Solutions. Chapter 1. Problem 1.1. Solution. Simulate free response of damped harmonic oscillator

Transcription

1 Chapter Solutions Problem. Simulate free response of damped harmonic oscillator ẍ + ζẋ + x = for different values of damping ration ζ and initial conditions. Plot the response x(t) and ẋ(t) versus t and the corresponding trajectories in the phase space for underdamped, critically damped, and overdamped systems. Do the same for the harmonically forced system: ẍ + ζẋ + x = f cos ωt for some choice of f and ω and plot the solutions including transient response and some part of the steady state. Solution MATLAB code for the simulations and figures: function [] = Problem MCE 567 Problem. Solution integrate the equations for the free response [t,y] = ode45(@harmonic,:.:,[ ],[],.,,); [t,y] = ode45(@harmonic,:.:,[ ],[],,,); [t,y3] = ode45(@harmonic,:.:,[ ],[],,,); plot the results figure() figure( PaperType, <custom>, PaperSize,[8 4], PaperPosition,[ ]); subplot() plot(t,y,t,y,t,y3) xlabel( $t$, FontSize,4, Interpreter, latex ); ylabel( $x(t), \dot x(t)$, FontSize,4, Interpreter, latex ); axis([ ]) box on 3

2 4 CHAPTER. SOLUTIONS legend( underdamped, x, underdamped, v, critically-damped, x,... critically-damped, v, overdamped, x, overdamped, v ) legend boxoff subplot() tau = ; plot(y(:,),y(:,),y(:,),y(:,),y3(:,),y3(:,)) xlabel( $x$, FontSize,4, Interpreter, latex ); ylabel( $\dot x$, FontSize,4, Interpreter, latex ); axis([ ]) box on legend( underdamped, critically-damped, overdamped, location, northwest ) legend boxoff print -dpdf Problem a integrate the equations for the forced response [t,y] = ode45(@harmonic,:.:,[ ],[],.,.,.8); [t,y] = ode45(@harmonic,:.:,[ ],[],,.,.8); [t,y3] = ode45(@harmonic,:.:,[ ],[],,.,.8); plot the results figure() figure( PaperType, <custom>, PaperSize,[8 4], PaperPosition,[ ]); subplot() plot(t,y,t,y,t,y3) xlabel( $t$, FontSize,4, Interpreter, latex ); ylabel( $x(t), \dot x(t)$, FontSize,4, Interpreter, latex ); axis([ -.5.5]) box on legend( underdamped, x, underdamped, v, critically-damped, x,... critically-damped, v, overdamped, x, overdamped, v ) legend boxoff subplot() tau = ; plot(y(:,),y(:,),y(:,),y(:,),y3(:,),y3(:,)) xlabel( $x$, FontSize,4, Interpreter, latex ); ylabel( $\dot x$, FontSize,4, Interpreter, latex ); axis([ ]) box on legend( underdamped, critically-damped, overdamped, location, northwest ) legend boxoff print -dpdf Problem b function dy=harmonic(t,y,zeta,f,omega) duffing equation for integration dy=[y(); -y()*zeta*y()+f*cos(omega*t)]; end end

3 underdamped, x underdamped, v critically-damped, x critically-damped, v overdamped, x overdamped, v.5 underdamped critically-damped overdamped Figure.: Output of the above script for the free response The figure above shows the free response of the damped harmonic oscillator. We observe that critically damped system reaches zero faster than over-damped. However, the over-damped has smaller maximum amplitude. The figure bellow shows the forced response of the system. We observe that the damping determines the magnitude of the steady state response: larger damping causes smaller steady state response amplitudes. The rate at which the steady state is reached is the same as for the unforced system underdamped, x underdamped, v critically-damped, x critically-damped, v overdamped, x overdamped, v.5 underdamped critically-damped overdamped Figure.: Output of the above script for the forced response (f =. and ω =.8)

