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LAPLACETRANSFORMEE602 : CIRCUIT ANALYSISSARIATI DALIBJABATAN KEJURUTERAAN ELETRIK,POLITEKNIK MERLIMAU27.11.2014
Laplace TransformChapter 2: Laplace Transform2.1 Understand the use of Laplace Transform and Inverse Laplace Transform insolving network analysis2.1.1 Define the Laplace Transform of an expression by using the integraldefinition. πΉ(π ) 0 π π π‘π(π‘) ππ‘2.1.2 Determine the Laplace Transform of a function by using table.Table Laplace Transform2.1.3 Use the linearity property.a) πΏ{π π } πΏ{π } πΏ{π }b) πΏ {ππ} π πΏ {π} if k is constant2.1.42.1.5Use the first shift theorem.π{π ππ‘ π(π‘)π } πΉ(π π)Use the Laplace Transform of derivatives.ππΏ {π‘ π π(π‘)}π ( 1) πΉ π (π )2.1.6Use the Laplace Transform of integrals.πΏ {π‘ π(π‘)π } πΉ β² (s)2.1.7Determine the Inverse Laplace Transforms of some standard functions.Table Inverse Laplace Transform 1πΏLinearity Property :{πΉ(π )} π(π‘)πΏ{π π } πΏ{π } πΏ{π }πΏ {ππ} π πΏ {π}First Shift Theorem:πΏ 1 {πΉ (π π)} π ππ‘ π(π‘)}2.1.82.1.92.1.102.1.112.1.12Compute the Inverse Laplace Transforms using partial fractions.Compute the Inverse Laplace Transforms by completing the square.Compute the Laplace Transform of first and second derivatives.Apply the Laplace Transform to solve differential equations.Apply the Laplace Transform in RLC circuit analysis:a.RL Series circuitb.RC Series circuitc.LC Series circuitd.RLC Series circuite.RLC Parallel circuitSARIATI DALIBPage 1
Laplace TransformChapter 2 : Laplace TransformIntroduction:The main idea behind the Laplace Transformation is that we can solve anequation (or system of equations) containing differential and integral termsby transforming the equation in "t-space" to one in "s-space". This makes theproblem much easier to solve. The kinds of problems where the LaplaceTransform is invaluable occur in electronics.If needed we can find the inverse Laplace transform, which gives us thesolution back in "t-space".Definition of Laplace TransformLetbe a given function which is defined forfunction. If there exists aso that,Thenis called the Laplace Transform of. Notice the integratorwhere, and will be denoted byis a parameter which may bereal or complex.Thus,The symbolwhich transformintois called the Laplace transformoperator.SARIATI DALIBPage 2
Laplace TransformLaplace Transform by Direct IntegrationTo get the Laplace transform of the given functionand integrate with respect to, multiplybyfrom zero to infinity. In symbol,Example 1 : Find the Laplace transform ofwhenby using DirectIntegration MethodSolution 01 β{π(π‘)} 0 π π π‘ π(π‘)ππ‘ β{1} 0 π π π‘ (1)ππ‘ β{1} π π π‘ ππ‘01 β{1} [ π π π‘ ]π 01 1 [ π π‘ ]π π 1 11[ 0]π π11 (0 1) π π Thus,SARIATI DALIBPage 3
Laplace TransformExample 2: Find the Laplace transform ofby using Direct IntegrationMethodSolution 02Thus,SARIATI DALIBPage 4
Laplace TransformExample 3: Find the Laplace transform of. by using DirectIntegration MethodSolution 03For.Using integration by parts:. LetUsing integration by parts again. LetSARIATI DALIBPage 5
