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序列 x(n)的奇偶分解的公式为:' x5 {5 i( e* G% f. _9 t
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编写一个序列 x(n) 的奇偶分解式 xe(n) 和 xo(n),需要考虑的问题是序列长度,下标的变化。1 q) d0 S5 f( d/ j1 D3 W) ]
4 Q/ W/ F/ V9 K( B2 S- g这里必须做个声明,下面的程序中用到了前几篇博客中的几个函数,这里给贴出来:. o: A5 c( ~- Y, Q* j
) w. u1 f7 E6 J$ R3 J) j$ M信号相加:
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: A7 ~; E2 t. [! A" Ffunction [y,n] = sigadd(x1,n1,x2,n2)
) @2 H% V' w. F+ M* d- i% implements y(n) = x1(n) + x2(n)' S3 P @* B: Z: S3 `4 ]
% [y,n] = sigadd(x1,n1,x2,n2)! l2 G( U o& n7 @8 g( C
%____________________________________2 A/ E- \& t6 U9 @# [& W! O3 d7 a8 |
% y = sum sequence over n, which includes n1 and n2
2 P7 D+ z- \. x9 ?+ z; _; C6 F; D F% x1 = first sequence over n10 m, Y& T0 A& q1 Q% o3 F
% x2 = second sequence over n2( n2 can be different from n1)
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6 R7 z$ i1 U* l* M9 y1 Dn = min( min(n1), min(n2) ):max( max(n1), max(n2) ); %duration of y(n)
1 Z! Z9 I: ^! q3 C- w0 Ty1 = zeros(1,length(n)); y2 = y1; %initialization2 q m9 d4 ~3 m0 ^" R3 H
y1( find( ( n >= min(n1) )&( n <= max(n1) ) == 1 ) ) = x1; %x1 with duration of y1, _/ K& _% |4 B3 S* p3 b
y2( find( ( n >= min(n2) )&( n <= max(n2) ) == 1 ) ) = x2; %x2 with duration of y2
4 Q7 _! I" a' m, b2 Qy = y1 + y2;( ]& C9 ?: h: t9 a* c4 ^$ T) R
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( C( a7 O/ j3 w& Y. v5 c信号移位:/ V4 u+ `# i3 ~0 c8 G' M6 w' L
+ W1 U" Y4 f, z7 ^" n Wfunction [y,n] = sigshift(x,m,k)8 R8 ~1 n1 q. _# F+ k u
%implements y(n) = x(n - k)
% \+ q9 u1 K8 }- Z3 R%_________________________! \& [3 M; m/ |+ |) u
%[y,n] = sigshift(x,m,k)6 e$ M9 L- p! n/ j* \9 W6 R" d8 C
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n = m+k;
- z% f1 Z# B( @( a* ?y = x;+ e0 ~6 I0 A% G0 w9 M* D
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: t/ j) ^" b, d5 z# N1 m. e单位阶跃序列:7 ~ F9 }. ]6 V0 h6 p& L
8 ]2 J# [2 X# g1 u" |% j9 s8 efunction [x,n]=stepseq(n0,n1,n2);
, q' r3 v8 H6 n( p) a( }% generate x(n) = u(n - n0); n1 <= n <= n2
0 J6 n9 ]+ z# Z" Y( h%_____________________________________________
4 T8 I6 G) x; C! E- p% l%[x,n] = stepseq(n0, n1, n2);
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: s( B$ q% g9 d5 I) k( Kn = [n1:n2];
7 C3 Q% b5 o% e: Q% _8 [+ E: n1 |x = [(n-n0) >= 0];
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L- s! _* G: V1 _: u下面给出函数程序:
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function [xe, xo, m] = evenodd(x, n)7 f9 c" C) H( r) g/ a
% Real signal decomposition into even and odd parts# d$ ]2 @- a' K9 K& m4 T' B
%__________________________________________________! c u4 b( d7 T6 {6 V( f. _7 Q
%[xe, xo, m] = evenodd(x, n)8 M \$ z* {( I3 I+ Y S' s
%$ {! X1 U# j2 b# g* Z& L
if any( imag(x) )" H2 C0 y9 K) P! t0 f
error('x is not a real sequence!');
; x1 m. K: _7 ^, g- P Fend
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% Ensure m of xe and xo
( J5 h2 S( v( H- P3 [3 l9 x7 s8 Gm = - fliplr(n);
7 S, ]/ M8 s+ S5 P+ Wm1 = min([m,n]);& Y ^0 y5 R! n+ L) j
m2 = max([m,n]);
+ n: \( E& {8 N' Y9 gm = m1:m2;
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% Ensure x over m* f9 \4 T m8 [8 J3 \# a9 Q
nm = n(1) - m(1);" `6 V% P0 s1 V
