0
$\begingroup$

The FT of a unit step function is taken as:

$$ X(\omega) = \int_0^\infty e^{-j\omega t}dt = \frac{-1}{jw}e^{-j\omega t} \Biggr |_{0}^{\infty} = \frac{j}{\omega} $$

The transform only has the imaginary component and nearly (actually) all literature that I have read deem this transform as non-existent because of the missing real component.

Unfortunately, I have not found an intuitive explanation that describes why this transform is invalid/non-existent.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

The integral doesn't exist in the conventional sense. How did you evaluate the limit $\lim_{t\to\infty}e^{-j\omega t}$?

If one allows distributions (generalized functions), such as the Dirac impulse (and its derivatives), as Fourier transforms then the class of functions that can be transformed becomes much wider, and the unit step is one of the functions that has a Fourier transform in that sense. However, its transform is not given by the expression in your question. The Fourier transform of $u(t)$ is

$$\mathcal{F}\{u(t)\}=\pi\delta(\omega)+\frac{1}{j\omega}\tag{1}$$

Take a look at this answer for a derivation of $(1)$.

A simple sanity check also shows that the expression in your question cannot be the transform of $u(t)$. A purely imaginary Fourier transform corresponds to an odd time domain function, which $u(t)$ is not. The expression in your question is actually a valid Fourier transform, but of a different function. Its inverse transform is $x(t)=-\frac12\textrm{sgn}(t)$, which is obviously an odd function.

Improve this answer
$\endgroup$
5
  • $\begingroup$ did you mean sgn(t) - 1/2? $\endgroup$
    – Dan Boschen
    Apr 21, 2019 at 22:02
  • $\begingroup$ @DanBoschen: No, that function isn't odd, so it wouldn't have a purely imaginary Fourier transform. $\endgroup$
    – Matt L.
    Apr 22, 2019 at 9:04
  • $\begingroup$ It is -1/2 for all negative t and +1/2 for all positive t, so isn’t that an odd function (or more likely what am I not seeing?) $\endgroup$
    – Dan Boschen
    Apr 22, 2019 at 14:53
  • $\begingroup$ @DanBoschen: That would indeed be an odd function, but $\textrm{sgn}(t)-\frac12$ is $\frac12$ for positive $t$, and $-\frac32$ for negative $t$. Are you confusing $\textrm{sgn}(t)$ with $u(t)$? The sign function is $-1$ for $t<0$ and $1$ for $t>0$, so it is an odd function. If you subtract a non-zero constant from an odd function it's not odd anymore. $\endgroup$
    – Matt L.
    Apr 22, 2019 at 17:00
  • 1
    $\begingroup$ that is exactly what I was doing! All makes sense now, thanks $\endgroup$
    – Dan Boschen
    Apr 22, 2019 at 19:41
0
$\begingroup$

A Fourier transform splits strictly real input input into an even (strictly symmetric) and odd (strictly antisymmetric) components (around the origin or center or location where the basis cosine == 1 and the basis sine function == 0).

Therefore: If the FT result of strictly real input is strictly imaginary (real part == 0), the input is strictly odd, and can be decomposed into only sine functions.

It also has no real-valued DC offset.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.