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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. x) \! Y- c9 A$ s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：
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& a( Y7 z* o, Q; Zfunction [y,n] = sigadd(x1,n1,x2,n2)
% implements y(n) = x1(n) + x2(n)
3 Y& E& y/ _7 l  G9 X* o3 h% [y,n] = sigadd(x1,n1,x2,n2)
- j$ t+ P" k! v5 f/ }%——————————————————————————————
% y = sum sequence over n, which includes n1 and n2
% x1 = first sequence over n1
% x2 = second sequence over n2( n2 can be different from n1)
%
n = min( min(n1), min(n2) ):max( max(n1), max(n2) ); %duration of y(n)
9 v, C' W! ?) d+ Iy1 = zeros(1,length(n)); y2 = y1; %initialization
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" R3 f  a3 S" r2 N! Sy = y1 + y2;

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直接给出MATLAB脚本：

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clear
$ T6 Z7 o, }- W5 ?- k/ Zclose all

  Z& {6 v& U. \0 q% help stepseq
%   generate x(n) = u(n - n0); n1 <= n <= n2
%  _____________________________________________
%  [x,n] = stepseq(n0, n1, n2);
! [& s) C# _: ]' }9 R* g, L$ e5 S[u1,n1] = stepseq(0,-5,45);
[u2,n2] = stepseq(10,-5,45);

% generate signal x(n)
7 |3 r$ R/ {% x[x,n] = sigadd(u1,n1,-u2,n2);

4 @, N. H9 b. ~- N" G% generate signal h(n)
) y/ W6 C# W3 g, g# Lm = -5:45;
6 `) D; G8 N) W  _h = ( (0.9).^m ).* u1;


& U/ f# K0 n5 F& w$ P3 l+ S1 n% the convolution of x(h) and h(n)
$ g! X7 \; B& X* V# ]# @y = conv(x,h);
: x, L! P6 }4 B- m9 x) S9 {% ensure the index
+ R+ L. K: B: g2 J5 V$ fnyb = n(1)+ m(1);
& e+ M) a2 D5 v* J$ ]nye = n(length(x)) +n(length(h));
ny = nyb:nye;


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  _" W$ }- f% j" Gstem(n,x);
( o0 L+ [7 ?9 }- ititle('x(n)');
" z9 y* _) F  o' Q1 S  zxlabel('n')
# w3 r: |( b2 R6 a/ o( k. E9 b
subplot(3,1,2);
# _9 H( d& u9 Y; E# V$ Fstem(m,h);
# G' `' e3 l0 e2 E9 c0 }title('h(n)');
xlabel('n')
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subplot(3,1,3);
stem(ny,y);
title('the conv of x(n) and h(n)');
xlabel('n')
xlim([-5,45]);
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Heaven_1

subplot这个函数经常使用
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