hw8-sol

hw8-sol - Homework 8 Solutions Chapter 14 25. Over a time...

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Homework 8 Solutions Chapter 14 25. Over a time range of 0 400 tm s << , signal ( )3 c o s ( 2 0)2 s i n ( 3 0) x tt t π = is shown in following figures (dashed line), together with sampled by different sampling intervals: 1/120s, 1/60s, 1/30s, 1/15s. 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 -5 0 5 t x(t) T s =1/120s 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 -5 0 5 t T s =1/60s 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 -5 0 5 t T s =1/30s
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0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 -5 0 5 t x(t) T s =1/15s From four figures shown above, this signal can be reconstructed when sampled by 1/120 s Ts = , 1/ 60 s = and cannot be reconstructed for 1/ 30 s = , 1/15 s = . Analytically, we can determine if the signal can be reconstructed by finding its Nyquist rate. ( )3 c o s ( 2 0)2 s i n ( 3 0) x tt t π =− 3 [ ] [ ( 10) ( 10)] [ ( 15) ( 15)] 2 Xf f f j f f δδ + + + + So, 15Hz m f = , 23 0 H z Nyq m ff == . In order to reconstruct the signal, sampling frequency should satisfy: 30Hz sN y q >= 1/30 s < CODE: clear all; t = 0:1e-3:400e-3; y0 = 3*cos(20*pi*t)-2*sin(30*pi*t); figure(1), subplot(2,1,1),plot(t,y0,'--'); xlabel('t');ylabel('x(t)'),hold on subplot(2,1,2),plot(t,y0,'--'); xlabel('t');ylabel('x(t)'),hold on figure(2), subplot(2,1,1),plot(t,y0,'--'); xlabel('t');ylabel('x(t)'),hold on subplot(2,1,2),plot(t,y0,'--'); xlabel('t');ylabel('x(t)'),hold on t1 = 0:1/120:400e-3; % (a) Ts = 1/120s; y1 = 3*cos(20*pi*t1)-2*sin(30*pi*t1); figure(1) subplot(2,1,1),stem(t1,y1,'fill'); title('T_s=1/120s'),hold off t2 = 0:1/60:400e-3; % (b) Ts = 1/60s;
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y2 = 3*cos(20*pi*t2)-2*sin(30*pi*t2); figure(1) subplot(2,1,2),stem(t2,y2,'fill'); title('T_s=1/60s'),hold off t3 = 0:1/30:400e-3; % (c) Ts = 1/30s; y3 = 3*cos(20*pi*t3)-2*sin(30*pi*t3); figure(2) subplot(2,1,1),stem(t3,y3,'fill'); title('T_s=1/30s'),hold off t4 = 0:1/15:400e-3; % (d) Ts = 1/15s; y4 = 3*cos(20*pi*t4)-2*sin(30*pi*t4); figure(2) subplot(2,1,2),stem(t4,y4,'fill'); title('T_s=1/15s'),hold off 32. (a) 4 ( ) 15rect(300 )cos(10 ) x tt t π = 15 1 [ ] sinc( )* [ ( 5000) ( 5000)] 300 300 2 f Xf f f δδ =− + + 1 5000 5000 sinc sinc 40 300 300 ff ⎡− + ⎛⎞ =+ ⎜⎟ ⎢⎥ ⎝⎠ ⎣⎦ -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 x 10 4 -0.01 0 0.01 0.02 0.03 f X[f] From the frequency domain analysis, we will see this signal is not band limited, meaning m f
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This note was uploaded on 01/15/2012 for the course MAE 143A taught by Professor Miroslavkrstic during the Summer '08 term at UCSD.

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hw8-sol - Homework 8 Solutions Chapter 14 25. Over a time...

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