MIT18_03S10_c27

MIT18_03S10_c27 - 18.03 Class 27, April 12, 2010 Laplace...

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18.03 Class 27 , April 12, 2010 Laplace Transform II 1. Delta signal 2. t-derivative rule 3. Inverse transform 4. Unit impulse response 5. Partial fractions 6. L[f'_r] Laplace Transform: F(s) = int_0^\infty f(t) e^{-st} dt We saw that this improper integral may only converge for Re(s) > a, for some a depending upon f(t) . The smallest such a gives the "region of convergence." Computations so far: L[1] = 1/s L[t^n] = n!/s^{n+1} L[e^{rt}] = 1/(s-r) L[cos(t)] = s / (s^2 + omega^2) L[sin(t)] = omega / (s^2 + omega^2) We also have Rule 1 (Linearity): af(t) + bg(t) -----> aF(s) + bG(s). Rule 2 (s-shift): L[e^{rt}f(t)] = F(s-r) Warm-up: What is the Laplace transform of f(t) = e^{-t} cos(3t) ? We could do this by writing it as (1/2)( e^{(-1+3i)t} + e^{(-1-3i)t} ) but it's a bit easier to use the s-shift : cos(3t) ----> s / (s^2 + 9) so e^{-t} cos(3t) ----> (s+1)/( (s+1)^2 + 9 ) If you like you can "uncomplete the square" and write this as = (s+1) / (s^2 + 2s + 10) [1] The delta function: For b > 0 , L[delta(t-b)] = integral_0^infty delta(t-b) e^{-st} dt What could this mean? If f(t) is continuous at t = b , delta(t-b) f(t) = f(b) delta(t-b) .
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so integral delta(t-b) f(t) dt = f(b) u(t-b) + const or, if a < b < c , integral_a^c delta(t-b) f(t) dt = f(b) . In some accounts, this is the DEFINITION of the delta function. In our situation, you get L[delta(t-b)] = e^{-bs} In this case, the region of convergence is the entire complex plane: the limit you take to get the improper integral is constant as soon as t > b. There is a problem when b = 0 . delta(t) e^{-st} = delta(t) for any s , but int_0^infty delta(t) dt = u(infty) - u(0) and u(0) is indeterminate. We want the formula that worked for
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MIT18_03S10_c27 - 18.03 Class 27, April 12, 2010 Laplace...

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