Laplace TransformThus,Therefore,Exercise 1Find the Laplace transform of the following whenby using DirectIntegration Method1. π(π‘) 42. π(π‘) π 2π‘3. π(π‘) π 4π‘SARIATI DALIBPage 6
Laplace TransformTable of Laplace Transforms of Elementary FunctionsBelow are some functionse-atSARIATI DALIBand their Laplace transforms.1π πPage 7
Laplace TransformExercise 2Use table to find the laplace transform of the following;a) π(π‘) 4d) π(π‘) πππ 6π‘b) π(π‘) π‘ 3e) π(π‘) π ππ 3π‘c) π(π‘) π 2π‘f)π(π‘) π 3π‘SARIATI DALIBPage 8
Laplace TransformProperties of Laplace Transform1. Constant MultipleIfis a constant andis a function of, thenExample:1. π(π‘) 4 πππ π‘πΉ(π ) 4 [ π 2π 2π ] 124π 12. π(π‘) 5π‘πΉ(π ) 5 β t1 5 [ 2]π 5π 2Linearity Property2.Linearity PropertyIfandare constants whileandare functions ofwhoseLaplace transform exists, thenProof of Linearity PropertySARIATI DALIBPage 9
Laplace TransformThis property can be easily extended to more than two functions as shownfrom the above proof. With the linearity property, Laplace transform can alsobe called the linear operator.Example 01: LinearityFind the Laplace transform of.Solution 01πβπ’π πΉ(π ) 5 2π π 2Exercise : LinearityDetermine the Laplace Transform of following functions by using table andtheorems1. π(π‘) 6 π 5π‘ π3π‘ 5π‘3 92. π(π‘) 4πππ 4π‘ 9π ππ4π‘ 2πππ 10π‘33. π(π‘) 3 sin 2 π‘4. π(π‘) 5 π 2π‘ 4 sin 3π‘5. π(π‘) 3 3π 2π‘6. π(π‘) 6π‘ 5π‘ 37. π(π‘) 2π ππ4π‘ β 7πππ 3π‘8. π(π‘) 3 2π‘ 6π‘ 2SARIATI DALIBPage 10
Laplace Transform3.First Shifting PropertyIf, whenIn words, the substitutionthen,forin the transform corresponds tomultiplication of the original function bythe.Proof of First Shifting PropertyExample 01:First Shifting PropertyFind the Laplace transform of.Solution 01 (1)βπ 2π‘ 1 . (2)π 2Replace s in (1) with (s-2)SARIATI DALIBThus,Page 11
Laplace TransformExample 02: First Shifting PropertyFind the Laplace transform of π(π‘) π 5π‘ π ππ3π‘Solution 02Using table : .(1)βπ 5π‘ 1 . (2)π 5Replace S in (1) with (S 5)Thus,β(π 5π‘ sin 3π‘) 33 (π 2 10π 25) 9 π 2 10π 34Example 03 : First Shifting PropertyFind the Laplace transform of.Solution 03Using table :βπππ π‘ π‘ π π 2 12βπ 3π‘ 1π 3 .(1) . (2)Replace S in (1) with (S 3)Thus,SARIATI DALIBPage 12
Laplace TransformExercise: First Shifting PropertyDetermine the Laplace Transform of following functions by using table andtheorems1. π(π‘) π 2π‘ πππ 2π‘2. π(π‘) π‘ π 2π‘3. π(π‘) π 3π‘ (2π‘ 3)4. π(π‘) π π‘ π‘5. π(π‘) π π‘ sin 2π‘.SARIATI DALIBPage 13
Laplace TransformLaplace Transform of DerivativesFor first-order derivative:For second-order derivative:For third-order derivative:For nth order derivative:Proof of Laplace Transform of DerivativesUsing integration by parts,Thus,Apply the limits from 0 to :SARIATI DALIBPage 14
Laplace TransformExample 01: Laplace Transform of DerivativesFind the Laplace transform ofusing the transform of derivatives.Solution 01.Example 02: Laplace Transform of DerivativesFind the Laplace transform ofusing the transform of derivatives.Solution 02.SARIATI DALIBPage 15
