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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
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) $
