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Laplace Transform of Step Function

 
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Jun21-12, 02:20 PM   #1
 

Laplace Transform of Step Function


1. The problem statement, all variables and given/known data

Solve

y'' + y = f(t), y(0)=0, y'(0)=1,

f(t)=
(0 for 0<t<pi)
(1 for pi<t<2pi)
(0 for t>2pi)

3. The attempt at a solution

y'' + y = upi(t)-u2pi(t)

s2L{y} -sy(0) -y'(0) +L{y} = L{upi(t)} -Lu2pi(t)}

L{y}(s2+1) -1 = (e-pi*s/s) -(e-2pi*s/s)

L{y} = (e-pi*s/s(s2+1)) -(e-2pi*s/s(s2+1)) +1/(s2+1)

This is where I get stuck... I'm assuming that I can factor out the e terms separately, then use decomposition of partial fractions to separate 1/s(s2+1), but when I do that I get meaningless values for A and B.

1/s(s2+1) = A/s + B/s2+1

1= A(s2+1) +Bs

1 = As2 +Bs +A

From that I can infer that A = 1, but also that A=0, and B=0.

What am I doing wrong?
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Jun21-12, 02:33 PM   #2
 
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The partial fraction expansion should be
$$\frac{1}{s(s^2+1)} = \frac{A}{s} + \frac{Bs+C}{s^2+1}$$ because the second term has a quadratic in the denominator.
Jun21-12, 03:27 PM   #3
 
Ah, thank you, I think I got it.

y = upi(t) -upi(t)cos(t-pi) -u2pi(t) +u2pi(t)cos(t-2pi) +sin(t)
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