# Fourier Transform of a Derivative

Discussion in 'Math' started by Fraser_Integration, Dec 12, 2010.

1. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
Hello.

Quick question about a F.T. problem that is bugging me. See attached, it's part b) that I am a little unsure of

I know that the Fourier transform of exp(-kt) = 1 / (k + jw)

Then the F.T. of derivative of f(t) - which is (-k)*exp(-kt) will simply be the above multiplied by -k: (-k) / (k + jw)

But now using the derivative property of fourier transforms:

d/dt[f(t)] <==> jw[F(w)]

the transform of the derivative of f(t) should be jw multiplied by the transform of the original function, or in other words: (jw) / (k + jw)

These two answers are supposed to be equal but they're not! I cant re-arrange one into the other, and when I put dummy values in for k and w, I am left with a result of -1/1 when I take one away from the other

Can anyone spot where I went wrong please?

Thanks.

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2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
I don't believe you are allowed to multiply by -k both ends of the Transformation and say that's a derivative.
You see, you have written "exp(-kt) = 1 / (k + jw)". This is not exactly true. Those expressions are not equal. They are connected by a transformation. The correct notation would be $e^{-kt} \Leftrightarrow^{\mathcal{L}} \frac{1}{k + jw}$.

You can multiply both ends with a constant because the transformation is linear, but that doesn't mean you are differentiating.

So, it is invalid to think that (-k) / (k + jw) is the derivative of the Fourier transformation.

3. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
but the derivative of e^-kt is -k*e^-kt

and when you take that to be the new f(t), you do end up with the -k / k + jw result if you just do another direct calculation i.e...

F(w) = Int [ -k*e(-kt)*e(-jwt) ]

sorry I dont know Latex, but if you do the sum, you will see that you come out with -k / k + jw

4. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6

I preceded it with a "The fourier transform of...."

5. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
I have the result now.

The correct expression for f'(t) should be -k*e(-kt) +sigma(t)

this is because f(t) has an instantaneous jump in it so when you differentiate that you get a sigma function, and the derivative of that is 1 which clears up the 1 I had left over!

6. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
You got me there. Can you write all that with more numbers and formulas? You know, strict math.

I 'd be in your debt.

7. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
hope you can read this.

the top line was obviously just f(t) = e^(-kt)

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8. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
the previous step was the easy proof of F.T. of e^(-kt) which is 1 / (k + jw)

and the point of the question is to say you don't need to do all that extra integration if you know the transform of the original function, because the transform of the derivative of that original function is simply 'jw' times the original transform.

9. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Nice one. Just correct the details: It's $e^{-kt}$ for $0 \leq t < \infty,\text{ not }0. (I know you wrote it sideways).

10. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
I can't do infinities very well, I know!