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From the notes, we also know the relationship between <math>\,\! X(w)</math> and <math>\,\! X_s(f)</math>
 
From the notes, we also know the relationship between <math>\,\! X(w)</math> and <math>\,\! X_s(f)</math>
  
*<math>\,\! X(w) = X_s((\frac{w}{2\pi})F_s)</math>
+
<p><math>\,\! X(w) = X_s((\frac{w}{2\pi})F_s)</math></p>
  
 
Rewriting <math>\,\! X_s(f)</math>
 
Rewriting <math>\,\! X_s(f)</math>
  
*<math>\,\! X_s(f) = FsX(f)*\sum_{-\infty}^{\infty}\delta(f-F_sk)</math>
+
<math>\,\! X_s(f) = FsX(f)*\sum_{-\infty}^{\infty}\delta(f-F_sk)</math>
  
Substituting known relation  
+
Substituting known relation
*<math>\,\! X(w) = FsX((\frac{w}{2\pi})F_s)*\sum_{-\infty}^{\infty}\delta((\frac{w}{2\pi})F_s-F_sk)</math>
+
<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
 
Using LTI, rearrange the equation
*<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><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
 
Re-arrange the delta function
*<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><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>
 
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>
*<math>\,\! X(w) =Fs(\frac{2\pi}{F_s}) \sum_{-\infty}^{\infty}X((\frac{w}{2\pi})F_s)*\delta((w-k2\pi))</math>
+
  
 
The <math>F_s</math> will cancel and employ sifting to get
 
The <math>F_s</math> will cancel and employ sifting to get
*<math>\,\! X(w) =2\pi \sum_{-\infty}^{\infty}X((\frac{w-k2\pi}{2\pi})F_s)</math>
+
<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
 
Now you can see how your X(f) is being scaled and shifted

Revision as of 12:54, 16 February 2009


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

Alumni Liaison

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

Buyue Zhang