4 6 CHAPTER. SOLUTIONS Problem. Simulate two-well Duffing oscillator ẍ +.5 ẋ + x(x ) =.3 cos t using MATLAB s ode45 algorithm and using initial conditions x() = and ẋ() = over the interval t [, ]. Sample the trajectory with the uniform sampling interval t s =.. Now assume that our measurement from this system is u i = x i, and do the following: Plot the original systems phase space trajectory, by plotting x i versus ẋ i. Now plot our measurements versus their -delayed version: x i versus x i+. How do these two plot compare? Can you say that our measurement can be used to reconstruct the original system s trajectory? Solution MATLAB code for the simulation and plots: function [] = Problem MCE 567 Problem. Solution integrate the equation [t,y] = ode45(@duffing,:.:,[ , ],[],,,.5,.3,); plot the results clf figure( PaperType, <custom>, PaperSize,[7 3], PaperPosition,[ 7 3]); subplot() plot(y(:,),y(:,)) xlabel( $x$, FontSize,4, Interpreter, latex ); ylabel( $\dot x$, FontSize,4, Interpreter, latex ); axis([ ]) box on subplot() tau = ; plot(y(:end-tau,),y(+tau:end,)) xlabel( $x_n$, FontSize,4, Interpreter, latex ); ylabel( $x_{n+}$, FontSize,4, Interpreter, latex ); axis([ ]) box on print the results print -dpdf Problem duffing oscillator function dy=duffing(t,y,alpha,beta,gamma,f,omega)

5 7 duffing equation for integration dy=[y(); alpha*y()-beta*y()^3-gamma*y()+f*cos(omega*t)]; end end Looking at the above figures: Figure.3: Output of the above script. The delay coordinate plot seems similar to the original phase portrait rotated by 45 degrees counter clockwise.. Since both plots share striking resemblance they seem geometrically equivalent, we can speculate that the measurements can be used to reconstruct the original trajectory in certain cases.

6 8 CHAPTER. SOLUTIONS

7 Chapter Solutions Problem. In Matlab, generate M trials of random processes with the forms. x(t) = cos(t) + n(t);. x(t) = cos(t + θ) + n(t); and 3. x(t) = A cos(t + θ) + n(t), where t T for some sample T, n(t) is a normally-distributed (Gaussian) random process with zero mean and variance σ, θ is a uniformly-distributed constant random variable on [; π], and A is a normally-distributed constant with zero mean and unit variance. In each case, generate uniformly-sampled series {x ; x ;... ; x N } where x i = x(t i ). Use the above-generated series to examine the stationarity and ergodicity of the different processes. Experiment with different values of M, N, T, and σ to make sure you get good results. Discuss your findings. Note: Think carefully about your sampling interval and the length of the series (relationship between N and T ). Solution MATLAB code for the simulation and plots: function [] = Problem (T,N,M,sigma) MCE 567: Assignment, Problem. Can test for different values of: T -- window N -- number of samples in this window M -- numbers of trial functions sigma -- standard deviation of additive noise generate 9

8 CHAPTER. SOLUTIONS t = linspace(,t,n); define random variables n = sigma*randn(m,n); theta = *pi*rand(m,n); A = randn(m,n); generate random process trials x = ones(m,)*cos(t)+n; x = cos(ones(m,)*t+theta)+n; x3 = A.*cos(ones(M,)*t+theta)+n; test for stationarity indx = floor(linspace(,n,)); uniformly spaced points figure( PaperType, <custom>, PaperSize,[ 5.3], PaperPosition,[ ]); test for signal (a) subplot(,3,), errorbar(t(indx),mean(x(:,indx)),std(x(:,indx)), o ) xlabel( ), ylabel( ) title( stationarity of x(t) = sin(t) +n(t) ) thus this signal is nonstationary test for signal (b) subplot(,3,), errorbar(t(indx),mean(x(:,indx)),std(x(:,indx)), o ) xlabel( ), ylabel( ) title( stationarity of x(t) = sin(t-\theta) +n(t) ) this seems to be stationary test for signal (c) subplot(,3,3), errorbar(t(indx),mean(x3(:,indx)),std(x3(:,indx)), o ) xlabel( ), ylabel( ) title( stationarity of x(t) = A sin(t-\theta) +n(t) ) this also seems to be stationary Now test for ergodicity signal (a) subplot(,3,4), plot(mean(x )), hold on, plot(mean(x)) xlabel( ), ylabel( ) legend( average, space average ),title( Signal (a) ) this was obvious: it s not ergodic signal (b) subplot(,3,5), plot(mean(x )), hold on, plot(mean(x)) xlabel( ), ylabel( ),title( Signal (b) ) this seems to be ergodic signal (c) subplot(,3,6), plot(mean(x3 )), hold on, plot(mean(x3)) xlabel( ), ylabel( ),title( Signal (c) )

