How to obtain the CTFT of a impulse train in terms of f in hertz (from the formula in terms of $ \omega $)
Recall:
$ x(t)=\sum^{\infty}_{n=-\infty} \delta(t-nT) $
$ \mathcal{X}(\omega)=\frac{2\pi}{T}\sum^{\infty}_{k=-\infty}\delta(w-\frac{2\pi k}{T}) $
To obtain X(f), use the substitution
$ \omega= 2 \pi f $.
More specifically
$ \begin{align} X(f) &=\mathcal{X}(2\pi f) \\ &=\frac{2\pi}{T}\sum^{\infty}_{k=-\infty}\delta(2\pi f-\frac{2\pi k}{T}) \\ &=\frac{1}{T}\sum^{\infty}_{k=-\infty}\delta(f-\frac{k}{T}) \end{align} $
$ Since\ k\delta (kt)=\delta (t),\forall k\ne 0 $
