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==Scaling of the Dirac Delta (Impulse Function)== | ==Scaling of the Dirac Delta (Impulse Function)== | ||
<math>\displaystyle\delta(\alpha f)=\frac{1}{\alpha}\delta(f)</math> <math>\textstyle{ for }\alpha>0</math> | <math>\displaystyle\delta(\alpha f)=\frac{1}{\alpha}\delta(f)</math> <math>\textstyle{ for }\alpha>0</math> | ||
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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> | ||
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| + | [[ECE438_(BoutinFall2009)|Back to ECE438 course page]] | ||
Revision as of 07:28, 20 September 2009
Scaling of the Dirac Delta (Impulse Function)
$ \displaystyle\delta(\alpha f)=\frac{1}{\alpha}\delta(f) $ $ \textstyle{ for }\alpha>0 $
Mini Proof
$ \int_{-\infty}^{\infty}\delta(x)dx = 1 $
Let $ \displaystyle 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} $
