IFFT 的实现
前些天给出了FFT的实现,现在给出IFFT(inverse FFT)的实现
基于IFFT 算法,对FFT的核心程序稍作修改即可 : )
%%*************************************************************************************** % code writer : EOF % code date : 2014.09.17 % e-mail : [email protected] % 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:-1:1 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
对比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
以后给出相应版本的C 语言实现
wait for update : )
时间: 2024-10-10 06:50:17