MATLAB —— 信号处理工具箱之 ifft 的案例分析

mytomorrow

案例分析

       Inverse Transform of Vector

       Padded Inverse Transform of Matrix

       Conjugate Symmetric Vector

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案例分析
7 @4 m$ x) E+ q5 ?+ S) _Inverse Transform of Vector

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• % The Fourier transform and its inverse convert between data sampled in time and space and data sampled in frequency.
• %
• % Create a vector and compute its Fourier transform.
• X = [1 2 3 4 5];
• Y = fft(X)
• % Y = 1×5 complex
• %
• %   15.0000 + 0.0000i  -2.5000 + 3.4410i  -2.5000 + 0.8123i  -2.5000 - 0.8123i  -2.5000 - 3.4410i ⋯
• %
• % Compute the inverse transform of Y, which is the same as the original vector X.
• ifft(Y)
• % ans = 1×5
• %
• %      1     2     3     4     5
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" A7 Y2 `4 S$ {1 r结果如下：

ifft_vector

Y =

  1 至 4 列

. p7 {1 Q! v0 u* T# w& {1 L  15.0000 + 0.0000i  -2.5000 + 3.4410i  -2.5000 + 0.8123i  -2.5000 - 0.8123i
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  5 列

  -2.5000 - 3.4410i
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     1     2     3     4     5
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. E. \4 p- @" l) v) s7 A' z% _: EPadded Inverse Transform of Matrix

• clc
• clear
• close all
• % The ifft function allows you to control the size of the transform.
• %
• % Create a random 3-by-5 matrix and compute the 8-point inverse Fourier transform of each row.
• % Each row of the result has length 8.
• Y = rand(3,5)
• n = 8;
• X = ifft(Y,n,2)
• size(X)
  

- K! Q2 [& y, i结果如下：

Y =

4 ^) l; q. Q( D7 c  ?% i    0.8147    0.9134    0.2785    0.9649    0.9572
0 t2 l7 Z; Y& Z$ R. Q4 Y    0.9058    0.6324    0.5469    0.1576    0.4854
" A2 |* K1 h4 B' C& I4 x  }1 Z    0.1270    0.0975    0.9575    0.9706    0.8003

X =
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  1 至 4 列

1 m  O5 Y1 v, ?) Z* X   0.4911 + 0.0000i  -0.0224 + 0.2008i   0.1867 - 0.0064i  -0.0133 + 0.1312i
   0.3410 + 0.0000i   0.0945 + 0.1382i   0.1055 + 0.0593i   0.0106 + 0.0015i
0 L8 C+ f( M. R9 Z   0.3691 + 0.0000i  -0.1613 + 0.2141i  -0.0038 - 0.1091i  -0.0070 - 0.0253i
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  5 至 8 列

3 S* [) C$ L, p. a   0.0215 + 0.0000i  -0.0133 - 0.1312i   0.1867 + 0.0064i  -0.0224 - 0.2008i
   0.1435 + 0.0000i   0.0106 - 0.0015i   0.1055 - 0.0593i   0.0945 - 0.1382i
; J- C: q: v  d4 E' Z0 U   0.1021 + 0.0000i  -0.0070 + 0.0253i  -0.0038 + 0.1091i  -0.1613 - 0.2141i

ans =
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9 \7 l! E& b7 F, p* s! j+ J8 b上面的程序是计算矩阵每一行的8点ifft，故结果是每一行的ifft有8个元素，而计算矩阵 Y 每一行的 ifft，关键语句为：

X = ifft(Y,n,2)，里面的2，如果去掉2，则是对矩阵Y的每一列计算ifft，测试如下：
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• clc
• clear
• close all
• % The ifft function allows you to control the size of the transform.
• %
• % Create a random 3-by-5 matrix and compute the 8-point inverse Fourier transform of each row.
• % Each row of the result has length 8.
• Y = rand(3,5)
• n = 8;
• X = ifft(Y,n)
• size(X)
  

  

Y =

" n9 q$ @9 e3 v. k7 o) ]) {    0.1419    0.7922    0.0357    0.6787    0.3922
! V2 T$ `) m- g4 g; |% {    0.4218    0.9595    0.8491    0.7577    0.6555
    0.9157    0.6557    0.9340    0.7431    0.1712

