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 January 9th, 2017, 08:27 AM #1 Member   Joined: Nov 2014 From: SG Posts: 55 Thanks: 1 Laplace transform property Could anyone explain to me why d/dt = s when initial conditions are zero ? If possible , please show me an example why when initial condition are zero , d/dt =s. Thank you so much for your help Last edited by KaiL; January 9th, 2017 at 08:33 AM. January 9th, 2017, 10:48 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,584 Thanks: 1430 $\mathscr{L}\{f^{(n)}\} = s^n \mathscr{L}\{f\} - \displaystyle{\sum_{k=1}^n}~ s^{n-k}f^{(k-1)}(0)$ so all the $f^{(n-k)}(0),~k=1,n$ must equal zero. January 9th, 2017, 05:32 PM   #3
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Quote:
 Originally Posted by romsek $\mathscr{L}\{f^{(n)}\} = s^n \mathscr{L}\{f\} - \displaystyle{\sum_{k=1}^n}~ s^{n-k}f^{(k-1)}(0)$ so all the $f^{(n-k)}(0),~k=1,n$ must equal zero.
Hi
Thank you for your reply but I still don't really get it.

$f^{(k-1)}(0)=0, k=1$ since initial conditions are zero --> I understand this part.

Since k = 1 and subbing it to this equation

$\mathscr{L}\{f^{(n)}\} = s^n \mathscr{L}\{f\} - \displaystyle{\sum_{k=1}^n}~$$s^{(n-1)} f^{(n-1)}(0) Then I am not sure why is n equal to 0 ? and why d/dt = s ? January 9th, 2017, 08:04 PM #4 Senior Member Joined: Sep 2015 From: USA Posts: 2,584 Thanks: 1430 Quote:  Originally Posted by KaiL Hi Thank you for your reply but I still don't really get it. f^{(k-1)}(0)=0, k=1 since initial conditions are zero --> I understand this part. Since k = 1 and subbing it to this equation \mathscr{L}\{f^{(n)}\} = s^n \mathscr{L}\{f\} - \displaystyle{\sum_{k=1}^n}~$$s^{(n-1)}$ $f^{(n-1)}(0)$ Then I am not sure why is n equal to 0 ? and why d/dt = s ?
Oh so you don't get it at all, not just about the initial conditions.

Consider the transform of $f^\prime(t)$

$\mathscr{L}\{f^\prime(t)\} = \displaystyle{\int_0^\infty}~f^\prime(t) e^{-s t}~dt$

$u = e^{-s t},~du = -se^{-st}$

$v = f^\prime(t)~dt,~v = f(t)$

$\left . e^{-s t}f(t) \right|_0^\infty + \displaystyle{\int_0^\infty}~f(t)s e^{-s t}~dt =$

$-f(0) + s F(s)$

You can use induction to prove the general formula.

Last edited by skipjack; January 10th, 2017 at 02:47 AM. January 9th, 2017, 10:30 PM   #5
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Quote:
 Originally Posted by romsek Oh so you don't get it at all, not just about the initial conditions. Consider the transform of $f^\prime(t)$ $\mathscr{L}\{f^\prime(t)\} = \displaystyle{\int_0^\infty}~f^\prime(t) e^{-s t}~dt$ $u = e^{-s t},~du = -se^{-st}$ $v = f^\prime(t)~dt,~v = f(t)$ $\left . e^{-s t}f(t) \right|_0^\infty + \displaystyle{\int_0^\infty}~f(t)s e^{-s t}~dt =$ $-f(0) + s F(s)$ You can use induction to prove the general formula.
I understand that
$\mathscr{L}\{f^\prime(t)\} = -f(0) + s F(s)$

Given that initial conditions are zero, $f(0) = 0$
So
$\mathscr{L}\{f^\prime(t)\} = 0 + s F(s)$
$\mathscr{L}\{f^\prime(t)\} = s F(s)$

$\mathscr{L}\{f(t)\} = F(s)$

$\mathscr{L}\{f^\prime(t)\} = s \mathscr{L}\{f(t)\}$

However, I still can't see the reason why $d/dt = s ?$

So sorry for asking this question again. Last edited by skipjack; January 10th, 2017 at 02:47 AM. January 9th, 2017, 11:37 PM   #6
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Quote:
 Originally Posted by KaiL I understand that $\mathscr{L}\{f^\prime(t)\} = -f(0) + s F(s)$ Given that initial conditions are zero, $f(0) = 0$ So $\mathscr{L}\{f^\prime(t)\} = 0 + s F(s)$ $\mathscr{L}\{f^\prime(t)\} = s F(s)$ $\mathscr{L}\{f(t)\} = F(s)$ $\mathscr{L}\{f^\prime(t)\} = s \mathscr{L}\{f(t)\}$ However, I still can't see the reason why $d/dt = s ?$ So sorry for asking this question again. It's not that they are equal, it's that they are Laplace transform pairs

$\dfrac{d}{dt} f(t) \overset{\mathscr{L}}{\Leftrightarrow} s F(s)$

Last edited by skipjack; January 10th, 2017 at 02:44 AM. Tags laplace, property, transform Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post pass92 Differential Equations 0 November 28th, 2015 03:51 AM szz Differential Equations 1 November 2nd, 2014 03:18 AM capea Complex Analysis 2 August 23rd, 2013 10:43 AM Deiota Calculus 1 April 28th, 2013 10:28 AM gelatine1 Calculus 5 November 2nd, 2012 12:17 PM

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