...


Check if s=2 is a solution to 1+H(s)=0. Same for s=-2+3j, s=-1+j, s=-1-j. Plot these s values in the complex plane(s-plane). Plot the corresponding value of H(s) in another complex plane plot (H(s) plane).

For the controlled mechanical system of earlier pages, take M=1 Kg, kp=2 N/m andkv=2 Ns/m.


Exercise sExercise s→→→→→→→→H(s) mapH(s) map

M

x

kp

kv

pv

pv

kskMsksk

HH

+++

=+ 21sxx =Closed loop:

)(2 sHMs

kskex pv =

+=Open-loop:

15


s plane H(s) plane

r>0

r

w

Re(H)

Im(H)




For the controlled mechanical system of earlier pages, take M=1 Kg, kp=2 N/m andkv=2 Ns/m.



M

x

kp

kv

pv

pv

kskMsksk

HH

+++

=+ 21sxx =Closed loop:

)(2 sHMs

kskex pv =

+=Open-loop:

1.Assume that we consider sinus-types of signals, i.e. s=j� .Make the H(s) plot for � =0….∞.

2. Is the system stable?

16


H(s)+-

xexs

)(1)(

)(sH

sHsHc +

=

Stability: check RHP poles of Hc(s)

1+H(s)=0 for Re(s)>0

Question:Given a frequency response H(j� ), how to check stability?

Answer:Use Nyquist plot of H(j� ) in the complex plane, and evaluate with respect to the (-1,0)


NyquistNyquist stability criterionstability criterion


Graphical evaluation of stability

For increasing frequency along the curve of H(j� ) in the complex plane, the point (-1,0)

should stay at the left hand side of the curve

s plane H(s) plane

r

w

Re(H)

Im(H)

r>0 -1

r>0

w

H(s) with s=r+j�



17


-1.5 -1 -0.5-2 1.510.50 2

-1.5

-1

-0.5

-2

1.5

1

0.5

0

2

-1000

-500

r=0

-1500

s=r+jw

REAL

IMAG




-1.5 -1 -0.5-2 1.510.50 2

-1.5

-1

-0.5

-2

1.5

1

0.5

0

2

1000

500

r=0

1500

s=r+jw

REAL

IMAG



18


frequency [Hz]101 102 103 104-50

0

50amplitude [dB]

101 102 103 104-200

0

200phase [deg]


Measurement of open loop reticle stage

Example of experimental systemExample of experimental system


frequency [Hz]101 102 103 104

-50

0

50amplitude [dB]

101 102 103 104-200

0

200phase [deg]



Measurement of closed loop reticle stage

19


-4 -3 -1-5 3210 4

-3

-2

-1

-4

4

3

2

0

5

Re(Ho)

Im(Ho)

-2 5

1

-5


Nyquist plot of open loop reticle stage



-1.5 -1 -0.5-2 1.510.50 2

-1.5

-1

-0.5

-2

1.5

1

0.5

0

2

Re(Ho)

Im(Ho)



Nyquist plot of open loop reticle stage (zoomed)

20


Stability marginsStability margins

Margin = Distance of H(j� ) to the critical point

Phase Margin (PM) = allowable phase shift up to instability (rotation of H(j� ) diagram)

Or: check phase of H(j� ) where H(j� ) =1

Gain Margin (GM) = allowable gain increase up to instability (enlargement of H(j� ) diagram)

Or: check gain of H(j� ) where ∠H(j� ) = -180º



-1.5 -1 -0.5-2 1.510.50 2

-1.5

-1

-0.5

-2

1.5

1

0.5

0

2

Re(Ho)

Im(Ho)


Stability marginsStability marginsNyquist plot of open loop reticle stage

PM

1/GM

21


SummarySummary

• Stability in time domain• Poles / Zeros• Stability in frequency domain

– Poles of closed loop should be in RHP

• The Nyquist plot• The Nyquist stability criterion

– Leave (-1,0) at left hand of open loop H(s)– Leave sufficient stability margins (gain/phase)


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