2y ago

27 Views

2 Downloads

832.59 KB

38 Pages

Transcription

Class Notes 12:Laplace Transform (1/3)Intro & Differential Equations82 – Engineering Mathematics

Laplace Transform – Introduction

Laplace Transform – Introduction

Pierre Simon Laplace (Marquis) French mathematician and astronomer whose work was pivotal tothe development of mathematical astronomy. He summarized andextended the work of his predecessors in his five volumeMécanique Céleste (Celestial Mechanics) (1799-1825). Thisseminal work translated the geometric study of classicalmechanics to one based on calculus, opening up a broader rangeof problems. He formulated Laplace's equation, and invented the Laplacetransform. In mathematics, the Laplace transform is one of thebest known and most widely used integral transforms. It iscommonly used to produce an easily solvable algebraic equationfrom an ordinary differential equation. It has many importantapplications in mathematics, physics, optics, electricalengineering, control engineering, signal processing, andprobability theory. He restated and developed the nebular hypothesis of the origin ofthe solar system and was one of the first scientists to postulate theexistence of black holes and the notion of gravitational collapse.Pierre-Simon, marquis de Laplace23 April 1749 - 5 March 1827Video Clip

Laplace Transform – Motivation Transform between the time domain (t) to the frequency domain (s) Lossless Transform – No information is lost when the transform andthe inverse transform is appliedFrequencyTimeDomain (t)DifferentialDomain (S)L( f (t )) F (s)EquationTime DomainSolutionAlgebraicEquationf (t ) L 1 ( F ( s))Frequency DomainSolution

Improper Integral – Definition An improper integral over an unbounded interval is definedas the limit of an integral over a finite interval aAf (t )dt lim f (t )dtA a where A is a positive real number. If the integral from a to A exists for each A a the limit as A exists then the improper integral is said to converge to thatlimiting value. Otherwise, the integral is said to diverge or fail toexist.

Improper Integral – Example 0 e dt lim ctAA 0ct Aee dt limA cct1 c 0c c 0 c 0 01 ct lim (e 1)A cconvergediverge 1

Integral Transform - Laplace Transform – Definition Tool for solving linear diff. eq. – Integral transform F ( s ) k ( s, t ) f (t )dt k(s,t) - The kernel of the transformationGiven , ( ; )fTransformFKernel k ( s, t ) e st Laplace transform L f (t ) F (s) e st f (t )dt0whenever this improper integral converges

Laplace Transform – Theorem 6.1.2 –Sufficient Conditions for Existence Suppose that f is a function for which the following hold:(1) f is piecewise continuous on [0, b] for all b 0.(2) f(t) Meat when t T, with T, M 0. Then the Laplace Transform of f exists for s a. L f (t ) F (s) e st f (t )dt finite0 Note: A function f that satisfies the conditions specifiedabove is said to have exponential order as t .

Laplace Transform – Theorem 6.1.2 –Condition No. 1 - Piecewise Continuous A function f is piecewise continuous on an interval [a,b] if this interval can be partitioned by a finite number ofpointsa t0 t1 tn b such that(1) f is continuous on each (tk, tk 1)(2) lim f (t ) , k 0, , n 1t t k(3) lim f (t ) , k 1, , nt t k 1 In other words, f is piecewise continuous on [a, b] if it iscontinuous there except for a finite number of jumpdiscontinuities.

Laplace Transform – Theorem 6.1.2 –Condition No. 1 - Piecewise Continuous t1 f (t )dt f (t )dt t2t1 f (t )dt f (t )dtt2

Laplace Transform – Theorem 6.1.2 –Condition No. 2 - Exponential Order If– f is an increasing function Then– f(t) Meat when t T, with T, M 0.

