My Math Forum Laplace transform of a forced square wave

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 December 4th, 2016, 12:33 PM #1 Newbie   Joined: Dec 2016 From: United States Posts: 5 Thanks: 0 Laplace transform of a forced square wave Hi, I am working on a problem involving an initial value problem with a forced square wave. I am trying to take the laplace inverse of [e^(-(pi)s)] / [s^3+s] From what I can tell, I can't do the partial fractions method to split the denominator and it doesn't match anything in my table of transforms. Any help is appreciated. Thanks!
December 4th, 2016, 12:47 PM   #2
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 Originally Posted by lcollins19 Hi, I am working on a problem involving an initial value problem with a forced square wave. I am trying to take the laplace inverse of [e^(-(pi)s)] / [s^3+s] From what I can tell, I can't do the partial fractions method to split the denominator and it doesn't match anything in my table of transforms. Any help is appreciated. Thanks!
1) note that $u(t-t_0)f(t-t_0) \Leftrightarrow e^{-s t_0} ~F(s)$

2) use partial fractions to break up $\dfrac{1}{s(s^2+1)}$ into functions that will appear in your table.

3) use your table to find the appropriate functions in the time domain

4) apply the time delay appropriate for the $e^{-\pi s}$ factor.

Last edited by romsek; December 4th, 2016 at 01:03 PM.

 December 4th, 2016, 12:51 PM #3 Global Moderator   Joined: May 2007 Posts: 6,759 Thanks: 696 $\displaystyle \frac{1}{s^3+s}=\frac{1}{s}-\frac{s}{s^2+1}$ Does that help?
 December 4th, 2016, 02:54 PM #4 Newbie   Joined: Dec 2016 From: United States Posts: 5 Thanks: 0 Thanks for the replies. I should have said I had already done the partial fractions with numerator = 1, and I know that the inverse is 1 - cos(t). I guess what I don't understand is the time shift. Could you explain taking "the time delay appropriate for the e^(-pis) factor"? Last edited by lcollins19; December 4th, 2016 at 03:09 PM.
December 4th, 2016, 03:22 PM   #5
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 Originally Posted by lcollins19 Thanks for the replies. I should have said I had already done the partial fractions with numerator = 1, and I know that the inverse is 1 - cos(t). I guess what I don't understand is the time shift. Could you explain taking "the time delay appropriate for the e^(-pis) factor"?
suppose you have a Laplace transform pair

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

then

$f(t-t_0) \overset{\mathscr{L}}{\Leftrightarrow} \large e^{{-s t}_0}~\normalsize F(s)$

Last edited by romsek; December 4th, 2016 at 03:25 PM.

December 4th, 2016, 05:50 PM   #6
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 Originally Posted by romsek suppose you have a Laplace transform pair $f(t) \overset{\mathscr{L}}{\Leftrightarrow} F(s)$ then $f(t-t_0) \overset{\mathscr{L}}{\Leftrightarrow} \large e^{{-s t}_0}~\normalsize F(s)$

I seem to have been overthinking this problem.

Is it correct that the answer is just u(t-pi)[1-cos(t-pi)] ?

Thanks for all the help!

December 4th, 2016, 06:48 PM   #7
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 Originally Posted by lcollins19 I seem to have been overthinking this problem. Is it correct that the answer is just u(t-pi)[1-cos(t-pi)] ? Thanks for all the help!
you can simplify it a bit further

$\cos(t-\pi) = \cos(t)\cos(\pi) + \sin(t)\sin(\pi) = -\cos(t)$

so you end up with

$(1+\cos(t))u(t-\pi)$

 December 5th, 2016, 01:22 PM #8 Newbie   Joined: Dec 2016 From: United States Posts: 5 Thanks: 0 I have a second question that may deserve its own thread, but thought I'd ask here first as it is related and you were very helpful. Similar IVP, but this time with different coefficients leading to a third term in LHS of ODE: $\displaystyle y'' + 0.1y' + y = u(t) + 2 * sum[ (-1)^k * u(t-k*pi) ]$ where sum is from k=1 to inf (sorry for lack of formatting) I have worked it out as before and have taken the Laplace transform and done partial fractions from there. I end with: $\displaystyle 1/s - [0.1] / [(s+.05)^2 + (sqrt(.9975))^2] - s / [(s+.05)^2 + (sqrt(.9975))^2]$ I have split the numerators like this in an effort to match one of the table values. Am I going about this in the correct manner? I can't see how to manipulate the numerators to get the correct values to get the Laplace inverse to be either the cos(at) or sin(at) functions. I assume that once I have done this I will again apply the time delay as before for the second half of my equation involving the summation of step functions? Thanks for the help!
December 5th, 2016, 03:50 PM   #9
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 Originally Posted by lcollins19 I have a second question that may deserve its own thread, but thought I'd ask here first as it is related and you were very helpful. Similar IVP, but this time with different coefficients leading to a third term in LHS of ODE: $\displaystyle y'' + 0.1y' + y = u(t) + 2 * sum[ (-1)^k * u(t-k*pi) ]$ where sum is from k=1 to inf (sorry for lack of formatting) I have worked it out as before and have taken the Laplace transform and done partial fractions from there. I end with: $\displaystyle 1/s - [0.1] / [(s+.05)^2 + (sqrt(.9975))^2] - s / [(s+.05)^2 + (sqrt(.9975))^2]$ I have split the numerators like this in an effort to match one of the table values. Am I going about this in the correct manner? I can't see how to manipulate the numerators to get the correct values to get the Laplace inverse to be either the cos(at) or sin(at) functions. I assume that once I have done this I will again apply the time delay as before for the second half of my equation involving the summation of step functions? Thanks for the help!
The left hand side should be staightforward.

Taking the Laplace Transform and collecting terms I get

$\mathscr{L}\left(y'' + 0.1y' + y \right) =s^2 Y(s)+s (0.1 Y(s)-y(0))+Y(s)-y'(0)-0.1 y(0)$

The right hand side is a bit more complicated

$\mathscr{L}\left(u(t)\right) = \dfrac 1 s$

then we have

$\mathscr{L}\left(2 \displaystyle{\sum_{k=1}^\infty}~(-1)^k u(t - k \pi)\right)= 2\displaystyle{\sum_{k=1}^\infty}\dfrac{(-1)^k e^{k \pi s}}{s} = \dfrac {2}{s}\displaystyle{\sum_{k=1}^\infty}(-1)^k e^{-k \pi s} = \dfrac 2 s \dfrac{1}{e^{\pi s}+1}$

so in total for the right hand side we get

$\mathscr{L}\left(u(t) + 2 \displaystyle{\sum_{k=1}^\infty}~(-1)^k u(t - k \pi)\right) = \dfrac 1 s\left(1 + \dfrac{2}{e^{\pi s}+1}\right)$

I leave it to you to fill in the initial conditions if you have them and finish up the algebra.

 Tags forced, laplace, square, transform, wave

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