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：
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function [y,n] = sigadd(x1,n1,x2,n2)
4 j- M' L: K2 I# u8 O% implements y(n) = x1(n) + x2(n)
% [y,n] = sigadd(x1,n1,x2,n2)
! ?' l! [6 U- Q2 w  L$ X" O+ k6 K%——————————————————————————————
! S8 q' A0 H, y& r% E  A9 ]% y = sum sequence over n, which includes n1 and n2
  v! E. ]. V4 q2 ~# Y% x1 = first sequence over n1
% x2 = second sequence over n2( n2 can be different from n1)
%
* B1 V( `1 j( sn = min( min(n1), min(n2) ):max( max(n1), max(n2) ); %duration of y(n)
; i% Z  X) A& }, q4 Ey1 = zeros(1,length(n)); y2 = y1; %initialization
y1( find( ( n >= min(n1) )&( n <= max(n1) ) == 1  )  ) = x1; %x1 with duration of y1
, H6 H/ e" F& y6 N, ny2( 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：

, `2 p* G  p) G6 q) k  y0 j. Tclc
clear
! @' n. X' r: Xclose all
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7 S7 k6 s3 S+ @  y% help stepseq
4 g6 n( e" x* y! h7 c%   generate x(n) = u(n - n0); n1 <= n <= n2
%  _____________________________________________
%  [x,n] = stepseq(n0, n1, n2);
[u1,n1] = stepseq(0,-5,45);
[u2,n2] = stepseq(10,-5,45);

  M$ q9 a0 N$ P. u" H& C% generate signal x(n)
[x,n] = sigadd(u1,n1,-u2,n2);
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% generate signal h(n)
( \& d6 N6 j. o' a, _m = -5:45;
8 C' h. \% H! l5 C7 r  Y  ih = ( (0.9).^m ).* u1;
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8 ]: P2 b$ Z6 _: X% the convolution of x(h) and h(n)
8 L) k6 U" b. Ay = conv(x,h);
4 v. d2 u2 i+ V% ensure the index
nyb = n(1)+ m(1);
) _" O9 Q* ^% s! x: vnye = n(length(x)) +n(length(h));
7 ^! \8 l0 Z7 Q. a0 D/ ^ny = nyb:nye;
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subplot(3,1,1);
0 z- P3 q+ U! I7 y! bstem(n,x);
  j; k# H  W0 P  Rtitle('x(n)');
xlabel('n')

8 x, ]7 _6 J5 d1 f1 Qsubplot(3,1,2);
7 B8 p( j! a' @2 Q( ystem(m,h);
7 u" k2 O7 X  k! ~% q" M4 Mtitle('h(n)');
3 _& v) K: x1 j, Lxlabel('n')

subplot(3,1,3);
stem(ny,y);
; ?% r8 h- D0 u  @' A9 @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