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MCE693/793: Analysis and
Control of Nonlinear Systems
Introduction to Describing Functions
Hanz Richter
Mechanical Engineering Department
Cleveland State University
Introduction
2 / 31
Frequency domain methods are essential for analysis and design of linearsystems. Stability, robustness and performance can be studied using classicaland modern tools such as
1. Bode, Nyquist and Nichols plots
2. Singular values and sensitivity plots (multivariable systems)
3. H∞ control for performance
4. µ-synthesis analysis and control for robust performance
5. Quantitative feedback theory (frequency domain design on the Nicholschart)
Describing functions provide an approximate method for frequency-domainanalysis of nonlinear systems.
We will focus on limit cycle detection/creation or avoidance.
Basic Idea
3 / 31
In linear systems, transfer functions become frequency response functionswhen the Laplace variable s is restricted to the imaginary axis.
That is, the Fourier transform is the Laplace transform restricted to theimaginary axis. Frequency response functions in linear systems are functions offrequency only:
G(s) =s− 1
s+ 1, G(jw) =
1− w2
1 + w2+
2w
1 + w2j
In control systems containing nonlinearities, we seek to find a frequencyresponse function for the nonlinear block. Due to nonlinearity, this functionwill be amplitude-dependent.
G(s)φ(e)
e
0
−
{
N(A,w)
{G(jw)
Memoryless vs. Memory, Soft vs. Hard Nonlinearities
4 / 31
A memoryless nonlinearity produces an output which depends only on thecurrent value of the input. The nonlinearity must then be a function of itsargument: φ = φ(e(t)). Example: saturation.
The output of a nonlineariry with memory depends on the current and pastvalues of the inputs. The nonlinearity is not a function. Sometimes theterminology “multiple-valued function” is used. Example: hysteresis.
A soft nonlinearity is continuous. The terminology is vague, sometimes softnonlinearities are also required to be differentiable. Example: φ(e) = e3.
A hard nonlinearity has discontinuities. In some cases, a continuous functionwhich is not differentiable everywhere is called a hard nonlinearity. Example:saturation.
Describing Function Concept
5 / 31
When a sinuosoidal function is applied to a nonlinearity, harmonic distortionoccurs and the output is no longer a sinusoid. Assuming the output remainsperiodic (always true for static nonlinearities, still true for some others), thereis a Fourier (trigonometric) decomposition.
The decomposition contains sinuosoidal components with frequencies whichare integer multiples of the input frequency, and possibly a bias term (zero forodd functions). In particular, the fundamental component will have anamplitude which depends nonlinearly on the input amplitude.
The principle behind describing functions is to retain only this fundamentaloutput component as an approximation. A “transfer function” is formed bytaking the ratio of the output component (expressed as a complexexponential) to the input sinusoid (also expressed as a complex exponential).
This defines the sinusoidal input describing function (SIDF).
Calculating a simple SIDF
6 / 31
As a first example, consider the cubic nonlinearity: φ(u) = u3. Let the inputbe u(t) = A sin(wt) and express this as the complex exponentialU(A,w) = Aejwt = A cos(wt) +Aj sin(wt).
The output of the nonlinearity is φ(u) = A3 sin3(wt). We must find thefundamental component of the Fourier series expansion of this function. Recallthat the expansion has the form:
φ(t) =a02
+∞∑
n=1
an cos(2πntT ) + bn sin(
2πntT ), where
an =2
T
∫ x0+T
x0
φ(t) · cos(2πntT ) dt
bn =2
T
∫ x0+T
x0
φ(t) · sin(2πntT ) dt
and a0/2 is the mean of φ(t) in one period.
Calculating a simple SIDF...
7 / 31
Because φ is odd, an = 0 for all n. We only need to calculate b1. WithT = 2π/w, the result is
b1 =3
4A3
The output sinusoid is then Φ = 3A3
4 ejwt Therefore the describing function forthe cubic nonlinearity is
N(A,w) = N(A) =3A2
4e0
All odd static nonlinearities will have a purely real and frequency-independentSIDF.
Example: Frequency-dependent SIDF
8 / 31
Following the introductory example in S&L on application of the DF, considerthe nonlinearity φ(x, x) = −x2x associated with the van der Pol equation.
