Matlab 基4 FFT的实现

如果我逐个设置4 xp值的幂，下面的倍频程基4 FFT代码工作正常

$ octave fft4.m

ans = 1.4198e-015

但是，如果取消对循环代码的注释，则会出现以下错误

$ octave fft4.m

error: `stage' undefined near line 48 column 68
error: evaluating argument list element number 1
error: evaluating argument list element number 2
error: called from:
error:   r4fftN at line 48, column 22
error:   c:\Users\david\Documents\Visual Studio 2010\Projects\mv_fft\fft4.m at line 80, column 7

the "error" refers to a line the in fft function code which otherwise works correctly when xp is not set by a loop ... very strange.


    function Z = radix4bfly(x,segment,stageFlag,W)
      % For the last stage of a radix-4 FFT all the ABCD multiplers are 1.
      % Use the stageFlag variable to indicate the last stage
      % stageFlag = 0 indicates last FFT stage, set to 1 otherwise

      % Initialize variables and scale to 1/4
      a=x(1)*.25;
      b=x(2)*.25;
      c=x(3)*.25;
      d=x(4)*.25;

      % Radix-4 Algorithm
      A=a+b+c+d;
      B=(a-b+c-d)*W(2*segment*stageFlag + 1);
      C=(a-b*j-c+d*j)*W(segment*stageFlag + 1);
      D=(a+b*j-c-d*j)*W(3*segment*stageFlag + 1);

      % assemble output
      Z = [A B C D];
    end % radix4bfly()


    % radix-4 DIF FFT, input signal must be floating point, real or complex
    %
    function S = r4fftN(s)

      % Initialize variables and signals: length of input signal is a power of 4
      N = length(s);
      M = log2(N)/2;

      % Initialize variables for floating point sim
      W=exp(-j*2*pi*(0:N-1)/N);
      S = complex(zeros(1,N));
      sTemp = complex(zeros(1,N));

      % FFT algorithm
      % Calculate butterflies for first M-1 stages
      sTemp = s;
      for stage = 0:M-2
        for n=1:N/4
          S((1:4)+(n-1)*4) = radix4bfly(sTemp(n:N/4:end), floor((n-1)/(4^stage)) *(4^stage), 1, W);
        end
        sTemp = S;
      end

      % Calculate butterflies for last stage
      for n=1:N/4
        S((1:4)+(n-1)*4) = radix4bfly(sTemp(n:N/4:end), floor((n-1)/(4^stage)) * (4^
    stage), 0, W);
      end
      sTemp = S;

      % Rescale the final output
      S = S*N;
    end % r4fftN(s)


    % test FFT code
    %
    xp = 2;

    % ERROR if I uncomment loop!

    %for xp=1:8
      N = 4^xp; % must be power of: 4 16 64 256 1024 4086 ....
      x = 2*pi/N * (0:N-1);
      x = cos(x);
      Y_ref = fft(x);
      Y = r4fftN(x);
      Y = digitrevorder(Y,4);
      %Y = bitrevorder(Y,4);
      abs(sum(Y_ref-Y)) % compare fft4 to built-in fft
    %end

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问题是指数xp的循环边界应该从2开始，因为fft4代码假设至少有两个基4蝴蝶阶段

对不起，各位：

---

-David

请在下面找到一个在频率FFT算法中进行基数4抽取的完整Matlab实现。我还提供了复杂矩阵乘法和加法的总体运算计数。事实上可以证明，每个基数为4的蝴蝶包含3个复数乘法和8个复数加法。由于存在log_4N=log_2N/2个阶段，并且每个阶段涉及N/4个蝴蝶，因此操作计数为

complex multiplications = (3 / 8) * N * log2(N)
complex additions       = N * log2(N)

代码如下：

% --- Radix-2 Decimation In Frequency - Iterative approach

clear all
close all
clc

% --- N should be a power of 4
N = 1024;

% x = randn(1, N);
x = zeros(1, N);
x(1 : 10) = 1;
xoriginal = x;
xhat = zeros(1, N);

numStages = log2(N) / 2;

W = exp(-1i * 2 * pi * (0 : N - 1) / N);
omegaa = exp(-1i * 2 * pi / N);

mulCount = 0;
sumCount = 0;

M = N / 4;
for p = 1 : numStages;
    for index = 0 : (N / (4^(p - 1))) : (N - 1);
        for n = 0 : M - 1;   
            a =  x(n + index + 1) +      x(n + index + M + 1) + x(n + index + 2 * M + 1) +      x(n + index + 3 * M + 1);
            b = (x(n + index + 1) -      x(n + index + M + 1) + x(n + index + 2 * M + 1) -      x(n + index + 3 * M + 1)) .* omegaa^(2 * (4^(p - 1) * n));
            c = (x(n + index + 1) - 1i * x(n + index + M + 1) - x(n + index + 2 * M + 1) + 1i * x(n + index + 3 * M + 1)) .* omegaa^(1 * (4^(p - 1) * n));
            d = (x(n + index + 1) + 1i * x(n + index + M + 1) - x(n + index + 2 * M + 1) - 1i * x(n + index + 3 * M + 1)) .* omegaa^(3 * (4^(p - 1) * n));
            x(n + 1 + index) = a;
            x(n + M + 1 + index) = b;
            x(n + 2 * M + 1 + index) = c;
            x(n + 3 * M + 1 + index) = d;
            mulCount = mulCount + 3;
            sumCount = sumCount + 8;
        end;
    end;
    M = M / 4;
end

xhat = bitrevorder(x);

tic
xhatcheck = fft(xoriginal);
timeFFTW = toc;

rms = 100 * sqrt(sum(sum(abs(xhat - xhatcheck).^2)) / sum(sum(abs(xhat).^2)));

fprintf('Theoretical multiplications count \t = %i; \t Actual multiplications count \t = %i\n', ...
         (3 / 8) * N * log2(N), mulCount);
fprintf('Theoretical additions count \t\t = %i; \t Actual additions count \t\t = %i\n\n', ...
         N * log2(N), sumCount);
fprintf('Root mean square with FFTW implementation = %.10e\n', rms);

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原始FFT代码的源代码是我的测试代码，它只是针对matlab/octave内置FFT进行测试