MEC702_Wk_9_Lect_2 - MEC702 Wk 9 Lect 2 Laplace Transforms...

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Unformatted text preview: MEC702 Wk 9 Lect 2 Laplace Transforms Laplace transforms are invaluable for any engineer's mathematical toolbox as they make solving linear ODEs and related IVP, as well as systems of linear ODEs, much easier. Applications abound: electrical networks, springs, mixing problems, signal processing, and other areas of engineering and physics. Solving ODE using Laplace transform 1. The given ODE is transformed into an algebraic equation, called the subsidiary equation. 2. The subsidiary equation is solved by purely algebraic manipulations. 3. The solution in Step 2 is transformed back, resulting in the solutions of the given problem. Advantages of Laplace Transform 1. Problems are solved more directly. IVPs are solved without going through a general solution. Nonhomogeneous ODEs are solved without rst solving the corresponding homogeneous ODE. 2. More importantly, the use of the unit step function and Dirac's delta make the method particularly powerful for problems with inputs (driving forces) that have discontinuities or represent short impulses or complicated periodic functions. 1 Background on Integrals and Limits Proper Integral A proper integral on a function f (t) over the interval [a, b] is of the form ˆ b f (t) dt. a This type of integral is easily solved using the integration methods. Improper Integral A improper integral on a function g(t) is of the form ´∞ a ´∞ or f (t) dt −∞ or f (t) dt ´b −∞ f (t) dt. Improper integrals can not be solved directly using integration methods, but there it can be evaluated according to the theorem below. Theorem. Let f (t) be a function over the interval can be solved in the following way: 1. 2. 3. ´∞ a ´b b→+∞ a f (t) dt = lim ´b −∞ ´∞ −∞ f (t) dt. ´b a→−∞ a f (t) dt = lim f (t) dt = ´ (−∞, ∞). Then the integrals f (t) dt. 0 −∞ f (t) dt + ´∞ 0 f (t) dt. Limits as Innity The limits at ininity for the exponential function lim t t→∞ 1 t→∞ t lim ex lim t→∞ lim e−x t→∞ Similarly, we have = +∞ = 0 = +∞ = lim t→∞ 1 = 0. ex lim e−st = e−∞ = 0. t→+∞ 2 Laplace Transform Denition. Let transform f (t) be a function dened for all t ≥ 0. Then its Laplace L (f ) is given by ˆ ∞ F (s) = L {f (t)} = f (t) e−st dt 0 Example. Let f (t) = 1 for t ≥ 0. Then ˆ ˆ +∞ L {f } = L {1} = T e−st dt = 1 . s te−st dt = 1 . s2 e−st dt = lim T →∞ 0 0 Example. Let f (t) = t for t ≥ 0. Then ˆ ˆ +∞ L {f } = L {t} = −st te T dt = lim T →∞ 0 0 Example. Let f (t) = t2 for t ≥ 0. Then ˆ ˆ +∞ L (f ) = L (t2 ) = T t2 e−st dt = lim t2 e−st dt = T →∞ 0 0 2 . s3 Example. Let f (t) = et for t ≥ 0. Then ˆ ˆ +∞ L {f } = L {et } = T et e−st dt = lim T →∞ 0 e−st dt = 0 1 . s−1 Example. Let f (t) = e for t ≥ 0. Then at ˆ ˆ +∞ L {f } = L {e } = at −st at e e T →∞ 0 T e−st dt = dt = lim 0 Theorem. 1 . s−a The Laplace transform is a linear operation; that is, it preserves addition, scalar multiplication and linear combination of the Laplace transform. 1. L {a · f (t)} = a · L {f (t)}. 2. L {f (t) + g(t)} = L {f (t)} + L {g(t)}. 3. L {a · f (t) + b · g(t)} = a · L {f (t)} + b · L {g(t)}. 3 Table of Laplace Transforms f (t) L {f } f (t) L {f } 1.) 1 1 s 7.) cos ωt s s2 +ω 2 2.) t 1 s2 8.) sin ωt ω s2 +ω 2 3.) t2 2 s3 9.) cosh at s s2 −a2 4.) tn , (n = 0, 1, 2, . . .) n! s(n+1) 10.) sinh at a s2 −a2 5.) ta , (a positive) Γ(a+1) s(a+1) 11.) eat cos ωt s−a (s−a)2 +ω 2 6.) eat 1 s−a 12.) eat sin ωt ω (s−a)2 +ω 2 Remark. The formula Γ(a + 1) is called the Gamma ˆ function which is given by ∞ tα−1 e−t dt Γ(α) = 0 for α > 0. Inverse Laplace Transform Note that the Laplace Transform is reversible, i.e., L −1 {L {f (t)}} = f (t) Hence and L (L −1 (F (s)) = F (s). f (t) = L −1 {F (s)} . We have to use the table to convert the function F (s) back to f (t). 4 First Shifting Theorem, s-Shifting Theorem. If f (t) has the transform F (s) (where Laplace transform of the function g(t) = f (t)eat s > k for some k ), then the L {f (t)eat } = F (s − a) or, if we take the inverse on both sides, eat f (t) = L −1 {F (s − a)}. Example. Find the Inverse Laplace for the function L {f } = 3s − 137 . s2 + 2s + 401 Solution. Applying the inverse transform, using linearity and completing the square, we obtain f =L −1  3s − 137 s2 + 2s + 401  −1  s+1 (s + 1)2 + 202  − 7L = 3L = 3(e−t cos 20t) − 7(e−t sin 20t). 5 −1  20 (s + 1)2 + 202  ...
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