Inverse Laplace transform Modeling systems and processes (11MSP) - PowerPoint PPT Presentation

Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transform Modeling systems and processes (11MSP) Bohumil Kov´ aˇ r, Jan Pˇ rikryl, Miroslav Vlˇ cek Department of Applied Mahematics CTU in Prague, Faculty of Transportation Sciences 5-th lecture 11MSP 2019 verze: 2019-04-01 11:22

Inverse Laplace transform Inverse Laplace transform - examples Obsah pˇ redn´ aˇ sky 1 Inverse Laplace transform Definition Partial fractions decomposition Heaviside cover-up method Multiple roots Heaviside cover-up rule 2 Inverse Laplace transform - examples

Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transform Definition We have already stated that the inverse Laplace transform has the form of an integral along a curve in a complex plane p � c + i ∞ 1 F ( p ) e pt d p ≡ L − 1 { F ( p ) } . f ( t ) = 2 π i c − i ∞ For rational fractions in complex variable p we will proceed differently.

Inverse Laplace transform Inverse Laplace transform - examples How? f ( t ) ⇒ ⇐ F ( p ) 1 e − α t p + α p + α e − α t cos ω t ( p + α ) 2 + ω 2 = e − ( α − i ω ) t + e − ( α +i ω ) t = 1 � � 1 1 p + α − i ω + p + α +i ω 2 2 ω e − α t sin ω t ( p + α ) 2 + ω 2 = e − ( α − i ω ) t − e − ( α +i ω ) t = 1 � 1 � 1 p + α − i ω − 2i 2i p + α + i ω

Inverse Laplace transform Inverse Laplace transform - examples Rational fraction function Partial fractions decomposition Laplace transform of the system output has the form of rational fraction function, N ( p ) = b m p m + b m − 1 p m − 1 + · · · + b 1 p + b 0 R ( p ) = Q ( p ) a n p n + a n − 1 p n − 1 + · · · + a 1 p + a 0 Fraction can be expressed as the sum of partial fractions which are simple fractions with a constant in the numerator and one root of N ( p ) in the denominator.

Inverse Laplace transform Inverse Laplace transform - examples Rational fraction function Partial fractions decomposition Rational fraction function Q ( p ) N ( p ) is said to have a zero points p 0 ν , if Q ( p 0 ν ) = 0 and roots p ∞ µ , if N ( p ∞ µ ) = 0. If the function Q ( p ) N ( p ) has distinct real roots, then n � N ( p ) = ( p − p ∞ µ ) = ( p − p ∞ 1 )( p − p ∞ 2 ) . . . ( p − p ∞ n ) . µ =1

Inverse Laplace transform Inverse Laplace transform - examples Rational fraction function I Example Example (Racion´ aln´ ı lomen´ a funkce) If N ( p ) = p 3 + 3 p 2 + 6 p + 4 = ( p + 1)( p 2 + 2 p + 4) then √ √ N ( p ) = ( p + 1)( p 2 + 2 p + 4) = ( p + 1)( p + 1 + i 3)( p + 1 − i 3) so 3 � N ( p ) = ( p − p µ ) = ( p − p 1 )( p − p 2 )( p − p 3 ) . µ =1

Inverse Laplace transform Inverse Laplace transform - examples Rational fraction function II Example Example (Racion´ aln´ ı lomen´ a funkce) The roots in this case are p 1 = − 1 √ p 2 = − 1 − i 3 √ p 3 = − 1 + i 3 and it is true that N ( p 1 ) ≡ N ( − 1) = 0 etc. From this example, the first step we have to do in the inverse Laplace transform is to: find the roots of the polynomial in the denominator of the rational function N ( p )

Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transform Decomposition of rational functions into partial fractions has the form n Q ( p ) k µ � N ( p ) = p − p ∞ µ µ =1 k 1 k 2 k n = + + · · · + p − p ∞ 1 p − p ∞ 2 p − p ∞ n k 1 k 2 k n ≡ + + · · · + , p − p 1 p − p 2 p − p n where k µ are called residue .

Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transform For residues k µ p → p ∞ µ ( p − p ∞ µ ) Q ( p ) k µ = lim N ( p ) 1 = Q ( p ∞ µ ) p → p ∞ µ ( p − p ∞ µ ) lim N ( p ) 1 = Q ( p ∞ µ ) lim N ( p ) p → p ∞ µ p − p ∞ µ 1 = Q ( p ∞ µ ) N ′ ( p ∞ µ )

Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transform For simplicity we will continue to write p ∞ µ → p µ . Because it’s true that � 1 � L − 1 = e α t , p − α we get   n n � Q ( p ) � k µ   � � L − 1 = L − 1 k µ e p µ t .  = N ( p ) p − p µ  µ =1 µ =1

Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transform We have proved the so-called Heaviside formula for the inverse transformation of a rational function with a simple roots � Q ( p ) � Q ( p µ ) L − 1 � N ′ ( p µ ) e p µ t = N ( p ) µ

Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transform I Example Example (Simple roots) Laplace transform of the system impulse response is 6 6 H ( p ) = p 3 + 3 p 2 + 6 p + 4 = ( p + 1)( p 2 + 2 p + 4) . Find h ( t ). Solution: First we decompose 6 H ( p ) = ( p + 1)( p 2 + 2 p + 4) k 1 k 2 k 3 √ √ = p + 1 + + . p + 1 + i 3 p + 1 − i 3

Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transform II Example Example (Simple roots) It’s true that 6 1 − 2 + 4 = 6 6 k 1 = lim p 2 + 2 p + 4 = 3 = 2 , p →− 1 6 k 2 = lim √ √ ( p + 1)( p + 1 − i 3) p →− 1 − i 3 6 √ √ √ = ( − 1 − i 3 + 1)( − 1 − i 3 + 1 − i 3) 6 √ √ = = − 1 , ( − i 3)( − i2 3)

Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transform III Example Example (Simple roots) 6 √ k 3 = lim √ ( p + 1)( p + 1 + i 3) p →− 1 − i 3 6 = √ √ √ ( − 1 + i 3 + 1)( − 1 + i 3 + 1 + i 3) 6 = √ √ = − 1 . (i 3)(i2 3)

Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transform What to do with multiple roots? If N ( p ) = ( p − p 1 ) β 1 ( p − p 2 ) β 2 . . . ( p − p n ) β n has repeated root with multiplicity β i , we need to modify the previous approach because 1 e − α t � � L = p + α 1! te − α t � � L = ( p + α ) 2 2! t 2 e − α t � � L = ( p + α ) 3 . . . n ! � t n e − α t � L = ( p + α ) n +1

Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transform What to do with multiple roots? Obviously, in the inverse transformation, the roots of a rational function play a privileged role. Therefore, in the next we can only deal with such rational functions, whose numerator is unitary 1 H ( p ) = N ( p ) . If so N ( p ) = ( p − p 1 ) β 1 ( p − p 2 ) β 2 . . . ( p − p n ) β n has multiple roots, then the inverse Laplace transform has a form

Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transform What to do with multiple roots? t β 1 − 1 � 1 � � t � k (1) + k (2) 1! + · · · + k ( β 1 ) L − 1 = e p 1 t 1 1 1 N ( p ) ( β 1 − 1)! t β 2 − 1 � � t k (1) + k (2) 1! + · · · + k ( β 2 ) + e p 2 t 2 2 2 ( β 2 − 1)! t β n − 1 . � � t . k (1) + k (2) 1! + · · · + k ( β n ) . + e p n t n n n ( β n − 1)!

Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transform I What to do with multiple roots? The coefficients k ( β m ) can be obtained as follows. µ Example (Inverse Laplace transform of function with multiple roots) Let, for example N ( p ) = ( p − 2) 2 ( p + 5)( p + 7) . We look for decomposition into partial fractions in the form ( p − 2) 2 + k (1) k (2) 1 k 2 k 3 1 1 ( p − 2) 2 ( p + 5)( p + 7) = p − 2 + p + 5 + p + 7

Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transform II What to do with multiple roots? Example (Inverse Laplace transform of function with multiple roots) Multiplying equation by ( p − 2) 2 ( p − 2) 2 ( p − 2) 2 ( p + 5)( p + 7) 1 ( p − 2) + k 2 ( p − 2) 2 + k 3 ( p − 2) 2 = k (2) + k (1) 1 p + 5 p + 7 and find the limit for p → 2, 1 (2 + 5)(2 + 7) = k (2) 1

Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transform III What to do with multiple roots? Example (Inverse Laplace transform of function with multiple roots) 1 If we subtract the expression 63( p − 2) 2 from both sides of the original equation, we get 63( p − 2) 2 = k (1) 1 1 k 2 k 3 1 ( p − 2) 2 ( p + 5)( p + 7) − p − 2 + p + 5 + p + 7 respectively, the equation = k (1) 1 � − ( p + 14) � k 2 k 3 1 p − 2 + p + 5 + p + 7 , 63 ( p − 2)( p + 5)( p + 7)

Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transform IV What to do with multiple roots? Example (Inverse Laplace transform of function with multiple roots) for which the calculation of k µ is reduced to a simple pole case and 2 4 k (1) = − 7 2 × 9 2 , 1 1 = k 2 2 × 7 2 , 1 = − k 3 2 × 9 2 .

Inverse Laplace transform Inverse Laplace transform - examples Heaviside cover-up rule for multiple roots I Example (Heaviside cover-up rule for multiple roots) We decompose rational fraction R ( p ) using Heaviside’s method 1 R ( p ) = ( p + 1) 2 ( p + 2) .