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<math>\displaystyle\int_{-\infty}^{\infty}\delta(\alpha x)dx=\int_{-\infty}^{\infty}\delta(y)\frac{dy}{\alpha}=\frac{1}{\alpha}</math>
 
<math>\displaystyle\int_{-\infty}^{\infty}\delta(\alpha x)dx=\int_{-\infty}^{\infty}\delta(y)\frac{dy}{\alpha}=\frac{1}{\alpha}</math>
  
Hence,
+
==Hence,==
  
 
<math>\displaystyle\delta(\omega)=\delta(2\pi f)=frac{1}{2\pi}\delta(f)</math>
 
<math>\displaystyle\delta(\omega)=\delta(2\pi f)=frac{1}{2\pi}\delta(f)</math>

Revision as of 08:46, 20 September 2009

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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} $

Hence,

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

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood