Difference between revisions of "Talk:Getting X(w) from X(f)" - Rhea

Latest revision as of 07:29, 18 February 2009

I think I see an error in his work. He says:

$ \,\! X_s(f) = FsRep_{Fs}[X(f)] $

And directly following that is the claim:

$ \,\! X_s(f) = FsX(f)*\sum_{-\infty}^{\infty}\delta(f-F_sk) $

I am pretty sure this is incorrect. From the notes and/or the posted equation sheet, the Rep function is not a summation of deltas. Rather, it is a summation of the X function, which produces copies, or repititions, hence the name. He has actually implemented a Comb function, which ultimately leads to his doom, as his answer is wrong. The equation should read:

$ \,\! X_s(f) = Fs*\sum_{-\infty}^{\infty}X(f-F_sk) $

The following equation that he put is correct:

$ \,\! X(w) = X_s((\frac{w}{2\pi})F_s) $

Thus, we can substitute, and we get:

$ \,\! X_s(f) = Fs*\sum_{-\infty}^{\infty}X((\frac{w}{2\pi})F_s-F_sk) $

Combining the summation and the shift into one Rep produces the following:

$ \,\! X_s(f) = FsRep_{Fs}[X((\frac{w}{2\pi})F_s)] $

Which is what is in the notes of Professor A. Alternatively, we can just substitute into the original equation with the Rep functions, without converting to a summation and then converting back. This would be a one-step process and would produce the same result:

$ \,\! X_s(f) = FsRep_{Fs}[X((\frac{w}{2\pi})F_s)] $

Which leaves me a bit curious. Is it really this simple? One step? Anyone have any ideas?

--Vhsieh 00:19, 18 February 2009

Based off of what Virgil said, I think the relationship we need to prove goes as follows:

We know:

$ \,\! X_s(f) = \frac{1}{T}Rep_\frac{1}{T}[X(f)] $

This can also be written as:

$ X_s(f) = \mathcal{F} [ \sum_{k}^{}x(kT)\delta(t-kT)] = \sum_{k}^{}x(kT)\mathcal {F}[\delta(t-kT)] = \sum_{k=-\infty}^{\infty}x(kT)e^{(-j2{\pi}fkT)} $

Now compare this with the Fourier transform of $ x_d(n) \! $:

$ X_d(w) = \sum_{n=-\infty}^{\infty}x_d(n)e^{-jwn} = \sum_{n=-\infty}^{\infty}x(nT)e^{-jwn} \! $

These formulas are identical when $ w=2{\pi}fT \! $

Thus $ X_d(2{\pi}Tf)=X_s(f)\! $

So if we know

$ \! X_s(f) = FsRep_{Fs}[X(f)] $ and

$ \! X(w) = X_s((\frac{w}{2\pi})F_s) $

then we just evaluate $ X_s(f) \! $ at $ f=\frac{w}{2\pi}F_s \! $ so:

$ \! X(w) = FsRep_{Fs}[X(\frac{w}{2\pi}F_s)] $

--Pjcannon 01:30, 18 February 2009 (UTC)

• Looks pretty good to me. Does it makes sense to everybody? Is it clear? --Mboutin 11:54, 18 February 2009 (UTC)
• Yes. Now it is the time to memorize. right? --Kim415 12:29, 18 February 2009 (UTC)