Block 2 Laplace Transform

Block 2 – Laplace Transforms

2. Laplace transforms

Contents

3.1 Introduction

3.2 The Laplace transform3.2.1 Definitions and notation3.2.2 Laplace transform of some simple functions3.2.3 Properties of the Laplace transform3.2.4 Table of Laplace transforms

3.3 The inverse Laplace transform3.3.1 Using partial fractions3.3.2 Finding inverses using the first shift theorem3.3.3 Completing the square

3.4 Solution of differential equations 13.4.1 Laplace transforms of derivatives3.4.2 Solution of 1st order differential equations3.4.3 Solution of 2nd order differential equations

3.5 Solution of differential equations3.5.1 Differential equations and Dirac functions3.5.2 Nonlinear differential equations

3.6 Systems of differential equations

3.7 Summary

Appendix A1 – Partial fractions

Appendix A2 – Completing the square

1


Prerequisite knowledge

• 1st Year Math Modules or A-level mathematics/ HND.

• An understanding of improper integrals would be useful

Learning objectives

By the end of this block, you should be able to:

• Evaluate Laplace transforms

• Evaluate inverse Laplace transforms

• Solve linear constant coefficient differential equations

• Apply Laplace transforms to problems in vibration, control systems and oscillations in electrical circuits.

• Solve systems of differential equations

2


3.1 Introduction

Laplace transforms are an interesting field of mathematics that can be used to solve problems involving differential equations and integro-differential equations. The technique is quite abstract in nature but it allows the solution of differential equations, without the need to do any integration or differentiation. The processes of integration and differentiation are replaced by algebraic manipulation, which is often considered easier to apply than concepts taken from calculus. A further advantage of the ‘Laplace transform method’ for solving initial value problems is the initial conditions are incorporated in an entirely natural way that does not require the solution of simultaneous equations.

The main reason for considering Laplace transforms at this stage is all engineering disciplines (except civil engineering) engage in a subject called control engineering, where Laplace transforms are used to analyse the response of engineering systems to changes in inputs to the system, whether it be a chemical reactor vessel or an auto-pilot in a plane.

Before we can attempt to solve differential equations using the Laplace transform, we need to introduce it and consider the Laplace transform and inverse Laplace transform for a number of simple functions and differential operators.

3.2 The Laplace transform

3.2.1 Definitions and notation

The Laplace transform of a function,

( )tf

is denoted as,

( ){ } ( )∫∞ −=

0dttfetfL st (3.1)

Alternative symbols used for the Laplace transform are given below,

( ){ } ( ) ( )sforsFtfL =

From (3.1) we see that the Laplace transform consists of an improper integral (one of the integration bounds is infinite). In the first instance with an improper integral you have to focus on whether the integral has a solution or not. For example the Laplace transform of the function,

( ) atetf =

only exists if

3


0<− sa

otherwise the Laplace transform is

( ) ∫∫∞∞ − =

00dtedttfe btst

and b=a-s is positive. This integral is unbounded. Returning to the terminology used to describe the Laplace transform (3.1)

ste−

is called the kernel of the integral. In the original function,

( )tf

The independent variable is t, which can be considered a time variable. The function f(t) is considered to exist in the time domain. The Laplace transform F(s) exists in a frequency domain. s is the independent variable of the Laplace transform and strictly speaking is a complex variable although we will for the most part only consider its real part.

3.2.2 Laplace transforms of some simple functions

Consider the function,

( ) ctf = (3.2)

where c is a constant. The Laplace transform of (3.2) is given below.

( ){ } ∫∞ −=0

dtectfL st

As this is an improper integral it should be considered in the limit of the upper bound being finite and the limit to infinity taken once the integral is evaluated.

