# Homework Help: Fourier transform of e^(-ltl)

1. Mar 1, 2012

### zezima1

1. The problem statement, all variables and given/known data
Find the fourier transform of f(t)=exp(-ltl)

2. Relevant equations
The expression for the fourier transform.

3. 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$\omega$t) - how do I do that?

2. Mar 1, 2012

### jbunniii

What is the expression you obtained?

3. Mar 1, 2012

### 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 + ∫0exp(-t-iωt)dt)
=
1/√2π([1/(1-iω)exp(t-iωt)]-∞0+[-1/(1+iω)exp(-t-iωt)]0

Last edited: Mar 1, 2012
4. Mar 1, 2012

### jbunniii

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

$$\lim_{t \rightarrow \infty} \exp(-i\omega t)$$
but rather
$$\lim_{t \rightarrow \infty} \exp(-|t|)\exp(-i\omega t)$$

5. Mar 1, 2012

### 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 :)

6. Mar 1, 2012

### zezima1

Either way I get:

1/√2π *(1/(1+ω2)

Do you also get that? :)

7. Mar 1, 2012

### zezima1

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

8. Mar 1, 2012

### jbunniii

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

$\exp(-|t|)\exp(-i\omega t)$ spirals endlessly around the origin, but the $\exp(-|t|)$ factor pushes it closer and closer to the origin as t increases toward $\infty$ (or decreases toward $-\infty$). 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:

$$|\exp(-|t|)\exp(-i\omega t)| = |\exp(-|t|)|\cdot |\exp(-i\omega t)| = |\exp(-|t|)| \cdot 1 = |\exp(-|t|)| = \exp(-|t|)$$

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