Describing Function Slides

7/29/2019 Describing Function Slides

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Lecture 6 Describing function analysis

Todays Goal: To be able to

Derive describing functions for static nonlinearities

Predict stability and existence of periodic solutions through

describing function analysis

Material:

Slotine and Li: Chapter 5

Chapter 14 in Glad & Ljung

Chapter 7.2 (pp.280290) in Khalil

(Chapter 8 in Adaptive Controlby strm & Wittenmark)

Lecture notes

FRTN05 Lecture 6 Automatic Control LTH, Lund University


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Course Outline

Lecture 1-3 Modelling and basic phenomena(linearization, phase plane, limit cycles)

Lecture 2-6 Analysis methods(Lyapunov, circle criterion, describing functions)

Lecture 7-8 Common nonlinearities(Saturation, friction, backlash, quantization)

Lecture 9-13 Design methods(Lyapunov methods, Backstepping, Optimal control)

Lecture 14 Summary



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Example: saturated sinusoidals

0 1 2 3 4 5 6 7 8 9 1 04

3

2

1

0

1

2

3

4

time [s]

inputsignal

0 1 2 3 4 5 6 7 8 9 1 02

1.5

1

0.5

0

0.5

1

1.5

2

time [s]

outputsignal

Sine Wave ScopeSaturation

The effective gain ( the ratiosat(A sint)

A sint) varies with the

input signal amplitude A.



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Motivating Example

r e u y

G(s)

0 5 10 15 20

2

1

0

1

2y

u

G(s) =4

s(s + 1)2and u = sat (e) gives stable oscillation for r = 0.

How can the oscillation be predicted?

Q: What is the amplitude/topvalue of u and y? What is the frequency?



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Motivating Example

r e u y

G(s)

0 5 10 15 20

2

1

0

1

2y

u

G(s) =4

s(s + 1)2and u = sat (e) gives stable oscillation for r = 0.

How can the oscillation be predicted?

Q: What is the amplitude/topvalue of u and y? What is the frequency?



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Recall the Nyquist Theorem

Assume G(s) stable, and k is positive gain.

The closed-loop system is unstable if the point 1/k isencircled by G(i)

The closed-loop system is stable if the point 1/k is not

encircled by G(i)

0 e u y

k G(s) 1/k

G(i)



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Motivating Example (contd)

Heuristic reasoning:

For what frequency and what amplitude is"the loop gain" f G = 1?

Introduce N(A) as an amplitude dependent approximation ofthe nonlinearity f().

0 e u y

f() G(s)

1/N(A)

A

G(i)

y = G(i)u G(i)N(A)y G(i) = 1

N(A)



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8 6 4 2 010

8

6

4

2

0

G(i)

1/N(A)

Introduce N(A) as an amplitude dependent gain-approximationof the nonlinearity f

(

).

Heuristic reasoning: For what frequency and what amplitude is"the loop gain" N(A) G(iw) = 1?

The intersection of the 1/N(A) and the Nyquist curve G(i)

predicts amplitude and frequency.FRTN05 Lecture 6 Automatic Control LTH, Lund University


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8 6 4 2 010

8

6

4

2

0

G(i)

1/N(A)


(

).





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8 6 4 2 010

8

6

4

2

0

G(i)

1/N(A)


(

).





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How do we derive the describing function N(A)?

Does the intersection predict a stable oscillation?

Are the estimated amplitude and frequency accurate?



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Fourier Series

Every periodic function u(t) = u(t + T) has a Fourier seriesexpansion

u(t) =a0

2+

n=1

(an cos nt + bn sin nt)

=a0

2 +

n=1

a2n + b2n sin[nt + arctan(an/bn)]where = 2/T and

an =2

T T0 u(t) cos nt dt bn = 2T T

0u(t) sin nt dt

.

Note: Sometimes we make the change of variable t /



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The Fourier Coefficients are Optimal

The finite expansion

uk(t) =

a0

2+

k

n=1(an cos nt + bn sin nt)

solves

minu

2

T

T0

u(t)

uk(t)

2dt

if{

an, b

n}are the Fourier coefficients.



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The Key Idea

0 e u y

N.L. G(s)

Assume e(t) = A sint and u(t) periodic. Then

u(t) =a0

2+

n=1

a2n + b

2n sin[nt + arctan(an/bn)]

If G(in) G(i) for n = 2,3,. . . and a0 = 0, then

y(t) G(i)

a21 + b21 sin[t + arctan(a1/b1) + argG(i)]

Find periodic solution by matching coefficients in y = e.



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Definition of Describing Function

The describing function is

N(A,) =b1() + ia1()

A

e(t) u(t)

N.L.

e(t)

u1(t)N(A,)If G is low pass and a0 = 0, then

u1(t) = N(A,)A sin[t + argN(A,)]can be used instead of u(t) to analyze the system.Amplitude dependent gain and phase shift!


