Chapter 7: The Laplace Transform Part 2 Department of Electrical - PowerPoint PPT Presentation

Translations and Scaling Summary Chapter 7: The Laplace Transform – Part 2 Department of Electrical Engineering National Taiwan University ihwang@ntu.edu.tw December 3, 2013 1 / 34 DE Lecture 11 王奕翔 王奕翔

Translations and Scaling Summary So far we have learned 1 Basic properties of Laplace Transform 2 Inverse transform (Memorize with Laplace transform in pairs!) 3 How to use Laplace Transform to solve an IVP End of story? 2 / 34 DE Lecture 11 王奕翔

Translations and Scaling Step 1 : Laplace-transform both sides: 3 / 34 Summary = = DE Lecture 11 Solving a Second-Order IVP with Laplace Transform Example Solve y ′′ − 2 y ′ + y = e 2 t , y (0) = 1 , y ′ (0) = 5 . e 2 t } { y ′′ } { y ′ } { − 2 L + L { y } = L L 1 s 2 Y ( s ) − sy (0) − y ′ (0) ( ) ⇒ − 2 ( sY ( s ) − y (0)) + Y ( s ) = s − 2 1 ⇒ ( s 2 − 2 s + 1) Y ( s ) = s + 3 + s − 2 s + 3 1 Step 2 : Solve Y ( s ) : Y ( s ) = ( s − 1) 2 + ( s − 1) 2 ( s − 2) . Step 3 : Compute the inverse Laplace transform of Y ( s ) : How to compute? { } 3 1 1 ⇒ y ( t ) = 3 L − 1 + e 2 t . Y ( s ) = ( s − 1) 2 + s − 2 = ( s − 1) 2 王奕翔

Translations and Scaling A 4 / 34 Summary C B DE Lecture 11 Partial fraction decomposition : ( s − 1) 2 = ( s − 1) + 4 s + 3 1 4 = s − 1 + ( s − 1) 2 . ( s − 1) 2 1 ( s − 1) 2 ( s − 2) = s − 2 + s − 1 + ( s − 1) 2 [ ] 1 A = = 1 ( s − 1) 2 ✘✘✘ ( s − 2) s =2 [ ] 1 C = = − 1 ( s − 1) 2 ( s − 2) ✘ ✘✘✘ s =1 1 = ( B ( s − 1) + C )( s − 2) + A ( s 2 − 2 s + 1) = ⇒ B = − A = − 1 . 王奕翔

Translations and Scaling Summary Need more properties of Laplace and its inverse transforms! 5 / 34 DE Lecture 11 We already know that L − 1 { 1 } = t . s 2 If we know what is the inverse transform of a function F ( s ) when it is translated by 1 in the s -axis, that is, L − 1 { F ( s − 1) } , we can solve! 王奕翔

Translations and Scaling Summary 1 Translations and Scaling 2 Summary 6 / 34 DE Lecture 11 王奕翔

Translations and Scaling e at 7 / 34 Proof : Summary Theorem DE Lecture 11 Translation on the s -Axis Recall that → 1 1 L L 1 − − → s , s − a . Multiplying 1 by e at in t -domain results in right-shift of a in s -domain. L Let f ( t ) − → F ( s ) . For any a , { } L e at f ( t ) = F ( s − a ) . ∫ ∞ ∫ ∞ e − ( s − a ) t f ( t ) dt = F ( s − a ) . { } L e at f ( t ) = e − st e at f ( t ) dt = 0 0 王奕翔

Translations and Scaling Summary 8 / 34 DE Lecture 11 F ( s − a ) F ( s ) F ( s − a ) s s = a, a < 0 s = a, a > 0 王奕翔

Translations and Scaling Summary 9 / 34 Hence, DE Lecture 11 Example Back to the Problem Solve y ′′ − 2 y ′ + y = e 2 t , y (0) = 1 , y ′ (0) = 5 . Step 3 : Compute the inverse Laplace transform of Y ( s ) : How to compute? { } 3 1 1 ⇒ y ( t ) = 3 L − 1 + e 2 t . Y ( s ) = ( s − 1) 2 + ( s − 1) 2 s − 2 = { 1 { 1 } } L − 1 = e t L − 1 = te t . ( s − 1) 2 s 2 3 1 y ( t ) = 3 te t + e 2 t . Y ( s ) = ( s − 1) 2 + ⇒ s − 2 = 王奕翔

