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

0 ^/ ?& X+ L! m' b6 e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：

ifft_vector

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  1 至 4 列

  15.0000 + 0.0000i  -2.5000 + 3.4410i  -2.5000 + 0.8123i  -2.5000 - 0.8123i

  5 列

  -2.5000 - 3.4410i
; k9 A9 L9 R4 S. q" ]4 X
6 n! G; C- q) kans =

8 `2 k0 K2 l& [" @4 z( h     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)
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结果如下：

Y =
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    0.8147    0.9134    0.2785    0.9649    0.9572
    0.9058    0.6324    0.5469    0.1576    0.4854
    0.1270    0.0975    0.9575    0.9706    0.8003
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6 m: c1 \: L9 t/ {X =

2 o" B; S# T% x3 D9 X  1 至 4 列

7 g! T& j- z" Y   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
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7 E- U; b: A2 T  l2 G  5 至 8 列
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( g0 U( @, a6 b7 r4 \   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
   0.1021 + 0.0000i  -0.0070 + 0.0253i  -0.0038 + 0.1091i  -0.1613 - 0.2141i

ans =

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4 w9 @1 Z& T! N6 `- I4 s# k* Z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：

• 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)
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3 C* V% K1 B' E$ c    0.1419    0.7922    0.0357    0.6787    0.3922
  N2 D' b# D7 K6 ^5 u    0.4218    0.9595    0.8491    0.7577    0.6555
    0.9157    0.6557    0.9340    0.7431    0.1712
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: z- R2 a8 V+ n& U$ l* m8 K  1 至 4 列

   0.1849 + 0.0000i   0.3009 + 0.0000i   0.2274 + 0.0000i   0.2725 + 0.0000i
0 v7 y' o# X% z, b, G! N   0.0550 + 0.1517i   0.1838 + 0.1668i   0.0795 + 0.1918i   0.1518 + 0.1599i
) Z  H* ^7 h9 Q9 x  -0.0967 + 0.0527i   0.0171 + 0.1199i  -0.1123 + 0.1061i  -0.0080 + 0.0947i
# ~9 E" D+ `( V( a- Y  -0.0195 - 0.0772i   0.0142 + 0.0028i  -0.0706 - 0.0417i   0.0179 - 0.0259i
$ v& S% ^9 ^$ X! V. V9 O, T   0.0795 + 0.0000i   0.0611 + 0.0000i   0.0151 + 0.0000i   0.0830 + 0.0000i
  -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
) C/ p* S3 O! [$ j2 k3 X8 {+ E   0.0550 - 0.1517i   0.1838 - 0.1668i   0.0795 - 0.1918i   0.1518 - 0.1599i
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: i' Z( }7 B- A) |- x" G) }  5 列

   0.1524 + 0.0000i
   0.1070 + 0.0793i
! V' O3 I8 a6 T& L8 U+ C6 v, d. h4 y  f   0.0276 + 0.0819i
  -0.0089 + 0.0365i
4 \1 w$ f: Y: H& v1 p  -0.0115 + 0.0000i
9 W' w. ]( C: {  -0.0089 - 0.0365i
   0.0276 - 0.0819i
% j0 |  H8 N+ B+ O% I. Z   0.1070 - 0.0793i
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& T) \5 I9 Z! z& }6 {ans =
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     8     5

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% v' q5 g! b' nConjugate Symmetric Vector
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/ t+ n( f* l5 U5 q; p# G# a4 e' |For nearly conjugate symmetric vectors, you can compute the inverse Fourier transform faster by specifying the 'symmetric' option,
8 R  t% j* U! `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.
Then, compute the inverse transform specifying the 'symmetric' option, which eliminates the nearly 0 imaginary parts.
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; F' o6 n) e& U6 ?/ P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。

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：

3 ]: x, ^) U8 p8 ?/ EY =
2 S' ~5 o* ?3 X4 z0 I1 v
7 T4 l( a" L" S0 O7 l4 c5 q3 c    1.0000    2.0000    3.0000    4.0000    4.0000    3.0000    2.0000
  X( Q5 \1 H2 ?0 X
X =
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  1 至 4 列
/ H2 I# b! y( _
   2.7143 + 0.0000i  -0.7213 + 0.0000i  -0.0440 - 0.0000i  -0.0919 + 0.0000i
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8 @5 j0 V5 ?6 v, ^' h2 m  5 至 7 列
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& w# k; H: C4 k. a2 Y; Y! |  -0.0919 - 0.0000i  -0.0440 + 0.0000i  -0.7213 - 0.0000i

1 I9 B- \! @4 VXsym =

# B# B7 F9 ^! v5 \. g6 |* P    2.7143   -0.7213   -0.0440   -0.0919   -0.0919   -0.0440   -0.7213
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