Overlap–add method

In signal processing, the overlap–add method is an efficient way to evaluate the discrete convolution of a very long signal with a finite impulse response (FIR) filter : where h[m] = 0 for m outside the region [1, M]. This article uses common abstract notations, such as or in which it is understood that the functions should be thought of in their totality, rather than at specific instants (see Convolution#Notation). The concept is to divide the problem into multiple convolutions of h[n] with short segments of : where L is an arbitrary segment length. Then: and y[n] can be written as a sum of short convolutions: where the linear convolution is zero outside the region [1, L + M − 1]. And for any parameter it is equivalent to the N-point circular convolution of with in the region [1, N]. The advantage is that the circular convolution can be computed more efficiently than linear convolution, according to the circular convolution theorem: where: DFTN and IDFTN refer to the Discrete Fourier transform and its inverse, evaluated over N discrete points, and L is customarily chosen such that N = L+M-1 is an integer power-of-2, and the transforms are implemented with the FFT algorithm, for efficiency. The following is a pseudocode of the algorithm: (Overlap-add algorithm for linear convolution) h = FIR_filter M = length(h) Nx = length(x) N = 8 × 2^ceiling( log2(M) ) (8 times the smallest power of two bigger than filter length M. See next section for a slightly better choice.) step_size = N - (M-1) (L in the text above) H = DFT(h, N) position = 0 y(1 : Nx + M-1) = 0 while position + step_size ≤ Nx do y(position+(1:N)) = y(position+(1:N)) + IDFT(DFT(x(position+(1:step_size)), N) × H) position = position + step_size end When the DFT and IDFT are implemented by the FFT algorithm, the pseudocode above requires about N (log2(N) + 1) complex multiplications for the FFT, product of arrays, and IFFT. Each iteration produces N-M+1 output samples, so the number of complex multiplications per output sample is about: For example, when M=201 and N=1024, equals 13.