Translations and Scaling We can obtain the inverse Laplace transform of 10 / 34 Summary DE Lecture 11 Laplace Transform of t n e at n ! { t n e at } L = n = 0 , 1 , 2 , . . . , s > a ( s − a ) n +1 , 1 ( s − a ) n : t n − 1 { 1 } L − 1 = n = 1 , 2 , . . . ( n − 1)! e at , ( s − a ) n 王奕翔

Translations and Scaling k 11 / 34 k We can obtain the corresponding inverse Laplace transforms: Summary DE Lecture 11 Laplace Transform of e at sin ( kt ) and e at cos ( kt ) { } L e at sin ( kt ) = ( s − a ) 2 + k 2 , s > a ( s − a ) { } L e at cos ( kt ) = ( s − a ) 2 + k 2 , s > a { } L − 1 = e at sin ( kt ) ( s − a ) 2 + k 2 { ( s − a ) } L − 1 = e at cos ( kt ) ( s − a ) 2 + k 2 王奕翔

Translations and Scaling = 12 / 34 = Summary = Step 1 : Laplace-transform both sides: Example Solving a Second-Order IVP with Laplace Transform DE Lecture 11 Solve y ′′ − 4 y ′ + 5 y = t 2 e 2 t , y (0) = 2 , y ′ (0) = 6 . t 2 e 2 t } { y ′′ } { y ′ } { L − 4 L + 5 L { y } = L 2 s 2 Y ( s ) − sy (0) − y ′ (0) ( ) ⇒ − 4 ( sY ( s ) − y (0)) + 5 Y ( s ) = ( s − 2) 3 2 ⇒ ( s 2 − 4 s + 5) Y ( s ) = 2 s − 2 + ( s − 2) 3 2 s − 2 2 Step 2 : Solve Y ( s ) : Y ( s ) = s 2 − 4 s + 5 + ( s 2 − 4 s + 5)( s − 2) 3 . Step 3 : Compute the inverse Laplace transform of Y ( s ) : 4( s − 2) 2 − 2 2 Y ( s ) = ( s − 2) 2 + 1 + ( s − 2) 2 + 1 + s − 2 + ( s − 2) 3 y ( t ) = 4 e 2 t cos t + 2 e 2 t sin t − 2 e 2 t + t 2 e 2 t . ⇒ 王奕翔

Translations and Scaling Summary 13 / 34 Tedious to calculate ... E D C DE Lecture 11 Partial fraction decomposition : 2 s − 2 2( s − 2) 2 s 2 − 4 s + 5 = ( s − 2) 2 + 1 + ( s − 2) 2 + 1 ( s 2 − 4 s + 5)( s − 2) 3 = A ( s − 2) + B 2 ( s − 2) 2 + 1 + s − 2 + ( s − 2) 2 + ( s − 2) 3 王奕翔

Translations and Scaling E 14 / 34 E Summary C F D D C DE Lecture 11 Tip 1 : Working in C makes life easier! ( s 2 − 4 s + 5)( s − 2) 3 = A ( s − 2) + B 2 ( s − 2) 2 + 1 + s − 2 + ( s − 2) 2 + ( s − 2) 3 F ∗ = s − 2 − i + s − 2 + i + s − 2 + ( s − 2) 2 + ( s − 2) 3 [ ] 2 2 F = = 2 i · i 3 = 1 ✘ ( ✘✘✘✘ ( s − 2 − i )( s − 2 + i )( s − 2) 3 s =2+ i A = F + F ∗ = 2 Re { F } = 2 , B = i ( F − F ∗ ) = − 2 Im { F } = 0 王奕翔

Translations and Scaling D 15 / 34 Summary d E DE Lecture 11 C Tip 2 : Taking derivatives 2 2( s − 2) ( s 2 − 4 s + 5)( s − 2) 3 = ( s − 2) 2 + 1 + s − 2 + ( s − 2) 2 + ( s − 2) 3 [ 2 ] E = = 2 s 2 − 4 s + 5 s =2 [ 2 ] [ − 2 ] D = = [( s − 2) 2 + 1] 2 2( s − 2) = 0 s 2 − 4 s + 5 d ( s − 2) s =2 s =2 [ 1 d 2 ] 2 C = s 2 − 4 s + 5 d ( s − 2) 2 2! s =2 ] 2 − 2( s − 2) 2 2 ( s − 2) 2 + 1 ( s − 2) 2 + 1 [[ ] [ ] = − 2 = − 2 [( s − 2) 2 + 1] 4 s =2 王奕翔

