# A9_SOL - Signals and Systems Fall 2013 Homework...

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Signals and Systems Fall 2013 Homework Assignment #9 1. 2 2 ) 95 . 0 ( ) 6 / cos( ) 95 . 0 ( 2 ) ( z z z z H (a) % Assignment #9 % Problem 1(a) w=0:pi/120:pi; %Create 120 points to evaluate H(e^jw) h=exp(j*w)./((exp(j*w)).^2 - 2*0.95*cos(pi/6)*(exp(j*w)) +0.95^2); subplot(2,1,1) plot(w,abs(h)) title( 'Magnitude' ) grid on subplot(2,1,2) plot(w,angle(h)) title( 'Phase' ) grid on (b) The corresponding difference equation is: 2 2 2 0.95cos 1 0.95 1 6 y k y k y k x k % Assignment #9 % Problem 1(b) N=50; k=[0:N]; a=pi/12; % for x2: a= pi/6, x3: a=pi/3 x=cos(a.*k); y=0*k; for ii=1:N-1 y(ii+2)=2*0.95*cos(pi/6)*y(ii+1)-(0.95^2)*y(ii)+x(ii+1); end subplot(2,1,1) stem(k,x) xlabel( 'k' ) ylabel( 'x_1(k)' ) subplot(2,1,2) stem(k,y) xlabel( 'k' ) 0 0.5 1 1.5 2 2.5 3 3.5 0 10 20 30 Magnitude 0 0.5 1 1.5 2 2.5 3 3.5 -4 -3 -2 -1 0 Phase /3 /6 M 2 =20.49 M 1 = 5.158 M 3 =1.43 /12

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ylabel( 'y_1(k)' ) ) ( ) 12 / cos( ) ( 1 k u k k x s , ) ( ) 6 / cos( ) ( 2 k u k k x s , ) ( ) 3 / cos( ) ( 3 k u k k x s . 0 5 10 15 20 25 30 35 40 45 50 -1 -0.5 0 0.5 1 k x 3 (k) 0 5 10 15 20 25 30 35 40 45 50 -4 -2 0 2 4 k y 3 (k) (c) + Input ) ( ) 12 / cos( ) ( 1 k u k k x s , frequency-response at 12 T . The output magnitude is 1 5.158 M . + Input 2 ( ) cos( / 6) ( ) s x k k u k , frequency-response at 6 T . The output magnitude is 2 20.49 M . 0 5 10 15 20 25 30 35 40 45 50 -1 -0.5 0 0.5 1 k x 1 (k) 0 5 10 15 20 25 30 35 40 45 50 -10 -5 0 5 k y 1 (k) 0 5 10 15 20 25 30 35 40 45 50 -1 -0.5 0 0.5 1 k x 2 (k) 0 5 10 15 20 25 30 35 40 45 50 -20 -10 0 10 20 k y 2 (k)
+ Input 3 ( ) cos( / 3) ( ) s x k k u k , frequency-response at 3 T . The output magnitude is 3 1.43 M (Please also seethe plot of frequency response). 2. (a) x 1 ( k ) cos( k /6). In this case, we can choose H ( z ) to be:   2 2 2cos 1 6 z z H z A z , and A is chosen so that   1 1 H , i.e. 2 2 1 1 1 2cos 1 1 3.73 6 1 2cos 1 6 A A   and   2 6 2 2cos 1 3.73 z z H z z

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