Solutions. Chapter 5. Problem 5.1. Solution. Consider the driven, two-well Duffing s oscillator. which can be written in state variable form as

Transcription

1 Chapter 5 Solutions Problem 5.1 Consider the driven, two-well Duffing s oscillator which can be written in state variable form as ẍ + ɛγẋ x + x 3 = ɛf cos(ωt) ẋ = v v = x x 3 + ɛ( γv + F cos(ωt)). In the above expressions, γ is the linear (viscous) damping coefficient, F is the forcing amplitude, and ω is the forcing frequency. 1 Develop a MATLAB code to simulate this system. You need to be able to generate uniformly sampled time series of any length, with any sampling time t s (recall that the sampling frequency is f s = 1/t s, and be very careful when choosing your sampling time). Experiment with the simulations, and by varying the system parameters and examples of: 1. a periodic, approximately simple harmonic response; 2. a periodic response that is not simple harmonic; and 3. a chaotic (i.e. aperiodic) response. Plot the steady state results in the time domain and in the x v phase plane. To help you get started, reasonable values for the parameters to use in the simulations are ɛ =.1, γ [.7; 2.], ω [.9; 1, 1], F [1.5; 2.8]. Solution %% Problem 5.1 figure(1), clf % set parameters for P-1 orbits epsilon =.1; gamma = 2; omega = 1.1; f = 1.5; 11 The parameter ɛ is included here only for convenience when comparing our results to certain theoretical treatments. In particular, the damping and forcing are scaled by ɛ to indicate that the oscillator can be thought of as a perturbation of of a conservative system, with the conservative system corresponding to the case ɛ =. See, for example, Guckenheimer and Holmes (1983). 29

2 Amplitude x Amplitude v 3 CHAPTER 5. SOLUTIONS Time Amplitude x 5 Approximatly simple harmonic response: =.1, = 2, = 1.1, f = Figure 5.1: Simple, close to harmonic response. % integrate 5 forcing cycles to get rid of transients [t,y]=ode45(@duff,[ 5*2*pi/omega],[;;],[],epsilon,gamma,f,omega); % now take the last point in y as new initial condition, get steady state [t,y1]=ode45(@duff,[:2*pi/omega/36:4*2*pi/omega],y(,:),[],epsilon,gamma,f,omega); % plot the results set(gcf, PaperSize,[11 7], PaperPosition, [ 11 7]); subplot(221), plot(t(1:94),y1(1:94,1), linewidth,1) xlabel( Time ),ylabel( Amplitude x ), ylim([ ]) subplot(222), plot(y1(:,1),y1(:,2), linewidth,1) xlabel( Amplitude x ),ylabel( Amplitude v ) subplot(2,2,[3,4]) pwelch(y1(1:1:,1),window(@hann,2*124),[],2*124,.48) set(get(gca, children ), linewidth,1) title([ Approximatly simple harmonic response: \epsilon = num2str(epsilon)..., \gamma = num2str(gamma), \omega = num2str(omega), f = num2str(f)]); print -dpdf Prob_5_1_1 %% Now let s try for periodic, not simple harmonic response figure(2),clf epsilon =.1; gamma = 2; omega = 1.1; f = 1.78; % integrate 1 forcing cycles to get rid of transients [t,y]=ode45(@duff,[ 1*2*pi/omega],[;;],[],epsilon,gamma,f,omega); % now take the last point in y as new initial condition, get steady state