Translations and Scaling Summary 16 / 34 DE Lecture 11 Translation on the t -Axis Let’s compute L { f ( t − a ) } , a > 0 , given that L { f ( t ) } = F ( s ) : ∫ ∞ ∫ ∞ τ := t − a f ( τ ) e − s ( τ + a ) d τ L { f ( t − a ) } = f ( t − a ) e − st dt = 0 − a ∫ 0 ∫ ∞ = e − as f ( τ ) e − s τ d τ + e − as f ( τ ) e − s τ d τ 0 − a ∫ 0 = e − as F ( s ) + e − as f ( τ ) e − s τ d τ − a = e − as F ( s ) , if f ( t ) = 0 when t < 0 王奕翔

Translations and Scaling Summary 17 / 34 DE Lecture 11 If f ( t ) = 0 for t < 0 , then L { f ( t − a ) } = e − as L { f ( t ) } , for a > 0 . f ( t ) f ( t − a ) t = a, a > 0 t 王奕翔

Translations and Scaling Summary 18 / 34 DE Lecture 11 How about functions f ( t ) that is non-zero for t < 0 ? f ( t ) f ( t − a ) t t = a, a > 0 王奕翔

Translations and Scaling Summary 19 / 34 DE Lecture 11 Unit Step Function Definition (Unit Step Function) { 1 , t ≥ 0 U ( t ) := t < 0 . 0 , Note : L { f ( t ) U ( t ) } = L { f ( t ) } . U ( t ) 1 t 王奕翔

Translations and Scaling Summary Theorem (Translation on the t -Axis) 20 / 34 DE Lecture 11 For a > 0 , L { f ( t − a ) U ( t − a ) } = e − as L { f ( t ) U ( t ) } = e − as L { f ( t ) } . f ( t ) U ( t ) f ( t − a ) U ( t − a ) t = a, a > 0 t 王奕翔

Translations and Scaling Example 21 / 34 s Summary DE Lecture 11 s Example Examples: Laplace Transforms Calculate L {U ( t − a ) } . L {U ( t − a ) } = L { 1 · U ( t − a ) } = e − as L { 1 } = e − as . Calculate L { cos t U ( t − π ) } . L { cos t U ( t − π ) } = L { cos ( t + π − π ) U ( t − π ) } = e − π s L { cos ( t + π ) } = e − π s L {− cos t } = − e − π s s 2 + 1 . 王奕翔

Translations and Scaling . 22 / 34 Summary DE Lecture 11 Examples: Inverse Laplace Transforms Example { 1 } Calculate L − 1 ( s − 4) e s Since L − 1 { } 1 = e 4 t , according to the translation property: s − 4 { 1 } { 1 } L − 1 = L − 1 = e 4( t − 1) U ( t − 1) . s − 4 e − s ( s − 4) e s 王奕翔

Translations and Scaling . 23 / 34 Due to linearity and the translation property, the inverse transform is Summary s A: First we organize the term as follows: DE Lecture 11 Example Examples: Inverse Laplace Transforms { se − 2 s + e − s } Calculate L − 1 ( s − 1)( s − 2) se − 2 s + e − s 1 ( s − 1)( s − 2) e − 2 s + ( s − 1)( s − 2) = ( s − 1)( s − 2) e − s { − 1 { − 1 2 } 1 } e − 2 s + = s − 1 + s − 1 + e − s . s − 2 s − 2 { − e ( t − 2) + 2 e 2( t − 2) } { − e ( t − 1) + e 2( t − 1) } U ( t − 2) + U ( t − 1) . 王奕翔

Translations and Scaling Summary 24 / 34 DE Lecture 11 Piecewise-Defined Function Unit step function is useful in representing piecewise-defined functions. Example : the following function can be rewritten in terms of U : { 1 , a ≤ t < b f ( t ) = otherwise = U ( t − a ) − U ( t − b ) . 0 , U ( t − a ) − U ( t − b ) 1 a b t 王奕翔