Assignment 2 Solutions Fourier Series

Transcription

1 Assignment 2 Solutions Fourier Series ECE 223 Signals and Systems II Version.2 Spring 26. DT Fourier Series. Consider the following discrete-time periodic signal n = +7l n =+7l x[n] = n =+7l Otherwise where l is any integer. a. What is the fundamental period of this signal. N =7 b. Does the signal have even or odd symmetry? Yes, it is even. c. Plot four fundamental periods of the signal Time Index (n)

2 d. Plot the real and imaginary parts of the Fourier series coefficients X[k]. Real X[k] 5 5 Imaginary X[k] 5 5 Frequency Index (k) e. Plot the magnitude and phase of the Fourier series coefficients X[k]. X[k] X[k] (degrees) 5 5 Frequency Index (k) f. How are real, negative coefficients represented in polar (magnitude & phase) form? The magnitude is equal to the absolute value of the real coefficient, X[k] = Re{X[k]}. The phase is if the coefficients are positive and ±8 if the coefficients are negative, X[k] =±8 sign(x[k]). g. What type of symmetry do the coefficients have? Complex conjugate symmetry: X[ k] =X [k]. Since the coefficients are real in this case, they also have even symmetry: X[ k] =X[k]. h. What is the fundamental period of the coefficients? K =7, X[k +7]=X[k]. 2

3 i. Plot the real and imaginary parts of the DT sinusoids that comprise the discrete-time signal. Re. k= Im. k= Re. k= 2 Im. k= Re. k= Im. k= Re. k= Im. k= Re. k= Im. k= Re. k=2 Im. k= Re. k=3 Im. k= j. Plot the real and imaginary parts of the sum of the DT sinusoids that comprise the discrete-time signal. 3

4 Real Part Imaginary Part Time Index (n) k. How does the sum of the complex sinusoids compare to the original signal? They are identical. function [] = DTFourierSeries N = 7; % Fundamental period n = -2*N:2*N; % Time indices x = -*(mod(n,n)==-+n) + *(mod(n,n)==) -*(mod(n,n)==); % Signal k = -2*N:2*N; % Frequency indices Omega = 2*pi/N; % Fundamental frequency for c=:length(k) X(c) = sum(x(:n).*exp(-j*k(c)*omega.*n(:n)))/n; end % Calculate the DT Fourier series coefficients % Plot of the Signal FigureSet(,5); h = stem(n,x, b ); xlim([n()-.5 n(end)+.5]); ylim([-.5.5]); xlabel( Time Index (n) ); print( DTFourierSeries-Signal, -depsc ); % Plot of the Fourier Series Coefficients (Rectangular form) FigureSet(,5); subplot(2,,); h = stem(k,real(x), r ); xlim([k()-.5 k(end)+.5]); ylim(.5*max(abs(x))*[- ]); h = get(gca, YLabel ); ylabel( Real $X[k]$ ); 4

5 subplot(2,,2); h = stem(k,imag(x), r ); xlim([k()-.5 k(end)+.5]); ylim(.5*max(abs(x))*[- ]); h = get(gca, YLabel ); h = get(gca, XLabel ); ylabel( Imaginary $X[k]$ ); xlabel( Frequency Index ($k$) ); print( DTFourierSeries-FourierSeriesRectangular, -depsc ); % Plot of the Fourier Series Coefficients (Rectangular form) FigureSet(,5); subplot(2,,); h = stem(k,abs(x), r ); xlim([k()-.5 k(end)+.5]); ylim([.5*max(abs(x))]); h = get(gca, YLabel ); ylabel( $ X[k] $ ); subplot(2,,2); h = stem(k,angle(x)*8/pi, r ); xlim([k()-.5 k(end)+.5]); ylim(9*[- ]); h = get(gca, YLabel ); h = get(gca, XLabel ); ylabel( $\angle X[k]$ (degrees) ); xlabel( Frequency Index ($k$) ); print( DTFourierSeries-FourierSeriesPolar, -depsc ); % Plot of the Fourier Series Components FigureSet(,5,9); cplots = ; % Plot counter s = zeros(size(n)); % Memory allocation for the sum for c=-(n-)/2:(n-)/2 ik = find(k==c); c = X(ik)*exp(j*c*Omega*n); % Component s = s + c; subplot(n,2,cplots); h = stem(n,real(c), b ); cplots = cplots + ; xlim([n()-.5 n(end)+.5]); ylim(.5*max(abs(x))*[- ]); ylabel(sprintf( Re. k=%d,c)); 5