Laplace TransformExample 03 : Laplace Transform of DerivativesFind the Laplace transform ofusing the transform of derivatives.Solution 03.SARIATI DALIBPage 16
Laplace TransformLaplace Transform of IntergralsTheoremIf,thenProofLetthen,andTaking the Laplace transform of both sides,From Laplace transform of derivative,and from theTheorem above,Thus,SARIATI DALIBPage 17
Laplace TransformExample 01: Laplace Transform of IntergralsFind the Laplace transform of.Solution 01Hence,SARIATI DALIBPage 18
Laplace TransformSummary of Laplace Transform PropertiesProperties of Laplace Transform1. Linearity Property: Constant MultipleIfis a constant andis a function of, then2. Linearity Property- addition/subtraction of function.Ifandare constants whileandare functions ofLaplace transform exists, thenwhoseExample:3. First Shifting PropertyIf, then,4. Transforms of DerivativesThe Laplace transform of the derivativeexists when, andIn general, the Laplace transform of nth derivative is5. Transforms of IntegralsTheoremIfSARIATI DALIB, thenPage 19
Laplace TransformThe InverseLaplace TransformSARIATI DALIBPage 20
Laplace TransformThe Inverse Laplace TransformDefinitionFrom, the valueis called the inverse Laplace transform of.In symbol,whereis called the inverse Laplace transform operator.Table of Inverse Laplace Transforms of Elementary FunctionsπΉ(π ) β{π(π‘)}SARIATI DALIBπ(π‘) β 1{F(s)}Page 21
Laplace TransformExercise : Inverse Laplace TransformFind the inverse Laplace transform of the following;a) πΉ(π ) 1π b) πΉ(π ) 2π c) πΉ(π ) 23π d) πΉ(π ) 1π 2e) πΉ(π ) 4π 2f)πΉ(π ) 1π g) πΉ(π ) 5π 3h) πΉ(π ) 12π 3i)πΉ(π ) 3π 4j)πΉ(π ) 22π 6k)πΉ(π ) 89 π 2l)πΉ(π ) 3π 6m) πΉ(π ) π π 2 16n) πΉ(π ) 3π π 2 9SARIATI DALIBPage 22
Laplace TransformTheorems on Inverse Laplace TransformationTheorem 1: Linearity TheoremIf a and b are constants,Example 01: : Linearity TheoremπΉ(π ) Find the inverse transform of8π 3 3π 2 1π Solution 01π(π‘) β 1 { 8β 183 1 2 }3π π π 111 1 1 3β βπ 3π 2π π‘ 3 1π‘ 2 1 8[] 3[] [π‘](3 1)!(2 1)! 8π‘2π‘ 3 121 4 π‘ 2 3π‘ 1Example 02: : Linearity TheoremFind the inverse transform of πΉ (π ) 6π 2 9Solution 02π(π‘) β 1 {β 1 {π 2π 26} 931} 6 β 1 { 2} 9π 32 6 13β { 2}3π 32 2π ππ3π‘SARIATI DALIBPage 23
Laplace TransformExample 03: : Linearity TheoremFind the inverse transform of.Solution 03π(π‘) β 1 {54π 2}π 2 π 9 5β 1 [1π ] 4β 1 [ 2]π 2π 32 5π 2π‘ 4πππ 3π‘Exercise : : Linearity TheoremUsing table ,find the inverse Laplace transform of :a) πΉ(π ) 2π 3 SARIATI DALIBπ 31π π 3b) πΉ(π ) c) πΉ(π ) 5π 3π 2 9Page 24
Laplace TransformTheorem 2:First Shift TheoremExample 1: First Shift TheoremFind the inverse transform of4π 4(π 1)2 9Solution 1:π(π‘) β 1 {4π 4}(π 1)2 9 β 1 {4(π 1)}(π 1)2 9π 4π π‘ β 1 { 2}π 32 4π π‘ πππ 3π‘Example 2: First Shift TheoremFind the inverse transform of2(π 3)4Solution 2:π(π‘) β 1 {21 3π‘ 1} 2πβ{}( π 3) 4π 43!1 2π 3π‘ β 1 { 3 1 π₯ }3!π 13! 2π 3π‘ π₯ β 1 { 3 1 }6π π 3π‘ 3 π‘3SARIATI DALIBPage 25
