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--[[User:Vhsieh|Vhsieh]] 00:19, 18 February 2009 | --[[User:Vhsieh|Vhsieh]] 00:19, 18 February 2009 | ||
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| + | ---- | ||
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| + | Based off of what Virgil said, I think the relationship we need to prove goes as follows: | ||
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| + | We know: | ||
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| + | <math>\,\! X_s(f) = \frac{1}{T}Rep_\frac{1}{T}[X(f)]</math> | ||
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| + | This can also be written as: | ||
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| + | <math>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)} </math> | ||
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| + | Now compare this with the Fourier transform of <math>x_d(n) \!</math>: | ||
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| + | <math>X_d(w) = \sum_{n=-\infty}^{\infty}x_d(n)e^{-jwn} = \sum_{n=-\infty}^{\infty}x(nT)e^{-jwn} \!</math> | ||
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| + | These formulas are identical when <math> w=2{\pi}fT \!</math> | ||
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| + | Thus <math>X_d(2{\pi}Tf)=X_s(f)\!</math> | ||
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| + | So if we know | ||
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| + | <math>\! X_s(f) = FsRep_{Fs}[X(f)]</math> and | ||
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| + | <math>\! X(w) = X_s((\frac{w}{2\pi})F_s)</math> | ||
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| + | then we just evaluate <math>X_s(f) \!</math> at <math>f=\frac{w}{2\pi}F_s \!</math> so: | ||
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| + | <math>\! X(w) = FsRep_{Fs}[X(\frac{w}{2\pi}F_s)] </math> | ||
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| + | --[[User:Pjcannon|Pjcannon]] 01:30, 18 February 2009 (UTC) | ||
| + | * Looks pretty good to me. Does it makes sense to everybody? Is it clear? --[[User:Mboutin|Mboutin]] 11:54, 18 February 2009 (UTC) | ||
| + | * Yes. Now it is the time to memorize. right? --[[User:Kim415|Kim415]] 12:29, 18 February 2009 (UTC) | ||
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)
