Chapter 7: The Laplace Transform Part 3 Department of Electrical - PowerPoint PPT Presentation

Inverse Transform of Derivatives and Product Laplace Transform of Periodic Functions and Dirac Delta Function Systems of Linear Differential Equations Summary Chapter 7: The Laplace Transform – Part 3 Department of Electrical Engineering National Taiwan University ihwang@ntu.edu.tw December 4, 2013 1 / 35 DE Lecture 12 王奕翔 王奕翔

Inverse Transform of Derivatives and Product Laplace Transform of Periodic Functions and Dirac Delta Function Systems of Linear Differential Equations Summary Properties of Laplace and its Inverse Transforms so far: 1 Laplace Transform of Polynomials, Exponentials, sin , cos , etc. 2 Laplace Transforms of Derivatives 3 Translation in s -Axis and t -Axis 4 Scaling End of story? 2 / 35 DE Lecture 12 王奕翔

Inverse Transform of Derivatives and Product Laplace Transform of Periodic Functions and Dirac Delta Function Systems of Linear Differential Equations Summary Questions : ? How to compute the Laplace transform of a periodic function? 3 / 35 DE Lecture 12 How to compute L { t n e at cos ( kt ) } ? How to compute L − 1 { } 1 (( s − a ) 2 + k 2 ) 2 王奕翔

Inverse Transform of Derivatives and Product Laplace Transform of Periodic Functions and Dirac Delta Function Systems of Linear Differential Equations Summary 1 Inverse Transform of Derivatives and Product 2 Laplace Transform of Periodic Functions and Dirac Delta Function 3 Systems of Linear Differential Equations 4 Summary 4 / 35 DE Lecture 12 王奕翔

Inverse Transform of Derivatives and Product ds 5 / 35 Theorem Applying the calculation repetitively, we obtain the following theorem: Laplace Transform of Periodic Functions and Dirac Delta Function dt DE Lecture 12 Systems of Linear Differential Equations Summary Derivatives of Laplace Transforms d Consider taking the derivative of the Laplace transform F ( s ) = L { f ( t ) } : (∫ ∞ ) ∫ ∞ ∂ ( f ( t ) e − st ) dsF ( s ) = d f ( t ) e − st dt = ∂ s 0 0 ∫ ∞ − tf ( t ) e − st dt = − L { tf ( t ) } . = 0 L Let f ( t ) − → F ( s ) and f ( t ) is of exponential order, { d n } L − 1 L { t n f ( t ) } = ( − 1) n d n ds n F ( s ) , ds n F ( s ) = ( − t ) n f ( t ) . 王奕翔

Inverse Transform of Derivatives and Product Laplace Transform of Periodic Functions and Dirac Delta Function 6 / 35 DE Lecture 12 Derivatives: Systems of Linear Differential Equations Summary n − 1 ∑ L f ( n ) ( t ) s k f ( n − 1 − k ) (0) − → s n F ( s ) − k =0 L − 1 F ( n ) ( s ) − → ( − t ) n f ( t ) 王奕翔

Inverse Transform of Derivatives and Product . 7 / 35 Laplace Transform of Periodic Functions and Dirac Delta Function s s DE Lecture 12 Summary Systems of Linear Differential Equations Examples Example { } t 2 cos t Evaluate L Solution 1 : Since L { cos t } = s 2 +1 , we have ( 1/2 = d 2 s 2 + 1 = d 2 ) s − i + 1/2 t 2 cos t { } L ds 2 ds 2 s + i 2 s 3 − 6 s 1 1 = ( s − i ) 3 + ( s + i ) 3 = ( s 2 + 1) 3 Solution 2 : Since e it = cos t + i sin t , we have 2 { t 2 e it } t 2 cos t t 2 sin t { } { } L = L + i · L = ( s − i ) 3 . { } 2 s 3 − 6 s t 2 cos t 2 { } Hence, L = Re = ( s − i ) 3 3 . ( s 2 +1 ) 王奕翔