Assume X = Aejwt. Then the output of the nonlinearity isφ(x, x) = −A3w sin2(wt) cos(wt). Using trigonometric identities this is thesame as
φ(x, x) = −A3w
4(cos(wt)− cos(3wt))
By the definition of SIDF, we retain only the fundamental component.Therefore
Φ =A3w
4ej(wt−π/2)
The SIDF is then N(A,w) = Φ/U = A2w4 e−jπ/2 = −A2w
4 j.
In S&L the input is taken as (−x), so their N(A,w) has the opposite sign
Example: Memory Relay with Deadzone
9 / 31
In class, we obtain the SIDF for a relay having the characteristic shown below:
δ−δ
D
−D
in
out
Show that N(A,w) = 4DπA
√
1− ( δA)
2 − j 4DδπA2 . Because the nonlinearity is not
dynamic, frequency is absent.
Limit Cycle Analysis with Describing Functions
10 / 31
If N(A,w) were a truly linear transfer function, we could investigate thepresence of oscillations by using the idea of marginal stability and the Nyquistcriterion. For example, linear analysis can provide the value of K at which thefollowing system oscillates:
16(s+2)(s2+2s+8)
0
−
k
-1 0 1 2-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1Nyquist Diagram
Real Axis
Imag
inar
y A
xis
-20 -10 0 10-15
-10
-5
0
5
10
15Root Locus
Real Axis (seconds -1 )
Imag
inar
y A
xis
(sec
onds
-1)
Review of the Nyquist Stability Criterion
11 / 31
Complex Mapping Theorem: Let F (s) : C 7→ C be a mapping defined on aregion of the complex plane and let C be a closed curve contained in thatregion. Assume:
1. F (s) has no more zeroes than poles and C does not pass through any ofthem.
2. F (∞) is finite.
3. C encloses Z zeroes and P poles of F (s)
Suppose C is parameterized with s = s(t) for t = [0, T ] so that the curve istraversed clockwise. Then the image curve F (s) has N = Z − P clockwiseencirclements of the origin.
(Animation)
Review of the Nyquist Stability Criterion...
12 / 31
Nyquist’s idea was to take C as an infinite half-circle encompassing the entireIm axis and the right half of the complex plane (where unstable poles lie).
Since F (∞) is a constant, only the Im axis portion determines how/if F (s)encircles the origin.
Also, the Complex Mapping Theorem can be “shifted”: if 1 + F (s) has Z − Pclockwise encirclements of the origin, then F (s) has the same number ofencirclements of (−1, 0).
Review of the Nyquist Stability Criterion...
13 / 31
In a unity negative feedback control system, the open-loop transfer function isL(s) = G(s)K(s) and the closed-loop transfer function is
T (s) =L(s)
1 + L(s)
The closed-loop system is stable whenever L(s) doesn’t have poles inside theNyquist contour.
Key idea: when L(s) is a rational function (L(s) = n(s)d(s) with n(s) and d(s)
polynomials in s), the characteristic equation for T (s) has the form:
L(s) =n(s)
d(s)= −1
*Note*: The closed-loop poles are the zeroes of 1 + L(s) and the open-looppoles (poles of L(s)) are also poles of 1 + L(s).
Review of the Nyquist Stability Criterion...
14 / 31
Apply the “shifted” Complex Mapping Theorem to 1 + L(s):
■ N is the number of clockwise origin encirclements of 1 + L(s), orequivalently (we actually use this form) the number of clockwiseencirclements of (-1,0). Note we evaluate L(s) on the Im axis only (this iscalled a Nyquist plot)
■ Z is the number of zeroes of 1+L(s) inside the Nyquist contour (unstableclosed-loop poles)
■ P is the number of poles of 1 + L(s) (equivalently of L(s) and we usethis) inside the Nyquist contour (unstable open-loop poles)
Then Z = N + P
Example
15 / 31
We use the Nyquist criterion to predict the stability of a unity feedback controlsystem with K(s) = k and G(s) = 8
(s−2)(s2+2s+8). Can the system be
stabilized with proportional control?
-1 -0.8 -0.6 -0.4 -0.2 0-0.25
-0.2
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25Nyquist Diagram
Real Axis
Imag
inar
y A
xis
Confirm analysis using the Routh-Hurwitz criterion.
Practical Significance of the Nyquist Stability Criterion
16 / 31
Many systems are open-loop stable, so P = 0 is known even without a mathmodel. We can use the Nyquist criterion to predict the critical proportionalgain for closed-loop stability without having to model the system.
Since P = 0, the system can be operate safely in open-loop. We can use adynamic signal analyzer or offline signal processing to obtain Nyquist plotdata.
