EL2620 2010 EL2620 2010
EL2620 Nonlinear Control Today’s Goal
You should be able to
Lecture 6
• Derive describing functions for static nonlinearities
• Describing function analysis • Analyze existence and stability of periodic solutions by describing
function analysis
Lecture 6 1 Lecture 6 2
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Motivating Example
y
r e u y
2
u A Frequency Response Approach
G(s) 1
− 0 Nyquist / Bode:
−1
A (linear) feedback system will have sustained oscillations
(center) if the loop-gain is 1 at the frequency where the phase lag
−2
0 5 10 15 20
is −180o
4 But, can we talk about the frequency response, in terms of gain and
G(s) = and u = sat e give a stable oscillation.
s(s + 1)2 phase lag, of a static nonlinearity?
• How can the oscillation be predicted?
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Fourier Series
A periodic function u(t) = u(t + T ) has a Fourier series expansion
The Fourier Coefficients are Optimal
a0 X
∞
u(t) = + (an cos nωt + bn sin nωt) The finite expansion
2 n=1
k
a0 X p 2
∞ a0 X
= + an + b2n sin[nωt + arctan(an /bn )] u
bk (t) = + (an cos nωt + bn sin nωt)
2 2 n=1
n=1
solves Z
where ω = 2π/T and T
2 2
Z Z min u(t) − u
bk (t) dt
2 T 2 T û T 0
an (ω) = u(t) cos nωt dt, bn (ω) = u(t) sin nωt dt
T 0 T 0
Note: Sometimes we make the change of variable t → φ/ω
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Key Idea Definition of Describing Function
r e u y
N.L. G(s) The describing function is
−
b1 (ω) + ia1 (ω)
N (A, ω) =
A
e(t) = A sin ωt gives e(t) u(t) e(t) u
b1 (t)
X N.L. N (A, ω)
∞
p
u(t) = a2n + b2n sin[nωt + arctan(an /bn )]
n=1 If G is low pass and a0 = 0, then
If |G(inω)|≪ |G(iω)| for n ≥ 2, then n = 1 suffices, so that u
b1 (t) = |N (A, ω)|A sin[ωt + arg N (A, ω)]
q
y(t) ≈ |G(iω)| a21 + b21 sin[ωt + arctan(a1 /b1 ) + arg G(iω)] can be used instead of u(t) to analyze the system.
That is, we assume all higher harmonics are filtered out by G Amplitude dependent gain and phase shift!
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Describing Function for a Relay
u 1
u Odd Static Nonlinearities
H 0.5
e
e 0 Assume f (·) and g(·) are odd (i.e. f (−e) = −f (e)) static
−0.5 nonlinearities with describing functions Nf and Ng . Then,
−H
−1
0 1 2 3 4 5 6
• Im Nf (A, ω) = 0
Z 2π
φ = 2πt/T
1
a1 = u(φ) cos φ dφ = 0
π 0 • Nf (A, ω) = Nf (A)
Z Z
1 2π 2 π 4H
b1 = u(φ) sin φ dφ = H sin φ dφ =
π 0 π 0 π • Nαf (A) = αNf (A)
The describing function for a relay is thus
b1 (ω) + ia1 (ω) 4H • Nf +g (A) = Nf (A) + Ng (A)
N (A) = =
A πA
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eplacements
Periodic Solutions in Relay System
Existence of Periodic Solutions
0.1 −1/N (A)
G(iω) r e u y 0
0 e u y G(s) −0.1
f (·) G(s) −
− −0.2
−0.3 G(iω)
−1/N (A) −0.4
A −0.5
−1 −0.8 −0.6 −0.4 −0.2 0
3
Proposal: sustained oscillations if loop-gain 1 and phase-lag −180o G(s) = with feedback u = −sgn y
(s + 1)3
G(iω)N (A) = −1 √
No phase lag in f (·), arg G(iω) = −π for ω = 3 = 1.7
The intersections of the curves G(iω) and −1/N (A) √
give ω and A for a possible periodic solution. G(i 3) = −3/8 = −1/N (A) = −πA/4 ⇒ A = 12/8π ≈ 0.48
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Describing Function for a Saturation
u
The prediction via the describing function agrees very well with the 1
true oscillations: H 0.5
e
1 −D D e u
0
u
y −0.5
0.5
−H −1
0
0 1 2
φ
3 4 5 6
Let e(t) = A sin ωt = A sin φ. First set H = D . Then for
−0.5
φ ∈ (0, π)
−1
A sin φ, φ ∈ (0, φ0 ) ∪ (π − φ0 , π)
0 2 4 6 8 10
u(φ) =
D, φ ∈ (φ0 , π − φ0 )
Note that G filters out almost all higher-order harmonics.
where φ0 = arcsin D/A.
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1
Hence, if H = D , then N (A) = 2φ0 + sin 2φ0 .
π
If H 6= D , then the rule Nαf (A) = αNf (A) gives
Z
1 2π
a1 = u(φ) cos φ dφ = 0 H
π 0 N (A) = 2φ0 + sin 2φ0
Z Z Dπ
1 2π 4 π/2
b1 = u(φ) sin φ dφ = u(φ) sin φ dφ 1.1
π 0 π 0 1
Z Z
4A φ0 2 4D π/2 0.9
N (A) for H = D = 1
= sin φ dφ + sin φ dφ 0.8
π 0 π φ0 0.7
0.6
A
= 2φ0 + sin 2φ0 0.5
π 0.4
0.3
0.2
0.1
0 2 4 6 8 10
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The Nyquist Theorem
G(iω)
KG(s)
− −1/K
5 minute exercise: What oscillation amplitude and frequency do the
describing function analysis predict for the “Motivating Example”?
Assume that G is stable, and K is a positive gain.
• If G(iω) goes through the point −1/K the closed-loop system
displays sustained oscillations
• If G(iω) encircles the point −1/K , then the closed-loop system
is unstable (growing amplitude oscillations).
• If G(iω) does not encircle the point −1/K , then the closed-loop
system is stable (damped oscillations)
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Stability of Periodic Solutions An Unstable Periodic Solution
Ω G(Ω)
G(Ω)
−1/N (A)
Assume that G(s) is stable. −1/N (A)
• If G(Ω) encircles the point −1/N (A), then the oscillation
amplitude is increasing.
• If G(Ω) does not encircle the point −1/N (A), then the An intersection with amplitude A0 is unstable if A < A0 leads to
oscillation amplitude is decreasing. decreasing amplitude and A > A0 leads to increasing.
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Automatic Tuning of PID Controller
Stable Periodic Solution in Relay System Period and amplitude of relay feedback limit cycle can be used for
autotuning:
0.2
0.15 G(iω)
r e u y 0.1 PID
G(s) 0.05
Σ A u y
− 0 T
Process
−0.05 −1/N (A) Relay
−0.1
−0.15
−0.2
−1
−5 −4 −3 −2 −1 0
(s + 10)2 1
y u
G(s) = with feedback u = −sgn y
(s + 1)3 0
gives one stable and one unstable limit cycle. The left most
intersection corresponds to the stable one. −1
0 5 10
Time
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Describing Function for a Quantizer
u 1 Z
1 0.8 e 1 2π
0.6 a1 = u(φ) cos φ dφ = 0
0.4 π 0
e 0.2 u Z Z
0
1 2π 4 π/2
D
−0.2
b1 = u(φ) sin φdφ = sin φdφ
−0.4
−0.6
π 0 π φ0
−0.8
4 4p
−1
0 1 2 3 4 5 6 = cos φ0 = 1 − D2 /A2
π π
Let e(t) = A sin ωt = A sin φ. Then for φ ∈ (0, π) (
0, A<D
0, φ ∈ (0, φ0 ) N (A) = 4 p
u(φ) = 1 − D2 /A2 , A ≥ D
1, φ ∈ (φ0 , π − φ0 ) πA
where φ0 = arcsin D/A.
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Plot of Describing Function Quantizer
0.7
0.6 Describing Function Pitfalls
0.5 N (A) for D = 1 Describing function analysis can give erroneous results.
0.4
• A DF may predict a limit cycle even if one does not exist.
0.3
• A limit cycle may exist even if the DF does not predict it.
0.2
• The predicted amplitude and frequency are only approximations
0.1 and can be far from the true values.
0
0 2 4 6 8 10
Notice that N (A) ≈ 1.3/A for large amplitudes
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Accuracy of Describing Function Analysis
Control loop with friction F = sgn y : 1
y z = 1/3 1.2
z = 4/3
F Friction
0
1
0.8
−1 z = 1/3
0.6
0 5 10 15 20 25 30
_ 0.4
yref y 0.4
C G 0.2 y z = 4/3 0.2
_ 0 0
−0.2 −0.2
−0.4 −0.4
0 5 10 15 20 25 30 −1 −0.5 0 0.5
Corresponds to DF gives period times and amplitudes (T, A) = (11.4, 1.00) and
G s(s − z) (17.3, 0.23), respectively.
= 3 with feedback u = −sgn y
1 + GC s + 2s2 + 2s + 1 Accurate results only if y is close to sinusoidal!
The oscillation depends on the zero at s = z.
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Harmonic Balance
e(t) = A sin ωt u(t)
f (·)
A few more Fourier coefficients in the truncation
2
2 minute exercise: What is N (A) for f (x) =x ? k
a0 X
u
bk (t) = + (an cos nωt + bn sin nωt)
2 n=1
may give much better result. Describing function corresponds to
k = 1 and a0 = 0.
Example: f (x) = x2 gives u(t) = (1 − cos 2ωt)/2. Hence by
considering a0 = 1 and a2 = 1/2 we get the exact result.
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Analysis of Oscillations—A Summary
Time-domain:
• Poincaré maps and Lyapunov functions Today’s Goal
• Rigorous results but only for simple examples You should be able to
• Hard to use for large problems • Derive describing functions for static nonlinearities
Frequency-domain: • Analyze existence and stability of periodic solutions by describing
• Describing function analysis function analysis
• Approximate results
• Powerful graphical methods
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Next Lecture
• Saturation and anti-windup compensation
• Friction modeling and compensation
Lecture 6 33