3 Amplitude x Amplitude v Time Amplitude x 5 Period-2 response: =.1, = 2, = 1.1, f = Figure 5.2: Period-2 orbit. [t,y2]=ode45(@duff,[:2*pi/omega/36:4*2*pi/omega],y(,:),[],epsilon,gamma,f,omega); % plot the results set(gcf, PaperSize,[11 7], PaperPosition, [ 11 7]); subplot(221), plot(t(1:94),y2(1:94,1), linewidth,1) xlabel( Time ),ylabel( Amplitude x ), % ylim([.7 1.3]) subplot(222), plot(y2(:,1),y2(:,2), linewidth,1) xlabel( Amplitude x ),ylabel( Amplitude v ) subplot(2,2,[3,4]) pwelch(y2(1:1:,1),window(@hann,2*124),[],2*124,.48) set(get(gca, children ), linewidth,1) title([ Period-2 response: \epsilon = num2str(epsilon)..., \gamma = num2str(gamma), \omega = num2str(omega), f = num2str(f)]); print -dpdf Prob_5_1_2 %% Finally, let s try for chaotic response figure(3),clf epsilon =.1; gamma = 2; omega = 1.1; f = 2; % integrate 1 forcing cycles to get rid of transients [t,y]=ode45(@duff,[ 1*2*pi/omega],[;;],[],epsilon,gamma,f,omega); % now take the last point in y as new initial condition, get steady state [t,y3]=ode45(@duff,[:2*pi/omega/36:4*2*pi/omega],y(,:),[],epsilon,gamma,f,omega); % plot the results set(gcf, PaperSize,[11 7], PaperPosition, [ 11 7]);

4 Amplitude x Amplitude v 32 CHAPTER 5. SOLUTIONS Time Amplitude x 4 Chaotic response: =.1, = 2, = 1.1, f = Figure 5.3: Chaotic orbit. subplot(221), plot(t(1:94),y3(1:94,1), linewidth,1) xlabel( Time ),ylabel( Amplitude x ), % ylim([.7 1.3]) subplot(222), plot(y3(:,1),y3(:,2), linewidth,1) xlabel( Amplitude x ),ylabel( Amplitude v ) subplot(2,2,[3,4]) pwelch(y3(1:1:,1),window(@hann,2*124),[],2*124,.48) set(get(gca, children ), linewidth,1) title([ Chaotic response: \epsilon = num2str(epsilon)..., \gamma = num2str(gamma), \omega = num2str(omega), f = num2str(f)]); print -dpdf Prob_5_1_3

5 33 Problem 5.2 Using the time series {x i } generated in Exercise 1, make an observable via {y i } = {x i + n i }, where the n i are indepent identically-distributed (IID) mean zero Gaussian random variable with variance σ 2 (be careful when choosing your parameters we want to approximate real-life situations). Find the digital Power Spectral Density (PSD) for the three types of response found in Exercise 1, using the MATLAB functions pwelch or periodogram (e.g., at MATLAB prompt type help pwelch ). By experimentation, examine the effect of σ, t s, the number of points used in the FFT, and windowing (including no window, which is called rectwin in MATLAB). A good window to use is the Hanning, however you can experiment to see the differences. Type help window at the command line for more information. Plot your results using a semilog or Decibel (db) scale. Discuss your results: what do you see? Solution Results of the simulation and analysis are shown in P521P523. As seen from these figures with the introduction of even level our PSD acquires a noise floor at approximately -65 db. With the increase in the noise level the noise floor rises, and at high noise levels can obscure the important information (observe the loss of third peak in the top rows). We explored the effects of three different windowing functions: (1) Rectangular, (2) Hanning, and (3) Hamming. With just rectangular windowing the effects of the finite size of data manifest itself in a rising energy levels at low frequencies even for values where there is no observed frequency content, since the peaks are wide and spread widely (spectral leakage). Adding a Hamming window helps a situation considerably, since we have more clear peaks, however there is still some spread at the base of the peaks. With Hanning window we see that situation improves considerably with the thickness of the peaks at P-1 signal, Rectangular Window 2 2 P-1 signal, Hanning Window 2 P-1 signal, Hamming Window -2 5% noise -2 5% noise -2 5% noise P-2 signal, Rectangular Window P-2 signal, Hanning Window P-2 signal, Hamming Window Chaotic Signal, Rectangular Window Chaotic Signal, Hanning Window Chaotic Signal, Hamming Window Figure 5.4: PSDs of signal in Problem 5.1 with different noise level and different windowing function.

