Inverse Laplace transform - cvut.cz · Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transformII Example Example (Continuous-time system of second

Inverse Laplace transform Inverse Laplace transform - examples

Inverse Laplace transformModeling systems and processes (11MSP)

Bohumil Kovar, Jan Prikryl, Miroslav Vlcek

Department of Applied MahematicsCTU in Prague, Faculty of Transportation Sciences

5-th lecture 11MSP2019

verze: 2019-04-01 11:22


Obsah prednasky

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 transformDefinition

We have already stated that the inverse Laplace transform has theform 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 proceeddifferently.


How?

f (t)⇒ ⇐ F (p)

e−αt1

p + α

e−αt cosωtp + α

(p + α)2 + ω2

=e−(α−iω)t + e−(α+iω)t

2=

1

2

(1

p+α−iω + 1p+α+iω

)e−αt sinωt

ω

(p + α)2 + ω2

=e−(α−iω)t − e−(α+iω)t

2i=

1

2i

(1

p+α−iω −1

p + α + iω

)


Rational fraction functionPartial fractions decomposition

Laplace transform of the system output has the form of rationalfraction function,

R(p) =Q(p)

N(p)=

bmpm + bm−1p

m−1 + · · ·+ b1p + b0anpn + an−1pn−1 + · · ·+ a1p + a0

Fraction can be expressed as the sum of partial fractions which aresimple fractions with a constant in the numerator and one root ofN(p) in the denominator.


Rational fraction functionPartial fractions decomposition

Rational fraction functionQ(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 functionQ(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).


Rational fraction function IExample

Example (Racionalnı lomena funkce)

IfN(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).


Rational fraction function IIExample

Example (Racionalnı lomena 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 inverseLaplace transform is to: find the roots of the polynomial in thedenominator of the rational function N(p)


Inverse Laplace transform

Decomposition of rational functions into partial fractions has theform

Q(p)

N(p)=

n∑µ=1

kµp − p∞µ

=k1

p − p∞1+

k2p − p∞2

+ · · ·+ knp − p∞n

≡ k1p − p1

+k2

p − p2+ · · ·+ kn

p − pn,

where kµ are called residue.



For residues kµ

kµ = limp→p∞µ

(p − p∞µ)Q(p)

N(p)

= Q(p∞µ) limp→p∞µ

(p − p∞µ)1

N(p)

= Q(p∞µ) limp→p∞µ

1

N(p)

p − p∞µ

= Q(p∞µ)1

N ′(p∞µ)



For simplicity we will continue to write p∞µ → pµ. Because it’strue that

L−1{

1

p − α

}= eαt ,

we get

L−1{Q(p)

N(p)

}= L−1

n∑

µ=1

kµp − pµ

=n∑

µ=1

kµepµt .



We have proved the so-called Heaviside formula for the inversetransformation of a rational function with a simple roots

L−1{Q(p)

N(p)

}=∑µ

Q(pµ)

N ′(pµ)epµt


Inverse Laplace transform IExample

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 IIExample


It’s true that

k1 = limp→−1

6

p2 + 2p + 4=

6

1− 2 + 4=

6

3= 2,

k2 = limp→−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 IIIExample


k3 = limp→−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 transformWhat to do with multiple roots?

If N(p) = (p−p1)β1(p−p2)β2 . . . (p−pn)βn has repeated root withmultiplicity 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



Obviously, in the inverse transformation, the roots of a rationalfunction play a privileged role. Therefore, in the next we can onlydeal with such rational functions, whose numerator is unitary

H(p) =1

N(p).

If soN(p) = (p − p1)β1(p − p2)β2 . . . (p − pn)βn

has multiple roots, then the inverse Laplace transform has a form



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 IWhat to do with multiple roots?

The coefficients k(βm)µ can be obtained as follows.

Example (Inverse Laplace transform of function with multipleroots)

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+

k2p + 5

+k3

p + 7


Inverse Laplace transform IIWhat to do 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 IIIWhat to do with multiple roots?


If we subtract the expression1

63(p − 2)2from both sides of the

original equation, we get

1

(p − 2)2(p + 5)(p + 7)− 1

63(p − 2)2=

k(1)1

p − 2+

k2p + 5

+k3

p + 7

respectively, the equation

1

63

[−(p + 14)

(p − 2)(p + 5)(p + 7)

]=

k(1)1

p − 2+

k2p + 5

+k3

p + 7,


Inverse Laplace transform IVWhat to do 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.


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).


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 functionhaving 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–upmethod to the simple rootfraction.

=1

(p + 1)2− 1

(p + 1)(p + 2)Multiply.

=1

(p + 1)2− 1

p + 1+

1

p + 2

Apply the cover–upmethod to the last fractionon the right.


Obsah prednasky

1 Inverse Laplace transform

2 Inverse Laplace transform - examples

Differential equation


Inverse Laplace transform IExample

Example (Continuous-time system of second order)

Consider a linear continuous-time system described by a differentialequation

y ′′(t) + 3y ′(t) + 2y(t) = 5u(t),

where u(t) = 1(t) is a step function and the initial state of thesystem 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 IIExample


After the Laplace transformation of the differential equation, weget an algebraic equation

p2Y (p)− py(0)− y ′(0) + 3pY (p)− 3y(0) + 2Y (p) = 51

p.

Using the initial conditions, we find a solution of the algebraicequation in the form

Y (p) =5− p − p2

p(p + 1)(p + 2).


Inverse Laplace transform IIIExample


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 twofractions describe the transient state.

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Inverse Laplace transform - cvut.cz · Inverse Laplace transform Inverse Laplace transform - examples Inverse Laplace transformII Example Example (Continuous-time system of second