Laplace Transform – Theorem 6.1.2 –Condition No. 2 - Exponential Ordert ete t et2 cos(t ) 2et A positive integral power of t is always of exponential order sincefor c 0t n Mecttn M for t cte

Laplace Transform (Function) – Example L 1 F (s) e st 1 dt0L 1 0 st A e1 e st dt limA s0 e sA 1 1 lim A ss

Laplace Transform (Function) – Example F ( s) Le Leat e st at0 1s ae dt at0 0 e ( s a )te st eatdt ( s a )t edt limA s aA0 e ( s a ) A 1 limA s a

Laplace Transform (Function) – Example uv dt uv vu dtL t 0L t F (s) e st t dt tee t dt sv' u st0 st 011 1 1 L 1 2ss s s st e 1 st (1 ) dt 0 0 1e0s su v

Laplace Transform (Function) – Example stL sin( at ) F ( s ) e sin( at )dt 0uv'u v e st cos(at ) A s A st lim e cos(at )dt 0A a a 01 s A st1 s 2 stF ( s ) e cos(at )dt 2 e sin( at )dt 0a a 0 u aa v F (s)second integration by parts1 s2F ( s) 2 F ( s)a aaF ( s) 2s a2

Laplace Transform – Discontinues Function - Example Transformation of piecewise continuous function 0 0 t 3f (t ) t 3 2 L f (t ) e0 st3 f (t )dt e (0)dt e st 2dt st0 st 2e 0 s02e 3s s3s 0

Operation Properties – Translation on the t-Axis (Time)Second Translation TheoremL f (t a)u (t a) e as F ( s) L 1 e as F ( s ) L 1 F ( s ) t t au (t a) f (t a)u (t a)

Operation Properties – Translation on the S-Axis (Freq.)First Translation Theorem L e at f (t ) F ( s a) L f (t ) s s a L 1 F ( s a ) L 1 F ( s ) s a s e at f (t ) Example: L e 5t t 3 L t 3s s 5 3!sn s s 56( s 5) 4

Laplace Transform – Linearity Suppose f and g are functions whose Laplacetransforms exist for s a1 and s a2, respectively. Then, for s greater than the maximum of a1 and a2, theLaplace transform of c1 f (t) c2g(t) exists. That is,L c1 f (t ) c2 g (t ) e st c1 f (t ) c2 g (t ) dt is finite 0with L c1 f (t ) c2 g (t ) c1 e0 st f (t )dt c2 e st g (t )dt0 c1 L f (t ) c2 L g (t )

Laplace Transform – Linearity – Example L 5e 2t 3 sin(4t ) 5 L e 2t 3L sin(4t ) 512 2s 2 s 16

Inverse Laplace Transform – Example 1 1 4! 1L 1 5 L 1 5 t 4 S 4! S 24

Inverse Laplace Transform – Example 1 1 1 7 1L 2L 2sin 7t 77 S 7 S 7 1

Inverse Laplace Transform – Example6 2 S 6 1 2 SL 1 2 L 2 2S 4S 4S 4 S 6 1 2 2 L 1 2 L 2 (linearity) S 4 2 S 4 2 cos 2t 3 sin 2t

Inverse Laplace Transform – Partial Fraction s 2 6s 9L (s 1)(s 2)(s 4) 1Partial Fraction:s 2 6s 9ABC ( s 1)(s 2)(s 4) s 1 s 2 s 4A( s 2)(s 4) B( s 1)(s 4) C ( s 1)(s 2) ( s 1)(s 2)(s 4)s 2 6s 9 A( s 2)(s 4) B( s 1)(s 4) C ( s 1)(s 2)Option 1:3 equations with 3 unknown variables A, B, Cs 2 6 s 9 f 2 ( A, B, C ) s 2 f1 ( A, B, C ) s f 0 ( A, B, C )f 2 ( A, B, C ) 1f1 ( A, B, C ) 6f 0 ( A, B, C ) 9

Inverse Laplace Transform – Partial Fractions 2 6s 9 A( s 2)(s 4) B( s 1)(s 4) C ( s 1)(s 2)Option 2:Plug (s 1)Plug (s 2)Plug (s -4)16 A(-1)(5)25 B(1)(6)1 C(-5)(-6)A -16/5B 25/6C 1/30s 2 6s 916 / 5 25 / 6 1 / 30 ( s 1)(s 2)(s 4)s 1 s 2 s 4 s 2 6s 916 1 1 25 1 1 1 1 1 L L L L (s 1)(s 2)(s 4)5s 16s 230s 4 16251 et e 2t e 4t5630 1

