Chapter 7: The Laplace Transform Part 1 Department of Electrical - PowerPoint PPT Presentation

Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary Chapter 7: The Laplace Transform – Part 1 Department of Electrical Engineering National Taiwan University ihwang@ntu.edu.tw November 26, 2013 1 / 34 DE Lecture 10 王奕翔 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary Solving an initial value problem associated with a linear differential equation: 2 Plug in the initial conditions to specify the undetermined coefficients. Question : Is there a faster way? 2 / 34 DE Lecture 10 1 General solution = complimentary solution + particular solution. 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms 3 / 34 Question : How to solve the current? How to deal with discontinuity? Square voltage input: Periodic, Discontinuous . DE Lecture 10 For example: But in real applications, sometimes this is not true. with continuous, differentiable, or analytic coefficients. In Chapter 4, 5, and 6, we majorly deal with linear differential equations Summary E ( t ) L R E ( t ) t C 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary In this lecture we introduce a powerful tool: Laplace Transform Invented by Pierre-Simon Laplace (1749 - 1827). 4 / 34 DE Lecture 10 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms 5 / 34 DE Lecture 10 Summary Overview of the Method Find unknown y ( t ) Transformed DE Apply Laplace that satisfies DE becomes an algebraic transform and initial conditions equation in Y ( s ) Solution y ( t ) Apply inverse Laplace Solve transformed − 1 transform of original IVP equation for Y ( s ) 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary 1 Laplace and Inverse Laplace Transform: Definitions and Basics 2 Solve Initial Value Problems using Laplace Transforms 3 Summary 6 / 34 DE Lecture 10 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics given that the improper integral converges. 7 / 34 Solve Initial Value Problems using Laplace Transforms Note : Use capital letters to denote transforms. DE Lecture 10 Summary Definition Definition of the Laplace Transform For a function f ( t ) defined for t ≥ 0 , its Laplace Transfrom is defined as ∫ ∞ e − st f ( t ) dt , F ( s ) := L { f ( t ) } := 0 L L L f ( t ) − → F ( s ) , g ( t ) − → G ( s ) , y ( t ) − → Y ( s ) , etc. Note : The domain of the Laplace transform F ( s ) (that is, where the improper integral converges) depends on the function f ( t ) 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms 8 / 34 s s DE Lecture 10 Summary Examples of Laplace Transform Example Evaluate L { 1 } . ∫ ∞ ∫ T L { 1 } = e − st (1) dt = lim e − st dt T →∞ 0 0 [ − e − st ] T 1 − e − sT = lim = lim . T →∞ T →∞ 0 When does the above converge? s > 0 ! Hence, the domain of L { 1 } is s > 0 , and L { 1 } = 1 s . 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics td 9 / 34 s s Solve Initial Value Problems using Laplace Transforms s DE Lecture 10 Examples of Laplace Transform Summary Example Evaluate L { t } . ∫ ∞ ( − e − st ) ∫ T L { t } = te − st dt = lim T →∞ 0 0 [ − te − st ] T ∫ T 1 − Te − sT + 1 s e − st dt = lim = lim + s L { 1 } . T →∞ T →∞ 0 0 When does the above converge? s > 0 ! Hence, the domain of L { t } is s > 0 , and L { t } = 1 s 2 . 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary Laplace Transform of t n Proof : One way is to prove it by induction. We will show another proof after discussing the Laplace transform of the derivative of a function. 10 / 34 DE Lecture 10 n ! L { t n } = s n +1 , n = 0 , 1 , 2 , . . . , s > 0 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Proof : 11 / 34 Solve Initial Value Problems using Laplace Transforms DE Lecture 10 Laplace Transform of e at Summary 1 { e at } = L s − a , s > a ∫ ∞ ∫ T { e at } e − ( s − a ) t dt = e at e − st dt = lim L T →∞ 0 0 [ − e − ( s − a ) t 1 − e − ( s − a ) T ] T = lim = lim s − a s − a T →∞ T →∞ 0 When does the above converge? s − a > 0 ! 1 Hence, the domain of L { e at } is s > a , and L { e at } = s − a . 