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).

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：

function [y,n] = sigadd(x1,n1,x2,n2)
7 o; {  i4 A4 w; s) x; Q% implements y(n) = x1(n) + x2(n)
: D8 u) p- o& S0 r) g& c% [y,n] = sigadd(x1,n1,x2,n2)
8 `0 L) f( l9 U& {* \8 B. G8 A%——————————————————————————————
3 b% P9 W$ G0 Q4 B' a# R) Z% y = sum sequence over n, which includes n1 and n2
% x1 = first sequence over n1
9 t) b+ ?3 W) o5 S. V8 \$ L3 N% 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)
2 {5 y. B; Q8 t! {5 q& Yy1 = zeros(1,length(n)); y2 = y1; %initialization
! W2 Z% e- O& C! r$ _- ]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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直接给出MATLAB脚本：

# _' {! o! I7 a( xclc
clear
5 q1 s% u6 A  C% L8 \2 @& q6 Q% }close all

% help stepseq
%   generate x(n) = u(n - n0); n1 <= n <= n2
%  _____________________________________________
5 N2 T1 D+ H. r; P( \%  [x,n] = stepseq(n0, n1, n2);
[u1,n1] = stepseq(0,-5,45);
[u2,n2] = stepseq(10,-5,45);
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; W- m1 t. A+ u' G% ~2 ?# Z% generate signal x(n)
1 k: G+ B7 F3 z7 w[x,n] = sigadd(u1,n1,-u2,n2);

% generate signal h(n)
m = -5:45;
6 x7 A+ q4 `* K  Z# N$ eh = ( (0.9).^m ).* u1;


% the convolution of x(h) and h(n)
1 q6 _5 e8 ~( F2 ]1 Ky = conv(x,h);
9 f/ ^# q. Z" [3 J/ m4 N% ensure the index
nyb = n(1)+ m(1);
nye = n(length(x)) +n(length(h));
ny = nyb:nye;
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3 l9 e; E1 T" |# H& ksubplot(3,1,1);
" J( r: Y/ w" l  Z$ Tstem(n,x);
& [+ a1 r+ T) z& ktitle('x(n)');
xlabel('n')

4 g! f# o% ]- X3 Y9 C/ V4 w8 S4 csubplot(3,1,2);
stem(m,h);
title('h(n)');
/ a* L1 @- o" e3 M# Txlabel('n')

; D. r. Y- @2 V# u+ Fsubplot(3,1,3);
stem(ny,y);
' R; U, D: ~' ]! rtitle('the conv of x(n) and h(n)');
xlabel('n')
xlim([-5,45]);

% k4 _2 P4 X" N9 f- e


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Heaven_1

subplot这个函数经常使用