question about sampling

Hello everyone,

I read somewhere that if we want to resolve two frequencies using the
fourier transform, one signal must have evolved an extra 2 pi relative
to the other. I am trying to picturise this, but am having trouble
understanding this relationship. Can someone explain why this is so?

Cheers,
Anja

On May 3, 8:16 am, Anja <anja.e...@googlemail.com> wrote:
> Hello everyone,
>
> I read somewhere that if we want to resolve two frequencies using the
> fourier transform, one signal must have evolved an extra 2 pi relative
> to the other. I am trying to picturise this, but am having trouble
> understanding this relationship. Can someone explain why this is so?
>
> Cheers,
> Anja

I think what you are describing is the frequency resolution of a
sampled waveform.  It actually has nothing to do with the Fourier
transform.  It is a fundamental property of a sampled signal of finite
length.  A time sampled signal can be considered as a sum of signals
each of which are integer multiples of the frequency of the sine wave
with a wavelength equal to the time duration of the sample size.  If
Fs is the sample rate and N is the total number of samples, then N/Fs
is the time duration of the sample set.  The fundamental frequency of
this sample set is Fs/N.  It can be considered to be composed of sine
waves with frequencies , 0*Fs/N, 1*Fs/N, 2*Fs/N,... up to (N/2)*(Fs/N)
or Fs/2.  In reality the sampled signal could have been produced from
an analog signal with different frequencies, but there is no way to
tell the difference in the sampled domain.  When converted back into
an analog signal, the result will be the same if filtered by the
"perfect" analog filter.

This frequency spacing of Fs/N is, I think, what you are talking about
when you say, "evolved an extra 2 pi relative", no?

The point is that this is a property of a finite, sampled signal, not
the Fourier transform.  If you are in a class on digital signal
processing, you need to review your early chapters to figure out why
you missed this.  If you are learning it on your own, then you are a
better man than I am.  Even *after* taking a class in DSP, it took me
years to actually grasp what a lot of this is about.  (I'll tell you a
secret, I *still* don't really understand the Z transform, but don't
tell anyone, it's our little secret!)

On May 3, 5:16 am, Anja <anja.e...@googlemail.com> wrote:
> I read somewhere that if we want to resolve two frequencies using the
> fourier transform, one signal must have evolved an extra 2 pi relative
> to the other. I am trying to picturise this, but am having trouble
> understanding this relationship. Can someone explain why this is so?

If you transform a finite length signal into the frequency
domain, two sinusoids which are periodic but of different
frequencies will be orthogonal and thus completely separable
in the frequency domain.  To be both periodic and different
requires a different integer multiple of full cycles, or of
2 pi of phase, within the window.

You can note that the two periodic sinusoids are orthogonal
by seeing that if you multiply them together, the positive
and negative contributions cancel each other out.



IMHO. YMMV.
--
rhn A.T nicholson d.0.t C-o-M