c27 - 18.03 Class 27, April 17, 2006 Laplace Transform II:...

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18.03 Class 27, April 17, 2006 Laplace Transform II: inverse transform, t-derivative rule, use in solving ODEs; partial fractions: cover-up method; s-derivative rule. Definition: F(s) = L[f(t)] = integral_{0-}^infty f(t) e^{-st} dt , Re(s) >> 0 Rules: L is linear: L[af(t)+bg(t)] = aF(s) + bG(s) F(s) essentially determines f(t) s-shift: L[ e^{at}f(t) = F(s-a) t-derivative: L[f'(t)] = s F(s) - f(0+) if we omit the singularity of f'(t) at t = 0 . Computations: L[1] = 1/s L[e^{as}] = 1/(s-a) L[cos(omega t)] = s/(s^2+omega^2) L[sin(omega t)] = omega/(s^2+omega^2) L[delta(t-a)] = e^{-as} [1] The t-derivative rule: Compute: L[f'(t)] = integral_{0-}^infty f'(t) e^{-st} dt u = e^{-st} du = -s e^{-st} dt dv = f'(t) dt v = f(t) ... = e^{-st} f(t) |_{0-}^infty + s integral f(t) e^{-st} dt We continue to assume that f(t) doesn't grow too fast with t (so that the integral defining F(s) converges for Re(s) sufficiently large). This means that for s sufficiently large, the evaluation of the first term at infinity becomes zero. Since we are always assuming rest initial conditions, the evaluation at zero is also zero. Thus ... = s F(s)
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Now, what is f'(t) ? If f(t) has discontinuities, we must mean the generalized derivative. There is one discontinuity in f(t) that we can't just wish away: f(0-) = 0 , while we had better let f(0+) be whatever it wants to be. We have to expect a discontinuity at t = 0 . Just to keep the notation in bounds, let's suppose that
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This note was uploaded on 01/31/2011 for the course MAT 17A taught by Professor Staff during the Winter '08 term at UC Davis.

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c27 - 18.03 Class 27, April 17, 2006 Laplace Transform II:...

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