X =

  1 至 4 列
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   0.1849 + 0.0000i   0.3009 + 0.0000i   0.2274 + 0.0000i   0.2725 + 0.0000i
   0.0550 + 0.1517i   0.1838 + 0.1668i   0.0795 + 0.1918i   0.1518 + 0.1599i
  -0.0967 + 0.0527i   0.0171 + 0.1199i  -0.1123 + 0.1061i  -0.0080 + 0.0947i
- o' Z5 A, D& J8 _4 J$ x  -0.0195 - 0.0772i   0.0142 + 0.0028i  -0.0706 - 0.0417i   0.0179 - 0.0259i
, m6 \: p  |% t& H: n   0.0795 + 0.0000i   0.0611 + 0.0000i   0.0151 + 0.0000i   0.0830 + 0.0000i
7 k' C$ i- V% i( c2 p. [# b  -0.0195 + 0.0772i   0.0142 - 0.0028i  -0.0706 + 0.0417i   0.0179 + 0.0259i
: H& D% e6 G8 A; @1 ?+ z% l% y7 m" r  -0.0967 - 0.0527i   0.0171 - 0.1199i  -0.1123 - 0.1061i  -0.0080 - 0.0947i
4 u, W% c+ g' S- g% X" a7 R1 j9 F% U   0.0550 - 0.1517i   0.1838 - 0.1668i   0.0795 - 0.1918i   0.1518 - 0.1599i
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6 U* m9 R% c" g7 H% V" T& x, `   0.1524 + 0.0000i
   0.1070 + 0.0793i
   0.0276 + 0.0819i
7 {5 q. k- [" J# M! O  -0.0089 + 0.0365i
9 C* a- y8 l7 K6 U4 j1 j  -0.0115 + 0.0000i
4 R. j0 k. K8 L6 H8 y4 f7 B. r/ x  -0.0089 - 0.0365i
   0.0276 - 0.0819i
   0.1070 - 0.0793i

ans =
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Conjugate Symmetric Vector
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3 K& d6 e4 i6 t4 }# w* T8 fFor nearly conjugate symmetric vectors, you can compute the inverse Fourier transform faster by specifying the 'symmetric' option,
3 q5 v' E! w* M5 _' Twhich also ensures that the output is real. Nearly conjugate symmetric data can arise when computations introduce round-off error.
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Create a vector Y that is nearly conjugate symmetric and compute its inverse Fourier transform.
Then, compute the inverse transform specifying the 'symmetric' option, which eliminates the nearly 0 imaginary parts.

! h! r/ w* b6 ^7 k对于近似共轭对称矢量，您可以通过指定“symmetric”选项来更快地计算逆傅里叶变换，这也确保了输出是真实的。 当计算引入舍入误差时，可能出现几乎共轭的对称数据。

3 c1 g6 X( r, [创建几乎共轭对称的向量Y并计算其逆傅里叶变换。然后，计算指定'对称'选项的逆变换，它消除了近0虚部。
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• clc
• clear
• close all
• % For nearly conjugate symmetric vectors, you can compute the inverse Fourier transform faster by specifying the 'symmetric' option,
• % which also ensures that the output is real. Nearly conjugate symmetric data can arise when computations introduce round-off error.
• %
• % Create a vector Y that is nearly conjugate symmetric and compute its inverse Fourier transform.
• % Then, compute the inverse transform specifying the 'symmetric' option, which eliminates the nearly 0 imaginary parts.
• Y = [1 2:4+eps(4) 4:-1:2]
• % Y = 1×7
• %
• %     1.0000    2.0000    3.0000    4.0000    4.0000    3.0000    2.0000 ⋯
• X = ifft(Y)
• % X = 1×7 complex
• %
• %    2.7143 + 0.0000i  -0.7213 + 0.0000i  -0.0440 - 0.0000i  -0.0919 + 0.0000i  -0.0919 - 0.0000i  -0.0440 + 0.0000i  -0.7213 - 0.0000i ⋯
• Xsym = ifft(Y,'symmetric')
• % Xsym = 1×7
• %
• %     2.7143   -0.7213   -0.0440   -0.0919   -0.0919   -0.0440   -0.7213 ⋯
  3 L- m& C- q- S! z2 X

   
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- U: |5 d" g8 T- J; d: u/ B结果如下：

: @0 m* M9 |# o" g% VY =
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) L+ @# G2 h- [) A  D    1.0000    2.0000    3.0000    4.0000    4.0000    3.0000    2.0000
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/ y2 U, V( h7 B- SX =
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  1 至 4 列
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4 q0 B3 b. z6 p: i   2.7143 + 0.0000i  -0.7213 + 0.0000i  -0.0440 - 0.0000i  -0.0919 + 0.0000i

  5 至 7 列
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- f" [, K3 b$ c" A  -0.0919 - 0.0000i  -0.0440 + 0.0000i  -0.7213 - 0.0000i
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Xsym =

    2.7143   -0.7213   -0.0440   -0.0919   -0.0919   -0.0440   -0.7213
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