Idea: Use the describing function to approximate the part of


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Idea: Use the describing function to approximate the part ofthe signal coming out from the nonlinearity which will survivethrough the low-pass filtering linear system.

e(t) = A sint = Im (Aeit)

e(t) u(t)N.L. u(t) =

a0

2+

n=1(an cos nt + bn sin nt)

e(t) u1(t)N(A,)

u1(t) = a1 cos(t) + b1 sin(t)

= Im (N(A,)Aeit)

where the describing function is defined as

N(A,) =b1() + ia1()

A= U(i) N(A,)E(i)



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Existence of Limit Cycles

0 e u y

f() G(s)

1/N(A)

A

G(i)

y = G(i)u G(i)N(A)y G(i) = 1

N(A)

The intersections of G(i) and 1/N(A) give and A forpossible limit cycles.


D ibi F i f R l


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Describing Function for a Relay

H

H

u

e

0 1 2 3 4 5 61

0.5

0

0.5

1

eu

a1 =1

20

u() cosd= 0

b1 =1

20 u() sind= 2

0Hsind=

4H

The describing function for a relay is thus N(A) =4H

A.


D ibi F ti f Odd St ti N li iti


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Describing Function for Odd Static Nonlinearities

Assume f() and () are odd static nonlinearities (i.e.,f(e) = f(e)) with describing functions Nf and N. Then,

Im Nf(A,) = 0, coeff. (a1 0) Nf(A,) = Nf(A)

N f(A) = Nf(A)

Nf+(A) = Nf(A) + N(A)


Li it C l i R l F db k S t


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Limit Cycle in Relay Feedback System

r e u y

G(s)

1 0.8 0.6 0.4 0.2 0

0.5

0.4

0.3

0.2

0.1

0

0.1

G(i)

1/N(A)

G(s) =3

(s + 1)3with feedback u = sgny

3/8 = 1/N(A) = A/4 A = 12/8 0.48G(i) = 3/8 1.7, T= 2/= 3.7


Li it C l i R l F db k S t ( td)


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Limit Cycle in Relay Feedback System (contd)

The prediction via the describing function agrees very well with

the true oscillations:

0 2 4 6 8 10

1

0.5

0

0.5

1u

y

G filters out almost all higher-order harmonics.


Describing Function for a Saturation


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Describing Function for a Saturation

H

H

DD

u

e

0 1 2 3 4 5 6 71

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

u

e

Let e(t) = A sint = A sin. First set H= D. If A D thenN(A) = 1, if A > D then for (0,)

u() = A sin, (0,0) ( 0,)D, (0, 0)

where 0 = arcsin D/A.


Describing Function for a Saturation (contd)


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Describing Function for a Saturation (cont d)

a1 =1

20

u() cosd= 0

b1 =

1

2

0 u() sind=

4

/2

0 u() sind

=4A

00

sin2 d+4D

/20

sind

=

A

20 + sin 20


Describing Function for a Saturation (contd)


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Describing Function for a Saturation (cont d)

If H= D

N(A) =

1

20 + sin20, A DFor H= D the rule N f(A) = Nf(A) gives

N(A) =H

D20 + sin20, A D

0 2 4 6 8 100.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

N(A) for H= D = 1

NOTE: dependance of A showsup in 0 = arcsin D/A


3 minute exercise:


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3 minute exercise:

What oscillation amplitude and frequency do the describing

function analysis predict for the Motivating Example?


Solution: Find and A such that G(i) N(A) = 1;


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As N(A) is positive and real valued, find s.t.argG(i) = =

2 2arctan

1= = = 1.0

which gives a period time of 6.28 seconds.

G(1.0i) N(A) = 2 N(A) = 1 = N(A) = 0.5.To find the amnplitude A, either

Alt. 1 Solve (numerically) N(A) = 11

20 + sin20

= 0.5,

where 0 = arcsin(1/A)

Alt. 2 From the diagram of N(A) one can find A 2.47

0 2 4 6 8 100.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

N(A) for H= D = 1


The Nyquist Theorem


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The Nyquist Theorem

Assume G(s) stable, and k is positive gain.

The closed-loop system is unstable if the point 1/k isencircled by G(i)

The closed-loop system is stable if the point 1/k is notencircled by G(i)

0 e u y

k G(s) 1/k

G(i)


How to Predict Stability of Limit Cycles


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Assume G(s) stable. For a given A = A0:

A increases if the point 1/Nf(A0) is encircled by G(i)

A decreases otherwise

0 e u y

f G(s)

1/N(A)

G(i)

A stable limit cycle is predicted




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1/N(A)

G()

An unstable limit cycle is predicted

An intersection with amplitude A0 is unstable if A < A0 givesdecreasing amplitude and A > A0 gives increasing.


