Fourier Transform using duality property?

1. Feb 25, 2016

Jake 7174

1. The problem statement, all variables and given/known data
Find the Fourier transform of

x(t) = 4 / (4 - i*t)^2

where i is imaginary

2. Relevant equations

Duality Property F(t) ↔ 2πf(-ω) when f(t) ↔ F(ω)

3. The attempt at a solution

I am not sure if duality property is the way to solve this. I look at a list of properties and this seems to hold most promise. The issue is the book simply states the property and gives no example. I am not even completely sure how to interpret this property. Perhaps this is my issue. Can someone please help explain this property to me?

2. Feb 25, 2016

alivedude

What you have is a Fourier transform of something, can you see what? When you figure that out you can take a look at the formula for the inverse Fourier transform, try to manipulate it so you get an expression that works in the other direction.

3. Feb 25, 2016

Jake 7174

I have this table of pairs I see a similarity with the following

e^( -a * |t| ) ; where a∈ℝ > 0 ↔ 2a / (a^2 + I * ω)

Would I be correct in saying I can rewrite my original function as f(-ω)

4 / (4 - i * t)^2 = 4 / (4 + i * ω)^2 = (2 * 2) / (2^2 + i * ω)^2

Is this is correct?

Last edited: Feb 25, 2016
4. Feb 25, 2016

alivedude

No, not quite. But take a look at the transformation you found in your table, in your case $a=2$ right? Now, plug this into the inverse Fourier transform here

$f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i \omega t} d \omega$

and see if you can come up with something.

Hint: The Inverse transform looks pretty similar to the transform itself, doesn't it?

5. Feb 25, 2016

Jake 7174

I am a bit confused. The transform looks very similar to the original function with the exception of the denominator being squared. But what do I plug in for F(ω) is it

∫ (2 * 2) / (2^2 + i * ω) eiωt

6. Feb 25, 2016

Jake 7174

There is a pair that looks like it matches much nicerit is

t e ^(- at) u(t) ↔ ( 1 / (a + i * ω) )^2

should I be focusing on this pair?

7. Feb 25, 2016

Ray Vickson

What do YOU think?

8. Feb 25, 2016

alivedude

Yes! This is what we want to use here.

What I ment was that the formula for the inverse transform looks very similar to the formula for the transform itself, i'm sorry if I confused you.

But ok, in this case what you have is this (you forgot the $2 \pi$)

$f(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty}\frac{1}{(a+i\omega)^2}e^{i \omega t}d \omega = u(t)te^{-at}$

Now I claim that you can get your wanted Fourier transform out of this by doing only three things, one is putting your $a = 4$, the next thing is multiply both sides by $2 \pi$ and then by $4$, the last and most important thing you will have to figure it out yourself. But do these two steps first and then compare it with the formula for the Fourier transform. I think you got this.

Last edited: Feb 25, 2016
9. Feb 25, 2016

Jake 7174

I think I am very confused and looking for help. I am not trying to get someone to give me an answer. I am looking to understand. Thanks for YOUR help.

10. Feb 25, 2016

Jake 7174

This is a nasty integral. I'm grinding it out.

11. Feb 25, 2016

alivedude

It looks like I made a mistake when I looked at your problem, you had it right when you asked me if you should focus on that second pair. I have changed my post above now and I hope I didn't screw up somewhere else.

And you don't need to compute any integrals at all, all you need to do is manipulate this expression.

Last edited: Feb 25, 2016