( ){ } ∫∫ −∞

∞→

− ==b st

b

st dtcedtectfL00

lim

−−−=

−= −−−∫ 0

00

es

ce

s

ce

s

cdtce sb

bstb st

( )sbes

c −−= 1

We can now let b tend to infinity,

( )s

ce

s

c sb

b=− −

∞→1lim

4


To summarise,

{ }s

ccL =

This can be considered as a Laplace transform pair,

( ) ( )

==

s

csFctf ,

Worked Examples (Evaluating Laplace transforms by direct integration)

Let us consider another example in detail,

( ) ttf =

Solution

Consider the Laplace transform of t,

{ } ∫ ∫∞ −

∞→

− ==0 0

limb st

b

st dttedtettL

This can be integrated by parts,

∫0

bu

dvdt

dt=uv−∫0

bv

dudt

dt

u= t ⇒ du /dt=1

dv /dt= e−st ⇒ v=−1s

e−st

∫∫ −−− +

−=

b stb

stb st dtes

es

tdtte

00

0

1

∫∫ −−− +−=b stsbb st dte

se

s

bdtte

00

1

We could evaluate the integral on the right hand side but we can let b tend to infinity now and avoid the integration,

{ }200

11

11lim

sL

sdte

sdtte stb st

b==+= ∫∫

∞ −−

∞→

So the Laplace transform pair is,

5


( ) ( )

==

2

1,

ssFttf

Exercise

Use the same approach (i.e. one integration by parts) to show that,

( ) ( )

==

32 2,

ssFttf

The main point of this block is not the integration process. In this section we are interested in growing our collection of functions that we ‘know’ the Laplace transform for although we will not do this by integrating the Laplace transform for every function we come across.

6


For example, inspecting the three Laplace transform pairs given above a pattern is emerging for algebraic terms. The general rule is,

General Rule for Laplace transforms of algebraic terms

Consider the function,

( ) nttf =

Its Laplace transform is,

( )1

!+=

ns

nsF

One more example evaluated by direct integration,

Worked Example

Evaluate the Laplace transform of the function,

( ) ktetf =

Solution

{ } ( )∫ ∫∞ −−

∞→

− ==0 0

limb tks

b

stktkt dtedteeeL

( ) ( ) ( )( )0

00

11ee

kse

ksdte bks

btksb tks −

−−=

−−= −−−−−−∫

( )( )ks

eeks

bks

b −=−

−−= −−

∞→

11lim 0

provided

ksorks >>− 0

So the Laplace transform pair is,

( ) ( )

−==

kssFetf kt 1

,

As you will appreciate from the above examples deriving Laplace transforms by direct integration is quite a tedious process. To avoid direct integration many ingenious mathematical tricks and theorems are often used to find Laplace transforms. For example consider the following worked example.

7


Worked Example

Evaluate the Laplace transform of the function,

( ) iatetf =

Where a is some real number and

1−=i

Solution

At first glance this looks almost the same as the preceding worked example and indeed the solution is the same.

( ) ( )

−==

iassFetf iat 1

,

So why have we done two worked examples that are so very similar? The answer is the second leads onto an additional useful result giving us two extra Laplace transforms. Three Laplace Transforms for the price of one!

Complex numbers given in exponential form can be represented using Euler’s formula

( ) atiatetf iat sincos +==

Reconsidering the Laplace transform of the previous example

{ } { } { } { }atiLatLatiatLeL iat sincossincos +=+= (3.3)

The separation of the Laplace transform into an application of the Laplace transform to the two terms is possible as the Laplace transform is a linear operator. This will be discuss further below, for now just accept it.

Again from the previous worked example,

{ }22

11

as

ias

ias

ias

iasiaseL iat

++=

++×

−=

−=

So from (3.3)

{ } { }2222

sincosas

ai

as

satiLatL

++

+=+ (3.4)

Therefore equating real and imaginary parts of (3.4) gives the following results.

8


Laplace Transforms of Trigonometric Functions

The Laplace transform pair for sin at

( ) ( )

+==

22,sin

as

asFattf

The Laplace transform pair for cos at,

( ) ( )

+==

22,cos

as

ssFattf

3.2.3 Properties of the Laplace transform

The Laplace transform is a linear operator which means it has the property that,

( ) ( ){ } ( ){ } ( ){ }tgLtfLtgtfL βαβα +=+

where α and β are constants and f(t) and g(t) are functions.