Then the stability of the closed-loop system can be studied using data fromthe open-loop system only. Specifically, we can find the critical proportionalgain that scales the Nyquist plot to touch the (-1,0) point.
Also, since pure phase lags (such as those introduced by delays) rotate theNyquist plot, we can determine the maximum amount of delay that the systemcan tolerate before going unstable.
Nyquist-like analysis with Describing Functions
17 / 31
Several conditions have to be satisfied for a DF-based frequency analysis toproduce correct results (see S&L Sect. 5.1.3):
1. There’s a single nonlinearity in the loop (unless 2 or more can be blendedinto one)
2. The nonlinearity is time-invariant (a relay with a time-dependent deadzonewouldn’t)
3. Higher harmonics produced by the nonlinearity are attenuated by the lineardynamics following the nonlinearity (the linear TF must be low-pass).
4. The nonlinearity is odd, so that the Fourier expansion of its output haszero bias.
Notes:
■ Even when these conditions hold, the DF technique may produceerroneous results, particularly when predicting limit cycles.
■ There are extended DF techniques relaxing assumptions 1,2 and 4, butthey are cumbersome.
Example: van der Pol oscillator
18 / 31
See S&L Sect. 5.1.1. The van der Pol equation is re-arranged as a negativeunity feedback loop, where the nonlinear block is x2x, followed by the unstabletransfer function
G(s) =µ
s2 − µs+ 1
We had determined that the SIDF of the nonlinear block is N(A,w) = jA2w4 .
We can solve the characteristic equation for A and w:
1 +N(A,w)G(jw) = 0
The solution is A = 2 and w = 1. Remarkably, the solution is independent ofµ.
The characteristic equation solution corresponds to a Nyquist plot ofN(2, w)G(jw) crossing through (−1, 0) at a frequency of w = 1, as we canverify by obtaining a computer plot.
Observations
19 / 31
We have solved the characteristic equation directly, but in the general case itwon’t be easy to find solutions for both A and w. A more exciting graphicalmethod will be used.
The existence of the limit cycle was correctly predicted (by the existence ofsolutions for A and w), but the accuracy in predicting w is reduced as µ isincreased.
Likewise, the limit cycle predicted by the DF is sinusoidal (circular phaseorbit), while the shape of the actual cycle is not circular unless µ = 0.
Observations...
20 / 31
The stability of the limit cycle can also be studied using the DF method. InS&L, the poles of the closed-loop characteristic equation are shown to be
−1
8µ(A2 − 4)±
√
1
64µ2(A2 − 4)2 − 1
If A > 2, the poles are stable, indicating L.C. stability (approaching the L.C.).If A < 2, the poles are unstable, indicating that trajectories grow towards theL.C. This indicates the L.C. is stable.
Alternatively, from the Nyquist diagram, A < 2 causes the plot to enclose(−1, 0) (trajectories grow towards A = 2), while A > 2 stays clear of (−1, 0)(trajectories decrease towards A = 2).
Extended Nyquist Criterion
21 / 31
To allow non-unity loops (for instance with sensor dynamics H(s)), theNyquist criterion is simply applied to L(s) = K(s)G(s)H(s)
Note that S&L uses L(s) = G(s)H(s), takes K(s) = K (constant) and usesthe critical point −1/K, but K is allowed to be a complex number. ButFigure 5.22 does not include K or H anymore. Our version:
■ The characteristic equation is 1 +N(A,w)L(jw) = 0.
■ Include everything linear in L(s) (L = KGH)
■ Limit cycle detection is based on L(jw) = −1/N(A,w)
■ Follow S&L Sect. 5.4.2 by thinking of their G as our L.
Limit Cycle Detection with DFs
22 / 31
Since N(A,w) is both amplitude and frequency dependent in general, therewill be a family of Nyquist plots of −1/N(A,w), where each curve has fixed wand is parameterized by A.
Graph L(jw) and the family of −1/N(A,w) and look for intersections wherefrequencies coincide.
When N(A,w) = N(A) (frequency-independent), there is only one curveparameterized by A, so the process is simpler.
Example
23 / 31
Consider the nonlinear 3rd-order diff.eq.:
d3x
dt3+ x+ 2(1 + kx2)x+ 3(1 + x2)x = u
with u = u0 (constant). There is an additional condition for a periodicsolution to be possible with u0 6= 0 (harmonic balance).
Represent the equation as a unity negative feedback loop with a cascade oflinear dynamics and nonlinearity:
1
s3+s
2+2s+3
2kxx2 + 3x3
u0
−
x
φ(x)
Example...
