西安电子科技大学-数字信号处理-试卷C答案.doc

Answer to “Digital Signal Processing of 2005” Problem 1 even part: odd part: 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 Plot subplot(2,1,1); plot(w/pi,magX);grid; xlabel(frequency in pi units); ylabel(|X|); title(Magnitude Response); % Phase response plot subplot(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)， ROC : ， Problem 4 y(n)={50,44,34,52}; y(n)={5,16,34,52,45,28,0}; N=6; MATLAB Program: Function y=circonv(x1,x2,N) If (length(x1)N) error(“N must not be smaller than the length of sequence”) else x1=[x1,zeros(1,N-length(x1))]; end if(length(x2)N) error(“N must not be smaller than the length of sequence”) else x2=[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