Revision as of 19:15, 17 February 2009 by Vhsieh (Talk | contribs)

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?

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett