西安电子科技大学-数字信号处理-试卷A答案.doc

Answer to “Digital Signal Processing”Problem 1 (a). x1(n)=3,7,-2,-1,5,-8, n1=-2:3 x2(n)=3,7,-2,-1,5,-8, n2=1:6 x3(n)=-8,5,-1,-2,7,3 , n3=-2:3 3 points(b) y(n)=14,9,-3,21,-11,-17,-5,-40,0, n=-2:6 3 points(c).MATLAB program:clear,close all;n=0:5; x=3,7,-2,-1,5,-8;x1,n1=sigshift(x,n,-2); x2,n2=sigshift(x,n,1); x3,n3=sigfold(x,n); x3,n3=sigshift(x3,n3,3); y1,yn1=sigadd(2*x1,n1,x2,n2);y1,yn1=sigadd(y1,yn1,-x3,n3); 3 pointsstem(yn1,y1);title(sequenceY(n) Problem 2 (a)is periodic in with period 2. 5 points(b) Matlab program:clear; close all;n = 0:6; x = 4,3,2,0,2,3,4;w = 0:1:100*pi/100;X = x*exp(-j*n*w); magX = abs(X); phaX = angle(X);% Magnitude Response Plotsubplot(2,1,1); plot(w/pi,magX);grid;xlabel(frequency in pi units); ylabel(|X|);title(Magnitude Response);% Phase response plotsubplot(2,1,2); plot(w/pi,phaX*180/pi);grid;xlabel(frequency in pi units); ylabel(Degrees);title(Phase Response); axis(0,1,-180,180) 5 points(c) Because the given sequence x (n)=4,3,2,0,2,3,4 (n=0,1,2,3,4,5,6) is symmetric about ,the phase response satisfied the condition so the phase response is a linear function in . 5 points(d) 5 points(e) The difference of amplitude and magnitude response:Firstly, the amplitude response is a real function, and it may be both positive and negative. The magnitude response is always positive. Secondly, the phase response associated with the magnitude response is a discontinuous function. While the associated with the amplitude is a continuous linear function. 5 pointsProblem 3 (a) Pole-zero plot MATLAB script: b=1,1,0;a=1,0.5,-0.24;zplane(b,a) Figure : Pole-zero plot in Problem A2Because all the poles and zero are all in the unit circle,so the system is stable. 5 points(b) Difference equation representation.:;After cross multiplying and inverse transforming; 5 points(c) impulse response sequence h(n) using partial fraction representation: 5 points(d) MATLAB verification: (1) b=1,1;a=1,0.5,-0.24;delta,n=impseq(0,0,9)x=filter(b,a,delta) (2) n=0:4;x=(-0.1818.*(-0.8).n+1.1818.*(0.3).n).*stepseq(0,0,9) 5 pointsProblem 4 (a) figure 4.1 figure 4.2 sequence x1(n) sequence x2 (n) The plots of x1(n) and x2(n) is shown in figure 4.1 and figure 4.2 4 points (b) figure 4.3 After calculate, we find circular convolution is equal when N is 7 or 8. 4 points(c)If circular convolution is equal to linear convolution ，the minimum N is m+n-1=7.The plots of circular convolution of x1(n) and x2(n) is shown in figure 4.3. 4 points(d) MATLAB program:clear,close all;n=0:7;m=0:6;l=0:3;x1=1,-1,1,-1;x2=3,1,1,3;figure,stem(l,x1),title(sequence x1),box off;figure,stem(l,x2),title(sequence x2),box off;%8-point circular convolutionx1_fft8=fft(x1,8);x2_fft8=fft(x2,8);y_fft8=x1_fft8.*x2_fft8;y8=real(ifft(y_fft8);figure,subplot(2,1,1),stem(n,y8),title(8-pointcircular convolution ),axis(0 7 -3 3),box off;%7-point circular convolutionx1_fft7=fft(x1,7);x2_fft7=fft(x2,7);y_fft7=x1_fft7.*x2_fft7;y7=real(ifft(y_fft7);subplot(2,1,2),stem(m,y7),title(7-point circular convolution ),axis(0 7 -3 3),box off; 4 points Problem 5 (a) Block diagrams are shown as under: 4 points 4 points(b)The advantage of the linear-phase form:1. For frequency-selective filters, linear-phase structure is generally desirable to have a phase-response that is a linear function of frequency. 2. This structure requires 50% fewer multiplications than the direct form. 2 points Problem 6 (a) we use Hamming window to design the bandpass filter because it can provide us attenuation exceed 50dB 5 points (b) MATLAB verification:% Specifications:ws1 = 0.3*pi; % lower stopband edgewp1 = 0.4*pi; % lower passband edgewp2 = 0.5*pi; % upper passband edgews2 = 0.6*pi; % upper stopband edgeRp = 0.5; % passband rippleAs = 50; % stopband attenuation%tr_width = min(wp1-ws1),(ws2-wp2);M = ceil(6.6*pi/tr_width); M = 2*floor(M/2)+1, % choose odd Mn = 0:M-1;w_ham = (hamming(M);wc1 = (ws1+wp1)/2; wc2 = (ws2+wp2)/2;hd = ideal_lp(wc2,M)-ideal_lp(wc1,M);h = hd .* w_ham;db,mag,pha,grd,w = freqz_m(h,1);delta_w = pi/500;Asd = floor(-max(db(1:floor(ws1/delta_w)+1), % Actual AttnRpd = -min(db(ceil(wp1/delta_w)+1:floor(wp2/delta_w)+1), % Actual passband ripple （5）% Filter Response Plotssubplot(2,2,1); stem(n,hd); title(Ideal Impulse Response: Bandpass);axis(-1,M,min(hd)-0.1,max(hd)+0.1); xlabel(n); ylabel(hd(n)set(gca,XTickMode,manual,XTick,0;M-1,fontsize,10)subplot(2,2,2); stem(n,w_ham); title(Hamming Window);axis(-1,M,-0.1,1.1); xlabel(n); ylabel(w_ham(n)set(gca,XTickMode,manual,XTick,0;M-1,fontsize,10)set(gca,YTickMode,manual,YTick,0;1,fontsize,10)subplot(2,2,3); stem(n,h); title(Actual Impulse Response: Bandpass);axis(-1,M,min(hd)-0.1,max(hd)+0.1); xlabel(n); ylabel(h(n)set(gca,XTickMode,manual,XTick,0;M-1,fontsize,10)subplot(2,2,4); plot(w/pi,db); title(Magnitude Response in dB);axis(0,1,-As-30,5); xlabel(frequency in pi units); ylabel(Decibels)set(gca,XTickMode,manual,XTick,0;0.3;0.4;0.5;0.6;1)set(gca,XTickLabelMode,manual,XTickLabels,0;0.3;0.4;0.5;0.6;1,.fontsize,10)set(gca,TickMode,manual,YTick,-50;0)set(gca,YTickLabelMode,manual,YTickLabels,-50;0);grid 10 pointsProblem