Fourier Transform of a Derivative

Thread Starter

Fraser_Integration

Joined Nov 28, 2009
142
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.
 

Attachments

Georacer

Joined Nov 25, 2009
5,182
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.
 

Thread Starter

Fraser_Integration

Joined Nov 28, 2009
142
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
 

Thread Starter

Fraser_Integration

Joined Nov 28, 2009
142
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!
 

Thread Starter

Fraser_Integration

Joined Nov 28, 2009
142
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.
 

Georacer

Joined Nov 25, 2009
5,182
Nice one. Just correct the details: It's \(e^{-kt}\) for \(0 \leq t < \infty,\text{ not }0<t<8\). (I know you wrote it sideways).
 
Top