speeding up ifft with conjugate symmetry

In my MATLAB program I convert a time signal to the frequency domain and then back to time domain via the overlap add method.

According to the documentation, ifft(X) is faster if X is conjugate symmetric.

going from frequency to time I've tried

      Yn = Xn;
      yn = ifft(Yn, N);

versus the following, where I've tried to implement it as I understand the documentation

      Yn(1:N) = Xn;
      Yn = [Yn(1:N) fliplr(conj(Yn(1:N)))];
      yn = ifft(Yn, N);

N = length(Xn), which is the same in both cases.

In both cases I preallocate the needed memory, but the first one is always faster than the second.

What am I doing wrong?

The full code is downloadable from my dropbox: https://dl.dropbox.com/u/51552385/overlap_add.m

Thanks for your help

Full code:

L = 64;             %Blocklength/Number of bands
M = L/2;            %filterlength/Decimation factor
N = L + M - 1;      %fft- and ifft length

fs = 20e3;
x = randn(1, 2*fs);       %generate full input signal

y = zeros(1, length(x));  %allocate memory for y

y = [y zeros(1, M-1)];    %add extra space for last overlap, eventually to be cut away

complex = false;          %choose ifft method

if(complex)
    Yn = [y y];
end

tic
for n = 1:length(x)/L
    blockn = [x((n-1)*L+1:n*L) zeros(1,M-1)];   % blockn is equal to nth block
    Xn = fft(blockn, N);
    if(complex)
        Yn(1:N) = Xn;
        Yn = [Yn(1:N) fliplr(conj(Yn(1:N)))];
        yn = ifft(Yn, N);
    else
        Yn = Xn;
        yn = ifft(Yn, N);
    end
    %overlap and add current block to output signal y
    y((n-1)*L+1:N+(n-1)*L) = y((n-1)*L+1:N+(n-1)*L) + yn;
end
toc
y = y(1:length(x));     %cut away last overlap
error = norm(y-x)

Answer 1

[Wayne King]

Have you tried just using the 'symmetric' flag to treat the input as conjugate symmetric?

    t = 0:0.001:1-0.001;
    x = cos(2*pi*100*t)+randn(size(t));
    xdft = fft(x);
    xhat = ifft(xdft,'symmetric');

[ro bader]

Is there a way other thann the option 'symmetric'?