代码：

function [y] = ovrlpadd_v3(x, h, N)%% Overlap-Add method of block convolution%% ------------------------------------------------------%% [y] = ovrlpadd(x, h, N)%% y = output sequence%% x = input sequence%% h = impulse response%% N = block length >= 2*length(h)-1N = 2^(ceil(log10(N)/log10(2)))  Lx = length(x);                        % ML    L = length(h);                         % length of impulse responseHk_FFT = fft(h, N);M = floor((Lx)/(L))                   % number of bolckY = zeros(1, Lx+N-L)% convolution with successive blocksfor m = 0:M-1	xm = x(m*L+1 : m*L+L);                                     % length is L	XMk_FFT = fft(xm, N);	YMk_FFT = XMk_FFT .* Hk_FFT;	ym = real(ifft(YMk_FFT))                                   % length is 2L-1	%Y(m*L+1 : m*L+(2*L-1)) = Y(m*L+1 : m*L+(2*L-1)) + ym	Y(m*L+1 : m*L+N) = Y(m*L+1 : m*L+N) + ymendy = Y;

　　

%% ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++%%            Output Info about this m-filefprintf('\n***********************************************************\n');fprintf('        
     
       Problem 5.32 \n\n');banner();%% ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++% -------------------------------------------------------------%           Overlap-Add                 % -------------------------------------------------------------n1 = [0:8-1];x1 = cos(pi*n1/500);N1 = length(x1);nh = [0:3]; h = [1, -1, 1, -1]; L = length(h);M = N1/L;N = 2*L-1; y = ovrlpadd_v3(x1, h, N);            % FFT
     

　　运行结果：

       原题中x(n)长4000，不好看图说明，这里假设长度只有8，便于上图说明道理。