Difference between revisions of "Talk:Getting X(w) from X(f)" - Rhea

Revision as of 19:16, 17 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?

-Virgil Hsieh