Laplace TransformExample 3: First Shift TheoremFind the inverse transform ofπ 4(π 2)2 4Solution 3:π(π‘) β 1 {π 4π 2 2 1} β{}(π 2)2 4(π 2)2 22π 22 β 1 {} }22(π 2) 2(π 2)2 22π 2 π 2π‘ β 1 2 π 2π‘ β 1 22π 2π 22 π 2π‘ πππ 2π‘ π 2π‘ π ππ2π‘Exercise: First Shift TheoremFind the inverse laplace transform of the following by using table andtheorems :π) πΉ(π ) 3π 1π) πΉ(π ) 3(π 1)6π) πΉ(π ) 3(π 1)6 9π) πΉ(π ) π 2(π 2)6 16SARIATI DALIBPage 26
Laplace TransformInverse Laplace Transform by Partial Fraction ExpansionThis technique uses Partial Fraction Expansion to split up a complicatedfraction into forms that are in the Laplace Transform table. As you readthrough this section, you may find it helpful to refer to the section on partialfraction expansion techniques.1.Distinct Real RootsConsider first an example with distinct real roots.Example: Distinct Real RootsQ: Find the inverse Laplace Transform of:Solution:We can find the two unknown coefficients using the "cover-up" method orresidue method:π΄ π 11 (π 2) π 0 2π΅ π 1 1 1 π π 2 2 2SoπΉ(π ) 1/21/2 ππ 2And the inverse laplaceπ(π‘) SARIATI DALIB1 1 2π‘ π2 2Page 27
Laplace Transform2.Repeated Real RootsConsider next an example with repeated real roots (in this case at the origin,s 0).Example: Repeated Real RootsQ: Find the inverse Laplace Transform of the function F(s).Solution 1:We can find two of the unknown coefficients ( A and C)using the "coverup" method.π΄ π 2 15 2π 4π 2πΆ π 2 11 (π 2) π 0 2We find B using cross-multiplication:π 2 1 π 2 π΄ π (π 2)π΅ (π 2)πΆπ 2 1 π 2 π΄ π΅π 2 2π΅π πΆπ 2πΆEquating like powers of "s" gives us:power of "s"left side right sidecoefficient coefficients21A Bs102B Cs012Cπ΄ π΅ 151 445/41/4 1/2πΉ(π ) 2π 2π π π΅ 1 π΄ 1 And511inverse laplace : π(π‘) 4 π 2π‘ 4 2 π‘SARIATI DALIBPage 28
Laplace TransformSolution 2:We can find two of the unknown coefficients ( A and C)using cross βmultiplication.π΄π΅ πΆπ 2 1 π 2 ( π 2) ( 2)π 2 π π π 2 1 π 2 π΄ π (π 2)π΅ (π 2)πΆπΏππ‘ π 2, 4 1 4π΄πβπ’π ,πΏππ‘ π 0π΄ 54, 0 1 2πΆπβπ’π ,π πΏππ‘ π ππ πππ¦ ππ’ππππ, π 15412, 1 1 π΄ 3π΅ 3πΆ122 3π π₯3323π΅ 2 5433π΅ 4πβπ’π ,π΅ 14511 π 2 144πΉ(π ) 2 22π (π 2) π 2π π 51 1π(π‘) π 2π‘ π‘44 2Many texts use a method based upon differentiation of the fraction whenthere are repeated roots.SARIATI DALIBPage 29
Laplace Transform3.Complex RootsAnother case that often comes up is that of complex conjugateroots. Consider the fraction:The second term in the denominator cannot be factored into real terms. Thisleaves us with two possibilities - either accept the complex roots, or find away to include the second order term.Simplify the function F(s) so that it can be looked up in the LaplaceTransform table.Solution:Another way to expand the fraction without resorting to complex numbersis to perform the expansion as follows.π΄ π 3 π 2 4π 5 π 5 5 3( 5)2 4( 5) 5 2 25 20 521 105 We can find the quantities B and C from cross-multiplication.π 3 π΄(π 2 4π 5) (π΅π πΆ)(π 5)SARIATI DALIBPage 30
Laplace Transformπ 3 (π΄π 2 4π΄π 5π΄) (π΅π 2 5π΅π πΆπ 5πΆ)π 3 π 2 (π΄ π΅) π (4π΄ 5π΅ πΆ) (5π΄ 5πΆ)If we equate like powers of "s" we getorder ofleft side right sidecoefficient coefficient coefficient2nd (s2)0A B1st (s1)14A 5B C0th (s0)35A 5CπΈππ1. π΄ π΅ 0π΄ π΅ πβπ’π π΅ π΄ , π΅ 1/5Eqn 3.5A 5C 35π 3 5π΄13 5 ( ) 3 1 45 πΆ 555ThusFinally, we get114π 555πΉ(π ) 2π 5π 4π 5141( )π 555πΉ(π ) π 5 π 2 4π 5 The inverse Laplace Transform is given below :1π 45 (1)π₯πΉ(π ) 2π 55 π 4π 5 SARIATI DALIBPage 31
Laplace TransformThen use method : Completing the square15 (1)π₯ π 2 2πΉ(π ) π 55 (π 2)2 1 1π 225 (1) {[πΉ(π ) ] []}2π 55 (π 2) 1(π 2)2 1 The inverse laplace transform :11π(π‘) π 5π‘ π 2π‘ [πππ π‘ 2π πππ‘ ]55π(π‘) 0.2π 5π‘ 0.2π 2π‘ [πππ π‘ 2π πππ‘ ]SARIATI DALIBPage 32
Laplace TransformHow To Complete The SquareGiven a term of the formΞ±s2 bs cit is often useful to express it as(s d)2 eThis is often useful when solving certain quadratic equations. In our case itlets us more easily use Laplace Transform Tables (in the table the form(s a)2 Ο02 comes up frequently).Without loss of generality, we will only consider the case where Ξ± 1. If wedivide the original equation by through by Ξ± and let b Ξ²/Ξ± and c Ξ³/Ξ±, wegets2 bs cand our task is to express it as(s d)2 eWe start by setting the two terms to be equal to each otherx2 bx c (x d)2 eExpand the right hand sidex2 bx c x2 2dx d2 eEquating the coefficients of like powers of x we getb 2d,c d2 eSARIATI DALIBor d b/2, and e c - d2Page 33
Laplace TransformExample 1: Completing the squareComplete the square for the expression: s2 2s 10Solution:The original function is of the form "s2 bs c", so b 2, c 10, andd b/2 1e c - d2 10 - 1 9.The desired expression is "(s d)2 e" or (s 1)2 9Thus :s2 2s 10 (s 1)2 9Example 2: Completing the squareComplete the square for the expression: x2 4x 29Solution:The original function is of the form "x2 bx c", so b 4, c 29, andd b/2 2e c - d2 29 - 4 25.The desired expression is "(x d)2 e" or (x 2)2 25Thus ,x2 4x 29 (x 2)2 25SARIATI DALIBPage 34
Laplace TransformExercise 1:Perform the indicated operation:Solution 1Forπ΄ π 5 3 58 π 2 π 3 3 25π΅ π 52 53 π 3 π 2 2 35Thus,SARIATI DALIBPage 35
Laplace TransformExercise 2:Find the inverse transform ofSolution 2Forπ΄ 2π 2 5π 62 5 61 (π 3)(π 5) π 14( 4) 162π 2 5π 62( 3)2 5( 3) 6 18 15 63π΅ (π 1)(π 5) π 3( 3 1)( 3 5)( 4 )( 8)32π 2π 2 5π 62(5)2 5(5) 650 25 6 69 (π 1)(π 3) π 5(5 1)(5 3)(4)(8)32Therefore,SARIATI DALIBPage 36
Laplace TransformUsing Inverse Laplace Transforms to Solve Differential EquationsThe Laplace Transform Of A DerivativesLet f(t) be a function of f and let F(s) be the laplace transform of f. The valueof f and its derivatives when t 0 are denoted by f(0) , fβ(0) , fβββ(0) and so on.The nth derivatives of f is denoted by fn(t). Then it can be shown that thelaplace transform of fn(t) is given by:fn(t). Sn F(s) β Sn-1 f(0) β Sn-2 fβ(0) .f(n-1)(0)Two common cases are when n 1 and n 2:f1 (t) SF(s) β f(0)f2 (t) S2 F(s) β S1 f(0) β f β(0)Example 1: DerivativesQ: The laplace transform of f(t) is F(s). Given f(0) 2 and fβ(0) -3. Writeexpression for the laplace transform of :a) f β (t)b) f 11 (t)Solution:a) πΉ(π ) π πΉ(π ) π(0) π πΉ(π ) 2b) πΉ(π ) π 2 πΉ(π ) π π(0) π β² (0) π 2 πΉ(π ) 2π 3SARIATI DALIBPage 37