6 subplot(n,2,cplots); h = stem(n,imag(c), b ); cplots = cplots + ; xlim([n()-.5 n(end)+.5]); ylim(.5*max(abs(x))*[- ]); ylabel(sprintf( Im. k=%d,c)); end print( DTFourierSeries-FourierSeriesComponents, -depsc ); % Sum of Components FigureSet(,5); subplot(2,,); h = stem(n,real(s), b ); xlim([n()-.5 n(end)+.5]); ylim([-.5.5]); ylabel( Real Part ); subplot(2,,2); h = stem(n,imag(s), b ); xlim([n()-.5 n(end)+.5]); ylim([-.5.5]); xlabel( Time Index (n) ); ylabel( Imaginary Part ); print( DTFourierSeries-SumOfComponents, -depsc ); 2. CT Fourier Series. Consider the following continuous-time periodic signal { t +7l t +7l x(t) = Otherwise where l is any integer. a. What is the fundamental period of this signal. T =7 b. Does the signal have even or odd symmetry? Yes, the signal is odd. c. Plot four fundamental periods of the signal. 6

7 Time (s) d. Solve for the Fourier series coefficients. Simplify your expression as much as possible. Hint: te at dt = a 2 e at (at ) + C. X[k] = T T x(t)e jkωt dt = te jkωt dt T = T ( jkω) 2 e jkωt ( jkωt ) [ = e jkω ( jkω ) e jkω (jkω ) ] T (jkω) 2 [ ( = jkω e jkω +e jkω) + ( e jkω e jkω)] T (jkω) 2 = [ j2kω cos(kω)+j2sin(kω)] T (kω) 2 j2kω cos(kω) j2sin(kω) = T (kω) 2 = = j2k 2π T cos(k 2π j4πk cos(k 2π T T ) j2sin(k 2π T ) T (k 2π T )2 ) j2t sin(k 2π T ) (2πk) 2 e. Plot the real and imaginary parts of the Fourier series coefficients X[k]. 7

8 Real X[k] Imaginary X[k] Frequency Index (k) f. Plot the magnitude and phase of the Fourier series coefficients X[k].. X[k] X[k] (degrees) Frequency Index (k) g. What type of symmetry do the coefficients have? Complex conjugate symmetry: X[ k] =X [k]. In this case the coefficients are imaginary, so they also have odd symmetry: X[ k] = X[k]. h. What is the fundamental period of the coefficients? The coefficients are not periodic and do not have a fundamental period. i. Plot the real and imaginary parts of the CT complex sinusoids that comprise the continuous-time signal for k = 5 tok =5. 8

9 Re. k=5 Re. k=4 Re. k=3 Re. k=2 Re. k= Re. k= Re. k= Re. k= 2 Re. k= 3 Re. k= 4 Re. k= Im. k=5 Im. k=4 Im. k=3 Im. k=2 Im. k= Im. k= Im. k= Im. k= 2 Im. k= 3 Im. k= 4 Im. k= j. Plot the partial sums of the analysis equation, K ˆx K (t) = X[k]e jkωt k= K of the CT complex sinusoids that comprise the continuous-time signal for K =, 2, 3, 4, 5, 25, 5,. 9

10 Re. K= Im. K= Re. K=2 Im. K= Re. K=3 Im. K= Re. K=4 Im. K= Re. K=5 Im. K= Re. K=25 Im. K= Re. K=5 Im. K= Re. K= 5 5 Im. K= 5 5 k. Plot the mean square error of the partial sums for K =,...,.

11 Mean Square Error Partial Sum Limit (K) l. How do the partial sums of the complex sinusoids compare to the original signal? They are approximations of the original signal. The approximations with a greater number of terms are more accurate. m. In the limit, does the following equality hold? lim ˆx K(t) =x(t) K No. This signal contains a discontinuity. At the times of the discontinuities the signal and the Fourier series approximations will not be equal, in general. However, the mean square error will approach zero as K. function [] = CTFourierSeries T = 7; % Fundamental period omega = 2*pi/T; % Fundamental frequency (rad/sec) t = linspace(-2*t,2*t,); % Time indices tmod = mod(t,t); x = -(T-tmod).*(-+T<=tmod & tmod<=+t) + tmod.*(tmod<=); % Signal kn = -:-; Xn = (j*4*pi*kn.*cos(kn*2*pi/t)-j*2*t*sin(kn*2*pi/t))./(2*pi*kn).^2; kp = :; Xp = (j*4*pi*kp.*cos(kp*2*pi/t)-j*2*t*sin(kp*2*pi/t))./(2*pi*kp).^2; k = [kn kp]; X = [Xn Xp]; % Plot of the Signal FigureSet(,5); h = plot(t,x, b ); xlim([t() t(end)]); ylim([-.5.5]); xlabel( Time (s) );

12 print( CTFourierSeries-Signal, -depsc ); % Plot of the Fourier Series Coefficients (Rectangular form) FigureSet(,5); subplot(2,,); h = stem(k,real(x), r ); xlim([k()-.5 k(end)+.5]); ylim(.5*max(abs(x))*[- ]); h = get(gca, YLabel ); ylabel( Real $X[k]$ ); subplot(2,,2); h = stem(k,imag(x), r ); xlim([k()-.5 k(end)+.5]); ylim(.5*max(abs(x))*[- ]); h = get(gca, YLabel ); h = get(gca, XLabel ); ylabel( Imaginary $X[k]$ ); xlabel( Frequency Index ($k$) ); print( CTFourierSeries-FourierSeriesRectangular, -depsc ); % Plot of the Fourier Series Coefficients (Rectangular form) FigureSet(,5); subplot(2,,); h = stem(k,abs(x), r ); xlim([k()-.5 k(end)+.5]); ylim([.5*max(abs(x))]); h = get(gca, YLabel ); ylabel( $ X[k] $ ); subplot(2,,2); h = stem(k,angle(x)*8/pi, r ); xlim([k()-.5 k(end)+.5]); ylim(9*[- ]); h = get(gca, YLabel ); h = get(gca, XLabel ); ylabel( $\angle X[k]$ (degrees) ); xlabel( Frequency Index ($k$) ); print( CTFourierSeries-FourierSeriesPolar, -depsc ); 2

13 % Plot of the Fourier Series Components FigureSet(,5,9); cplots = ; % Plot counter s = zeros(size(t)); % Memory allocation for the sum for c=-5:5, ik = find(k==c); c = X(ik)*exp(j*c*omega*t); % Component s = s + c; subplot(5*2+,2,cplots); h = plot(t,real(c), b ); cplots = cplots + ; xlim([t() t(end)]); ylim(.5*max(abs(x))*[- ]); ylabel(sprintf( Re. k=%d,c)); subplot(5*2+,2,cplots); h = plot(t,imag(c), b ); cplots = cplots + ; xlim([t() t(end)]); ylim(.5*max(abs(x))*[- ]); ylabel(sprintf( Im. k=%d,c)); end print( CTFourierSeries-FourierSeriesComponents, -depsc ); % Plot of the Partial Sums FigureSet(,5,9); K = [,2,3,4,5,25,5,]; cplots = ; % Plot counter s = zeros(size(t)); % Memory allocation for the sum mse = zeros(max(k),); for c=:max(k), ip = find(k== c); % Index of positive coefficient in = find(k==-c); % Index of negative coefficient c = X(ip)*exp(j*c*omega*t) + X(in)*exp(-j*c*omega*t); % Component s = s + c; % Partial sum, k=-c to c mse(c+) = mean(abs(x-s).^2); if any(c==k), subplot(length(k),2,cplots); h = plot(t,real(s), b ); cplots = cplots + ; xlim([t() t(end)]); ylim([-.5.5]); ylabel(sprintf( Re. K=%d,c)); subplot(length(k),2,cplots); h = plot(t,imag(s), b ); cplots = cplots + ; xlim([t() t(end)]); ylim([-.5.5]); ylabel(sprintf( Im. K=%d,c)); end end print( CTFourierSeries-PartialSums, -depsc ); % Plot of the Mean Square Error of the Partial Sums 3

14 FigureSet(,5); h = stem(:max(k),mse, k ); xlim([-.5 max(k)+.5]); ylim([ max(mse)*.5]); xlabel( Partial Sum Limit (K) ); ylabel( Mean Square Error ); print( CTFourierSeries-MSE, -depsc ); 4