The Fourier Transform of a complex exponential is an impulse (or Dirac) function. Depending on how frequency is written ($ \omega $ or $ f $) we have two identical definitions of the same FT:
$$ \mathcal{F} \{ e^{j n \omega_0 t} \} = 2 \pi \cdot \delta(\omega – n \omega_0) $$
$$ \mathcal{F} \{ e^{j n 2 \pi f_0 t} \} = \delta(f – n f_0) $$
That implies
$$ \implies 2 \pi \cdot \delta(\omega – n \omega_0) = \delta(f – n f_0) $$
In my attempt to verify this, I expanded $ \omega $ into $ 2 \pi f $ and there was this $ 2 \pi $ factor inside and out of $ \delta $ that don't seem to cancel out:
$$ \begin{align} 2 \pi \cdot \delta(\omega – n \omega_0) &= 2 \pi \cdot \delta(2\pi f – n 2 \pi f_0) \\ &= 2 \pi \cdot \delta \big[2 \pi (f-nf_0) \big] \neq \delta(f-nf_0) \; ??? \end{align} $$
Can someone help to proof these two equations and explain the intuition behind it? What does it mean to have a $ 2 \pi $ scaling factor on the x-axis if I transform $ f $ to $ \omega $?
Best Answer
The Dirac $\delta$ distribution is better defined by $\int \delta(x) \, \phi(x) \, dx = \phi(0)$ for every continuous function $\phi.$
With the distribution $a\delta(x),$ where $a\in\mathbb{R},$ we then mean that $$\int (a\delta)(x) \, \phi(x) \, dx = a \int \delta(x) \, \phi(x) \, dx = a\phi(0).$$
And $\delta(ax)$ we define by making a variable substitution: $$ \int \delta(ax) \, \phi(x) \, dx = \{ y := ax \} = \int \delta(y) \, \phi(y/a) \, \frac{dy}{|a|} = \frac{1}{|a|} \phi(0/a) = \frac{1}{|a|} \phi(0) = \int \frac{1}{|a|} \delta(x) \, \phi(x) \, dx, $$ i.e. $\delta(ax) = \frac{1}{|a|} \delta(x)$ since the above is valid for all continuous functions $\phi.$