MCE693/793: Analysis and Control of Nonlinear Systems

MCE693/793: Analysis and Control of Nonlinear Systems Introduction to Describing Functions Hanz Richter Mechanical Engineering Department Cleveland State University

Introduction Frequency domain methods are essential for analysis and design of linear systems. Stability, robustness and performance can be studied using classical and 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 Nichols chart) Describing functions provide an approximate method for frequency-domain analysis of nonlinear systems. We will focus on limit cycle detection/creation or avoidance. 2 / 31

Basic Idea In linear systems, transfer functions become frequency response functions when the Laplace variable s is restricted to the imaginary axis. That is, the Fourier transform is the Laplace transform restricted to the imaginary axis. Frequency response functions in linear systems are functions of frequency only: G(s) = s 1 1 w2, G(jw) = s+1 1+w 2 + 2w 1+w 2j In control systems containing nonlinearities, we seek to find a frequency response function for the nonlinear block. Due to nonlinearity, this function will be amplitude-dependent. 0 φ(e) e G(s) { N(A,w) { G(jw) 3 / 31

Memoryless vs. Memory, Soft vs. Hard Nonlinearities A memoryless nonlinearity produces an output which depends only on the current value of the input. The nonlinearity must then be a function of its argument: φ = φ(e(t)). Example: saturation. The output of a nonlineariry with memory depends on the current and past values of the inputs. The nonlinearity is not a function. Sometimes the terminology multiple-valued function is used. Example: hysteresis. A soft nonlinearity is continuous. The terminology is vague, sometimes soft nonlinearities are also required to be differentiable. Example: φ(e) = e 3. A hard nonlinearity has discontinuities. In some cases, a continuous function which is not differentiable everywhere is called a hard nonlinearity. Example: saturation. 4 / 31

Describing Function Concept When a sinuosoidal function is applied to a nonlinearity, harmonic distortion occurs and the output is no longer a sinusoid. Assuming the output remains periodic (always true for static nonlinearities, still true for some others), there is a Fourier (trigonometric) decomposition. The decomposition contains sinuosoidal components with frequencies which are integer multiples of the input frequency, and possibly a bias term (zero for odd functions). In particular, the fundamental component will have an amplitude which depends nonlinearly on the input amplitude. The principle behind describing functions is to retain only this fundamental output component as an approximation. A transfer function is formed by taking the ratio of the output component (expressed as a complex exponential) to the input sinusoid (also expressed as a complex exponential). This defines the sinusoidal input describing function (SIDF). 5 / 31

Calculating a simple SIDF As a first example, consider the cubic nonlinearity: φ(u) = u 3. Let the input be u(t) = Asin(wt) and express this as the complex exponential U(A,w) = Ae jwt = Acos(wt)+Ajsin(wt). The output of the nonlinearity is φ(u) = A 3 sin 3 (wt). We must find the fundamental component of the Fourier series expansion of this function. Recall that the expansion has the form: φ(t) = a 0 2 + n=1 a n cos( 2πnt T )+b nsin( 2πnt T ), where a n = 2 T b n = 2 T x0 +T x 0 x0 +T x 0 φ(t) cos( 2πnt T ) dt φ(t) sin( 2πnt T ) dt and a 0 /2 is the mean of φ(t) in one period. 6 / 31

Calculating a simple SIDF... Because φ is odd, a n = 0 for all n. We only need to calculate b 1. With T = 2π/w, the result is b 1 = 3 4 A3 The output sinusoid is then Φ = 3A3 4 ejwt Therefore the describing function for the cubic nonlinearity is N(A,w) = N(A) = 3A2 4 e0 All odd static nonlinearities will have a purely real and frequency-independent SIDF. 7 / 31

Example: Frequency-dependent SIDF Following the introductory example in S&L on application of the DF, consider the nonlinearity φ(x,ẋ) = x 2 ẋ associated with the van der Pol equation. Assume X = Ae jwt. Then the output of the nonlinearity is φ(x,ẋ) = A 3 wsin 2 (wt)cos(wt). Using trigonometric identities this is the same as φ(x,ẋ) = A3 w 4 (cos(wt) cos(3wt)) By the definition of SIDF, we retain only the fundamental component. Therefore Φ = A3 w 4 ej(wt π/2) The SIDF is then N(A,w) = Φ/U = A2 w 4 e jπ/2 = A2 w 4 j. In S&L the input is taken as ( x), so their N(A,w) has the opposite sign 8 / 31

Example: Memory Relay with Deadzone In class, we obtain the SIDF for a relay having the characteristic shown below: out D δ δ in D Show that N(A,w) = 4D πa 1 ( δ A )2 j 4Dδ. Because the nonlinearity is not πa 2 dynamic, frequency is absent. 9 / 31

