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Laplace Transform of Step Function |
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| Jun21-12, 02:20 PM | #1 |
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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? |
| 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 |
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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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