# What does fourier transform of exponential function mean?

We just started with Fourier Transform in our course. Basically, we are doing this to break the function into its sine and cosine components. But for $$f(t) = e^{-\alpha |t|}$$ we get the transform to be $$g(\omega) = 2\alpha/({\alpha^2+\omega^2}).$$

What does this mean? Why aren't we getting sine or cosine wave expressions as answer?

Calculations are given on page 6 of https://web.stanford.edu/class/ee102/lectures/fourtran

• You can use {} to combine elements you want to include in an exponent. See my edit to demonstrate how.
– The Photon
Aug 29, 2019 at 18:21
• Why do you expect to get sine and cosines for $g(\omega)$? Aug 29, 2019 at 18:22
• @DanielSank I mean that's how we usually introduce Fourier Transform to a newbie. I am ready to accept the fact that answer is not a sine or a cosine function but what does this expression in $\omega$ mean then? Aug 29, 2019 at 18:26
• @user185887 Fourier transforms are rarely sine or cosine. If you have an arbitrary function $f(t)$, then you get the Fourier transform via $\tilde{f}(\omega) = \int f(t) \exp(-i \omega t) \, dt$. This new function $\tilde{f}$ would only be a sine or cosine in a few specific cases. Maybe you're thinking about the fact that the original function of time can be written as $f(t) = \int \tilde{f}(\omega) \exp(i \omega t) \, (d\omega / 2\pi)$, which can be rewritten in terms of sine and cosine if you want. Aug 29, 2019 at 18:31
• @DanielSank But what does that expression for $g(\omega)$ mean then? Is it something like the function $f(t)$ is composed of a continuous frequency distribution and the weight for a particular $\omega$ is given by $g(\omega)$ ? Aug 29, 2019 at 18:37

In using the Fourier transform, you are assuming your function $$f(t)$$ takes on the form (sometimes differing depending on the convention) $$f(t)=\frac1{2\pi}\int_{-\infty}^\infty g(\omega)e^{i\omega t}\,\text d\omega$$

In other words, you are thinking of your function as a sum of sine and cosine terms, where the "amount" of each term is determined by $$g(\omega)$$. This $$g(\omega)$$ function is the Fourier transform of your function $$g(\omega)=\int_{-\infty}^\infty f(t)e^{-i\omega t}\,\text dt$$

Just because we are expressing the entire function $$f(t)$$ as a sum of sines and cosines does not mean its Fourier transform is sines and cosines. The Fourier transform $$g(\omega)$$ just tells us how much of each term makes up our function.

An analogy I like is that the Fourier transform tells us the recipe, i.e. the amount of ingredients you need. You need to know how much you need of each ingredient to make your final product. If you asked for a recipe and all they gave you were the ingredients (the sines and cosines) you would not know how to make the product.

If it is easier, just think about discrete Fourier sine series $$f(t)=\sum_{n=1}^\infty a_n\sin(nt)$$

The "Fourier transform" of this sum is your function $$a(n)=a_n$$ that is a function that goes from the natural numbers to the real numbers. The "Fourier transform" is not the entire sum. The entire sum is just your function. The "Fourier transform" is not the sine terms. Those are just the terms we are using to make up our function. Don't mix up the recipe for the ingredients, or even the whole product.

So, for your example, we have $$e^{-\alpha|t|}=\frac1{2\pi}\int_{-\infty}^\infty \frac{2\alpha}{\alpha^2+\omega^2}e^{i\omega t}\,\text d\omega$$

This means that there is an "amount" $$\frac1{2\pi}\, g(\omega)=\frac{2\alpha}{2\pi(\alpha^2+\omega^2)}$$

of each term $$e^{i\omega t}$$ in your function $$f(t)=e^{-\alpha|t|}$$.