Chapter 6 Slides Complete ppt

Chapter 6 Slides Complete ppt - Chapter 6: LAPLACE...

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Unformatted text preview: Chapter 6: LAPLACE TRANSFORM Solution of Linear Differential Equations with constant coefficients by Integral Transform Methods Example: Mechanical Vibration due to Impulsive Forces 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 1 IMPULSIVE PROBLEM TREATED AS INITIAL VALUE PROBLEM WITH NO EXTERNAL LOADING 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 2 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 3 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 4 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 5 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 6 Laplace Transform of Derivatives of a Function 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 7 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 8 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 9 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 10 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 11 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 12 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 13 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 14 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 15 TRANSFORM OF GENERALIZED FUNCTION Dirac Delta Function 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 16 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 17 Bromwich Contour 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 18 MORE DETAILED DISCUSSION ON CONVOLUTION INTEGRAL FOLLOWS 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 19 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 20 WE OBTAINED THE SAME RESULT USING CONVENTIONAL METHOD 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 21 PROPERTIES OF THE SOLUTION IN THE TRANSFORMED DOMAIN 1. Reduces the differential Equation in u(t ) to an Algebraic Equation in u(s ) (A Simpler Problem to Solve) 2. Initial Conditions are Automatically Included in u(s ) 3. Solution u(s ) in the Transformed Domain is Unique When the Differential Equation is Given, and Initial Conditions are Prescribed 4. Solution u(t ) in the Physical Domain can be Found by Inverse Transform u(t ) = L−1 { u(s)} 5. There Exists an Unique one-to-one Mapping u(t ) ⇔ u(s) 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 22 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 23 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 24 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 25 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 26 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 27 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 28 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 29 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 30 GENERAL SOLUTION OF NONHOMOGENEOUS SLDE WITH CONSTANT COEFFICIENTS USING LAPLACE TRANSFORM & & && & Differential Equation: mu + cu + ku = F (t ), u(0) = u0 , u(0) = u0 Taking Laplace Transform of both sides & m −u(0) − su(0) − s 2u + c ( −u(0) − su ) + ku = F ( s ) Solving for u , we have u( s ) = ( ) & F ( s ) + mu0 + u0 ( ms + c ) ms 2 + cs + k & F ( s ) + mu0 + u0 ( ms + c ) u( s ) = m ( s − s1 ) ( s − s2 ) u( s ) = where s1 and s2 are the roots of ms 2 + cs + k = 0 i.e., s1,2 c2 − 4km c = − ± 2m 2m & u ( s + c / m) u0 F (s ) + +0 m ( s − s1 ) ( s − s2 ) ( s − s1 ) ( s − s2 ) ( s − s1 ) ( s − s2 ) Inverse transform yields the solution u(t ) in the time domain 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 31 INVERSION IN THE TIME DOMAIN 1 1 Write = ( s − s1 )( s − s2 ) s1 − s2 Recall L { e at } 1 1 − s − s1 s − s2 e s1t − e s2t 1 1 = . L−1 = s−a ( s − s1 )(s − s2 ) s1 − s2 When s1 = s2 (equal roots) , differentiating at ∞ 1 1 st L{ e } = = ∫ e −( s−a )t dt , we get L−1 = te 1 , Re( s) > s1 2 s−a 0 ( s − s1 ) Underdamping: c 2 < 4km , c k c s1,2 = λ ± iµ , λ = − , µ = ωn 1 − ζ 2 , ωn = , ζ = 2m m 2 km where s1 and s2 are the roots of ms 2 + cs + k = 0 i.e., c 2 − 4km c s1,2 = − ± 2m 2m 01/17/11 Inverse transform yields the solution u(t ) in the time domain MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 32 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 33 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 34 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 35 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 36 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 37 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 38 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 39 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 40 01/17/11 MAE-182A: Lecture Notes: Laplace Tansform: Chatterjee 41 ...
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