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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...

$ \displaystyle \delta(\omega)=\delta(2\pi f)=\frac{1}{2\pi}\delta(f) $

$ \displaystyle 2\pi\delta(\omega)=\delta(f) $

To convert $ \delta(f) $ to radians, simply replace $ \delta(f) $ with $ 2\pi\delta(\omega) $

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} $

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Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin