Thread Subject: IFFT of symmetric vector: must I input the redundant coefficients?

I want to do the IFFT of a conjugate symmetric vector. Is it possible to do the IFFT of the first (non-redundant) coefficients, using the "half-complex" functions mentioned in the FFTW docs?

"Knut" wrote in message <j4j2l3$1kf$1@newscl01ah.mathworks.com>...
> I want to do the IFFT of a conjugate symmetric vector. Is it possible to do the IFFT of the first (non-redundant) coefficients, using the "half-complex" functions mentioned in the FFTW docs?

Hi Knut, Yes, you can just use the 'symmetric' option in ifft.

x = 1:8;
xdft = fft(x);
ifft([xdft(1:5) zeros(1,3)],'symmetric')
ifft(xdft)

Wayne

"Wayne King" wrote in message <j4jbi9$fi$1@newscl01ah.mathworks.com>...
> Hi Knut, Yes, you can just use the 'symmetric' option in ifft.
>
> x = 1:8;
> xdft = fft(x);
> ifft([xdft(1:5) zeros(1,3)],'symmetric')
> ifft(xdft)
>
> Wayne
Thank you, I just learned this myself :-)

But it still seems like a waste of bandwidth to have to copy a vector into something ~twice it size filled with zeros, just to have the underlying FFTW library ignore the latter half (assuming that MATLAB utilize the half-complex interface under the hood)?

-k