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 Mimi wants us to prove the relationship between $ X_s(f) and X_d(w)\! $:
I thought in class she said the proof we needed to know was done in the notes, which this one is. It 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)\! $
--Pjcannon 00:56, 18 February 2009 (UTC)