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics s 12 / 34 s s s Solve Initial Value Problems using Laplace Transforms s DE Lecture 10 Proof : Summary s k Laplace Transform of sin ( kt ) and cos ( kt ) L { sin ( kt ) } = s 2 + k 2 , L { cos ( kt ) } = s 2 + k 2 , s > 0 ∫ ∞ ∫ ∞ ( − e − st ) sin ( kt ) e − st dt = L { sin ( kt ) } = sin ( kt ) d 0 0 ∫ ∞ ] ∞ [ − sin ( kt ) e − st cos ( kt ) e − st dt = + k 0 0 ] ∞ [ − sin ( kt ) e − st = + k s L { cos ( kt ) } 0 ] ∞ [ − sin ( kt ) e − st When does the above converge? s > 0 ! = ⇒ 0 = 0 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics s 13 / 34 s . s s s Solve Initial Value Problems using Laplace Transforms s DE Lecture 10 Summary Proof : s k Laplace Transform of sin ( kt ) and cos ( kt ) L { sin ( kt ) } = s 2 + k 2 , L { cos ( kt ) } = s 2 + k 2 , s > 0 ∫ ∞ ∫ ∞ ( − e − st ) cos ( kt ) e − st dt = L { cos ( kt ) } = cos ( kt ) d 0 0 ∫ ∞ ] ∞ [ − cos ( kt ) e − st sin ( kt ) e − st dt = − k 0 0 ] ∞ [ − cos ( kt ) e − st = − k s L { sin ( kt ) } 0 ] ∞ [ − cos ( kt ) e − st 0 = 1 When does the above converge? s > 0 ! = ⇒ 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Proof : 14 / 34 s k = Solve Initial Value Problems using Laplace Transforms Solve the above, we get the result: DE Lecture 10 s Summary k Laplace Transform of sin ( kt ) and cos ( kt ) L { sin ( kt ) } = s 2 + k 2 , L { cos ( kt ) } = s 2 + k 2 , s > 0 { L { sin ( kt ) } = k s L { cos ( kt ) } L { cos ( kt ) } = 1 s − k s L { sin ( kt ) } s 2 − k 2 L { sin ( kt ) } = k s L { cos ( kt ) } = k s 2 L { sin ( kt ) } ⇒ s 2 + k 2 L { sin ( kt ) } = k ⇒ L { sin ( kt ) } = s 2 + k 2 s 2 s 2 = L { cos ( kt ) } = s k L { sin ( kt ) } = s 2 + k 2 . 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Proof : It can be proved by the linearity of integral. 15 / 34 s k Solve Initial Value Problems using Laplace Transforms . Hence Example DE Lecture 10 Theorem Laplace Transform is Linear Summary L L For any α, β , f ( t ) − → F ( s ) , g ( t ) − → G ( s ) , L { α f ( t ) + β g ( t ) } = α F ( s ) + β G ( s ) Evaluate L { sinh ( kt ) } and L { cosh ( kt ) } . e kt − e − kt ) e kt + e − kt ) A: sinh ( kt ) = 1 , cosh ( kt ) = 1 ( ( 2 2 → 1 ( 1 1 ) L sinh ( kt ) − s − k − = s 2 − k 2 , s > | k | 2 s + k → 1 ( 1 1 ) L cosh ( kt ) − s − k + = s 2 − k 2 , s > | k | . 2 s + k 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics e at 16 / 34 s k s Solve Initial Value Problems using Laplace Transforms k DE Lecture 10 Summary t n Laplace Transforms of Some Basic Functions f ( t ) F ( s ) Domain of F ( s ) n ! s > 0 s n +1 1 s > a s − a sin ( kt ) s > 0 s 2 + k 2 cos ( kt ) s > 0 s 2 + k 2 sinh ( kt ) s > | k | s 2 − k 2 cosh ( kt ) s > | k | s 2 − k 2 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics of exponential order , 17 / 34 lim Solve Initial Value Problems using Laplace Transforms Definition Theorem (Sufficient Conditions for the Existence of Laplace Transform) Existence of Laplace Transform Summary DE Lecture 10 If a function f ( t ) is piecewise continuous on [0 , ∞ ) , and then L { f ( t ) } exists for s > c for some constant c . A function f ( t ) is of exponential order if ∃ c ∈ R , M > 0 , τ > 0 such that | f ( t ) | ≤ Me ct , ∀ t > τ. Note : If f ( t ) is of exponential order, then ∃ c ∈ R such that for s > c , t →∞ f ( t ) e − st = 0 . 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics of exponential order , 18 / 34 exists. Solve Initial Value Problems using Laplace Transforms DE Lecture 10 Existence of Laplace Transform Summary Theorem (Sufficient Conditions for the Existence of Laplace Transform) If a function f ( t ) is piecewise continuous on [0 , ∞ ) , and then L { f ( t ) } exists for s > c for some constant c . Proof : For sufficiently large T > τ , we split the following integral: ∫ T ∫ τ ∫ T f ( t ) e − st dt f ( t ) e − st dt f ( t ) dt = + . 0 0 τ � �� � � �� � I 1 I 2 We only need to prove that I 2 converges as T → ∞ : ∫ T ∫ T ∫ T | f ( t ) e − st | dt = | f ( t ) | e − st dt ≤ Me ct e − st dt , | I 2 | ≤ τ τ τ { e ct } which converges as T → ∞ for s > c since L 王奕翔