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 f (t) = 1 2πi c+i∞

c−i∞

F(p) ept dp ≡ L−1 {F(p)} . For rational fractions in complex variable p we will proceed differently.

Inverse Laplace transform Inverse Laplace transform - examples

How?

f (t) ⇒ ⇐ F(p) e−αt 1 p + α e−αt cos ωt p + α (p + α)2 + ω2 = e−(α−iω)t + e−(α+iω)t 2 = 1 2

• 1

p+α−iω + 1 p+α+iω

• e−αt sin ωt

ω (p + α)2 + ω2 = e−(α−iω)t − e−(α+iω)t 2i = 1 2i

• 1

p+α−iω −

1 p + α + iω

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, R(p) = Q(p) N(p) = bmpm + bm−1pm−1 + · · · + b1p + b0 anpn + an−1pn−1 + · · · + a1p + a0 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 p0ν, if Q(p0ν) = 0 and roots p∞µ, if N(p∞µ) = 0. If the function Q(p) N(p) has distinct real roots, then N(p) =

n

• µ=1

(p − p∞µ) = (p − p∞1)(p − p∞2) . . . (p − p∞n).

Inverse Laplace transform Inverse Laplace transform - examples

Rational fraction function I

Example

Example (Racion´ aln´ ı lomen´ a funkce) If N(p) = p3 + 3p2 + 6p + 4 = (p + 1)(p2 + 2p + 4) then N(p) = (p + 1)(p2 + 2p + 4) = (p + 1)(p + 1 + i √ 3)(p + 1 − i √ 3) so N(p) =

3

• µ=1

(p − pµ) = (p − p1)(p − p2)(p − p3).

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 p1 = −1 p2 = −1 − i √ 3 p3 = −1 + i √ 3 and it is true that N(p1) ≡ 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 Q(p) N(p) =

n

• µ=1

kµ p − p∞µ = k1 p − p∞1 + k2 p − p∞2 + · · · + kn p − p∞n ≡ k1 p − p1 + k2 p − p2 + · · · + kn p − pn , where kµ are called residue.

Inverse Laplace transform Inverse Laplace transform - examples

Inverse Laplace transform

For residues kµ kµ = lim

p→p∞µ(p − p∞µ)Q(p)

N(p) = Q(p∞µ) lim

p→p∞µ(p − p∞µ)

1 N(p) = Q(p∞µ) lim

p→p∞µ

1 N(p) p − p∞µ = Q(p∞µ) 1 N′(p∞µ)

Inverse Laplace transform Inverse Laplace transform - examples

Inverse Laplace transform

For simplicity we will continue to write p∞µ → pµ. Because it’s true that L−1

• 1

p − α

• = eαt,

we get L−1 Q(p) N(p)

• = L−1

  

n

• µ=1

kµ p − pµ    =

n

• µ=1

kµepµt.

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 L−1 Q(p) N(p)

• =
• µ

Q(pµ) N′(pµ)epµt

Inverse Laplace transform Inverse Laplace transform - examples

Inverse Laplace transform I

Example

Example (Simple roots) Laplace transform of the system impulse response is H(p) = 6 p3 + 3p2 + 6p + 4 = 6 (p + 1)(p2 + 2p + 4). Find h(t). Solution: First we decompose H(p) = 6 (p + 1)(p2 + 2p + 4) = k1 p + 1 + k2 p + 1 + i √ 3 + k3 p + 1 − i √ 3 .

Inverse Laplace transform Inverse Laplace transform - examples

Inverse Laplace transform II

Example

Example (Simple roots) It’s true that k1 = lim

p→−1

6 p2 + 2p + 4 = 6 1 − 2 + 4 = 6 3 = 2, k2 = lim

p→−1−i √ 3

6 (p + 1)(p + 1 − i √ 3) = 6 (−1 − i √ 3 + 1)(−1 − i √ 3 + 1 − i √ 3) = 6 (−i √ 3)(−i2 √ 3) = −1,

Inverse Laplace transform Inverse Laplace transform - examples

Inverse Laplace transform III

Example

Example (Simple roots) k3 = lim

p→−1−i √ 3

6 (p + 1)(p + 1 + i √ 3) = 6 (−1 + i √ 3 + 1)(−1 + i √ 3 + 1 + i √ 3) = 6 (i √ 3)(i2 √ 3) = −1.

