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January 9th, 2017, 09:27 AM   #1
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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 09:33 AM.
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January 9th, 2017, 11:48 AM   #2
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$\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.
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January 9th, 2017, 06:32 PM   #3
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Quote:
Originally Posted by romsek View Post
$\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 ?
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January 9th, 2017, 09:04 PM   #4
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Quote:
Originally Posted by KaiL View Post
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 03:47 AM.
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January 9th, 2017, 11:30 PM   #5
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Quote:
Originally Posted by romsek View Post
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 03:47 AM.
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January 10th, 2017, 12:37 AM   #6
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Quote:
Originally Posted by KaiL View Post
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 03:44 AM.
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