Limit Cycle Analysis with Describing Functions If N(A,w) were a truly linear transfer function, we could investigate the presence of oscillations by using the idea of marginal stability and the Nyquist criterion. For example, linear analysis can provide the value of K at which the following system oscillates: 1 Nyquist Diagram 15 Root Locus 0.8 0.6-1 ) 10 0 k 16 (s+2)(s 2 +2s+8) Imaginary Axis 0.4 0.2 0-0.2-0.4-0.6-0.8 Imaginary Axis (seconds 5 0-5 -10-1 -1 0 1 2 Real Axis -15-20 -10 0 10 Real Axis (seconds -1 ) 10 / 31

Review of the Nyquist Stability Criterion Complex Mapping Theorem: Let F(s) : C C be a mapping defined on a region of the complex plane and let C be a closed curve contained in that region. Assume: 1. F(s) has no more zeroes than poles and C does not pass through any of them. 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 is traversed clockwise. Then the image curve F(s) has N = Z P clockwise encirclements of the origin. (Animation) 11 / 31

Review of the Nyquist Stability Criterion... Nyquist s idea was to take C as an infinite half-circle encompassing the entire Im 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 P clockwise encirclements of the origin, then F(s) has the same number of encirclements of ( 1,0). 12 / 31

Review of the Nyquist Stability Criterion... In a unity negative feedback control system, the open-loop transfer function is L(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 the Nyquist 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-loop poles (poles of L(s)) are also poles of 1+L(s). 13 / 31

Review of the Nyquist Stability Criterion... Apply the shifted Complex Mapping Theorem to 1+L(s): N is the number of clockwise origin encirclements of 1+L(s), or equivalently (we actually use this form) the number of clockwise encirclements of (-1,0). Note we evaluate L(s) on the Im axis only (this is called a Nyquist plot) Z is the number of zeroes of 1 + L(s) inside the Nyquist contour (unstable closed-loop poles) P is the number of poles of 1+L(s) (equivalently of L(s) and we use this) inside the Nyquist contour (unstable open-loop poles) Then Z = N +P 14 / 31

Example We use the Nyquist criterion to predict the stability of a unity feedback control 8 system with K(s) = k and G(s) =. Can the system be (s 2)(s 2 +2s+8) stabilized with proportional control? 0.25 Nyquist Diagram 0.2 0.15 0.1 Imaginary Axis 0.05 0-0.05-0.1-0.15-0.2-0.25-1 -0.8-0.6-0.4-0.2 0 Real Axis Confirm analysis using the Routh-Hurwitz criterion. 15 / 31

Practical Significance of the Nyquist Stability Criterion Many systems are open-loop stable, so P = 0 is known even without a math model. We can use the Nyquist criterion to predict the critical proportional gain 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 a dynamic signal analyzer or offline signal processing to obtain Nyquist plot data. Then the stability of the closed-loop system can be studied using data from the open-loop system only. Specifically, we can find the critical proportional gain that scales the Nyquist plot to touch the (-1,0) point. Also, since pure phase lags (such as those introduced by delays) rotate the Nyquist plot, we can determine the maximum amount of delay that the system can tolerate before going unstable. 16 / 31

Nyquist-like analysis with Describing Functions Several conditions have to be satisfied for a DF-based frequency analysis to produce correct results (see S&L Sect. 5.1.3): 1. There s a single nonlinearity in the loop (unless 2 or more can be blended into one) 2. The nonlinearity is time-invariant (a relay with a time-dependent deadzone wouldn t) 3. Higher harmonics produced by the nonlinearity are attenuated by the linear dynamics following the nonlinearity (the linear TF must be low-pass). 4. The nonlinearity is odd, so that the Fourier expansion of its output has zero bias. Notes: Even when these conditions hold, the DF technique may produce erroneous results, particularly when predicting limit cycles. There are extended DF techniques relaxing assumptions 1,2 and 4, but they are cumbersome. 17 / 31

Example: van der Pol oscillator See S&L Sect. 5.1.1. The van der Pol equation is re-arranged as a negative unity feedback loop, where the nonlinear block is x 2 ẋ, followed by the unstable transfer function µ G(s) = s 2 µs+1 We had determined that the SIDF of the nonlinear block is N(A,w) = j A2 w 4. 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 of µ. The characteristic equation solution corresponds to a Nyquist plot of N(2,w)G(jw) crossing through ( 1,0) at a frequency of w = 1, as we can verify by obtaining a computer plot. 18 / 31

Observations We have solved the characteristic equation directly, but in the general case it won t be easy to find solutions for both A and w. A more exciting graphical method will be used. The existence of the limit cycle was correctly predicted (by the existence of solutions for A and w), but the accuracy in predicting w is reduced as µ is increased. Likewise, the limit cycle predicted by the DF is sinusoidal (circular phase orbit), while the shape of the actual cycle is not circular unless µ = 0. 19 / 31

Observations... The stability of the limit cycle can also be studied using the DF method. In S&L, the poles of the closed-loop characteristic equation are shown to be 1 8 µ(a2 4)± 1 64 µ2 (A 2 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 the L.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). 20 / 31

Extended Nyquist Criterion To allow non-unity loops (for instance with sensor dynamics H(s)), the Nyquist 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 uses the critical point 1/K, but K is allowed to be a complex number. But Figure 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. 21 / 31

Limit Cycle Detection with DFs Since N(A,w) is both amplitude and frequency dependent in general, there will be a family of Nyquist plots of 1/N(A,w), where each curve has fixed w and is parameterized by A. Graph L(jw) and the family of 1/N(A,w) and look for intersections where frequencies coincide. When N(A,w) = N(A) (frequency-independent), there is only one curve parameterized by A, so the process is simpler. 22 / 31

Example Consider the nonlinear 3rd-order diff.eq.: d 3 x dt 3 +ẍ+2(1+kx2 )ẋ+3(1+x 2 )x = u with u = u 0 (constant). There is an additional condition for a periodic solution to be possible with u 0 0 (harmonic balance). Represent the equation as a unity negative feedback loop with a cascade of linear dynamics and nonlinearity: u 0 1 s 3 +s 2 +2s+3 x φ(x) 2kẋx 2 +3x 3 23 / 31

Example... We note that the nonlinearity is odd and the linear transfer function has low-pass properties. Let x = x 0 +Asin(wt). Work with trigonometric identities and leave out second and higher harmonics to find: ẋx 2 Awcos(wt)(x 2 0 + A2 4 ) x 3 3(x 2 0 + A2 4 )Asin(wt)+x 0(x 2 0 + 3 2 A2 ) The output of the non-bias portion of nonlinearity is then φ(x,ẋ) = 9(x 2 0 + A2 4 )A[(2k/9)wcos(wt)+sin(wt)] As an operator on y = sin(wt) this is: [ 9(x 2 0 + A2 4 )A (2k/9) dy ] dt +y 24 / 31

Example... That is Φ(s) = 9(x 2 0 + A2 4 )A[(2k/9)s+1] In the frequency domain: N(A,w) = 9(x 2 0 + A2 4 )A[(2k/9)jw+1] Suppose that u 0 = 0. A DC harmonic balance will show that x 0 must be zero. To find potential limit cycles we solve (graphically or numerically): G(jw) = 1/N(A,w) Taking x 0 = 0 and k = 6, a solution is found: w = 2.4495 and A = 1.1006. 25 / 31

Example... 1 Nyquist Diagram 0.8 w=3 0.6 0.4 Imaginary Axis 0.2 0-0.2 w=0-0.4-0.6-0.8-1 -1-0.5 0 0.5 1 1.5 Real Axis 26 / 31

Example... 0.3 Nyquist Diagram 0.25 0.2 Imaginary Axis 0.15 0.1 System: sysg Real: -0.0283 Imag: 0.0925 Frequency (rad/s): 2.45 0.05 0-0.06-0.05-0.04-0.03-0.02-0.01 0 Real Axis 27 / 31

Example... DC Harmonic Balance Once the nonlinearity has been approximated with the SIDF, substitution of the candidate periodic solution into the diff. eq. gives: Aw 3 cos(wt) Aw 2 sin(wt)+2awcos(wt)+3(x 0 +Asin(wt)) = u 0 2kAw(x 2 0 + A2 4 ) 9(x2 0 + A2 4 )Asin(wt) 3x 0(x 2 0 + 3 2 A2 ) In particular, the DC components must be matched: u 0 = 3x 0 (1+x 2 0 + 3 2 A2 ) This provides information about the possible combinations of u 0 and x 0 associated 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 frequency domain analysis with the Nyquist criterion. We obtain a system of nonlinear equations. 28 / 31

Example: Limit Cycle Stability and Routh-Hurwitz Analysis The nonlinearity was replaced by the amplitude-dependent linear operator N(A, s). A characteristic equation can be written for the closed-loop system: [ s 3 +s 2 +2 1+k(x 2 0 + 1 ] [ 4 A2 ) s+3 1+3(x 2 0 + 1 ] 4 A2 ) = 0 The classical Routh-Hurwitz criterion can provide an equation for the onset of oscillations. Using the critical stability condition and the DC harmonic balance leads to the system of nonlinear equations: (2k 9)(x 2 0 + 1 4 A2 ) = 1 3x 0 (1+x 2 0 + 3 2 A2 ) = u 0 From the first equation, it s clear that k > 4.5 is necessary for a limit cycle to exist. The R-H criterion can also indicate the stability of any limit cycles. 29 / 31

DF Analysis of Forced Sinusoidal Input Response - Jump Resonance Consider the nonlinear system with sinusoidal input: ẋ = f(x)+u(t) u(t) = Ue jwt Use SIDF as necessary to represent f as f = N(X,w)x with x = Xe jwt and X i = A i e jφ i, also U = [u 1 u 2...u n ] 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 be solved for the A i and φ i. When more than one solution exists for each frequency, the system exhibits jump resonance. 30 / 31

Example Consider the mass-spring-damper system with nonlinear stiffness: ẍ+2ζẋ+x(1+µx 2 ) = u Let the input be u(t) = U sin(wt). In class, we show that the output-input amplitude ratio satisfies X U = 1 (1+ 3µ 4 X2 w 2 )+(2ζw) 2 Up to 3 solutions are found for X in a frequency range. Amplitude ratio A B D C frequency 31 / 31