Starting with some $ \,\! X(f) $, we want to derive a mathematical expression for $ \,\! X(w) $
Though we already know that it's just some shift/scale version with period 2*pi, here is the math behind it.
We know $ \,\! X_s(f) = FsRep_{Fs}[X(f)] $ from the discussion of $ \,\!x_s(t) = comb_t(x(t)) $
From the notes, we also know the relationship between $ \,\! X(w) $ and $ \,\! X_s(f) $
$ \,\! X(w) = X_s((\frac{w}{2\pi})F_s) $
Rewriting $ \,\! X_s(f) $
$ \,\! X_s(f) = FsX(f)*\sum_{-\infty}^{\infty}\delta(f-F_sk) $
Substituting known relation
$ \,\! X(w) = FsX((\frac{w}{2\pi})F_s)*\sum_{-\infty}^{\infty}\delta((\frac{w}{2\pi})F_s-F_sk) $
Using LTI, rearrange the equation
$ \,\! X(w) = Fs\sum_{-\infty}^{\infty}X((\frac{w}{2\pi})F_s)*\delta((\frac{w}{2\pi})F_s-F_sk) $
Re-arrange the delta function
$ \,\! X(w) = Fs\sum_{-\infty}^{\infty}X((\frac{w}{2\pi})F_s)*\delta((\frac{F_s}{2\pi})(w-k2\pi)) $
Using delta properties, you can take out the $ (\frac{F_s}{2\pi}) $
$ \,\! X(w) =Fs(\frac{2\pi}{F_s}) \sum_{-\infty}^{\infty}X((\frac{w}{2\pi})F_s)*\delta((w-k2\pi)) $
The $ F_s $ will cancel and employ sifting to get
$ \,\! X(w) =2\pi \sum_{-\infty}^{\infty}X((\frac{w-k2\pi}{2\pi})F_s) $
Now you can see how your X(f) is being scaled and shifted
- I think what Myron is trying to show here is the relationship between the FT of a signal and the FT of a sampling of that signal. Anybdy sees a mistake? Perhaps one can rewrite this so it becomes a bit clearer. --Mboutin 16:22, 17 February 2009 (UTC)
- Note that the answer is given in Prof. Allebach's table. But somehow, the table's answer has a factor Fs in front of the summation. So most likely a factor Fs was dropped somewhere: can anybody find where? --Mboutin 16:35, 17 February 2009 (UTC)
