View source for Getting X(w) from X(f) - Rhea

[[Category:ECE438Spring2009mboutin]] 

Starting with some <math>\,\! X(f)</math>, we want to derive a mathematical expression for <math>\,\! X(w)</math>


Though we already know that it's just some shift/scale version with period 2*pi, here is the math behind it.


We know <math>\,\! X_s(f) = FsRep_{Fs}[X(f)]</math> from the discussion of <math> \,\!x_s(t) = comb_t(x(t))</math>


From the notes, we also know the relationship between <math>\,\! X(w)</math> and <math>\,\! X_s(f)</math>

<p><math>\,\! X(w) = X_s((\frac{w}{2\pi})F_s)</math></p>

Rewriting <math>\,\! X_s(f)</math>

<math>\,\! X_s(f) = FsX(f)*\sum_{-\infty}^{\infty}\delta(f-F_sk)</math>

Substituting known relation
<p><math>\,\! X(w) = FsX((\frac{w}{2\pi})F_s)*\sum_{-\infty}^{\infty}\delta((\frac{w}{2\pi})F_s-F_sk)</math></p>

Using LTI, rearrange the equation
<p><math>\,\! X(w) = Fs\sum_{-\infty}^{\infty}X((\frac{w}{2\pi})F_s)*\delta((\frac{w}{2\pi})F_s-F_sk)</math></p>

Re-arrange the delta function
<p><math>\,\! X(w) = Fs\sum_{-\infty}^{\infty}X((\frac{w}{2\pi})F_s)*\delta((\frac{F_s}{2\pi})(w-k2\pi))</math></p>

Using delta properties, you can take out the <math>(\frac{F_s}{2\pi})</math>
<p><math>\,\! X(w) =Fs(\frac{2\pi}{F_s}) \sum_{-\infty}^{\infty}X((\frac{w}{2\pi})F_s)*\delta((w-k2\pi))</math></p>

The <math>F_s</math> will cancel and employ sifting to get
<p><math>\,\! X(w) =2\pi \sum_{-\infty}^{\infty}X((\frac{w-k2\pi}{2\pi})F_s)</math></p>

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. --[[User:Mboutin|Mboutin]] 16:22, 17 February 2009 (UTC)