Inverse Laplace transform Inverse Laplace transform - examples

Inverse Laplace transform

What to do with multiple roots?

If N(p) = (p −p1)β1(p −p2)β2 . . . (p −pn)βn has repeated root with multiplicity βi, we need to modify the previous approach because L

• e−αt

= 1 p + α L

• te−αt

= 1! (p + α)2 L

• t2e−αt

= 2! (p + α)3 . . . L

• tne−αt

= n! (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 H(p) = 1 N(p). If so N(p) = (p − p1)β1(p − p2)β2 . . . (p − pn)βn 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?

L−1

• 1

N(p)

• = ep1t
• k(1)

1

+ k(2)

1

t 1! + · · · + k(β1)

1

tβ1−1 (β1 − 1)!

• + ep2t
• k(1)

2

+ k(2)

2

t 1! + · · · + k(β2)

2

tβ2−1 (β2 − 1)!

• .

. . + epnt

• k(1)

n

+ k(2)

n

t 1! + · · · + k(βn)

n

tβn−1 (βn − 1)!

Inverse Laplace transform Inverse Laplace transform - examples

Inverse Laplace transform I

What to do with multiple roots?

The coefficients k(βm)

µ

can be obtained as follows. 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 1 (p − 2)2(p + 5)(p + 7) = k(2)

1

(p − 2)2 + k(1)

1

p − 2 + k2 p + 5 + k3 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) = k(2)

1

+ k(1)

1 (p − 2) + k2(p − 2)2

p + 5 + k3(p − 2)2 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) If we subtract the expression 1 63(p − 2)2 from both sides of the

• riginal equation, we get

1 (p − 2)2(p + 5)(p + 7) − 1 63(p − 2)2 = k(1)

1

p − 2 + k2 p + 5 + k3 p + 7 respectively, the equation 1 63

• −(p + 14)

(p − 2)(p + 5)(p + 7)

• = k(1)

1

p − 2 + k2 p + 5 + k3 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 kµ is reduced to a simple pole case and k(1)

1

= − 24 72 × 92 , k2 = 1 2 × 72 , k3 = − 1 2 × 92 .

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 R(p) = 1 (p + 1)2(p + 2).

Inverse Laplace transform Inverse Laplace transform - examples

Heaviside cover-up rule for multiple roots II

Example (Heaviside cover-up rule for multiple roots) We proceed as follows: R(p) = 1 (p + 1)2(p + 2) A sample rational function having repeated roots = 1 (p + 1) · 1 (p + 1)(p + 2) Factor-out the repeats. = 1 (p + 1)

• 1

p + 1 + −1 p + 2

• Apply

the cover–up method to the simple root fraction. = 1 (p + 1)2 − 1 (p + 1)(p + 2) Multiply. = 1 (p + 1)2 − 1 p + 1 + 1 p + 2 Apply the cover–up method to the last fraction

• n the right.

Inverse Laplace transform Inverse Laplace transform - examples

Obsah pˇ redn´ aˇ sky

1 Inverse Laplace transform 2 Inverse Laplace transform - examples

Differential equation

Inverse Laplace transform Inverse Laplace transform - examples

Inverse Laplace transform I

Example

Example (Continuous-time system of second order) Consider a linear continuous-time system described by a differential equation y′′(t) + 3y′(t) + 2y(t) = 5u(t), where u(t) = 1(t) is a step function and the initial state of the system is given by the state of output and speed over time t = 0: y(0) = −1 a y′(0) = 2. We have to find a solution y(t).

Inverse Laplace transform Inverse Laplace transform - examples

Inverse Laplace transform II

Example

Example (Continuous-time system of second order) After the Laplace transformation of the differential equation, we get an algebraic equation p2Y (p) − py(0) − y′(0) + 3pY (p) − 3y(0) + 2Y (p) = 5 1 p. Using the initial conditions, we find a solution of the algebraic equation in the form Y (p) = 5 − p − p2 p(p + 1)(p + 2).

Inverse Laplace transform Inverse Laplace transform - examples

Inverse Laplace transform III

Example

Example (Continuous-time system of second order) The partial fractions decomposition is Y (p) = 5 2p − 5 p + 1 + 3 2 1 (p + 2). The solution we are looking for is for t ≥ 0 y(t) = 5 2 − 5e−t + 3 2e−2t. The first fraction corresponds to the steady state, the other two fractions describe the transient state.