Question

I am stuck for 2 days on this... Fourier transform of f(t) = sin(2t) / e^|t|

Answer 1

f(t) = sin(2t) / e^t * U(t) + sin(2t) / e^(-t) * U(-t)

= sin(2t) * e^-t * U(t) + sin(2t) * e^t * U(-t)

Answer 2

Here's the definition I have
\[\large F(\xi)=\mathcal{F}\left\{f(t)\right\}(\xi)=\int_{-\infty}^\infty f(t)e^{-i\xi t}~dt\]
Yours may be different, so make the necessary corrections.

You're given \(f(t)=\dfrac{\sin(2t)}{e^{|t|}}\).
This may make the integration easier:
\[\begin{align*}
\sin(2t)&=2\sin t\cos t\\
&=2\left(\frac{e^{it}-e^{-it}}{2}\right)\left(\frac{e^{it}+e^{-it}}{2}\right)\\
&=\frac{e^{2it}-e^{-2it}}{2}
\end{align*}\]
So, the Fourier transform of the given function is
\[\large{ \begin{align*}F(\xi)&=\frac{1}{2}\int_{-\infty}^\infty \frac{e^{2it}-e^{-2it}}{e^{|t|}}~dt\\
&=\frac{1}{2}\left[\int_{-\infty}^0 \frac{e^{2it}-e^{-2it}}{e^{-t}}~dt +\int_0^\infty \frac{e^{2it}-e^{-2it}}{e^t}~dt \right]\\

&=\frac{1}{2}\left[\int_{-\infty}^0 \left(e^{(1+2i)t}-e^{(1-2i)t}\right)~dt +\int_0^\infty \left(e^{-(1-2i)t}-e^{-(1+2i)t}\right)~dt \right]
\end{align*}}
\]

Answer 3

let me check

Answer 4

Hmm, I got total fourier transform = 0

Answer 5

Oh wait, I left out the \(\xi\) term... Just a sec

Answer 6

\[\large{ \begin{align*}F(\xi)&=\frac{1}{2}\int_{-\infty}^\infty \frac{e^{2it}-e^{-2it}}{e^{|t|}}~e^{i\xi t}~dt\\
\end{align*}}\]

Answer 7

So the integral should be
\[\large \frac{1}{2} \Bigg[ \int_{-\infty}^0 \left(e^{(1+2i)t}-e^{(1-2i)t}\right)e^{i\xi t}~dt\\
~~~~~~~~~~~~~~~~~~~~\large+\int_0^\infty \left(e^{-(1-2i)t}-e^{-(1+2i)t}\right)e^{i\xi t}~dt \Bigg]\]

Answer 8

I'll work out the first integral:
\[\large\int_{-\infty}^0 \left(e^{(1+2i)t}-e^{(1-2i)t}\right)e^{i\xi t}~dt\\
\large\int_{-\infty}^0 \left(e^{(1+(2+\xi)i)t}-e^{(1-(2-\xi)i)t}\right)~dt\\
\large\left[ \frac{e^{(1+(2+\xi)i)t}}{1+(2+\xi )i}-\frac{e^{(1-(2-\xi)i)t}}{1-(2-\xi )i } \right]_{-\infty}^0\\

\large \left[ \frac{1}{1+(2+\xi )i}-\frac{1}{1-(2-\xi )i } \right] - \left[ 0-0 \right]\\
\large \frac{-2(2-\xi )i}{1+(2+\xi )^2}
\]
I hope I'm doing this right, it's been a while...

Answer 9

The second integral would be done similarly. Finding a common denominator among the resulting terms should get you something like what WA is giving me:
http://www.wolframalpha.com/input/?i=FourierTransform%5BSin%5B2*t%5D%2FExp%5B%7Ct%7C%5D%2Ct%2Cw%5D

Answer 10

F(sin 2t) = pi/j * [$(w-2) - $(w+2) ]

F ( sin(2t) * e^-t * U(t) ) = pi/j * [$(w-1) - $(w+3) ]

similarily,

F ( sin(2t) * e^t * U(-t) ) = -pi/j * [$(w-1) - $(w+3) ]

Answer 11

we sure have different ways about this.. :)