西安电子科技大学-数字信号处理-试卷A答案

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1、真诚为您提供优质参考资料，若有不当之处，请指正。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

2、(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 =

3、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;xlab

4、el('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、 . 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 discontinuou

6、s 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(

7、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

8、+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 conv

9、olution 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),

10、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_f

11、ft7=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

12、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 atte

13、nuation 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-wp

14、2);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 =

15、 -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

16、','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',&#

17、39;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',

18、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','X

19、TickLabels','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 7Firstly, we use the given specifications of to design an analog lowpass IIR filter. Secondly, we change the analog lowpass IIR filter into the analog highpass IIR filte. Thirdly, we change the analog highpass IIR filter into the digital highpass IIR filte. 5 points 6 / 6