Stable Periodic Solution in Relay System


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Stable Periodic Solution in Relay System

r e u y

G(s)

5 4 3 2 1 00.2

0.15

0.1

0.05

0

0.05

0.1

0.15

0.2

G(i)

1/N(A)

G(s) =(s + 10)2

(s + 1)3with feedback u = sgny

gives one stable and one unstable limit cycle. The left mostintersection corresponds to the stable one.


Periodic Solution in Relay System


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Periodic Solution in Relay System

The relay gain N(A) is higher for small A:

orbit

Growing amplitudes

Shrinking relay gain

Stable

periodic

orbit

Unstable

periodic

Growing relay gain

One encirclement

Shrinking amplitudes

Growing relay gain

Small amplitudes

High relay gain

No encirclement

Shrinking amplitudes

Big amplitudes

Small relay gain

No encirclement


Automatic Tuning of PID Controller


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Automatic Tuning of PID Controller

Period and amplitude of relay feedback limit cycle can be usedfor autotuning.

ProcessPID

Relay

A

T

u y

1


Describing Function for a dead-zone relay


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b g y

D

D

u

e

0 1 2 3 4 5 6

1

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

e

u

Let e(t) = A sint = A sin. Then for (0,)

u() = 0, (0,0)D, (0, 0)where 0 = arcsin D/A (if A D)


Describing Function for a dead-zone relaycontd.


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g y

a1 =1

20

u() cosd= 0

b1 =1

2

0

u() sind=4

/2

0

D sind

=4D

cos0 =

4D

1 D2/A2

N(A) = 0, A < D4A

1 D2/A2, A D


Plot of Describing Function for dead-zone relay


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g y

0 2 4 6 8 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

N(A) for D = 1

Notice that N(A) 1.3/A for large amplitudes


Pitfalls


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Describing function analysis can give erroneous results.

DF analysis may predict a limit cycle, even if it does notexist.

A limit cycle may exist, even if DF analysis does not predictit.

The predicted amplitude and frequency are onlyapproximations and can be far from the true values.


Example


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p

The control of output power x(t) from a mobile telephone is critical forgood performance. One does not want to use too large power sinceother channels are affected and the battery length is decreased.

Information about received power is sent back to the transmitter and

is used for power control. A very simple scheme is given by

x(t) = u(t)

u(t) = sign y(t L), ,> 0

y(t) = x(t).

Use describing function analysis to predict possible limit cycle

amplitude and period of y. (The signals have been transformed so

x = 0 corresponds to nominal output power)y(t) = x(t)

x(t)

y(t L)

s e

sL

u(t)

x(t) y(t) y(t L)


The system can be written as a negative feedback loop with


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G(s) =esL

s

and a relay with amplitude 1. The describing function of a relaysatisfies 1/N(A) = A/4 hence we are interesting in G(i)on the negative real axis. A stable intersection is given by

= arg G(i) = /2Lwhich gives = /(2L).This gives

A

4= G(i) =

=

2L

and hence A = 8L/2. The period is given byT= 2/= 4L. (More exact analysis gives the true valuesA = L and T= 4L, so the prediction is quite good.)


Accuracy of Describing Function Analysis


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Control loop with friction F= sgny:

_

_

GC

Friction

yref u

F

y

Corresponds to

G1+ G C

= s(s b)s3 + 2s2 + 2s + 1

with feedback u = sgny

The oscillation depends on the zero at s = b.


Accuracy of Describing Function Analysis


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1 0.5 0 0.50.4

0.2

0

0.2

0.4

0.6

0.8

1

1.2

b = 4/3

b = 1/3

DF predicts period times andampl. (T,A)b=4/3 = (11.4,1.00)and (T,A)b=1/3 =(17.3,0.23)

0 5 10 15 20 25 30

1

0

1

0 5 10 15 20 25 300.4

0.2

0

0.2

0.4

y

y

b = 4/3

b = 1/3

Simulation:

(T,A)b=4/3 = (12, 1.1)

(T,A)b=1/3 = (22, 0.28)

Accurate results only if y is sinusoidal!


Analysis of OscillationsA summary


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There exist both time-domain and frequency-domain methodsto analyze oscillations.

Time-domain:

Poincar maps and Lyapunov functions

Rigorous results but hard to use for large problems

Frequency-domain:

Describing function analysis

Approximate resultsPowerful graphical methods


Todays Goal


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To be able to

Derive describing functions for static nonlinearities

Predict stability and existence of periodic solutions through

describing function analysis


Next Lecture


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Saturation and antiwindup compensation

Lyapunov analysis of phase locked loopsFriction compensation


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Describing Function Slides