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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> | ||
| + | |||
| + | ==Therefore...== | ||
| + | |||
| + | <math>\delta(\omega)=\delta(2\pi f)=frac{1}{2\pi}\delta(f)</math> | ||
==Which also means that..== | ==Which also means that..== | ||
Revision as of 16:09, 21 October 2009
Contents
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...
$ \delta(\omega)=\delta(2\pi f)=frac{1}{2\pi}\delta(f) $
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} $
