MATLAB —— 信号处理工具箱之 ifft 的案例分析

mytomorrow

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案例分析

       Inverse Transform of Vector

       Padded Inverse Transform of Matrix

       Conjugate Symmetric Vector

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

0 `( Z' O) v; [6 ~2 @8 c1 IY =
8 d$ @' o2 T; D) X
  1 至 4 列

6 E2 e: @# }; f  15.0000 + 0.0000i  -2.5000 + 3.4410i  -2.5000 + 0.8123i  -2.5000 - 0.8123i
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  5 列

8 ]0 u2 l9 `: r0 F& g  -2.5000 - 3.4410i

* ~; e* n6 o2 A8 m' j+ i* \/ g* Bans =

     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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4 h0 K. s6 B1 \9 ^结果如下：

Y =
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    0.8147    0.9134    0.2785    0.9649    0.9572
2 b# I) L0 I) D# Z9 @8 f    0.9058    0.6324    0.5469    0.1576    0.4854
8 E7 P8 z* _6 ?- y2 g    0.1270    0.0975    0.9575    0.9706    0.8003

" ~/ d- ^6 D+ ]$ v3 J  y" y1 CX =
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8 f  U: Q2 l8 T) \4 m  1 至 4 列

  b: z- q; [; A) k- H8 H* r3 Z   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
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   0.0215 + 0.0000i  -0.0133 - 0.1312i   0.1867 + 0.0064i  -0.0224 - 0.2008i
% t  H& @! p  N4 q( L8 M$ k, O   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
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ans =

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, `; j5 m# j  Q- a& q- t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：

" h4 I! ^% G# w" a2 NX = 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)
  

  

1 I! D- p+ G8 r. ]4 ]3 _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
# V% D$ Z5 U* J! f6 A$ ?; e+ U    0.9157    0.6557    0.9340    0.7431    0.1712

' D. T( g8 [3 \X =

  1 至 4 列
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9 p4 A- S/ I" C! W/ @5 [   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
9 d6 U9 Z- g4 k' M  -0.0967 + 0.0527i   0.0171 + 0.1199i  -0.1123 + 0.1061i  -0.0080 + 0.0947i
% E  r. `+ S  n* ?& _4 y  -0.0195 - 0.0772i   0.0142 + 0.0028i  -0.0706 - 0.0417i   0.0179 - 0.0259i
   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
5 C1 {; i0 e+ n/ R   0.0550 - 0.1517i   0.1838 - 0.1668i   0.0795 - 0.1918i   0.1518 - 0.1599i

! {* A% b% D% Y" `8 L, b# l  5 列
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   0.1524 + 0.0000i
   0.1070 + 0.0793i
, s9 s$ V0 u; j% X   0.0276 + 0.0819i
1 S1 \5 R& I% i) B6 I( t  -0.0089 + 0.0365i
; R. @, y9 Q+ t  -0.0115 + 0.0000i
; A  v: i& P3 G! `. m  -0.0089 - 0.0365i
   0.0276 - 0.0819i
6 L! y, E) \/ j   0.1070 - 0.0793i
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# n# x( Z: k  K( J% A" `2 Tans =
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Conjugate Symmetric Vector

1 ]+ [# `4 ?+ }1 D3 n# R) pFor 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.
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% q' h9 g, k/ z' ]! n. tCreate 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.

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。

6 e+ `- r) P' e  X2 ^! S) p  y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。

• 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 ⋯
  - L5 P" x3 |! k5 a" J; V

   

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结果如下：
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. w8 C4 k% s7 `" D: X    1.0000    2.0000    3.0000    4.0000    4.0000    3.0000    2.0000
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- |+ ^1 X% [! ~7 s& t) q$ [7 |X =
7 M* ~, O) f6 G# e- _
( n- O6 ?/ T+ L, q4 N  1 至 4 列

+ O, \1 m* y$ l5 ]   2.7143 + 0.0000i  -0.7213 + 0.0000i  -0.0440 - 0.0000i  -0.0919 + 0.0000i

  5 至 7 列

/ B( t$ B" a+ z$ u9 M  -0.0919 - 0.0000i  -0.0440 + 0.0000i  -0.7213 - 0.0000i

7 H0 T. N+ a& ?. JXsym =
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    2.7143   -0.7213   -0.0440   -0.0919   -0.0919   -0.0440   -0.7213

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