6 34 CHAPTER 5. SOLUTIONS the base. However, the peaks themselves are not as defined (pointy) as before. These are really important for periodic data. However, for the chaotic data the effects are less pronounced. We know that increasing the sampling time period, causes the loss of information in high frequencies and may lead to the aliasing. As we see from plots in Fig. 5.5 and previous PSD plots, our systems bandwidth is approximately 1/2 of its normalized frequency range. Thus, by downsampling the signal by factor of 2 we retain important information. However the noise that exists at high frequencies is still folded back onto our signal. So the proper procedure would be to low-pass filter our signal with.5 cutoff frequency and then do the psd estimates. When signals are downsampled by factor of 4, we clearly see the appearance of peaks that were not present before as a result of aliasing (this is most clear in the top raw of plots). P-1 signal, Rectangular Window 2 2 P-1 signal, Hanning Window 2 P-1 signal, Hamming Window P-2 signal, Rectangular Window Chaotic Signal, Rectangular Window P-2 signal, Hanning Window Chaotic Signal, Hanning Window P-2 signal, Hamming Window Chaotic Signal, Hamming Window % noise % noise % noise Figure 5.5: Same as Fig. 5.4 but with data recorded with twice the sampling period. %% Problem 5.2 figure(4),clf set(gcf, PaperSize,[11 7], PaperPosition, [ 11 7]); % now add noise to see the effects n = randn(size(y1(:,1))); % generate noise signal nl = [.1.1.5]; % three distinct noise levels clas = { P-1 signal, P-2 signal, Chaotic Signal }; clas2 = { Rectangular Window, Hanning Window, Hamming Window }; for i = 1:3 eval([ x = y num2str(i) (:,1); ]); sd = std(x); % define signal and its std subplot(3,3,3*i-2)

7 35 P-1 signal, Rectangular Window 2 2 P-1 signal, Hanning Window 2 P-1 signal, Hamming Window P-2 signal, Rectangular Window Chaotic Signal, Rectangular Window P-2 signal, Hanning Window Chaotic Signal, Hanning Window P-2 signal, Hamming Window Chaotic Signal, Hamming Window % noise % noise % noise Figure 5.6: Same as Fig. 5.4 but with data recorded with four-times the sampling period. [Pxx,F] = pwelch(s(1:2:),window(@rectwin,124),[],124,1*.48/2); plot(f,1*log1(pxx), linewidth,1, color,co(4-j,:)) ylim([-7 2]);title([clas{i}, clas2{1}]); xlim([ 12]) xlabel( ) ylabel( ) leg(,, 5% noise ) subplot(3,3,3*i-1) [Pxx,F] = pwelch(s(1:2:),window(@hann,124),[],124,1*.48/2); plot(f,1*log1(pxx), linewidth,1, color,co(4-j,:)) ylim([-7 2]);title([clas{i}, clas2{2}]); xlim([ 12]) xlabel( ) ylabel( ) leg(,, 5% noise ) subplot(3,3,3*i) [Pxx,F] = pwelch(s(1:2:),window(@hamming,124),[],124,1*.48/2); plot(f,1*log1(pxx), linewidth,1, color,co(4-j,:))

8 36 CHAPTER 5. SOLUTIONS xlabel( ) ylabel( ) ylim([-7 2]);title([clas{i}, clas2{3}]); xlim([ 12]) leg(,, 5% noise ) print -dpdf Prob_5_2_1 %% figure(5),clf set(gcf, PaperSize,[11 7], PaperPosition, [ 11 7]); % now add noise to see the effects n = randn(size(y1(:,1))); % generate noise signal nl = [.1.1.5]; % three distinct noise levels clas = { P-1 signal, P-2 signal, Chaotic Signal }; clas2 = { Rectangular Window, Hanning Window, Hamming Window }; for i = 1:3 eval([ x = y num2str(i) (:,1); ]); sd = std(x); % define signal and its std subplot(3,3,3*i-2) [Pxx,F] = pwelch(s(1:4:),window(@rectwin,124),[],124,1*.48/4); plot(f,1*log1(pxx), linewidth,1, color,co(4-j,:)) ylim([-7 2]);title([clas{i}, clas2{1}]); xlabel( ) ylabel( ) leg(,, 5% noise ) subplot(3,3,3*i-1) [Pxx,F] = pwelch(s(1:4:),window(@hann,124),[],2*124,1*.48/4); plot(f,1*log1(pxx), linewidth,1, color,co(4-j,:)) ylim([-7 2]);title([clas{i}, clas2{2}]); xlabel( ) ylabel( ) leg(,, 5% noise ) subplot(3,3,3*i) [Pxx,F] = pwelch(s(1:4:),window(@hamming,124),[],124,1*.48/4); plot(f,1*log1(pxx), linewidth,1, color,co(4-j,:)) ylim([-7 2]);title([clas{i}, clas2{3}]); xlabel( ) ylabel( )