Laplace TransformExample 2: DerivativesQ; If the initial value, f(0) y(0) 2 and fβ(0) yβ(0) -3. Write expression for thelaplace transform of :a) 2fββ β 3fβ fb)πβ²β²π ππβ² πc)ππ π ππ ππ π ππ πSolution;β²π) 2π β²β² 3π β² π 2 [π 2 πΉ(π ) π π(0) π (0)] 3[π πΉ(π ) π(0)] πΉ(π ) 2[π 2 πΉ(π ) 2π 3] 3[π πΉ(π ) 2] πΉ(π ) 2π 2 πΉ(π ) 4π 6 3π πΉ(π ) 6 πΉ(π ) πΉ(π )[ 2π 2 3π 1] 4π 1211β²π) π β²β² 2π β² π [π 2 πΉ(π ) π π(0) π (0)] 2[π πΉ(π ) π(0)] πΉ(π )22 1 2[π πΉ(π ) 2π 3] 2[π πΉ(π ) 2] πΉ(π )213 [ π 2 πΉ(π )) π ] [2π πΉ(π ) 4] πΉ(π )2213 πΉ(π ) [ π 2 2π 1] π 42215πΉ(π ) [ π 2 2π 1] π 22π)3π2 π¦ ππ¦ 3π¦ β²β² π¦ππ‘ 2ππ‘ 3[π 2 π(π ) π π¦(0) π¦ β² (0)] π(π ) 3[π 2 π(π ) 2π 3] π(π ) 3π(π )(π 2 1) 6π 9SARIATI DALIBPage 38
Laplace TransformSolving Differential Equation using laplace Transform:Example 1: Solve DEQ: Consider the initial value problem.Solveππ¦ππ‘ 3π¦ 0 , π¦(0) 4Solution:ππ¦ 3π¦ 0ππ‘π π(π ) π¦(0) 3π(π ) 0π π(π ) 4 3π(π ) 0π(π )[π 3]π(π ) 44π 3Inverse laplaceπ¦(π‘) 4π 3π‘Exercise: Solve DE1. Solveππ₯ππ‘2. Solveπ2 π¦ππ‘ 23. Solve 2π₯ 2π 3π‘ , π₯(0) 2 π¦ 2 , π¦(0) π¦ β² (0) 0π¦ β²β² π¦ β² 2π¦ 4 , π¦(0) 2, π¦ β² (0) 1SARIATI DALIBPage 39
Laplace TransformExample 2:Solve DEWrite down the subsidiary equations for the following differential equationsand hence solve them.ππ¦ππ‘ π¦ π ππ3π‘,given that y 0 when t 0.Solution:Taking Laplace transform of both sides gives:π π(π ) π¦(0) π(π ) π(π )[ π 1] π¦(0) π(π )[ π 1] π 2π 23 32π 23 93π ππππ π¦(0) 0 9Solving for Y(s) and finding the partial fraction decomposition gives:π(π ) 3π΄π΅π πΆ 22(π 1)(π 9) π 1 (π 9)π΄ π 233 9 π 1 10Apply βcross multiplicationβ:3 π΄(π 2 9) (π΅π πΆ)(π 1)3 π΄π 2 9π΄ π΅π 2 π΅π πΆπ πΆleft sidecoefficientright sidecoefficients20A Bs10B Cs039A Cpower of "s"SARIATI DALIBPage 40
Laplace TransformπΈππ’ππ‘πππ 1 π΄ π΅ 0π΅ π΄ 310πΈππ’ππ‘πππ 2; π΅ πΆ 0πΆ π΅ [ Soπ(π ) 33] 10103π΄π΅π πΆ 22(π 1)(π 9) π 1 (π 9)333( 10) π 1010 π 1(π 2 9) 31 π 1[ 2]10 π 1 (π 3231 π 1[ 2 ]10 π 1 (π 32 ) (π 2 32 )Finding the inverse Laplace tranform gives us the solution for y as a functionof t:π¦(π‘) SARIATI DALIB3 π‘31π πππ 3π‘ π ππ3π‘101010Page 41
Laplace TransformExample 5 : Solve DESolveπ2π¦ππ‘ 2 2ππ¦ππ‘ 5π¦ 0 , given that π¦ β² (0) 0, π¦(0) 1when t 0Solution:Taking Laplace transform of both sides and applying initial conditions of y(0) 1 and y'(0) 0 gives:π 2 π(π ) π π¦(0) π¦ β² (0) 2(π π(π ) π¦(0) 5π(π ) 0(π 2 π(π ) π ) 2(π π(π ) 1) 5π(π ) 0(π 2 2π 5)π(π ) π 2Solving for Y and completing the square on the denominator gives:π(π ) π 2π 2π 2 (π 1)2 4 2π 5 π 1 1(π 1)2 4π 11 22(π 1) 2(π 1)2 22π 112 22(π 1) 22 (π 1)2 22Now, finding the inverse Laplace Transform gives us the solution for y as afunction of t:1π¦(π‘) π π‘ πππ 2π‘ π π‘ π ππ2π‘2SARIATI DALIBPage 42
Laplace TransformExample 6: Solve DESolvey'' y' - 2y 4y(0) 2y'(0) 1Solution[ yββ yβ β 2y] {4}A table of Laplace transforms is useful here. We get(s2{y} - 2s - 1) (s{y} - 2) - 2{y} 4/sNext, combine like terms to get(s2 s - 2){y} 4/s 2s 3Putting under a common denominator, dividing and factoring we getπΏ{π(π )} 2π 2 3π 4(π )(π 1)(π 2)Use partial fractions. We writeπΏ{π(π )} 2π 2 3π 4π΄π΅πΆ (π )(π 1)(π 2) π π 1 (π 2)which givesA(s - 1)(s 2) Bs(s 2) Cs(s - 1) 2s2 3s 4Letting s 0 gives-2A 4Letting s 1 gives3B 9Letting s -2 givesA -2B 36C 6C 1Now we solveπΏ{π(π )} 231 π π 1 (π 2)Now we can use the table to getπ¦(π‘) 2 3π π‘ π 2π‘SARIATI DALIBPage 43
