(10 intermediate revisions by one other user not shown)
Line 1: Line 1:
[[ECE438_(BoutinFall2009)|Back to ECE438 course page]]
+
[[Hw3ECE438F09boutin|Return to previous page]]
  
 
==Scaling of the Dirac Delta (Impulse Function)==
 
==Scaling of the Dirac Delta (Impulse Function)==
Line 16: Line 16:
 
x)dx=\int_{-\infty}^{\infty}\delta(y)\frac{dy}{\alpha}=\frac{1}{\alpha}</math>
 
x)dx=\int_{-\infty}^{\infty}\delta(y)\frac{dy}{\alpha}=\frac{1}{\alpha}</math>
  
==Hence,==
+
==Therefore...==
  
<math>\displaystyle\delta(\omega)=\delta(\frac{f}{2\pi})=2\pi\delta(f)</math>
+
<math>\displaystyle \delta(\omega)=\delta(2\pi f)=\frac{1}{2\pi}\delta(f)</math>
  
This may seem strange at first. I had the urge to simply replace \omega with 2\pi f as well. But that wouldn't be telling the same story. If you have an impulse located at 1 hz with some arbitrary magnitude, then the signal in radians would naturally be the same impulse located at 2\pi (because w = 2\pi f). We'll ignore the magnitude for now. Essentially all that is done going from <math>X(f)->X(w)</math> is a frequency scale where every frequency is multiplied by 2\pi to obtain the spectrum in radians.
+
<math>\displaystyle 2\pi\delta(\omega)=\delta(f)</math>
  
 +
To convert <math>\delta(f)</math> to radians, simply replace <math>\delta(f)</math> with <math>2\pi\delta(\omega)</math>
 
==Which also means that..==
 
==Which also means that..==
  
Line 27: Line 28:
  
 
<math>P_T(\omega)=\frac{2\pi}{T_s}\sum_{n=-\infty}^{\infty}\delta(w-n\frac{2\pi}{T_s})\;\;\;\;\;\;\;w_s=\frac{2\pi}{T_s}</math>
 
<math>P_T(\omega)=\frac{2\pi}{T_s}\sum_{n=-\infty}^{\infty}\delta(w-n\frac{2\pi}{T_s})\;\;\;\;\;\;\;w_s=\frac{2\pi}{T_s}</math>
 +
 +
[[Hw3ECE438F09boutin|Return to previous page]]
 +
 +
[[ECE438|Return to ECE438]]

Latest revision as of 07:37, 25 August 2010

Return to previous page

Scaling of the Dirac Delta (Impulse Function)

$ \displaystyle\delta(\alpha f)=\frac{1}{\alpha}\delta(f)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;for\;\;\alpha>0 $

Mini Proof

$ \int_{-\infty}^{\infty}\delta(x)dx = 1 $

$ \displaystyle Let\;\;\;y=\alpha x\;\;\;\;\;\;\;\;\;\;\;\;\;dx=\frac{dy}{\alpha} $

$ \displaystyle\int_{-\infty}^{\infty}\delta(\alpha x)dx=\int_{-\infty}^{\infty}\delta(y)\frac{dy}{\alpha}=\frac{1}{\alpha} $

Therefore...

$ \displaystyle \delta(\omega)=\delta(2\pi f)=\frac{1}{2\pi}\delta(f) $

$ \displaystyle 2\pi\delta(\omega)=\delta(f) $

To convert $ \delta(f) $ to radians, simply replace $ \delta(f) $ with $ 2\pi\delta(\omega) $

Which also means that..

$ P_T(f)=\frac{1}{T_s}\sum_{n=-\infty}^{\infty}\delta(f-\frac{n}{T_s})\;\;\;\;\;\;\;\;\;\;\;f_s=\frac{1}{T_s} $

$ P_T(\omega)=\frac{2\pi}{T_s}\sum_{n=-\infty}^{\infty}\delta(w-n\frac{2\pi}{T_s})\;\;\;\;\;\;\;w_s=\frac{2\pi}{T_s} $

Return to previous page

Return to ECE438

Alumni Liaison

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

Dr. Paul Garrett