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<math>P_T(f)=\frac{1}{T_s}\sum_{n=-\infty}^{\infty}\delta(f-\frac{n}{T_s})</math> | <math>P_T(f)=\frac{1}{T_s}\sum_{n=-\infty}^{\infty}\delta(f-\frac{n}{T_s})</math> | ||
| − | <math>P_T(\omega)\frac{2\pi}{T_s}\sum_{n=-\infty}^{\infty}\delta(w-n\frac{2\pi}{T_s})</math> | + | <math>P_T(\omega)=\frac{2\pi}{T_s}\sum_{n=-\infty}^{\infty}\delta(w-n\frac{2\pi}{T_s})</math> |
Revision as of 07:50, 20 September 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} $
Hence,
$ \displaystyle\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}) $
$ P_T(\omega)=\frac{2\pi}{T_s}\sum_{n=-\infty}^{\infty}\delta(w-n\frac{2\pi}{T_s}) $