Inverse Transform of Derivatives and Product Definition (Convolution) 8 / 35 Laplace Transform of Periodic Functions and Dirac Delta Function DE Lecture 12 We have seen the Laplace transform of derivatives. How about integrals? Convolution and its Laplace Transform Summary Systems of Linear Differential Equations The convolution of two functions f ( t ) and g ( t ) is defined as ∫ t ( f ∗ g )( t ) := f ( τ ) g ( t − τ ) d τ 0 Note : Convolution is exchangeable: f ∗ g = g ∗ f . (why?) Theorem (Convolution in t ⇐ ⇒ Multiplication in s ) L L Let f ( t ) − → F ( s ) and g ( t ) − → G ( s ) . Then, L { ( f ∗ g )( t ) } = F ( s ) G ( s ) . 王奕翔

Inverse Transform of Derivatives and Product Laplace Transform of Periodic Functions and Dirac Delta Function 9 / 35 DE Lecture 12 Proof of the Convolution Theorem Summary Systems of Linear Differential Equations ∫ ∞ ∫ ∞ Write F ( s ) = f ( τ 1 ) e − s τ 1 d τ 1 , G ( s ) = g ( τ 2 ) e − s τ 2 d τ 2 . Hence, 0 0 (∫ ∞ ) (∫ ∞ ) F ( s ) G ( s ) = f ( τ 1 ) e − s τ 1 d τ 1 g ( τ 2 ) e − s τ 2 d τ 2 0 0 ∫ ∞ ∫ ∞ f ( τ 1 ) g ( τ 2 ) e − s ( τ 1 + τ 2 ) d τ 2 d τ 1 = 0 0 ∫ ∞ ∫ ∞ f ( τ 1 ) g ( t − τ 1 ) e − st dt d τ 1 = ( t := τ 1 + τ 2 ) 0 τ 1 ∫ ∞ ∫ t = f ( τ 1 ) g ( t − τ 1 ) e − st d τ 1 dt ( exchange the order ) 0 0 ∫ ∞ (∫ t ) e − st dt = f ( τ 1 ) g ( t − τ 1 ) d τ 1 0 0 = L { ( f ∗ g )( t ) } 王奕翔

Inverse Transform of Derivatives and Product Laplace Transform of Periodic Functions and Dirac Delta Function 10 / 35 Hence, DE Lecture 12 Example (Use Laplace Transform to Compute Convolution) Examples Systems of Linear Differential Equations Summary Evaluate the convolution of e t and sin t . 1 1 Since L { e t } = s − 1 , L { sin t } = s 2 +1 , we have ( s − 1)( s 2 + 1) = 1/2 1 s − 1 − 1/2 s 1/2 { e t ∗ sin t } = s 2 + 1 − L s 2 + 1 . { } 1 = 1 2 e t − 1 2 cos t − 1 e t ∗ sin t = L − 1 2 sin t . ( s − 1)( s 2 + 1) 王奕翔

Inverse Transform of Derivatives and Product Write 11 / 35 k s By the convolution theorem, we have Laplace Transform of Periodic Functions and Dirac Delta Function s s s DE Lecture 12 . Examples Systems of Linear Differential Equations s Summary Example (Finding Inverse Transforms of Products) Evaluate L − 1 { } ( s 2 + k 2 ) 2 1 2 = s 2 + k 2 · ( s 2 + k 2 ) s 2 + k 2 . Note that { } { 1 } = 1 L − 1 L − 1 = cos ( kt ) , k sin ( kt ) . s 2 + k 2 s 2 + k 2 { } ∫ t = 1 L − 1 cos ( k τ ) sin ( k ( t − τ )) d τ ( s 2 + k 2 ) 2 0 = 1 ∫ t { sin ( kt ) − sin ( k (2 τ − t )) } d τ 2 k 0 [ ] t = 1 τ sin ( kt ) + 1 1 2 k cos ( k (2 τ − t )) = 2 kt sin ( kt ) . 2 k 0 王奕翔

Inverse Transform of Derivatives and Product s 12 / 35 s s Laplace Transform of Periodic Functions and Dirac Delta Function s . Example DE Lecture 12 Systems of Linear Differential Equations Summary Theorem Laplace Transform of Integrals L Let f ( t ) − → F ( s ) . By the convolution theorem, {∫ t } = F ( s ) f ( τ ) d τ L 0 Evaluate L − 1 { } 1 ( s 2 +1) 2 { } We know that L − 1 = 1 2 ( s 2 +1 ) 2 t sin t . By the theorem above, we have { 1 } [ sin τ − τ cos τ ] t ∫ t τ sin τ = sin t − t cos t L − 1 = d τ = . ( s 2 + 1) 2 2 2 2 0 0 王奕翔

