Fourier Transform using duality property?

[Jake 7174]

Homework Statement

Find the Fourier transform of

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

where i is imaginary

Homework Equations

Duality Property F_(t) ↔ 2πf_(-ω) when f_(t) ↔ F_(ω)

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?

 

[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.

 

[Jake 7174]

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.

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?

 

[alivedude]

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

If this is correct?

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?

 

[Jake 7174]

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?

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) e^iωt dω

 

[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) e^iωt dω

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?

 

[Ray Vickson]

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?

What do YOU think?

 

[alivedude]

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?

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.

 

[Jake 7174]

What do YOU think?

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.

 

[Jake 7174]

The one you got first is actually more neat in this situation, when you solve this you will understand why. But let's go back to the problemWhat 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 $2 \pi$)

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

Now I claim that you can get your wanted Fourier transform out of this by doing only three things, one is putting your $a = 2$, the next thing is multiply both sides by $2 \pi$ and 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.

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

 

[alivedude]

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

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.