I'd like to compute the response to a step function of a electrical/thermal system. Generally I can "easily" compute the transfer function $H$:

$$H(\omega) = \frac{V_{out}(\omega)}{V_{in}(\omega)}$$

Since the Fourier transform ($\mathcal{F}$) of the Heaviside function is (computed with WA):

$$\mathcal{F}(\theta(t)) = V_{in}(\omega) = \sqrt{\frac{\pi}{2}}\delta(\omega)+\frac{i}{\sqrt{2\pi}\omega}$$

Hence, noting $\mathcal{IF}$ the Inverse Fourier transform:

$$V_{out}(t) = \mathcal{IF} \left\{ \left( \sqrt{\frac{\pi}{2}}\delta(\omega)+\frac{i}{\sqrt{2\pi}\omega} \right) H(\omega) \right\}$$

To check my math I tried to compute the response for a simple RC system:

enter image description here

I should get the well known charge of the capacitor. The transfer function:

$$H(\omega) = \frac{1}{1+i\omega R C}$$

Computing the Inverse Fourier transform ($\mathcal{IF}$) with WA ($R=C=1$) I get:

enter image description here

This would be correct if we were going backward in time :/. So the question is... What am I doing wrong?

I did the same using Laplace Transforms and everything works fine... But I don't understand why.

P.S. I don't want another method, I just want to understand what's wrong in my approach.

P.S. the reason why I am using WA is that for my more complicated system I need to compute the Fourier transforms using WA.

  • $\begingroup$ This isn't the answer you are looking for, but this article on how to do a Discrete Inverse Laplace Transform for virtually any transfer function may be of interest to you. $\endgroup$ Commented Dec 23, 2015 at 15:36
  • $\begingroup$ Thank you for the interesting link! I'm still trying to understand why Laplace transforms are needed. Or better, why Fourier transforms don't work... $\endgroup$
    – Worldsheep
    Commented Dec 23, 2015 at 21:53
  • $\begingroup$ Are you familiar with Laplace Transforms? The Laplace and Fourier Transforms are quite similar, but I am not a good enough mathematician to describe the exact differences. EE's typically work in the s domain (Laplace transform) which would be the same as your H(w) equation if you replace replace jw with s. Also, you will probably get a better answer if you post this question on the dsp.stackexchange.com site. Those guys are in tune to this stuff. $\endgroup$ Commented Dec 23, 2015 at 22:47
  • $\begingroup$ Yes I noticed that EE always work with Laplace in these cases and when i've tried that, it worked fine! But intuitively, I would use Fourier. I'll follow your advice and I'll visit the other site! $\endgroup$
    – Worldsheep
    Commented Dec 24, 2015 at 8:44
  • 2
    $\begingroup$ You can find an answer to this question here: dsp.stackexchange.com/questions/27896/… $\endgroup$
    – Worldsheep
    Commented Dec 26, 2015 at 9:53

1 Answer 1


The main reason is likely due to Wolfram Alpha applying the inverse Fourier transform as a second Fourier transform. In fact, doing so "flips time" - as can be shown mathematically:

Defining the '''flip-time operator''' $\mathcal{P}$ that inverts time, $\mathcal{P}[f(t)] ↦ f(−t)$ $$\begin{align} \mathcal{F}^0 &= \mathrm{Id}, \quad \mathcal{F}^1 = \mathcal{F}, \\ \mathcal{F}^2 &= \mathcal{P}, \quad \mathcal{F}^4 = \mathrm{Id}, \\ \mathcal{F}^3 &= \mathcal{F}^{-1} = \mathcal{P} \circ \mathcal{F} = \mathcal{F} \circ \mathcal{P} \end{align}$$

Applying the fourier transform 3 times to the system will get you the version in normal time. Since waves are time consistent, it normally does not matter.


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