Revision as of 17:20, 11 February 2009 by Mlo (Talk | contribs)


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) $

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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang