describing function slides
TRANSCRIPT
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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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Motivating Example (contd)
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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Motivating Example (contd)
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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Motivating Example (contd)
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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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!
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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.
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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.
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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)
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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.
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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.
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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?
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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)
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How to Predict Stability of Limit Cycles
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How to Predict Stability of Limit Cycles
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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How to Predict Stability of Limit Cycles
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How to Predict Stability of Limit Cycles
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.
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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.
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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
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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
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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)
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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
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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
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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.
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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)
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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.)
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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.
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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!
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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
FRTN05 Lecture 6 Automatic Control LTH, Lund University
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
FRTN05 Lecture 6 Automatic Control LTH, Lund University
Next Lecture
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Saturation and antiwindup compensation
Lyapunov analysis of phase locked loopsFriction compensation
FRTN05 Lecture 6 Automatic Control LTH, Lund University