Question

Fourier Transform of f(t) = t h(t), where h(t) is the heaviside step function.

Answer 1

\[\int\limits_{0}^{\infty}t e ^{iwt}\]

Answer 2

But how do I solve it? Im not quite sure how to do it by hand

Answer 3

What's your definition for the step function?

Answer 4

The way we've been using it is 0 for t < 0, 1 for t > 0. Sorry for the late response

Answer 5

Okay, so it's the standard definition. Your integral setup is correct. Try integrating by parts.

Answer 6

Where \[u = t, du = dt; dv = e ^{iwt} dt, v = \frac{ 1 }{ iw } e ^{iwt}\] \[\left[ \frac{ t }{ iw }e ^{iwt} \right] 0 \to \infty - \int\limits_{0}^{\infty} \frac{ 1 }{ wt }e ^{iwt}dt\]

Answer 7

I feel like doing it the other way made it worse. These integrals go to infinity o_0

Answer 8

If I recall correctly, there's a formula for Fourier transforms that says
\[\large\mathcal{F}\left\{t~u(t)~e^{-\alpha t}\right\}=\frac{1}{(\alpha+i\omega)^2}\]
In this case, you'd have \(\alpha=0\), so it looks like that integral should converge...

Answer 9

In mathematica, I got something in terms of a delta'(w), one sec...

Answer 10

Yes I'm getting the same result,
\[-\frac{1}{\sqrt{2\pi}\omega^2}-i\sqrt\frac{\pi}{2}\delta'(\omega)\]

Answer 11

Yet integrating manually (via Mma, at least) yields \(-\dfrac{1}{\omega^2}\), which agrees with that formula.

Answer 12

Can you explain that part please? I wasnt sure how to get the complicated result but I see now how to get the result from the transform formula you showed me just now

Answer 13

I'm trying to figure out the difference myself ;)

Answer 14

Haha, alright, thanks for the help again

Answer 15

And where can I find that transform formula/others like it by chance?

Answer 16

Here's one table I just found (bottom of third page): http://uspas.fnal.gov/materials/11ODU/FourierTransformPairs.pdf

Answer 17

I think the discrepancy is that Mathematica uses a different definition for the transform.

Answer 18

Ah, just missed that one. Thank you

Answer 19

Yeah, it uses the different coefficient =/

Answer 20

What Mma uses: http://mathworld.wolfram.com/FourierTransform.html

What you seem to be using:
\[\large\mathcal{F}\{f(t)\}:=\int_{-\infty}^\infty f(t)e^{-i\omega t}~dt\]

Answer 21

I think, in your work above, you're missing the negative in the exponent. This should fix the result.

Answer 22

I blame the teacher. He has it opposite and I have been trying to figure out why. He uses positive for t-> frequencey and negative for backwards. Then he uses negative/positive for x -> k

Answer 23

Why does the MM answer equal the simpler answer?

Answer 24

Im not sure what to do with delta prime... or how to understand it anyways

Answer 25

I think the software uses a different algorithm for `FourierTransform[]` than it does for `Integrate[]`.

Answer 26

It might have to do with its definition of the step function, which seems to be zero for \(t<0\) and 1 for \(t>0\) as opposed to \(\le0\) or \(\ge1\).

Answer 27

Gotcha, ill try to look into some more. Thanks for everything!

Answer 28

Wait. Does the above Fourier Transform formula work if e isnt explicitly in the function?

Answer 29

f(t) = t h(t)

Answer 30

You can always write any function that doesn't explicitly contain an exponential function by tacking on a factor of \(e^0\).

Answer 31

True point

Answer 32

So what we could have done was multiply the transform of t by the transform of u(t) ? This gives me....