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January 14th, 2012, 06:31 AM   #11
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Re: solve differential lap ace where f(t)=square wave

How is this?
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 January 14th, 2012, 02:38 PM #12 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: solve differential lap ace where f(t)=square wave Still a problem... $\frac{e^{-s}}{s}-\frac{e^{-2s}}{s}=e^{-3s}$ ??????
January 15th, 2012, 01:45 AM   #13
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Re: solve differential lap ace where f(t)=square wave

yeah thought I may be going in the wrong direction there.
how is this for a strong sprint to the finish?
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January 15th, 2012, 07:44 AM   #14
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Re: solve differential lap ace where f(t)=square wave

Quote:
 Originally Posted by defunktlemon yeah thought I may be going in the wrong direction there. how is this for a strong sprint to the finish?
Completely wrong. A mix of s and t in one equation is not good.

 January 16th, 2012, 03:08 PM #15 Member   Joined: Dec 2011 From: Oxford, UK Posts: 48 Thanks: 0 Math Focus: Linear Algebra / 3d Math / Geometry & Trig / Topology Re: solve differential lap ace where f(t)=square wave any hints please - I'm not seeing this
 January 17th, 2012, 08:50 AM #16 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: solve differential lap ace where f(t)=square wave use the shifting property http://www.intmath.com/laplace-trans...properties.php
 January 24th, 2012, 04:26 PM #17 Senior Member   Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0 Re: solve differential laplace where f(t)=square wave Ok, seeing you're getting lots of errors along the solution, I'll try to go step by step, as clearly as possible. $y''- 3y' + 2y = S(t)$ where $S(t)= 1$ for $(1,2)$,$S(t)= 0$ otherwise. Then we can represent $S(t)$ by the Heaviside function $\theta(t)$ like this: $S(t)= \theta(t-1)-\theta(t-2)$ Thus we have $y''- 3y' + 2y = \theta(t-1)-\theta(t-2)$ We use the Laplce transform, setting $L \{ y \}(s)= x(s)$for an easier working. Using the rules of tranforms of derivatives and the transform of the Heaviside function, $L \{ \theta(t-a) \}(s)= \frac{e^{-sa}}{s}$ we get the following equation: $s^2 x(s) - sy(0) - y'(0) - 3(s x(s)-y(0)) + 2x(s)= \frac{e^{-s}-e^{-2s}}{s}$ Recalling y(0) = y'(0) = 0 we get $s^2 x(s) -3s x(s) + 2x(s)= \frac{e^{-s}-e^{-2s}}{s}$ Factor out $x(s)$ $x(s) (s^2 -3s + 2)= \frac{e^{-s}-e^{-2s}}{s}$ Note that $s^2 -3s + 2= (s-1)(s-2)$ and write $x(s)= \frac{e^{-s}-e^{-2s}}{s(s-1)(s-2)}$ Use partial fractions: $\frac{1}{s(s-1)(s-2)}= \frac{1}{2s} +\frac{1}{2(s-2)}-\frac{1}{s-1}$ Multiply the last expression through $e^{-s}-e^{-2s}$ to get $x(s)={e^{ - s}}\frac{1}{{2s}} + {e^{ - s}}\frac{1}{{2\left( {s - 2} \right)}} - {e^{ - s}}\frac{1}{{s - 1}} - {e^{ - 2s}}\frac{1}{{2s}} - {e^{ - 2s}}\frac{1}{{2\left( {s - 2} \right)}} + {e^{ - 2s}}\frac{1}{{s - 1}}$ Use the theorems $L \{ f(t-a) \}(s)= e^{-sa} L \{ f(t) \}(s)$ and $L \{ e^{ta} f(t) \}(s)= L \{f(t) \}(s-a)$ to obtain $y$. Remeber that $L \{ \theta(t-a) \}(s)= \frac{e^{-sa}}{s}$. You get $y= \frac{{\theta \left( {t - 1} \right)}}{2} + {e^{2\left( {t - 1} \right)}}\frac{{\theta \left( {t - 1} \right)}}{2} - {e^{t - 1}}\theta \left( {t - 1} \right) - \frac{{\theta \left( {t - 2} \right)}}{2} - {e^{2\left( {t - 2} \right)}}\frac{{\theta \left( {t - 2} \right)}}{2} + {e^{t - 2}}\theta \left( {t - 2} \right)$ You can group in terms of the Heaviside function to get $y= \theta \left( {t - 1} \right)\left[ {\frac{{{e^{2\left( {t - 1} \right)}}}}{2} - {e^{t - 1}} + \frac{1}{2}} \right] - \theta \left( {t - 2} \right)\left[ {\frac{{{e^{2\left( {t - 2} \right)}}}}{2} - {e^{t - 2}} + \frac{1}{2}} \right]$ So it is neater to read. You can see you actually have a function that is being shifted: $f(x)= \theta \left( x \right)\left[ {\frac{{{e^{2x}}}}{2} - {e^x} + \frac{1}{2}} \right]$ $f(x)= \frac{1}{2} \theta \left( x \right)\left( {e^x} -1 \right)^2$ So you actually have: $y= f(x - 1) - f(x - 2)$ Rings a bell? If you're up to it, check that indeed you have y''-3y'+2y = S(t) (You can use GeoGebra, setting $\theta(x)= \frac{1}{2}\left( \frac{|x|}{x}+1\right)$ as the heaviside function, $g(x)= {\frac{{{e^{2x}}}}{2} - {e^x} + \frac{1}{2}}$,$f(x)= \theta(x) g(x)$, and then $p(x)= f(x-1) - f(x-2)$. Input then $p''(x) - 3p'(x) +2p(x)$, and check what you get.)

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### laplace of square wave

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