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HW12sol_FA08 - UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN...

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UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN Department of Electrical and Computer Engineering ECE 410 Digital Signal Processing Homework 12 Solutions Wednesday, November 12, 2008 Prof. Bresler, Prof. Jones Due by November 19, 2008 Problem 1 (20 points) Design a length-7, symmetric ramp FIR filter h [ n ] , n = 0 , 1 , . . . , 6 with desired frequency re- sponse D ( ω ) = | ω | (before the linear phase shift) by hand. Use the window design method with a simple truncation ( i.e. , rectangular/boxcar) window and give the filter coefficients as your answer. Using Matlab, plot separately the magnitude and phase of H ( ω ), using at least 1000 points between - π and π . Also plot the ideal desired magnitude response | D ( ω ) | on the same plot as the actual response. Solution: Introducing the required linear phase-shift, the desired frequency response is G d ( ω ) = | ω | e - j 3 ω and the ideal (IIR) filter coefficients are its inverse DTFT, i.e. , g [ n ] = 1 2 π Z π - π | ω | e - j 3 ω e jωn = 1 2 π - Z 0 - π ωe j ( n - 3) ω + Z 0 - π ωe j ( n - 3) ω = 1 2 π •Z π 0 ωe - j ( n - 3) ω + Z 0 - π ωe j ( n - 3) ω = 1 π Z π 0 ω cos( n - 3) ω dω = ω π ( n - 3) sin( n - 3) ω fl fl fl π 0 + 1 π ( n - 3) 2 cos( n - 3) ω fl fl fl π 0 , n 6 = 3 π 2 , n = 3 = ( 1 π ( n - 3) 2 [cos( n - 3) π - 1] , n 6 = 3 π 2 , n = 3 = ( 1 π ( n - 3) 2 £ ( - 1) n - 3 - 1 / , n 6 = 3 π 2 , n = 3 With a length 7 rectangular window, h [ n ] = 1 π ( n - 3) 2 £ ( - 1) n - 3 - 1 / , 0 n 6 , n 6 = 3 π 2 , n = 3 0 , otherwise 1

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The magnitude and phase response for the ideal and FIR filter are computed in Matlab using the following code and shown in figure 1. % Problem 1 N = 7; M = (N-1)/2; n = 0:(N-1); h = zeros(1,N); fftlen = 1024; w =[-fftlen/2: fftlen/2-1]/fftlen*2*pi; % Filter coefficients for G_d(w) = |w|, using rectangular window h = 1./(pi*(n-3).^2).*( (-1).^(n-3) -1); h(4) = pi/2; % Estimated filter response H = fftshift(fft(h,fftlen)); mag = abs(H); ang = angle(H); % Ideal (desired) filter response mag_g = abs(w); ang_g = angle(exp(-j*3*w)); figure(1); plot(w, mag, w, mag_g,’r:’, ’LineWidth’, 2); title(’Magnitude Responses’); xlabel(’\omega (radian)’); legend(’|H_d(\omega)|’, ’|G_d(\omega)|’,-1); xlim([-pi pi]); ylim([0 pi]); grid on print -depsc P1_mag figure(2); plot(w, ang, w, ang_g,’r:’, ’LineWidth’, 2); title(’Phase Responses’); xlabel(’\omega (radian)’); legend(’\angle H_d(\omega)’, ’\angle G_d(\omega)’, -1); xlim([-pi pi]); ylim([-pi pi]); grid on print -depsc P1_phase 2
Problem 2 (20 points) Use the frequency sampling method respectively to design a length-101 type-1 GLP FIR high- pass filter having cutoff frequency ω c = 3 π/ 4 and generalized linear phase. (a) Find an expression for the filter coefficients { h n } 100 n =0 . (b) Use Matlab to evaluate H d ( 512 ), for n = - 512 , - 511 , . . . , 511. Plot the magnitude and phase responses, respectively, and analyze the plots to see if they are what you expected. Solution: (a) Since the filter is high-pass and odd-length, it needs to be a type-1 GLP filter with even symmetry. The desired filter response G d ( ω ) for ω [0 , 2 π ] is therefore given by: G d ( ω ) = ( e - (101 - 1) 2 , ω [ 3 π 4 , 5 π 4 ] 0 , else In order to use the frequency sampling method, we need to find inverse DFT of the sequence H [ m ] = G d ( 2 πm 101 ) , m = 0 , 1 , . . . , 100, i.e.

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HW12sol_FA08 - UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN...

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