Choosing FFT points

Hi All. It is not clear to me how to choose points in a FFT, and matlab Help, doesn't help on this. E.g., I have an L long array and I want an fft window of 32ms (which is 704 samples with a fs of 22kHz); fft(X(1:704)) gives me 704 frequency bins by default, but I want only a subset of frequency "lines", and specifically I need M=128 frequency bins. I tought I would just use the syntax fft(x(1:704),128), however:

x1=ifft( fft( x(1:704),128 ), 704) 
is not equal to
x1=ifft( fft( x(1:704) ) )

Why?? (thanks)

[Wayne King]

By the way:

x1=ifft( fft( x(1:704),128 ), 704)

does not take the DFT of x(1:704) as you think, it takes the DFT of x(1:128)

isequal(fft(x(1:128)),fft(x(1:704),128))

Answer 1

[Wayne King]

The bins in the DFT are Fs/N where N is the length of the input vector and Fs is the sampling frequency.

If you only want to evaluate the DFT at a subset of 128 DFT bins, then you can select those from the output of fft()

Also, see the help for goertzel(), which evaluates the DFT at specific bins.

[joeDiHare]

Thanks for your answer.

Goertzel algorithm looks very good, but there is no IDFT.

Is it correct to use ifft(X), where X is the output of the goertzel "zeropadded" in the frequency bins discarded by the DFT, so that I have back a time domain signal 704 samples long?.

something like this:

ind_freq=1:2:128*2;

G=goertzel(x(1:704),ind_freq);

magn=abs(G); phas=phase(G);

X=zeros(704,2);

X(ind_freq,:)=[magn phas];

x=ifft( X(:,1).*exp(1j*X(:,2) )