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
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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结果如下：
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ifft_vector
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Y =

  1 至 4 列

1 v; l/ }# t7 \) T$ Z" v1 o6 P4 j  15.0000 + 0.0000i  -2.5000 + 3.4410i  -2.5000 + 0.8123i  -2.5000 - 0.8123i
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. k" h7 }6 @  ]5 c1 I# C  S: a  5 列
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  -2.5000 - 3.4410i
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ans =
8 y0 S$ d8 V7 Q+ k
5 _5 w! T/ p1 A     1     2     3     4     5


Padded 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)
  

8 d6 S0 _) U# p" g* L: M% s5 ~结果如下：
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    0.8147    0.9134    0.2785    0.9649    0.9572
% w. f4 I# e# T  _/ L    0.9058    0.6324    0.5469    0.1576    0.4854
    0.1270    0.0975    0.9575    0.9706    0.8003
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X =
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: ~; _5 |8 [: s  1 至 4 列
  k1 H  t2 G4 \6 v' e0 I: [9 z$ ^2 i
( l$ w( }/ I# w   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
   0.3691 + 0.0000i  -0.1613 + 0.2141i  -0.0038 - 0.1091i  -0.0070 - 0.0253i

  5 至 8 列

) R- I, ?; v9 r4 z0 `   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
2 Q7 O( D0 s- _( [; _$ u+ q# B! i   0.1021 + 0.0000i  -0.0070 + 0.0253i  -0.0038 + 0.1091i  -0.1613 - 0.2141i

  x, f( m  h/ Y7 X" ~7 Hans =

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& p+ D% ^8 L* @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：

' k: G% ~: S/ z* t  |" iX = 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)
  4 p; l# b  d# a3 u' @# m

  

Y =
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    0.1419    0.7922    0.0357    0.6787    0.3922
    0.4218    0.9595    0.8491    0.7577    0.6555
6 i6 @9 F8 p5 U7 K    0.9157    0.6557    0.9340    0.7431    0.1712
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X =

: \; B. d) f' x" b  1 至 4 列

: m/ Q8 r% `4 d4 ~' c/ D   0.1849 + 0.0000i   0.3009 + 0.0000i   0.2274 + 0.0000i   0.2725 + 0.0000i
- x7 b/ {, V% _5 V   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
6 ^+ v4 i4 N) ^' O2 ^( p0 V) C( t3 r  -0.0195 - 0.0772i   0.0142 + 0.0028i  -0.0706 - 0.0417i   0.0179 - 0.0259i
! W, W1 l+ v* M+ ^6 f) ?0 S# E) v" d   0.0795 + 0.0000i   0.0611 + 0.0000i   0.0151 + 0.0000i   0.0830 + 0.0000i
+ o# b" e. |1 K0 y, r  -0.0195 + 0.0772i   0.0142 - 0.0028i  -0.0706 + 0.0417i   0.0179 + 0.0259i
  -0.0967 - 0.0527i   0.0171 - 0.1199i  -0.1123 - 0.1061i  -0.0080 - 0.0947i
) E3 f: A! H! \: N- C/ K9 z/ n+ {( @   0.0550 - 0.1517i   0.1838 - 0.1668i   0.0795 - 0.1918i   0.1518 - 0.1599i

! i5 _2 ]- R: ~7 u  5 列
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& l7 W7 b1 s* Z8 j' O   0.1524 + 0.0000i
" l) B: T  V3 X3 Z   0.1070 + 0.0793i
   0.0276 + 0.0819i
( p% K( m) d& s  -0.0089 + 0.0365i
  -0.0115 + 0.0000i
  -0.0089 - 0.0365i
   0.0276 - 0.0819i
   0.1070 - 0.0793i
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ans =

     8     5


Conjugate Symmetric Vector
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' }: B/ e0 b% f4 k  f& MFor nearly conjugate symmetric vectors, you can compute the inverse Fourier transform faster by specifying the 'symmetric' option,
( n9 `6 T6 T5 D( n0 J5 a4 _which 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.
- `& ], h! k' i8 [5 p  g+ c7 |Then, compute the inverse transform specifying the 'symmetric' option, which eliminates the nearly 0 imaginary parts.
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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。
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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 ⋯
  

   
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结果如下：

Y =
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- r& W$ b/ W, N3 b9 q$ a% P    1.0000    2.0000    3.0000    4.0000    4.0000    3.0000    2.0000
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; S& p  H2 U2 ~+ Q- m0 {X =

  1 至 4 列

$ v/ {5 Q" i" ~% j9 n, S4 J: f   2.7143 + 0.0000i  -0.7213 + 0.0000i  -0.0440 - 0.0000i  -0.0919 + 0.0000i

  5 至 7 列

" _5 c6 G5 z6 G) c  -0.0919 - 0.0000i  -0.0440 + 0.0000i  -0.7213 - 0.0000i
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% ~! A$ F& }  }& s: xXsym =
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    2.7143   -0.7213   -0.0440   -0.0919   -0.0919   -0.0440   -0.7213

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