两个序列的卷积和运算的MATLAB实现（1）

mytomorrow

设矩形脉冲 是脉冲响应 的LTI系统的输入，求输出 y(n).
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下面的脚本中用到了一个自定义的函数，也就是两个信号相加的函数：

! j- m7 j+ w7 C0 n+ Xfunction [y,n] = sigadd(x1,n1,x2,n2)
0 q* m7 {2 B) A$ P1 m. [# E% implements y(n) = x1(n) + x2(n)
% [y,n] = sigadd(x1,n1,x2,n2)
%——————————————————————————————
$ M* ?4 }$ j* d; c/ a6 m# `" L% y = sum sequence over n, which includes n1 and n2
% x1 = first sequence over n1
" k, A/ e3 M+ Q5 H" @) V" J% x2 = second sequence over n2( n2 can be different from n1)
1 ~( C* d' u, R; N" h5 j% {3 e$ a%
6 S* J+ T( M3 f$ ~n = min( min(n1), min(n2) ):max( max(n1), max(n2) ); %duration of y(n)
1 p; v3 E. A$ [, m4 \y1 = zeros(1,length(n)); y2 = y1; %initialization
" \4 g9 e/ F/ `4 j3 J- ~3 [y1( find( ( n >= min(n1) )&( n <= max(n1) ) == 1  )  ) = x1; %x1 with duration of y1
y2( find( ( n >= min(n2) )&( n <= max(n2) ) == 1  )  ) = x1; %x2 with duration of y2
y = y1 + y2;

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( X) d) X! j* q: d$ ~% R7 g直接给出MATLAB脚本：

clc
clear
4 m  Y: I) r5 A+ M, g* Z! Zclose all
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% help stepseq
%   generate x(n) = u(n - n0); n1 <= n <= n2
%  _____________________________________________
%  [x,n] = stepseq(n0, n1, n2);
) A, i$ {6 J3 v5 z- z& A& _/ y8 l[u1,n1] = stepseq(0,-5,45);
( R2 N, x9 G# M- p[u2,n2] = stepseq(10,-5,45);
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% generate signal x(n)
[x,n] = sigadd(u1,n1,-u2,n2);

) T7 i4 ^& O/ C  }) k3 I4 l% generate signal h(n)
: n9 o3 o7 I; W9 H: ?9 Km = -5:45;
3 F' R0 j8 d: s% ^; Th = ( (0.9).^m ).* u1;

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% the convolution of x(h) and h(n)
; H3 I) Y$ b; F! Zy = conv(x,h);
% ensure the index
nyb = n(1)+ m(1);
nye = n(length(x)) +n(length(h));
ny = nyb:nye;
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subplot(3,1,1);
stem(n,x);
/ E6 P6 H! U9 U; ~  {title('x(n)');
xlabel('n')
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subplot(3,1,2);
stem(m,h);
title('h(n)');
) k3 p6 _8 Q6 S1 l! Bxlabel('n')
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6 o9 K& \6 X  X5 Y  x% f  p" Q- Ysubplot(3,1,3);
stem(ny,y);
title('the conv of x(n) and h(n)');
xlabel('n')
& c* U' F- c/ C: x0 }! mxlim([-5,45]);

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Heaven_1

subplot这个函数经常使用
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