Question

(Fourier Transform) How do I take the Fourier Transform of a Heaviside/Unit Step function?

Answer 1

@dan815

Or how exactly would I start on a derivation of this?

Answer 2

If it's the typical definition of the step function, the integral can be written as
\[\large \int_0^1e^{-2\pi i\xi t}\,dt\]

Answer 3

I don't even remember how the unit step function is defined other than the symbol, one sec.

Answer 4

Yeah, I know how to define a piecewise function using a linear combination of unit step functions, but what's the integral definition/how did you figure out how to write it like that?

Answer 5

@SithsAndGiggles

Answer 6

I've seen the Heaviside function typically defined as
\[u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t<c\end{cases}\]
So if we take the most basic form, when \(c=0\), we have
\[\mathcal{F}\{u(t)\}=\int_{-\infty}^\infty u(t)e^{-2\pi i\xi t}\,dt=\int_0^11e^{-2\pi i\xi t}\,dt+\left\{\int_{-\infty}^0+\int_1^\infty\right\}0e^{-2\pi i\xi t}\,dt\]
so in fact we're left with computing
\[\mathcal{F}\{u(t)\}=\int_0^1e^{-2\pi i\xi t}\,dt\]

Answer 7

Oh, okay! That makes a lot more sense, thank you.

Answer 8

I don't want to complain/critique you, but I think what you said above is slightly wrong (and feel free to totally ignore that until I actually come back with a response, it has to do with the way you defined the Heaviside function in relation to the three integrals produced w/different bounds)

Answer 9

If there are any discrepancies between what I suggested and what you think the answer might be, it's probably due to different definitions of the transform. The Heaviside function's definition is standard, but the transform has many (admittedly confusing) forms.

Answer 10

The one I used here was
\[\mathcal{F}\{f(t)\}=\int_{-\infty}^\infty f(t)e^{-2\pi i\xi t}\,dt\]
which is equivalent to
\[\mathcal{F}\{f(t)\}=\frac{1}{2\pi}\int_{-\infty}^\infty f(t)e^{-i\xi t}\,dt\]
yet different from
\[\mathcal{F}\{f(t)\}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(t)e^{-i\xi t}\,dt\]
which is the definition used by W|A here: http://www.wolframalpha.com/input/?i=FourierTransform%5BHeavisideTheta%5Bt%5D%2Ct%2Cxi%5D

Answer 11

I feel like in order to get the three integrals that you had above, you would need a piecewise-defined function with three separate intervals. As I understand it, you have three integrals, one spanning from -infty<t<0, where the whole thing is presumed zero on that interval, and from 0 to 1, you have some nonzero expression or, the Heaviside is equal to one, and then, after that, you have an integral from 1 to infty which is also equal to zero, because another heviside cancels it out.

What this sounds like in my mind is:

\[f(x) = \left\{
\begin{array}{lr}
0, & -\infty <x<0\\
1, & 0<x<1\\
0, & 1<x<\infty
\end{array}
\right.\]

Answer 12

e.g. your piecewise-defined function as written, I think, would be nonzero forever after c. And as such, that integral from 1 to infinity would not be zero.

Answer 13

Oh sorry, I don't know what I was talking about :P I've been helping some students around with wavelet functions defined as
\[\phi(x)=\begin{cases}1&\text{for }0<x<1\\0&\text{otherwise}\end{cases}\]
and I mixed the two up.

(See the graph on the right of the second slide: http://www.cyut.edu.tw/teacher/ft00002/Wavelet1.pdf)

Answer 14

You're right, it should be
\[\int_{-\infty}^00\,dt+\int_0^\infty e^{-i\xi t}\,dt\]
or however your transform is defined.

Answer 15

Oh, okay! Thanks nonetheless, what you said made sense. I'll just make it a different question, but I'm trying to solve an integral equation.