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 Chapter 7: The Laplace Transform – Part 3 Department of Electrical Engineering National Taiwan University ihwang@ntu.edu.tw December 3, 2013 1 / 25 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 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 / 25 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 Questions : ? How to compute the Laplace transform of a periodic function? 3 / 25 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 1 Inverse Transform of Derivatives and Product 2 Laplace Transform of Periodic Functions and Dirac Delta Function 3 Systems of Linear Differential Equations 4 / 25 DE Lecture 12 王奕翔

Inverse Transform of Derivatives and Product dt 5 / 25 Theorem Applying the calculation repetitively, we obtain the following theorem: Laplace Transform of Periodic Functions and Dirac Delta Function DE Lecture 12 ds Derivatives of Laplace Transforms Systems of Linear Differential Equations d Consider taking the derivative of the Laplace transform F ( s ) = L { f ( t ) } : (∫ ∞ ∫ ∞ ) ∂ f ( t ) e − st dt ( f ( t ) e − st ) dsF ( s ) = d = ∂ 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 / 25 DE Lecture 12 Systems of Linear Differential Equations Derivatives: 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 / 25 Laplace Transform of Periodic Functions and Dirac Delta Function s s DE Lecture 12 Systems of Linear Differential Equations Example Examples { 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 / 25 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 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 / 25 DE Lecture 12 Systems of Linear Differential Equations Proof of the Convolution Theorem ∫ ∞ ∫ ∞ 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 ( τ 1 ) e − s τ 1 d τ 1 g ( τ 2 ) e − s τ 2 d τ 2 F ( s ) G ( s ) = 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 / 25 Hence, DE Lecture 12 Systems of Linear Differential Equations Example (Use Laplace Transform to Compute Convolution) Examples 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 { } L = s 2 + 1 − 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 s 11 / 25 k s By the convolution theorem, we have Laplace Transform of Periodic Functions and Dirac Delta Function s s DE Lecture 12 Write Example (Finding Inverse Transforms of Products) s Examples Systems of Linear Differential Equations . Evaluate L − 1 { } ( s 2 + k 2 ) 2 1 2 = s 2 + k 2 · s 2 + k 2 . Note that ( s 2 + k 2 ) { } { 1 } = 1 L − 1 L − 1 = cos ( kt ) , k sin ( kt ) . s 2 + k 2 s 2 + k 2 { } = 1 ∫ t 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 = 1 [ τ sin ( kt ) + 1 ] t 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 / 25 s s Laplace Transform of Periodic Functions and Dirac Delta Function s . Example DE Lecture 12 Theorem Laplace Transform of Integrals Systems of Linear Differential Equations L Let f ( t ) − → F ( s ) . By the convolution theorem, {∫ t } = F ( s ) L f ( τ ) d τ 0 Evaluate L − 1 { } 1 ( s 2 +1) 2 { } We know that L − 1 = 1 2 2 t sin t . By the theorem above, we have ( s 2 +1 ) { 1 } ∫ t [ sin τ − τ cos τ ] t = sin t − t cos t τ sin τ L − 1 = d τ = . ( s 2 + 1) 2 2 2 2 0 0 王奕翔

Inverse Transform of Derivatives and Product We can efficiently solve this kind of equation using Laplace transform. 13 / 25 = Hence, Laplace Transform of Periodic Functions and Dirac Delta Function Example DE Lecture 12 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 1 Inverse Transform of Derivatives and Product 2 Laplace Transform of Periodic Functions and Dirac Delta Function 3 Systems of Linear Differential Equations 14 / 25 DE Lecture 12 王奕翔

Inverse Transform of Derivatives and Product and periodic with period T , then 15 / 25 Laplace Transform of Periodic Functions and Dirac Delta Function For example, DE Lecture 12 Theorem Periodic Functions 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, 1 ∫ T L { f ( t ) } = f ( t ) e − st dt 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 16 / 25 Hence, Laplace Transform of Periodic Functions and Dirac Delta Function DE Lecture 12 Systems of Linear Differential Equations Proof : ∫ ∞ ∫ ∞ ∫ T L { f ( t ) } = f ( t ) e − st dt = f ( t ) e − st dt + f ( t ) e − st dt 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

Inverse Transform of Derivatives and Product Ldi 17 / 25 Taking the Laplace transform on both sides, we get Laplace Transform of Periodic Functions and Dirac Delta Function DE Lecture 12 Systems of Linear Differential Equations LR -Circuit with Square-Wave Driving Voltage Consider an LR -circuit with E ( t ) being a unit square wave, period of which is 2 T , and L E { 1 , 0 ≤ t < T E ( t ) = R 0 , T ≤ t < 2 T To determine its current i ( t ) with i (0) = 0 , we solve the following IVP: dt + Ri = E ( t ) , i (0) = 0 . ∫ 2 T 1 ( Ls + R ) I ( s ) = L { E ( t ) } = E ( t ) e − st dt 1 − e − 2 sT 0 1 1 − e − sT ∫ T = e − st dt = 1 − e − 2 sT s (1 − e − 2 sT ) 0 王奕翔