| Line 11: | Line 11: | ||
<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, | ||
| + | |||
| + | <math>\delta(\omega)=\delta(2\pi f)=frac{1}{2\pi}\delta(f)</math> | ||
Revision as of 07:46, 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,
$ \delta(\omega)=\delta(2\pi f)=frac{1}{2\pi}\delta(f) $
