hw6-sol

hw6-sol - Homework 6 Problem 33. Excitation of the system...

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Unformatted text preview: Homework 6 Problem 33. Excitation of the system and its response are given () ( ) () ( ( ) )( ) ( ( ) )( ) We know that response is equal to the convolution of excitation and impulse response () ( ) ( ) or in Fourier space ( ) ( ) ( ), where convolution in time domain is ‘equivalent’ to usual multiplication in frequency domain. We can calculate ( ) ( ). Then () ( ) ( ) and the impulse response ( ) will be the inverse Fourier transform of ( ). Step#1. ( ) () ( ( )) () * () ( ) (Formulas page 369-371) ( )) ( ( ) , here () () () () () ( + ( denotes Fourier transform. Then ( ) ) ( Problem 34b. ( ) ( ) ) Step#2. We need to find the inverse Fourier transform of pages 369-371 suggest ( ) ) () . Formulas on ( ), this is the impulse response of the system. ( ) ( ) . One can observe that ( ) ( ) and ( ) since period of each function is 4. However, let us go to Fourier space and check if we indeed get zero. () ( ( )) [ ∑ ( ( )[ )[ ∑[ ()[ ∑ So the magnitude and phase are constant zeros. Problem 34c. ( ) ( ) () ( ) First, we notice that ( ) ( ) and ( ) () () ( ( )) [ [ () [ () Clearly at the value of ( ) is infinite, one can observe this from the fact that the total area of the signal in the time-domain is infinite. Matlab code: clear all close all clc f=-5:0.1:5; % frequency range y=dirac(f)+1/2/i/pi./f.*(1+cos(2*pi*f)-i*sin(2*pi*f)); % Fourier image subplot(1,2,1) plot(f,abs(y),'-') title('|G(f)|') xlabel('f') ylabel('Magnitude of G(f)') subplot(1,2,2) plot(f,angle(y)) title('Phase of G(f)') xlabel('f') ylabel('Phase of G(f)') Problem 39a. () () () Problem 39b. () () ( According to this, () ( ), then from Formulas on page 369-371 Magnitude of the latter factor is one | () () () ( ). The maximum value of | | ( )| | | | ( )| | ( )| ), then from Formulas on page 369-371 ( )| is equal to the maximum value of | ( )| ( )|. Problem 39c. CTFT of ( ) has the value of at a frequency We can assume that signal ( ) is real-valued. Then from the definition of Fourier transform ( ) ∫ () ( ) (∫ ( ) () ) ( ) , here ( ) stands for conjugate. However, if we do not consider ( ) to be purely real, the answer cannot be given based only on the given data, more information is needed. Then we can express the function ( ) as Problem 47. () () ( ) and then ( ) ( Matlab code: clear all close all clc f=-5:0.1:5; y=4*sin(pi*4*f)./f/pi/4 -4*sin(pi*2*f)./f/pi/2; subplot(1,2,1) plot(f,abs(y),'-') title('|G(f)|') xlabel('f') ylabel('Magnitude of G(f)') subplot(1,2,2) plot(f,angle(y)) title('Phase of G(f)') xlabel('f') ) ( ). ylabel('Phase of G(f)') ( Problem 48. Then in Fourier space ( ) () () ) () () ( ( % defining function that computes rect(t) function y = rect(t) y=heaviside (t+0.5)-heaviside(t-0.5); end f1=linspace(-15,15); x1=1/2*rect(f1/20); plot(f1,abs(x1),'-o') title('|X1(f)|') xlabel('f') ylabel('Magnitude of X1(f)') () () () ( ). Again, using Formulas on pages 369-371 Matlab code: clear all close all clc ) ) ( ) () ( ∫ ( ) ( ) ( ) ( ) ) Locations of the rect functions are centered at Matlab code: function y = rect(t) y=heaviside (t+0.5)-heaviside(t-0.5); end clear all close all clc f1=[-1020:1:-980]; % location of the first rectangular y1=5/4*(rect(f1/20-50)+rect(f1/20+50)); subplot(1,2,1) plot(f1,abs(y1),'-o') title('Magintude of Y(f)') xlabel('f') ylabel('|Y(f)|') f2=[980:1020]; % location of the second rectangular y2=5/4*(rect(f2/20-50)+rect(f2/20+50)); subplot(1,2,2) with height 1.25. plot(f2,abs(y2),'-o') title('Magnitude of Y(f)') xlabel('f') In simple words, mixer is producing the signal with frequencies equal to the sum and difference of the corresponding frequencies of 2 input signals. Problem 49. ( ) () ( ) We want to reject 60Hz and all its harmonics. Let us look at the Fourier transform of the impulse response () ( ( )) [ We need ( ) words ( () )[ This is possible when [ where n is integer. Letting [ or in other we will exclude all desired harmonics. ...
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