Fourier Transform-1

Question

I am trying to solve a Fourier transform problem and I am stuck. The problem is:

$$f(t)= \frac{\sin(2t)}{e^{|t|}}.$$

I have used integration, but the answer that I come up with is different than that of mathematica, etc.

Answer 1

We want to evaluate

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\omega t}\sin(2t)e^{-|t|}\,dt.$$

The simplest approach - I think - is to introduce the complex exponential. Recognize that $\sin(x) = \text{im}(e^{ix})$, where $\text{im}$ means to take the imaginary part. So the equivalent problem is to evaluate

$$\text{im}\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\omega t}e^{2it}e^{-|t|}\,dt\right).$$

This is an easy integral to manipulate due to properties of exponentials! First let's break our integral into two parts because of the pesky absolute value. This gives us

$$\int_{-\infty}^0 e^{-i\omega t}e^{2it}e^{t}\,dt + \int_0^{\infty}e^{-i\omega t}e^{2it}e^{-t}\,dt.$$

Now we can combine our exponentials to get

$$\int_{-\infty}^0e^{(-i\omega+2i+1)t}\,dt + \int_0^{\infty} e^{(-i\omega+2i-1)t}\,dt.$$

Can you take it from here? Remember to take the imaginary part after doing the above computation.

Answer 2

With $f(t)=\sin(2t)e^{-|t|}$ one simple option to compute the Fourier transform is to use the Fourier transform $G(\omega)$ of $g(t)=e^{-|t|}$ (I leave this part up to you), and then use the modulation property of the Fourier transform:

$$F(\omega)=\frac{1}{2i}\left[G(\omega-2)-G(\omega+2)\right]$$

(because $\sin(2t)=\frac{1}{2i}(e^{2it}-e^{-2it})$)