Laplace transform property

Thread Starter

KaiL

Joined Aug 30, 2014
69
upload_2017-1-10_16-5-23.png

This is taken from a lecture note but I am not sure why d/dt = s

Assuming that I let n = 1

upload_2017-1-10_16-6-48.png

Here I am struck and I can't figure out why d/dt = s
 

MrAl

Joined Jun 17, 2014
11,389
Hi,

Try finding the Laplace Transform of dv/dt, and that brings you to a function that has one part that disappears as t goes to infinity, and you are left with s*V(s). So the transform of df(t)/dt is s*F(s).

You should also try this with a known function like 1-e^(-t) which after transforming and multiplying by 's' and then taking the inverse transform you should end up with the derivative of that original time function and so you can compare the two and observe what is happening while you do this.
 

Papabravo

Joined Feb 24, 2006
21,157
The statement is being slightly misconstrued. The double headed arrow does not mean "equal to"; at best it means equivalent to. Let me explain:
  1. d/dt is an operator on functions defined in the time domain
  2. s is an opeator on functions defined in the frequency domain
  3. What the statement says is that applying s to a function in the frequency domain, that is the Laplace transform of some time domain function, is the same as applying d/dt to that time domain function.
  4. As MrAl suggested this proposition is easy to demonstrate with an example. The formal proof is also reasonably accessible.
 
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