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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

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  • $\begingroup$ What is it that you don't understand, exactly? Any online resource would tell you, at least from the engineering/physics perspective where you do some hand-waving, that it's a shifted impulse (scaled by some constant). $\endgroup$
    – Metric
    Commented Jan 23, 2019 at 3:33
  • $\begingroup$ @Metric like do i integrate the euler form? $\endgroup$
    – aceminer
    Commented Jan 23, 2019 at 3:35
  • $\begingroup$ No, you integrate it directly, but this is probably from an engineering/physics class, so there's going to be some handwaving. $\endgroup$
    – Metric
    Commented Jan 23, 2019 at 3:38
  • $\begingroup$ To offer some assitance towards the classical approach of finding it, I suggest you look at the inverse fourier transform of the dirac delta. $\endgroup$
    – Metric
    Commented Jan 23, 2019 at 3:39

1 Answer 1

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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

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

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