9 37 leg(,, 5% noise )%, location, southwest ); print -dpdf Prob_5_2_2 % figure(6),clf set(gcf, PaperSize,[11 7], PaperPosition, [ 11 7]); % now add noise to see the effects n = randn(size(y1(:,1))); % generate noise signal nl = [.1.1.5]; % three distinct noise levels clas = { P-1 signal, P-2 signal, Chaotic Signal }; clas2 = { Rectangular Window, Hanning Window, Hamming Window }; for i = 1:3 eval([ x = y num2str(i) (:,1); ]); sd = std(x); % define signal and its std subplot(3,3,3*i-2) [Pxx,F] = pwelch(s(1:8:),window(@rectwin,124),[],124,1*.48/8); plot(f,1*log1(pxx), linewidth,1, color,co(4-j,:)) ylim([-7 2]);title([clas{i}, clas2{1}]); xlabel( ) ylabel( ) leg(,, 5% noise, location, southwest ); subplot(3,3,3*i-1) [Pxx,F] = pwelch(s(1:8:),window(@hann,124),[],2*124,1*.48/8); plot(f,1*log1(pxx), linewidth,1, color,co(4-j,:)) ylim([-7 2]);title([clas{i}, clas2{2}]); xlabel( ) ylabel( ) leg(,, 5% noise, location, southwest ); subplot(3,3,3*i) [Pxx,F] = pwelch(s(1:8:),window(@hamming,124),[],124,1*.48/8); plot(f,1*log1(pxx), linewidth,1, color,co(4-j,:)) ylim([-7 2]);title([clas{i}, clas2{3}]); xlabel( ) ylabel( ) leg(,, 5% noise, location, southwest );

10 38 CHAPTER 5. SOLUTIONS print -dpdf Prob_5_2_3 References John Guckenheimer and Philip Holmes (1983) Nonlinear Oscillations, Dynamical Systems, and Bifurcations of Vector Fields, Springer-Verlag, New York. See in particular Ch. 4.5, Eq. (4.5.18).

11 Chapter 6 Solutions Problem 6.1 [Bifurcation Diagram for Logistics Map] Consider a Logistics map equation: x n+1 = Ax n (1 x n ). (6.1) Run the simulations with the initial condition X =.1, for different values of A starting from A = 2.9 and going up to A = 4. Make increments in A small enough, e.g., A <.1 (e.g., A =.1). For each magnitude of A record long time series and retain only the last 1, (or more, e.g., 4,) points representing its steady state response. Now plot the bifurcation diagram, by plotting all the response points versus the corresponding value of A. What can you see in the diagram? Solution Results of the simulation are shown in Fig %% Problem 6.1 X=zeros(111:1); A = 2.9:.1:4; x=zeros(1,21); for j = 1:111 x(1) =.1; for k = 1:12 x(k+1)=a(j)*x(k)*(1-x(k)); X(j,:) = x(-999:); figure(7) set(gcf, PaperSize,[11 7], PaperPosition, [ 11 7]); plot(a,x,., markersize,.5, markeredgecolor,co(1,:)) axis([ ]) xlabel( $A$, Interpreter, Latex, FontSize,14) ylabel( $x_n$, Interpreter, Latex, FontSize,14) print -dpdf Prob_6_1 39