Laplace TransformApply Laplace Transform In RLCCircuit AnalysisSARIATI DALIBPage 44
Laplace TransformSARIATI DALIBPage 45
Laplace TransformLaplace Transform in RL CircuitExample 1: RL circuitBy using laplace transform, find i(t) when switch is close at t 0. The initialcurrent value i(0) 0.Solution:Derive the circuit equation in t-domain then transform these differentialequation in s-domain.By KVL:In t-domain;π ππ πΏ ππ/ππ‘ππ100 25π 0.1 ππ‘Applying lapl
Nov 27, 2014Β Β· If needed we can find the inverse Laplace transform, which gives us the solution back in "t-space". Definition of Laplace Transform Let be a given function which is defined for . If there exists a function so that , Then is called the Laplace Transform of , and will be denoted by
Laplace Transform in Engineering Analysis Laplace transforms is a mathematical operation that is used to βtransformβ a variable (such as x, or y, or z, or t)to a parameter (s)- transform ONE variable at time. Mathematically, it can be expressed as: L f t e st f t dt F s t 0 (5.1) In a laymanβs term, Laplace transform is used
Introduction to Laplace transform methods Page 1 A Short Introduction to Laplace Transform Methods (tbco, 3/16/2017) 1. Laplace transforms a) Definition: Given: ( ) Process: β( ) ( ) 0 Μ( ) Result: Μ( ), a function of the βLaplace tr
Using the Laplace Transform, differential equations can be solved algebraically. 2. We can use pole/zero diagrams from the Laplace Transform to determine the frequency response of a system and whether or not the system is stable. 3. We can tra
Objectives: Objectives: Calculate the Laplace transform of common functions using the definition and the Laplace transform tables Laplace-transform a circuit, including components with non-zero initial conditions. Analyze a circuit in the s-domain Check your s-domain answers using the initial value theorem (IVT) and final value theorem (FVT)
No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t) Transform domain (s) Original DE & IVP Algebraic equation for the Laplace transform Laplace transform of the solu
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
The Laplace transform technique is a huge improvement over working directly with differential equations. It is relatively straightforward to convert an input signal and the network description into the Laplace domain. However, performing the Inverse Laplace transform can be challeng
Fundamentals; Harmony; Jazz, Pop, and Contemporary Music Theory (including Twentieth-Century Music); and Form in Music. The format for each volume is consistent: 1. The left column lists terms to help you organize your study and find topics quickly. 2. Bold indicates key concepts. 3. Each volume ends with a Remember-Forever Review and More
