FFT and sin wave

A small doubt in signal processing. When you take fft of a signal -- k=fft(y,512); and plot it, we get 512 points. I want to represent the 'x' axis in Hz . How would you do that? For eg: if k=40 what is the the corresponding frequency in Hz for a 512 point FFT?

Answer 1

[Wayne King]

If the signal is real-valued:

t = 0:.001:1-0.001;
Fs = 1e3;
x = cos(2*pi*100*t)+randn(size(t));
xdft = fft(x);
xdft = xdft(1:length(x)/2+1);
freq = 0:Fs/length(x):Fs/2;
plot(freq,abs(xdft));
xlabel('Hz');

[Wayne King]

Frequencies are spaced at Fs/N where Fs is the sampling frequency and N is the length input signal.

Answer 2

[i Venky]

This is okay when the no of points in the fft is same as that of the length of the input signal. When I change the no of points in the fft. For eg: in your code if I change xdft=fft(x) into xdft=fft(x,512), what should be the change that I should make to get the frequency in Hz?

Answer 3

[i Venky]

I got it.

Frequency in Hz= k * Sampling frequency (in time domain)/ N

Where 'N' is the no of points in the FFT.

[Wayne King]

Yes, but just keep in mind that the frequency resolution of your DFT is determined by the length of your input signal, NOT by the value of any zero padding. Zero padding just interpolates your DFT, it does not give you any better resolution.