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Re: [Phys-l] some questions related to sampling

• From: Stefan Jeglinski <jeglin@4pi.com>
• Date: Fri, 9 Jan 2009 18:34:38 -0500

In the spirit of the correspondence principle, let's examine the
relationship between a discrete Fourier transform and a non-discrete
happens:

1. As previously discussed: When dt becomes small, the
frequency-range of the first period of the output becomes large. (The
output is always defined for all frequencies, so we can't talk about
the range of the output, only the range of one period.)

2. As previously discussed: When Ndt becomes large, the output
becomes nearly continuous.

3. If you do both together - small dt and large Ndt - the discrete
Fourier transform begins to look a whole lot like the old-fashioned
non-discrete Fourier transform.

OK?

Agreed, as long as we are referring in bullet 2 to Nf, as per your Eq 24. Increasing Nf leads to a higher resolution plot of Vf, for a given Ntdt. Granted, if I decrease dt in order to sample more of Vin, Nt will necessarily increase if I am to still cover the entire region of interest in Vin. But my choice of Nf is independent of that, used only to increase the resolution in frequency space. No?

I'm either being pedantic or still missing the point. Given a choice, I'd rather not miss the point.

You can't have a sampled input without having the output of the
transform be periodic. The proof is simple:

Understood. The way I had always looked at this was as follows (hoping I can make this clear in my non-UTF8 client, and hopefully without typos). Casting a sampled x(t) as

sx(t) = [Sum_over_all_n] x[n] delta-function[t - nT]

where T is the sampling interval. Then, fourier-transform sx(t):

X(f) = [Integral over +- infinity] sx(t) exp(-2pi*i*f*t) dt