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I have been given this problem to solve:

Define the function f(t) by $$ f(t) =\begin{cases} e^{-kt},& t \geq 0 \\ 0,& \text{otherwise}\end{cases} $$

where $k > 0$ is a real number.

Calculate the F.T. of df/dt in two ways: (i) by differentiating f(t) and then finding the F.T. of the result; and (ii) using the F.T. differentiation property.

I have managed to do part (i) and have come up with the answer of $\frac{jw}{jw+k}$.

I have come up with $F(w) = \frac{1}{k+jw}$ as the transform for f(t) but am struggling to complete part (ii) as I am unsure about the notation of the fourier differentiation property.

$$\frac{d(f(t))}{dt} \rightleftharpoons jwF(w)$$

Could someone explain the steps I need to take to solve part (ii), an explanation of the steps would suffice, I should then be able to finish the problem.

stuart194
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  • Your part (i) is incorrect. How did you obtain the $-1$? – Daniel Fischer Nov 10 '14 at 18:46
  • I differentiated f(t) to get $-ke^(-kt)$ then intergrated between infinity and zero – stuart194 Nov 10 '14 at 18:51
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    Two mistakes. First, the derivative is not that. You have a jump at $0$, so the function isn't classically differentiable. You must use the distributional derivative. And second, you must multiply with $e^{-jw}$ before you integrate to get the Fourier transform. You have the right Fourier transform later, so what you got would give you $-k\cdot F(w) = \frac{-k}{k+jw}$ for the Fourier transform of $\frac{df}{dt}$. That is a part of it, but something is still missing. Look at what part (ii) gave you, perhaps you can see what is missing from comparing to that. – Daniel Fischer Nov 10 '14 at 18:56
  • so my fourier transform of f(t) is okay? how are you changing that to the fourier transform for $\frac{df}{dt}$ using the differentiation property is what is really stumping me, I thought the equation I had for the property meant multiply by $jw$. I realise I made mistakes doing part (i) now. – stuart194 Nov 10 '14 at 19:16
  • Your Fourier transform of $f(t)$ is okay for one normalisation of the Fourier transform, $F(w) = \int f(t) e^{-jwt},dt$. There are other conventions, but that would only produce minor variations. Right, the differentiation property of the FT is that, when things are sensible, you have $$\mathscr{F}\left(\frac{df}{dt}\right)(w) = jw\cdot\mathscr{F}(f)(w).$$ Now, for the given $f$, things are sensible, but you can't get the Fourier transform of $\frac{df}{dt}$ by the Fourier integral. For $\frac{df}{dt}$ is not a function. It's a (tempered) distribution. – Daniel Fischer Nov 10 '14 at 19:39
  • So if I follow correctly you are saying the only way to actually get the F.T of $\frac{df}{dt}$ is by differentiating $f(t)$ then finding the F.T of that and that should give $\frac{-k}{k+jw}$? – stuart194 Nov 10 '14 at 19:54
  • No, it would give $1 + \frac{-k}{k+jw} = \frac{jw}{k+jw}$. You get the result from the differentiation property on the one hand. On the other, you can compute $\frac{df}{dt}$ (in the distributional sense!). Do you know what $\frac{df}{dt}$ is? – Daniel Fischer Nov 10 '14 at 19:58
  • I am not sure I follow you, I am looking to solve it firstly by differentiating $f(t)$. I was under the impression I could just do this normally, then finding the FT of that. Then for part two I am looking to use the differentiation property on whatever $\frac{df(t)}{dt}$ would be. – stuart194 Nov 10 '14 at 20:09
  • But $f(t)$ isn't differentiable in the classical sense. It has a jump at $0$. So when you differentiate $f$, you get a (tempered) distribution. You can then apply the Fourier transform to that tempered distribution, and if everything is done correctly, you obtain $$\mathscr{F}\left(\frac{df}{dt}\right)(w) = jw\cdot \mathscr{F}(f)(w) = \frac{jw}{k+jw}.$$ – Daniel Fischer Nov 10 '14 at 20:15
  • So you are telling me that the question I have been set is infact not actually possible? – stuart194 Nov 10 '14 at 20:18
  • No. It is doable if you have the necessary theory. You need the Fourier transform of tempered distributions. If you don't have that, $\mathscr{F}\left(\frac{df}{dt}\right)$ doesn't make sense. – Daniel Fischer Nov 10 '14 at 20:20
  • Okay, the fourier transform I want is actually $\frac{df(t)}{dt}$ and not $F(w)$ please can you relate what you are saying to the terms in my question as this is what I have to work with, it's not my choice unfortunately. Am I right in saying that to do part two all I have to do is multiply my original transform by $jw$? – stuart194 Nov 10 '14 at 20:29
  • I don't understand. You write you want the Fourier transform of $\frac{df}{dt}$. That's what I have denoted by $\mathscr{F}\left(\frac{df}{dt}\right)$. Just like $\mathscr{F}(f)$ denotes the Fourier transform of $f$, which you have denoted by $F$. – Daniel Fischer Nov 10 '14 at 20:33
  • I see so what I have called $\frac{d(f(t))}{dt} = \frac{df}{dt}$ I was confused because you have placed an $(w)$ after that and could not see how that fitted with what I had. – stuart194 Nov 10 '14 at 20:40
  • The Fourier transform of $\frac{df}{dt}$ is a function of $w$. So if we evaluate the function $\mathscr{F}\left(\frac{df}{dt}\right)$ at the point $w$, we obtain $\mathscr{F}\left(\frac{df}{dt}\right)(w)$. – Daniel Fischer Nov 10 '14 at 20:45
  • I see now, Thanks for your help. – stuart194 Nov 10 '14 at 20:55
  • For future reference differating something such as $f(t)$ in the question is simple providing you remember to add the delta dirac function when you do so. – stuart194 Nov 10 '14 at 22:24

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For future reference when differentiating something with a jump in it is important to add a delta dirac function

As for part (ii) it is a simple matter of multiplying the original transform by $jw$

stuart194
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