n1 = 1:length(n);# c/ j( w" q8 l9 i
x1 = zeros(1,length(m)); % initialization& f+ i' X9 z2 B/ T
x1(nm + n1) = x;8 J; o _9 J5 Q! [3 N
x = x1; % new x which enlarge index n+ N1 g! t; n% c* {* b
- s+ t8 u9 W; t3 Q2 F% xe and xo
1 G& B% E( z4 C; `5 K$ o D% `xe = 0.5*(x + fliplr(x));
5 N1 h" z. k* C' z, D" hxo = 0.5*(x - fliplr(x));' `8 e7 F- p& u- K2 Z
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! A: s: Z! ~" c t) u, G7 a序列和及其位置分别装入 x 和 n 数组。首先确认是否已知序列是实序列并在m数组中确定偶部和奇部分量的位置,最后将所得奇偶分量存入xe和xo数组中。
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: [( `! _3 h. n: x/ X/ A下面以一个实例来验证上述函数:
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6 E& r' j8 A8 E0 n7 C8 m将x(n)分解为奇偶分量。
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clc0 x+ }1 r: h- d) F) U& w, Q
clear* U& d$ d% ]4 d1 c; z5 ]7 D: t: e
close all
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3 o& b9 w, z& P* ]n = 0:10;2 f9 g! T% C* N+ K6 |: b6 e; b
x = stepseq(0,0,10) - stepseq(10,0,10);
1 j* b/ t8 u# z; H[xe, xo, m] = evenodd(x,n);+ C# Z' j8 ^2 A D" n. [4 B
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subplot(2,2,1);
" a$ o! F8 o' d! ^; g u1 V$ ~stem(n,x,'filled');
" C) o9 b! A; q- O0 ~5 C2 ]title('Rectangular pulse');( N. h. ?- S' Q7 W! y$ O8 [( F. j9 X
xlabel('n');ylabel('x(n)');
) M$ m+ P9 D' Raxis([-10,10,0,1.2]);% ^+ J( F. w3 _2 `) V
/ D0 R4 k1 u5 I1 N8 Y- ~0 O8 ^3 o) Msubplot(2,2,2);
0 G5 U8 N' A) @% Mstem(m,xe,'filled');. o1 W2 u5 j/ K* {: |9 H6 y
title('Even part');
, o/ P5 G1 K+ d& ]6 J0 Axlabel('n');ylabel('xe(n)');' t" C) S: f2 x! |
axis([-10,10,0,1.2]);
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subplot(2,2,4);
8 L2 m0 t) Q& U) G f- {stem(m,xo,'filled');
- U8 o7 u5 ~4 q6 p8 ~* d; [6 x& Htitle('Odd part');
8 W! M9 d) ?3 y# A& U% ~xlabel('n');ylabel('xo(n)');
) J; O" q+ C$ ~% M) @& K4 |axis([-10,10,-0.6,0.6]);+ E# b+ ?" ^' F- |
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事实上,这篇博文到这里已经结束了,那我还想看看序列x(n)= u(n) - u(n-10)的合成过程:/ Y# E# G) {5 {4 q0 [! \ h1 D
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. G6 e+ l B' M: wclear
5 a: m) F3 H$ J- Pclose all" P3 ~4 e! P+ ^7 v2 p- z
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[u1,n1] = stepseq(0,0,10);
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subplot(3,1,1); ]) f% O& m9 S8 R- @# J1 H, N; L
stem(n1,u1,'filled');. m& `* h( }. f
title('u(n)');
7 K# X3 X& P$ s' U' M: K5 U4 Dylabel('u(n)');xlabel('n');
7 ?" H4 e& t) g2 kaxis([-10,10,0,1.2]);
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3 ?; c+ U& I$ d4 i7 K! B[u2,n2] = sigshift(u1,n1,10);& r+ `- T! z; [
subplot(3,1,2)
1 R. \5 C* q# ^: a4 }' Bstem(n2,u2,'filled');
% I* V+ h: P9 c/ x* Ktitle('u(n-10)')
, F% I4 Y; e8 P$ M- uxlabel('n');ylabel('u(n - 10)');
/ \( M' {. U& O7 Q e8 A4 E; Gaxis([0,20,0,1.2]);
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[x,n] = sigadd(u1,n1,-u2,n2);& N$ m; X2 k6 ?" j
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subplot(3,1,3)
S) K4 \$ e5 N% R: [3 N" b0 fstem(n,x,'filled');, B3 U5 w5 K c; Z
title('Rectangular sequence');/ z' V0 ?+ E3 v2 s
xlabel('n');ylabel('x(n)= u(n) - u(n -10)');
, L# J6 V1 b9 I3 r& x0 E% o: vaxis([-10,10,0,1.2]);
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