Fourier transform of general complex signal

Question

I am trying to understand a sinusoid and its fourier transform.

Given an example sinusoid

$Ze^{iwt}$, how do i calculate the fourier transform of it?

How do i even represent this?

In euler form?

like $Zcost(wt) + iZsin(wt)$ ?

Been looking at online materials but still cannot understand

Answer 1

Since I think I know what you're asking for, I'm going to hand-wave a little bit here.

Suppose the fourier transform $F$ of a function $f$ is defined as

$$F(f)(\omega) = \int_{-\infty}^\infty f(t) e^{-i \omega t} dt$$

and the inverse-fourier transform $F^{-1}$ of a function $g$ is defined as

$$F^{-1}(g)(t) =\frac{1}{2\pi}\int_{-\infty}^\infty g(\omega) e^{i \omega t} d \omega$$

Since

$$ F(\delta)(\omega) = \int_{-\infty}^\infty \delta(t)e^{-i\omega t} d t = \int_{-\infty}^\infty \delta(t)e^{0} d t = \int_{-\infty}^\infty \delta(t) d t = 1$$

we have that

$$F^{-1}(1)(t) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{i\omega t} d\omega = \delta(t) $$

That is,

$$ \int_{-\infty}^\infty e^{i \omega t} d \omega = 2 \pi \delta(t) \tag 1$$

Now, if your function is $f(t) = ze^{i \omega_0 t}$, then using $(1)$ we have that its fourier transform is

$$F(f)(\omega) = \int_{-\infty}^\infty f(t) e^{-i \omega t} d t = z\int_{-\infty}^\infty e^{i (\omega_0-\omega) t} dt = 2 \pi z\delta(\omega_0 - \omega) = 2 \pi z\delta(\omega - \omega_0)$$