Inverse Transform of Derivatives and Product exists, then 13 / 35 t t dt s s Laplace Transform of Periodic Functions and Dirac Delta Function s Proof: s t DE Lecture 12 Theorem Summary t Systems of Linear Differential Equations Integral of Laplace Transform { } L f ( t ) Let f ( t ) − → F ( s ) . If L { f ( t ) } ∫ ∞ = F ( u ) du L ∫ ∞ ∫ ∞ ∫ ∞ ∫ ∞ (∫ ∞ ) f ( t ) e − ut dt du = e − ut du F ( u ) du = f ( t ) 0 0 ∫ ∞ { f ( t ) } f ( t ) e − st = dt = L 0 王奕翔

Inverse Transform of Derivatives and Product We can efficiently solve this kind of equation using Laplace transform. 14 / 35 = Hence, Laplace Transform of Periodic Functions and Dirac Delta Function Example DE Lecture 12 Summary Systems of Linear Differential Equations Integral Equation Volterra Integral Equation of y ( t ) : ∫ t y ( t ) = g ( t ) + ( h ∗ y )( t ) = g ( t ) + y ( τ ) h ( t − τ ) d τ. 0 Solve y ( t ) = 3 t 2 − e − t − ∫ t 0 y ( τ ) e t − τ d τ . s +1 − Y ( s ) 6 1 Taking Laplace transform on both sides, we get Y ( s ) = s 3 − s − 1 . Y ( s ) = 6( s − 1) s ( s + 1) = 6 s − 1 s 3 − 6 s 4 + 1 2 − s − s 4 s + 1 y ( t ) = 3 t 2 − t 3 + 1 − 2 e − t . ⇒ 王奕翔

Inverse Transform of Derivatives and Product Laplace Transform of Periodic Functions and Dirac Delta Function Systems of Linear Differential Equations Summary 1 Inverse Transform of Derivatives and Product 2 Laplace Transform of Periodic Functions and Dirac Delta Function 3 Systems of Linear Differential Equations 4 Summary 15 / 35 DE Lecture 12 王奕翔

Inverse Transform of Derivatives and Product and periodic with period T , then 16 / 35 Laplace Transform of Periodic Functions and Dirac Delta Function For example, DE Lecture 12 Theorem Periodic Functions Summary Systems of Linear Differential Equations A function f ( t ) is periodic with period T > 0 if f ( t ) = f ( t + T ) , for all t . If a function f ( t ) is piecewise continuous on [0 , ∞ ) , of exponential order, ∫ T 1 f ( t ) e − st dt L { f ( t ) } = 1 − e − sT 0 ∫ 2 π 1 sin te − st dt L { sin t } = 1 − e − 2 π s 0 [ − cos te − st − s sin te − st ] 2 π 1 = s 2 + 1 1 − e − 2 π s 0 1 − e − 2 π s 1 1 = = s 2 + 1 s 2 + 1 1 − e − 2 π s 王奕翔

Inverse Transform of Derivatives and Product T 17 / 35 Hence, Laplace Transform of Periodic Functions and Dirac Delta Function DE Lecture 12 Proof : Systems of Linear Differential Equations Summary ∫ ∞ ∫ T ∫ ∞ f ( t ) e − st dt = f ( t ) e − st dt + f ( t ) e − st dt L { f ( t ) } = 0 0 ∫ ∞ ∫ T f ( τ + T ) e − s ( τ + T ) d τ = f ( t ) e − st dt + ( τ := t − T ) 0 0 ∫ ∞ ∫ T = f ( t ) e − st dt + e − sT f ( τ ) e − s τ d τ 0 0 ∫ T = f ( t ) e − st dt + e − sT L { f ( t ) } 0 ( 1 − e − sT ) ∫ T L { f ( t ) } = 0 f ( t ) e − st dt . 王奕翔