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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.

Worldsheep
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  • This isn't the answer you are looking for, but this article on how to do a [Discrete Inverse Laplace Transform](http://iowahills.com/A9DiscreteInverseLaPlace.html) for virtually any transfer function may be of interest to you. – user5108_Dan Dec 23 '15 at 15:36
  • 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... – Worldsheep Dec 23 '15 at 21:53
  • 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. – user5108_Dan Dec 23 '15 at 22:47
  • 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! – Worldsheep Dec 24 '15 at 08:44
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    You can find an answer to this question here: http://dsp.stackexchange.com/questions/27896/response-of-a-system-to-a-step-function-heaviside/27903?noredirect=1#comment52176_27903 – Worldsheep Dec 26 '15 at 09:53
  • transfer functions are based on Laplace transform, not Fourier. This is by definition. – Nasser Dec 27 '15 at 12:04
  • Sounds strange that you can just define things like that. Do you have any reference? – Worldsheep Dec 27 '15 at 12:18
  • @Worldsheep Both transforms can be seen as the integral from $-\infty$ to $+\infty$. However for the Laplace transform it is assumed that the function is equal to zero for $t<0$, such that the part before zero will not contribute to the integral. When taking the inverse Laplace transform it is also assumed that the output is equal to zero before $t=0$, which can be seen as the system being at rest in the steady state (assuming it is stable). – fibonatic Dec 27 '15 at 14:58

1 Answers1

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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.

Mark
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