# Computing Large DFTs Using Small FFTs

It is possible to compute *N*-point discrete Fourier transforms (DFTs) using radix-2 fast Fourier transforms (FFTs) whose sizes are less than *N*. For example, let's say the largest size FFT software routine you have available is a 1024-point FFT. With the following trick you can combine the results of multiple 1024-point FFTs to compute DFTs whose sizes are greater than 1024.

The simplest form of this idea is computing an *N*-point DFT using two *N*/2-point FFT operations. Here's how the trick works for computing a 16-point DFT, of a 16-sample *x*(*n*) input sequence, using two 8-point FFTs; we execute the following steps:

- Perform an 8-point FFT on the
*x*(*n*) samples where*n*= 0, 2, 4, ..., 14. We'll call those FFT results*X*_{0}(*k*). - Store two copies of
*X*_{0}(*k*) in Memory Array 1 as shown in Figure 1. - Next we compute an 8-point FFT on the
*x*(*n*) samples where*n*= 1, 3, 5, ..., 15. We call those FFT results*X*_{1}(*k*). - Store two copies of
*X*_{1}(*k*) in Memory Array 3 in Figure 1. - In Memory Array 2 we have stored 16 samples of one cycle of the complex exponential
*e*^{-j2}^{π}^{m}^{/N}, where*N*= 16, and 0 ≤*m*≤ 15. - Finally we compute our desired 16-point
*X*(*m*) samples by performing the arithmetic shown in Figure 1 on the horizontal rows of the memory arrays. That is,

*X*(0) =*X*_{0}(0) +*e*^{-j2}^{π}^{0/16}·*X*_{1}(0),*X*(1) =*X*_{0}(1) +*e*^{-j2}^{π}^{1/16}·*X*_{1}(1),. . .

*X*(15) =*X*_{0}(7) +*e*^{-j2}^{π}^{15/16}·*X*_{1}(7).

The final desired *X*(*m*) DFT results are stored in Memory Array 4.

**Figure ****1:** 16-point DFT using two 8-point FFTs.

We describe the above process, algebraically, as

*X*(*k*) = *X*_{0}(*k*) + *e*^{-j2}^{π}^{k}^{/16} · *X*_{1}(*k*) (1)

and

*X*(*k* + 8) = *X*_{0}(*k*) + *e*^{-j2}^{π}^{(k+8)/16} · *X*_{1}(*k*) (1')

for *k* in the range 0 ≤ *k* ≤ 7.

Notice that we did nothing to reduce the size of Memory Array 2 due to redundancies in the complex exponential sequence *e*^{-j2}^{π}^{m}^{/N}. As it turns out, for an *N*-point DFT, only *N*/4 complex values need be stored in Memory Array 2. The reason for this is because

(2)

which involves a simple sign change on *e*^{-j2}^{π}^{m}^{/N}. In addition,

(2')

which is merely swapping the real and imaginary parts of *e*^{-j2}^{π}^{m}^{/N} plus a sign change of the resulting imaginary part. So Eqs. (2) and (2') tell us that only the values *e*^{-j2}^{π}^{m}^{/N} for 0 ≤ *m* ≤ *N*/4-1 need be stored in Memory Array 2. With that reduced storage idea aside, to be clear regarding exactly what computations are needed for our 'multiple-FFTs' technique we leave Memory Array 2 unchanged from what is in Figure 1.

The neat part of this 'multiple-FFTs' scheme is that our DFT length, *N*, is not restricted to be an integer power of two. We can use computationally efficient radix-2 FFTs to compute DFTs whose lengths are any integer multiple of an integer power of two. For example, we can compute an *N* = 24-point DFT using three 8-point FFTs. To do so, we perform an 8-point FFT on the *x*(*n*) samples, where *n* = 0, 3, 6, ..., 21, to obtain *X*_{0}(*k*). Next we compute an 8-point FFT on the *x*(*n*) samples, where *n* = 1, 4, 7, ..., 22 to yield *X*_{1}(*k*). And then we perform an 8-point FFT on the *x*(*n*) samples, where *n* = 2, 5, 8, ..., 23, to obtain an *X*_{2}(*k*) sequence. Finally we compute our desired 24-point DFT results using

*X*(*k*) = *X*_{0}(*k*) + *e*^{-j2}^{π}^{k}^{/24} · *X*_{1}(*k*) + *e*^{-j2}^{π}^{(2k)/24} · *X*_{2}(*k*) (3)

*X*(*k* + 8) = *X*_{0}(*k*) + *e*^{-j2}^{π}^{(k+8)/24} · *X*_{1}(*k*) + *e*^{-j2}^{π}^{[2(k+8)]/24} · *X*_{2}(*k*) (3')

and

*X*(*k* + 16) = *X*_{0}(*k*) + *e*^{-j2}^{π}^{(k+16)/24} · *X*_{1}(*k*) + *e*^{-j2}^{π}^{[2(k+16)]/24} · *X*_{2}(*k*). (3'')

for *k* in the range 0 ≤ *k* ≤ 7. The memory-array depiction of this process is shown in Figure 2, with our final 24-point DFT results residing in Memory Array 6. Memory Array 2 contains *N* = 24 samples of one cycle of the complex exponential *e*^{-j2}^{π}^{m}^{/24}, where 0 ≤ *m* ≤ 23. Memory Array 4 contains 24 samples of two cycles of the complex exponential *e*^{-j2}^{π}^{(2m)/24}.

**Figure ****2** 24-point DFT using three 8-point FFTs.

To conclude this blog, we mention that the larger the size of the FFTs the more computationally efficient is this 'multiple-FFTs' spectrum analysis technique. This behavior is illustrated in Figure 3 where we show the number of complex multiplies required by the 'multiple-FFTs' algorithm versus the desired DFT size (*N*). The top bold curve is the number of complex multiplies required by the standard (inefficient) DFT algorithm, and the bottom dashed curve is the number of complex multiplies required by a single *N*-point radix-2 FFT. The curves in the center of the figure show the number of complex multiplies required by the 'multiple-FFTs' algorithm when various FFT sizes (*P*) are used to compute an *N*-point DFT.

**Figure ****3** Number of complex multiplies versus *N*.

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function Y = fftx(X)

% Y = fftx(X)

%Extended fft by combining smaller fft's recursively.

%URL: http://www.mathworks.com/company/newsletters/articles/faster-finite-fourier-transforms-matlab.html

%Modified to end recursion at max fft size by Don Gateley

%Maximum allowed fft size (use whatever is appropriate for you.)

max = 512;

X = X(:);

n = length(X);

if(n <= max)

%Down to chunks and end of recursion. Use native fft() for chunks or less.

Y = fft(X);

else

% Recursive divide and conquer.

w = exp(2 * 1i * pi * (0: n / 2 - 1) / n)';

u = fftx(X(1: 2: n-1));

v = fftx(X(2: 2: n)) .* w;

Y = [u + v; u - v];

end

end

/////////////////////////////////////////////////////////////////////////////////////////////////////

And the inverse:

/////////////////////////////////////////////////////////////////////////////////////////////////////

function T = ifftx(X)

%ifftx() using fftx()

n = size(X, 1);

T = fftx(imag(X) + 1i * real(X));

T= imag(T) + 1i * real(T);

T = T / n;

end

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I'd like to ask if you know any way of combining two sequential FFTs into a larger one efficiently,

and how is it possible to do so.

Thanks,

Ben

As always great!

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