24 / 31
We note that the nonlinearity is odd and the linear transfer function haslow-pass properties.
Let x = x0 +A sin(wt). Work with trigonometric identities and leave outsecond and higher harmonics to find:
xx2 ≈ Aw cos(wt)(x20 +A2
4)
x3 ≈ 3(x20 +A2
4)A sin(wt) + x0(x
20 +
3
2A2)
The output of the non-bias portion of nonlinearity is then
φ(x, x) = 9(x20 +A2
4)A [(2k/9)w cos(wt) + sin(wt)]
As an operator on y = sin(wt) this is:
9(x20 +A2
4)A
[
(2k/9)dy
dt+ y
]
Example...
25 / 31
That is
Φ(s) = 9(x20 +A2
4)A [(2k/9)s+ 1]
In the frequency domain:
N(A,w) = 9(x20 +A2
4)A [(2k/9)jw + 1]
Suppose that u0 = 0. A DC harmonic balance will show that x0 must be zero.To find potential limit cycles we solve (graphically or numerically):
G(jw) = −1/N(A,w)
Taking x0 = 0 and k = 6, a solution is found: w = 2.4495 and A = 1.1006.
Example...
26 / 31
-1 -0.5 0 0.5 1 1.5-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1Nyquist Diagram
Real Axis
Imag
inar
y A
xis
w=0
w=3
Example...
27 / 31
-0.06 -0.05 -0.04 -0.03 -0.02 -0.01 00
0.05
0.1
0.15
0.2
0.25
0.3Nyquist Diagram
Real Axis
Imag
inar
y A
xis
System: sysGReal: -0.0283Imag: 0.0925Frequency (rad/s): 2.45
Example... DC Harmonic Balance
28 / 31
Once the nonlinearity has been approximated with the SIDF, substitution ofthe candidate periodic solution into the diff. eq. gives:
−Aw3 cos(wt)−Aw2 sin(wt) + 2Aw cos(wt) + 3(x0 +A sin(wt)) =
u0 − 2kAw(x20 +A2
4)− 9(x20 +
A2
4)A sin(wt)− 3x0(x
20 +
3
2A2)
In particular, the DC components must be matched:
u0 = 3x0(1 + x20 +3
2A2)
This provides information about the possible combinations of u0 and x0associated with a periodic solution of amplitude A.
Coefficient matching for sin(wt) and cos(wt) on both sides of the equation is essentially the same as the frequencydomain analysis with the Nyquist criterion. We obtain a system of nonlinear equations.
Example: Limit Cycle Stability and Routh-Hurwitz
Analysis
29 / 31
The nonlinearity was replaced by the amplitude-dependent linear operatorN(A, s). A “characteristic equation” can be written for the closed-loop system:
s3 + s2 + 2
[
1 + k(x20 +1
4A2)
]
s+ 3
[
1 + 3(x20 +1
4A2)
]
= 0
The classical Routh-Hurwitz criterion can provide an equation for the onset ofoscillations.
Using the critical stability condition and the DC harmonic balance leads to thesystem of nonlinear equations:
(2k − 9)(x20 +1
4A2) = 1
3x0(1 + x20 +3
2A2) = u0
From the first equation, it’s clear that k > 4.5 is necessary for a limit cycle toexist. The R-H criterion can also indicate the stability of any limit cycles.
DF Analysis of Forced Sinusoidal Input Response -
Jump Resonance
30 / 31
Consider the nonlinear system with sinusoidal input:
x = f(x) + u(t)
u(t) = Uejwt
Use SIDF as necessary to represent f as
f = N(X,w)x
with x = Xejwt and Xi = Aiejφi , also U = [u1 u2 ...un]
T .Then, for each frequency w, the following must hold:
X = [jwI −N(X,w)]−1 U
This is a system of 2n nonlinear equations with 2n unknowns that can besolved for the Ai and φi. When more than one solution exists for eachfrequency, the system exhibits jump resonance.
Example
31 / 31
Consider the mass-spring-damper system with nonlinear stiffness:
x+ 2ζx+ x(1 + µx2) = u
Let the input be u(t) = U sin(wt). In class, we show that the output-inputamplitude ratio satisfies
X
U=
1√
(1 + 3µ4 X2 − w2) + (2ζw)2
Up to 3 solutions are found for X in a frequency range.
frequency
Amplitude ratio
A
B
C
D
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