IFFT 的实现

前些天给出了FFT的实现，如今给出IFFT(inverse FFT)的实现

基于IFFT 算法。对FFT的核心程序稍作改动就可以 : )

%%***************************************************************************************
% code writer : EOF
% code date : 2014.09.17
% e-mail : jasonleaster@gmail.com
% code file : IFFT_EOF.m
% Version : 1.0
%
% code purpose :
%
% It's time to finish my demo for DFT. I would like to share my code with
% someone who is interesting in DSP. If there is something wrong with my code,
% please touche me by e-mail. Thank you!
%
%%***************************************************************************************

clear all;

%*************************************************
% The number of all the signal that our sensor got
%*************************************************
TotalSample = 8;

% We assume that the preiod of the signal we generated is 'circle';
circle = TotalSample/2;

%**************************************************************
% This varible is used for recording the signal which
% were processed by inverse-DFT in time domain
%**************************************************************
SignalInT = zeros(TotalSample,1);

SignalInT_reversed = zeros(TotalSample,1);
%This varible is used for recording the signal which were processed by inverse-DFT in time domain

OutPutSignal = zeros(TotalSample,1);

OriginalSignal = zeros(TotalSample,1);
%This varible is used for recording the original signal that we got.

%% initialize a square wave
for SampleNumber = -(TotalSample/2):(TotalSample/2)-1

if (mod(abs(SampleNumber),circle) < (circle/2))&&(SampleNumber>0)

    OriginalSignal((TotalSample/2)+1+SampleNumber) = 5;

elseif (mod(abs(SampleNumber),circle) >= (circle/2))&&(SampleNumber>0)

    OriginalSignal((TotalSample/2)+1+SampleNumber) = 0;

elseif (mod(abs(SampleNumber),circle) < (circle/2))&&(SampleNumber<0)

    OriginalSignal((TotalSample/2)+1+SampleNumber) = 0;   

elseif (mod(abs(SampleNumber),circle) >= (circle/2))&&(SampleNumber<0)

    OriginalSignal((TotalSample/2)+1+SampleNumber) = 5;  
end  

end

InPutSignal = fft(OriginalSignal); % for testing

TotalSample = size(InPutSignal,1);

OutPutSignal_to_time = zeros(TotalSample,1);

tmp = TotalSample - 1;

%%***********************************************************************
% @Bits : describe how many bits should be used to make up the TotalSample
%%***********************************************************************
Bits = 0;

while tmp > 0

%%  floor (X) Return the largest integer not greater than X.  
tmp = floor(tmp/2);

Bits = Bits + 1;  

end

%*******************************************************************
% | | | |
% input X(n) | layer 3 |layer 2 |layer 1 | x(n) output
% | | | |
%
%
% @layyer : the number of layyer
% @SampleNumber: the start number of point which
% is going to do butter-fly operation.
% @pre_half : the pre_half point of current butter-fly
%***********************************************************************
for layyer = Bits👎1

  for SampleNumber = 1 : 2^(layyer) : TotalSample

        for  pre\_half = SampleNumber:(SampleNumber+2^(layyer-1) -1)

             r = -get\_r\_in\_Wn(pre\_half-1,layyer,TotalSample,Bits);

             W\_rN = exp(-2\*pi\*j\*(r)/TotalSample) ;

             OutPutSignal\_to\_time(pre\_half) = ...  
             0.5\*(InPutSignal(pre\_half) +  ...  
             InPutSignal(pre\_half + 2^(layyer-1)));

             OutPutSignal\_to\_time(pre\_half  + 2^(layyer-1)) =  ...  
             0.5\*W\_rN \*(InPutSignal(pre\_half) -  ...  
             InPutSignal(pre\_half + 2^(layyer-1)));

       end  
  end

  InPutSignal = OutPutSignal\_to\_time;  

end

%******************************************
% Reverse the bits of output number
%******************************************
for SampleNumber = 1 : TotalSample

ret     = bit\_reverse(SampleNumber - 1,Bits);

OutPutSignal\_to\_time(SampleNumber) = InPutSignal(ret+1);  

end

对照原始时间领域的输入信号和最后IFFT的输出信号，一致。于是IFFT实现成功

以后给出对应版本号的C 语言实现

wait for update : )