How does the Fourier transform of f(t)=exp(-ltl) behave at infinity?

[zezima1]

Homework Statement

Find the Fourier transform of f(t)=exp(-ltl)

Homework Equations

The expression for the Fourier transform.

The Attempt at a Solution

Applying the Fourier transform I get an expression, where I have to take the limit of t->-∞ of exp(-i[itex]\omega[/itex]t) - how do I do that?

 

[jbunniii]

Homework Statement

Find the Fourier transform of f(t)=exp(-ltl)

Homework Equations

The expression for the Fourier transform.

The Attempt at a Solution

Applying the Fourier transform I get an expression, where I have to take the limit of t->-∞ of exp(-i[itex]\omega[/itex]t) - how do I do that?

What is the expression you obtained?

 

[zezima1]

Yes I better post it - just tried to avoid it since I'm not good at latex. I get:
1/√2π(∫_-∞⁰exp(t-iωt)dt + ∫₀^∞exp(-t-iωt)dt)
=
1/√2π([1/(1-iω)exp(t-iωt)]_-∞⁰+[-1/(1+iω)exp(-t-iωt)]₀^∞

 

[jbunniii]

Yes I better post it - just tried to avoid it since I'm not good at latex. I get:
1/√2π(∫_-∞⁰exp(t-iωt)dt + ∫₀^∞exp(-t-iωt)dt)
=
1/√2π([1/(1-iω)exp(t-iωt)]_-∞⁰+[-1/(1+iω)exp(-t-iωt)]₀^∞

OK, that looks right. But look more carefully: the limit you need is not

[tex]\lim_{t \rightarrow \infty} \exp(-i\omega t)[/tex]
but rather
[tex]\lim_{t \rightarrow \infty} \exp(-|t|)\exp(-i\omega t)[/tex]

 

[zezima1]

oh god yes :)

But still: What if exp(-iwt) tends to infinity? Okay I don't think it does since its a sum of a real cos and an imaginary sin but still I want to know how to calculate the limit of the term with the complex exponential :)

 

[zezima1]

Either way I get:

1/√2π *(1/(1+ω²)

Do you also get that? :)

 

[zezima1]

nvm that was almost correct but got the right one now :)

 

[jbunniii]

[itex]\exp(-i\omega t)[/itex] has no limit as [itex]t \rightarrow \infty[/itex]. It just spins endlessly around the unit circle in the complex plane.

[itex]\exp(-|t|)\exp(-i\omega t)[/itex] spirals endlessly around the origin, but the [itex]\exp(-|t|)[/itex] factor pushes it closer and closer to the origin as t increases toward [itex]\infty[/itex] (or decreases toward [itex]-\infty[/itex]). This is why it has a limit even though one of its factors does not.

You can see this more formally by looking at the magnitude:

[tex]|\exp(-|t|)\exp(-i\omega t)| = |\exp(-|t|)|\cdot |\exp(-i\omega t)| = |\exp(-|t|)| \cdot 1 = |\exp(-|t|)| = \exp(-|t|)[/tex]

which goes to zero as t goes to either [itex]+\infty[/itex] or [itex]-\infty[/itex]. And the magnitude of a function goes to zero if and only if the function itself goes to zero.