西安电子科技大学-数字信号处理-试卷C答案.doc

Digital Signal Processing of 2005Answer to “Digital Signal Processing of 2005”Problem 1(a) even part: odd part: (b)(c) MATLAB Program n=-4:2;x=1 -2 4 6 -5 8 10;x11,n11=sigshift(x,n,2);x12,n12=sigshift(x,n,-1);x13,n13=sigfold(x,n);x13,n13=sigshift(x13,n13,-2);x12,n12=sigmult(x,n,x12,n12);y,n=sigadd(2*x11,n11,x12,n12);y,n=sigadd(y,n,-1*x13,n13)Problem 2(a)，is periodic in with period 2(b) MATLAB Program:clear; close all;n = 0:6; x = 4,2,1,0,1,2,4;w = 0:1:1000*pi/1000;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) (c) Because the given sequence x (n)=4,2,1,0,1,2,4 (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 . (d) ;(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. Problem 3(a) Zero:0 and 1;Pole:-0.6 and 1.5;(b)， (c) ROC : ， Problem 4(a) y(n)=50,44,34,52;(b) y(n)=5,16,34,52,45,28,0;(c) N=6;(d) MATLAB Program:Function y=circonv(x1,x2,N) If (length(x1)N) error(“N must not be smaller than the length of sequence”)elsex1=x1,zeros(1,N-length(x1);endif(length(x2)N) error(“N must not be smaller than the length of sequence”)elsex2=x2,zeros(1,N-length(x2); end y1=dft(x1,N).*dft(x2,N);y=idft(y,N);(e) DTFT is discrete in time domain, but continuous in frequency domain. The DFT is discrete both in time and frequency domain.The FFT is a very efficient method for calculating DFT. Problem 5(a) Direct form II uses the little delay and it can decrease the space of the compute.(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.(c) Block diagrams are shown as under:Problem 6(a) we use Hamming or Blackman window to design the bandpass filter because it can provide us attenuation exceed 60dB .(b) According to Blackman window :first, Determine transition width = ;second, Determine the type of the window according to ;third, Compute M using the formula ;fourth, Compute ideal LPF ;fifth, design the window needed, multiply point by point;sixth, determine (c) MATLAB Program:% Specifications about Blackman window:ws1 = 0.2*pi; % lower stopband edgewp1 = 0.3*pi; % lower passband edgewp2 = 0.6*pi; % upper passband edgews2 = 0.7*pi; % upper stopband edgeRp = 0.5; % passband rippleAs = 60; % 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);gridProblem 7Firstly, we use the given speci