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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} $
Hence,
$ \displaystyle\delta(\omega)=\delta(\frac{f}{2\pi})=2\pi\delta(f) $
This may seem strange at first. I had the urge to simply replace $ \omega $ with $ 2\pi $ f as well. But that wouldn't be telling the same story. If you have an impulse located at 1 hz with some arbitrary magnitude, then the signal in radians would naturally be the same impulse located at $ 2\pi<\math>. We'll ignore the magnitude for now. Essentially all that is done going from <math>X(f) $ to $ X(\omega) $ is a frequency scale where every frequency is multiplied by $ 2\pi $ to obtain the spectrum in radians. However, when the impulse function is scaled, there is also an effect on the magnitude of the impulse function, which can be seen from the proof.
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} $
