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 December 30th, 2012, 12:07 PM #1 Newbie   Joined: Dec 2012 Posts: 3 Thanks: 0 Laplace transform Hi there, I just started to learn the laplace transform. I know that the transform for gamma(t)= 1/s But how do you perform a transformation on gamma(sin(?t)) ? Thanks in advance! Greetings, Frederic
 December 30th, 2012, 12:26 PM #2 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Laplace transform $\mathcal{L} (\Gamma (t))= \frac{1}{s}$ ?? I know that $\mathcal{L}(t^n)= \frac{\Gamma{(n+1)}}{s^{n+1}}$ or I think you mean that $\mathcal{L}(t)= \frac{1}{s^2}$ if you mean by gamma the laplace transformation ?
 December 30th, 2012, 12:32 PM #3 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Laplace transform $\mathcal{L}(\sin (wt))= \frac{w}{s^2+w^2}$
 December 31st, 2012, 06:05 AM #4 Newbie   Joined: Dec 2012 Posts: 3 Thanks: 0 Re: Laplace transform Im sorry, what I meant was the laplace transform of the unit step of sin(?t). So L( u(sin(?t))) = ? How can i deal with this problem?
 December 31st, 2012, 06:10 AM #5 Newbie   Joined: Dec 2012 Posts: 3 Thanks: 0 Re: Laplace transform Let me post it like this: $\lambda (\Gamma( sin(\pi t)))$
December 31st, 2012, 11:46 AM   #6
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Re: Laplace transform

Quote:
 Originally Posted by fredericP Im sorry, what I meant was the laplace transform of the unit step of sin(?t). So L( u(sin(?t))) = ? How can i deal with this problem?
Sorry never heard of unit step of sin (t) !

January 1st, 2013, 08:30 PM   #7
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Re: Laplace transform

Quote:
 Originally Posted by fredericP Im sorry, what I meant was the laplace transform of the unit step of sin(?t). So L( u(sin(?t))) = ? How can i deal with this problem?
I may be way off base here, but if the unit step is defined as

$U(x)=\left{0, \ x<0,
1, \ x \geq 0$

then $\ U(\sin(\pi t)) \$ is a periodic square wave with a period of 2. For example, it has a value of 1 from t = 0 to 1, 0 from t = 1 to 2, 1 from t = 2 to 3 and so on (I am being a bit sloppy here with overlap at the endpoints, but hopefully the waveform values are clear).

I will not try to take the LT of $\ U(\sin(\pi t)) \$ but I will take the LT of $\ U(t) \cdot U(\sin(\pi t)) \$ (at least according to how I defined it). Multiplying by U(t) makes the waveform equal to 0 for t<0. Otherwise, strange things tend to happen (convergence problems, I guess).

I had to dust off a (very) old textbook for this, but let's define a function, f, for the first period:

$f(t)=U(t)-U(t-1) \$ whose LT is $\ F(s)=\frac{1}{s}-\frac{e^{-s}}{s}=\frac{1-e^{-s}}{s}$

Now we add in the periodic nature of the waveform by noting that:

$f_p(t)=f(t)+f(t-2)+f(t-4)+f(t-6)+... \$ and the LT of $\ f_p(t) \$ is:

$F_p(s)=F(s)(1+e^{-2s}+e^{-4s}+e^{-6s}+...)$

Since $\ (1+e^{-2s}+e^{-4s}+e^{-6s}+...)=\frac{1}{1-e^{-2s}}$

$F_p(s)=\frac{F(s)}{1-e^{-2s}}=\frac{1-e^{-s}}{s(1-e^{-2s})}=\frac{1-e^{-s}}{s(1-e^{-s})(1+e^{-s})}=\frac{1}{s(1+e^{-s})}$

So $\mathcal{L}(U(t) \cdot U(\sin(\pi t)))=\frac{1}{s(1+e^{-s})}$

Do you know if this is anywhere close to the answer?

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