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How to obtain the CTFT of a periodic function in terms of f in hertz (from the formula in terms of $ \omega $)

Recall:

$ x(t)=\sum^{\infty}_{k=-\infty} a_{k}e^{ikw_{0}t} $

$ \mathcal{X}(\omega)=2\pi\sum^{\infty}_{k=-\infty}a_{k}\delta(w-kw_{0}) $

To obtain X(f), use the substitution

$ \omega= 2 \pi f $.

More specifically

$ \begin{align} X(f) &=\mathcal{X}(2\pi f) \\ &=2\pi\sum^{\infty}_{k=-\infty}a_{k}\delta(2\pi f-kw_{0}) \\ &=\sum^{\infty}_{k=-\infty}a_{k}\delta(f-\frac{kw_{0}}{2\pi}) \end{align} $

$ Since\ k\delta (kt)=\delta (t),\forall k\ne 0 $


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