12 4 CHAPTER 6. SOLUTIONS Problem 6.2 Figure 6.1: Bifurcation diagram of the Logistics map. Numerically integrate the Lorenz system described in Problem 3.2 and perform a Poincaré section. Namely, record (x, z) every time the y-coordinate equals zero and its derivative is negative. You can accomplish by providing options to ode45 MATLAB program by first setting it through options = odeset( Events,@events);, where events is the function that specifies the when to sample the flow (i.e., you need to specify y = and ẏ < conditions). I would suggest reading ODE Event Location on Mathworks site and following their example. Once you have the points, plot the Poincaré section by plotting z vs x as points. Solution Results of the simulation are shown in Fig %% Problem 6.2 sigma = 1; r = 28; b = 8/3; options = odeset( Events,@events); [~,~,~,ye] = ode45(@lorenz,[ 5],ones(3,1),options); figure(8) set(gcf, PaperSize,[11 7], PaperPosition, [ 11 7]); plot(ye(:,1),ye(:,3),., markersize,1) xlabel( x ) ylabel( z ) title( Poincare Section for y= & y < ) print -dpdf Prob_6_2

13 z 41 Poincare Section for y= & y'< x Figure 6.2: Poincaré section of the Lorenz equation obtained by sampling every time the y-coordinate equals zero and its derivative is negative. function [value,isterminal,direction] = events(~,y) % Locate the time when y passes through zero in a % decreasing direction and record x and z. value = y(2); % Detect y crossing isterminal = ; % Do not stop the integration direction = -1; % Negative direction only function dy = lorenz(~,y) % dy = [ -b*y(1) + y(2)*y(3); % sigma*(y(2) - y(3)); % r*y(2) - y(3) - y(1)*y(2)]; % dy = [-b*y(1) + y(2)*y(3); -sigma*(y(2) - y(3)); -y(1)*y(2) + r*y(2) - y(3)]; Problem 6.3 Numerically integrate the Lorenz system described in Problem 3.2 and collect all the maxima of y variable. Again, you can use events to accomplish this, or use findpeaks command in MATLAB

14 42 CHAPTER 6. SOLUTIONS on y time series that is sampled sufficiently. compare it to the attractor of Problem 6.2. Plot the series of the maxima Y n versus Y n+1 and Solution Results of the simulation are shown in Fig Poincare Section for dy/dt= & y''< 1 Reconstructed Poincare Section z n 5 y n x n y n Figure 6.3: Poincaré section of the Lorenz equation obtained by sampling every time the y-coordinate achieves its local maximum. We can see that the actual and reconstructed Poincaré sections are quite different from the previous problem. This is caused by the fact that on average there are five local maxima before zero crossing occurs for the y variable. options2 = odeset( Events,@events2); [~,~,~,ye] = ode45(@lorenz,[ 5],ones(3,1),options2); figure(9) set(gcf, PaperSize,[11 4.5], PaperPosition, [ ]); subplot(1,2,1) plot(ye(:,1),ye(:,3),., markersize,1) xlabel( x_n ) ylabel( z_n ) title( Poincare Section for dy/dt= & y < ) subplot(1,2,2) plot(ye(1:-1,2),ye(2:,2),., markersize,1) xlabel( y_n ) ylabel( y_{n+1} ) title( Reconstructed Poincare Section ) print -dpdf Prob_6_3 function [value,isterminal,direction] = events2(t,y) % Locate the time when dx/dt passes through zero in a % decreasing direction... dy = lorenz(t,y); value = sigma*(dy(2)-dy(1)); % Detect dx/dt crossing

15 43 isterminal = ; direction = -1; % Do not stop the integration % Negative direction only function dy = lorenz(~,y) % dy = [ -b*y(1) + y(2)*y(3); % sigma*(y(2) - y(3)); % r*y(2) - y(3) - y(1)*y(2)]; % dy = [-b*y(1) + y(2)*y(3); -sigma*(y(2) - y(3)); -y(1)*y(2) + r*y(2) - y(3)];