9 this seems to be ergodic print -dpdf Problem From the figures it is clear the the process (a) is not stationary since its mean varies with. Processes (b) and (c) seem to be stationary since their mean and variance stay approximately constant along. As for the ergodicity: process (a) is not ergodic since average is different from the space average. However, processes (b) and (c) seem stationary since their both averages seem to have similar behavior.

10 CHAPTER. SOLUTIONS stationarity of x(t) = sin(t) +n(t) stationarity of x(t) = sin(t- ) +n(t) stationarity of x(t) = A sin(t- ) +n(t) Signal (a) average space average Signal (b) Signal (c) stationarity of x(t) = sin(t) +n(t) stationarity of x(t) = sin(t- ) +n(t) stationarity of x(t) = A sin(t- ) +n(t) Signal (a) average space average Signal (b) Signal (c) Figure.: Problem. (Top: T =, N =, M =, σ =.5; Bottom: T =, N =, M =, σ =.)

11 3 stationarity of x(t) = sin(t) +n(t).5 stationarity of x(t) = sin(t- ) +n(t) stationarity of x(t) = A sin(t- ) +n(t) Signal (a) average space average Signal (b) Signal (c) stationarity of x(t) = sin(t) +n(t) stationarity of x(t) = sin(t- ) +n(t) stationarity of x(t) = A sin(t- ) +n(t) Signal (a) average space average Signal (b) Signal (c) Figure.: Problem. (Top: T =, N =, M =, σ =.5; Bottom: T =, N =, M =, σ =.)

12 4 CHAPTER. SOLUTIONS Problem. Generate one long series of the random process () in Exercise. Estimate the mean of the process many s from different subsets of the series, and show that the distribution of µ x is approximately Gaussian with mean and variance, as predicted by the Central Limit Theorem. Note: the way in which you make the samples used for each estimate can have a large impact on your results! Solution MATLAB code for the simulation and plots: function [mn, sd] = Problem MCE 567 Assignment, Problem. Solution define the parameters t = linspace(,48,48); generate random process trials x = cos(t)+.*randn(size(t)); mn=[]; sd=[]; sz = [ ]; for i = :, s = sz(i); for j = :, mun(j) = mean(x((j-)*s+:j*s)); end mn = [mn mean(mun)]; sd = [sd std(mun)]; subplot(4,3,i),hist(mun,5) xlabel( ),ylabel([ PSD (N = numstr(s) ) ]) end print -dpdf Problem a figure( PaperType, <custom>, PaperSize,[ 3.], PaperPosition,[. 3]); subplot(,,), semilogx(sz,mn, o- ) xlabel( data set size (N) ),ylabel( mean of the means ) subplot(,,), loglog(sz,sd, o- ) xlabel( data set size (N) ),ylabel( standard deviation in the mean ) print -dpdf Problem b The first figure here shows the clear trend of approaching a normal probability distribution as the number of points is increased in the data set. Also the mean of means converges to zero as data set size is increased and the standard deviation decreases according to the power law predicted by the CLT.

13 mean of the means standard deviation in the mean PSD (N = 5) PSD (N = 4) PSD (N = 48) PSD (N = 64) PSD (N = 8) PSD (N = 56) PSD (N = 8) PSD (N = 6) PSD (N = 3) PSD (N = ) PSD (N = ) PSD (N = 4) data set size (N) data set size (N) Figure.3: Problem. Plots