Laplace Transform of a Derivative (First) st st (t )dt e L f (t ) e f f (t ) ( s )e st f (t )dt 0uuv e stv0u f (t ) ( s ) e st f (t )dt0L f (t ) sL f (t ) f (0) sF ( s ) f (0)v

Laplace Transform of a Derivative (Second) stL f (t ) e f (t )dt e f (t ) s e f (t )dt 00uuu st stv v0vL f (t ) f (0) sL f (t ) s sF ( s ) f (0) f (0)L f (t ) s 2 F ( s) sf (0) f (0)

Laplace Transform of a Derivativefirst derivativeL f (t ) sF ( s) f (0)Initial Conditionssecond derivativeL f (t ) s 2 F ( s) sf (0) f (0)third derivativeL f (t ) s 3 F ( s ) s 2 f (0) sf (0) f (0) n-th derivative L f ( n ) (t ) s n F ( s) s n 1 f (0) s n 2 f (0) f ( n 1) (0)

Laplace Transform – Solving Linear ODEsdnyd n 1 yan n an 1 n 1 a0 y g (t )dtdty(0) y0 ; y (0) y1; y ( n 1) (0) yn 1 Initial Conditions:aii 0,1, , ny0 ; y1 ; yn 1- constants(I.C.s) - constantsApply Laplace Transform an s nY ( s ) s n 1 y (0) y ( n 1) (0) an 1 s n 1Y ( s ) s n 2 y (0) y ( n 2 ) (0) a0 Y ( s ) G ( s )

Laplace Transform – Solving Linear ODEs The Laplace Transform of a linear differential equation with constantcoefficients become an algebraic equation in Y(s)Time DomainDifferential EquationLaplace TransformFrequency DomainAlgebraic EquationP( s ) Y ( s) G(s) Q(s) PolynimialY ( s) Non-homoinputI.C.G ( s) Q( s) P( s) P( s) G ( s ) Q( s ) y(t ) L 1 Y ( s) L 1 P(s)P( s )

Laplace Transform – Example First Order Linear ODEs Example:dy 3 y 13sin 2tdty(0) 6L y 3 y L 13sin 2t sY ( s) 6 3Y ( s) s 3 Y (s) 26s2 426 62s 42666s 2 50Y ( s) 2 s 4 s 3 s 3 s 3 s 2 4ABs C 2s 3 s 4 As 2 4 A Bs 2 3Bs Cs 3C ( A B ) s 2 (3B C ) s (4 A 3C )

Laplace Transform – Example First Order Linear ODEs A B 6 A 3 3B C 0 B 2 4 A 3C 50 C 6 Y ( s) 8 2s 682s6 2 2 2s 3 s 4s 3 s 4 s 4s 2 1 1 1 y(t ) 8L 1 2L 3L 2 2 s 3 s 4 s 4 y (t ) 8e 3t 2 cos 2t 3 sin 2t

Laplace Transform – Example Second Order Linear ODEs Example:y 3 y 2 y e 4ty (0) 1; y (0) 5 L y 3 y 2 y L e 4t1s 2Y ( s ) sy(0) y (0) 3 sY ( s ) y (0) 2Y ( s ) s 4 L y L 3 y L 2 y L e 4 t s2 3s 2 Y ( s) s 2 1s 4s 21s 2 6s 9Y ( s) 2 s 3s 2 s 4 s 2 3s 2 s 1 s 2 s 4 (See previous example)y(t ) L 1 Y ( s) 16 t 25 2t 1 4te e e5630

Intro & Differential Equations . He formulated Laplace's equation, and invented the Laplace transform. In mathematics, the Laplace transform is one of the best known and most widely used integral transforms. It is commo

Related Documents: