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If we want to have the Fourier transform of a complex exponential $x(t) = e^{i\omega_0t}$ we could "guess" that it's $X(\omega)=2\pi\delta(\omega-\omega_0)$ and prove the equality: $$ x(t) = \frac{1}{2\pi}\int_{-\infty}^\infty 2\pi\delta(\omega-\omega_0)e^{i\omega t}d\omega = \int_{-\infty}^\infty \delta(\omega-\omega_0)e^{i\omega t}d\omega = e^{i\omega t} |_{\omega=\omega_0} = e^{i\omega_0t} $$

QUESTIONS:

1) What is the logic behind the above mentioned "guess"?

2) What is the correct way to get the Fourier transform of a complex exponential without "guessing"?

Thank you for your help.

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3 Answers 3

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In general, the fourier transform of a continuous time signal $x(t)$ is given by:

$$\begin{align*} X(\omega) &= \int_{-\infty}^{\infty} x(t) e^{-i\omega t} dt \end{align*}$$

But, please note that the signal $x(t)$ must be absolutely integrable over all time i.e.,

$$\begin{align*} \int_{-\infty}^{\infty} |x(t)| dt \space < \space \infty \end{align*}$$

The function $e^{i\omega_0 t}$ however, is not absolutely integrable and the fourier transform does not converge: $$\begin{align*} \int_{-\infty}^{\infty} |e^{i\omega_0 t}| dt \space = \space \int_{-\infty}^{\infty} 1. dt \space= \space \infty \end{align*}$$

However, to find the Fourier Transform (F.T) of the above function, we can use the duality property of F.T i.e., $$\begin{align*} if \space \space &F\{x(t)\} \space \space = \space \space X(\omega), \space then\\ & F\{X(t)\} \space = \space 2\pi x(-\omega) \end{align*}$$

Now, consider the Dirac delta function (it's not a function, really), $\delta(t)$. $$\begin{align*} &F\{\delta(t)\} \space \space = \space \space 1, \space then\\ & F\{1\} \space = \space 2\pi \delta(-\omega) \space = 2\pi\delta(\omega) \space \space \space (1) \end{align*}$$

Ok, so far so good. Duality alone will not fetch the desired result. We now use the modulation property of F.T. $$\begin{align*} if \space \space &F\{x(t)\} \space \space = \space \space X(\omega), \space then\\ & F\{x(t).e^{i\omega_0t}\} \space = \space X(\omega-\omega_0) \end{align*}$$

Therefore, from eqn (1), $$\begin{align*} & F\{1.e^{i\omega_0t}\} \space = \space F\{e^{i\omega_0t}\} \space = \space 2\pi\delta(\omega-\omega_0) \end{align*}$$

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Just apply the Fourier transformation to $e^{i\omega_0t}$:

$$\begin{align*} X(\omega) &= \int_{-\infty}^{\infty} e^{i\omega_0t} e^{-i\omega t} dt\\ &= \int_{-\infty}^{\infty} e^{-i(\omega-\omega_0)t} dt\\ &= 2\pi\delta(\omega - \omega_0) \end{align*}$$

Compare it with the Fourier transform of the constant function: (proof not given)

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

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Up to a factor of $2\pi$ the Fourier transformation can be seen as an expansion in terms of $e^{i\omega t}$. Clearly for $e^{i\omega_0 t}$ there is only one component in the expansion. In a discrete expansion this would mean that we have a Kronecker delta $\delta_{\omega_0}^{\omega}$ as component. But because we are doing a continuous transformation this becomes the Dirac delta. This (generalized) function will filter out that single component $e^{i\omega_0 t}$.

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  • $\begingroup$ I think this is the answer that best addresses the point raised. $\endgroup$ Jun 7, 2020 at 10:05

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