subplot(3,1,1);stem(n,x);axis([-10 20 0 2]);title('Input sequence'); ylabel('x[n]');
subplot(3,1,2);stem(n,h);axis([-10 20 0 6]);title('Inpulse Response'); ylabel('h[n+5]'); %output response [y2,ny]=conv_m(x,n,h,n);
subplot(3,1,3);stem(ny,y2);title('Output Sequence');xlabel('n'); ylabel('y_2[n]'); stepseq.m的源程序如下：
function [x,n]=stepseq(n0,n1,n2); if nargin ~=3
disp('Usage:Y=stepseq(n0,n1,n2)'); elseif((n0
error('arguments must satisfy n1<=n0<=2') end n=[n1:n2]; x=[(n-n0)>=0];
conv_m.m的源程序如下：
function[y,cy]=conv_m(x,nx,h,nh) if nargin~=4
disp('Usage:Y=conv_m(x,nx,h,nh)'); return; end;
nyb=nx(1)+nh(1);
nye=nx(length(x))+nh(length(x)); ny=[nyb:nye]; y=conv(x,h);
3.设h[n]?(0.9)nu[n],输入x[n]?u[n]?u[n?10]，求系统输出y[n]?x[n]*h[n]:
n=-5:50;
u1=stepseq(0,-5,50);u2=stepseq(10,-5,50);%u1=u[n];u2=u[n-10] %input x[n] x=u1-u2;
%impluse response h[n] h=((0.9).^n).*u1;
subplot(3,1,1);stem(n,x);axis([-5 20 0 2]);title('Input Sequence'); ylabel('x[n]');
subplot(3,1,2);stem(n,h);axis([-5 20 0 6]);title('Inpulse Response'); ylabel('h[n]'); %output response [y,ny]=conv_m(x,n,h,n);
subplot(3,1,3);stem(ny,y);title('Output Sequence');xlabel('n'); ylabel('y[n]');
stepseq.m的源程序如下：
function [x,n]=stepseq(n0,n1,n2); if nargin ~=3
disp('Usage:Y=stepseq(n0,n1,n2)'); elseif((n0
error('arguments must satisfy n1<=n0<=2') end n=[n1:n2]; x=[(n-n0)>=0];
conv_m.m的源程序如下：
function[y,cy]=conv_m(x,nx,h,nh) if nargin~=4
disp('Usage:Y=conv_m(x,nx,h,nh)'); return; end;
nyb=nx(1)+nh(1);
nye=nx(length(x))+nh(length(x)); ny=[nyb:nye]; y=conv(x,h);
思考题：在此问题中h[n]为无限长序列，在计算卷积和时应如何处理？ 解：取尽可能多的包含能量分布集中的具有代表性的点
4某离散时间线性时不变系统的差分方程为：y[n]?0.8[n?1]?x[n] a.计算n?[?10:50]时的系统冲激响应h[n] b.计算n?[?10:50]时的系统阶跃响应s[n]
a=[1,-0.8];b=1;n=[-10:50]; x=impseq(0,-10,50);
h=filter(b,a,x);%MATLAB固有函数filter表示系统方程ay=bx；x是冲击信号，则h就是冲击响应 subplot(2,1,1);stem(n,h);
title('Inpulse Response');xlabel('n');ylabel('h[n]'); x=stepseq(0,-10,50);
s=filter(b,a,x); %x是阶跃信号，则s就是阶跃响应 subplot(2,1,2);stem(n,s);
title('Step Response');xlabel('n');ylabel('s[n]');
impseq.m的源程序如下：
function[x,n]=impseq(n0,n1,n2)
%Generates x[n]=delta(n-n0)；n1<=n<=n2,n1<=n0<=n2 %[x,n]=impseq(n0,n1,n2)＇)； if nargin~=3
disp('Usage:Y=impseq(n0,n1,n2)'); return;
elseif((n0
error('arguments must satisfy n1<=n0<=n2') end
n=[n1:n2]; x=[(n-n0)==0];
stepseq.m的源程序如下：
function [x,n]=stepseq(n0,n1,n2); if nargin ~=3
disp('Usage:Y=stepseq(n0,n1,n2)'); elseif((n0
error('arguments must satisfy n1<=n0<=2') end n=[n1:n2]; x=[(n-n0)>=0];