Example

Determine,

{ } { } { }tt eLtLetL 33 2323 +=+

Solution

Using the linearity property of the Laplace transform,

{ } { } { }tt eLtLetL 33 2323 +=+

We can now apply the Laplace transforms derived in the previous section,

3

232 −

+=ss

{ }3

2323

23

−+=+

ssetL t

9


Example

Determine,

{ }tettL 462sin435 −+−

Solution

Using the linearity property of the Laplace transform,

{ } { } { } { } { }tt eLtLtLLettL 44 62sin431562sin435 −+−=−+−

Applying the Laplace transforms derived or presented in the previous section,

4

6

4

24

3522 −

−

++−=

ssss

{ }4

6

4

83562sin435

224

−−

++−=−+−

ssssettL t

10


Another mathematical tool that can be used to grow our collection of Laplace transforms is called the First Shift Theorem.

First Shift Theorem

If f(t) is a function having a Laplace transform, F(s) then the function,

( )tfeat

has a Laplace transform given by,

( ){ } ( )asFtfeL at −=

Sometimes,

( )asF − is denoted, ( )[ ] asssF −→

Here are a couple of worked examples using the first shift theorem to derive Laplace transforms.

Worked Example

Determine,

{ }tetL 2−

Solution

{ } ( )sFs

tL ==2

1

By the first shift theorem,

{ } ( )[ ] ( ) 222

2

1

+== +→

−

ssFetL ss

t

Worked Example

Determine,

{ }teL t 2sin3−

11


Solution

{ } ( )sFs

tL =+

=4

22sin

2

By the first shift theorem,

{ } ( )[ ] ( ) 43

22sin

233

++== +→

−

ssFteL ss

t

3.2.4 Table of Laplace transforms

Deriving a Laplace transform every time a function crops up is a time consuming process. Remembering the Laplace transforms for all of the functions given above is also not a realistic proposition for most students.

Therefore in the exam the table of Laplace transforms presented below is handed out.

The table together with the first Shift theorem and the linearity property of the Laplace transform allows you to determine the Laplace transform of many functions.

12


( )tf ( )sF

cs

c

t2

1

snt

1

!+ns

n

kteks −

1

atsin22 as

a

+atcos

22 as

s

+att sin

( ) 222

2

as

sa

+att cos

( ) 222

22

as

as

+−

( ) ( )tbgtaf + ( ) ( )sbGsaF +( )tf ′ ( ) ( )0fssF −

( )tf ′′ ( ) ( ) ( )002 fsfsFs ′−−( )at −δ sae−

( ) ( )

<≥−

=at

atatgtf

0

( )sGe sa−

Table of Laplace transforms.

The table above summarises the results presented in the previous two sections. In addition there are Laplace transforms for derivatives and something called the Dirac delta function. These are useful for the solution of differential equations and will be considered in detail in the coming sections.

13


Challenge Exercise

Confirm by applying the first shift theorem to the Laplace transform of

{ }iatetL

that

{ } ( )222

22

cosas

asattL

+−=

and

{ } ( )222

2sin

as

asattL

+=

3.3 The inverse Laplace transform

If,

( ){ } ( )sFtfL =

Then the inverse Laplace transform is denoted, L-1 and,

( ){ } ( )tfsFL =−1

Examples

a) As

{ }ks

eL kt

−= 1

kteks

L =

−− 11

b) As

{ }22

sinas

aatL

+=

atas

aL sin

221 =

+−

14


3.3.1 Using partial fractions

The most obvious way to find an inverse transformation is to use the table of Laplace transforms, however some manipulation of rational polynomials of the form,

( )( )sq

sp(3.5)

is often required. For example they might be represented as partial fractions, see Appendix A1 on partial fractions if you need your memory jogging.

Similar to the Laplace transformation the inverse Laplace transform is a linear operator, so

( ) ( ){ } ( ){ } ( ){ }sGLsFLsGsFL 111 −−− +=+ βαβα

Using partial fractions and the linear property given above we can calculate our first inverse Laplace transforms that do not appear in the Laplace transform table on page 11. See the worked example below.

Worked Example

Find

( )( )

−+−

23

11

ssL

Solution

We are ultimately going to use the first shift theorem and the table of Laplace transforms, but the first thing to do is represent the Laplace transform using partial fractions,

( )( )( ) ( )

( )( )23

32

2323

1

−+++−=

−+

+=

−+ ss

sBsA

s

B

s

A

ss

( )( )( )23

32

−++−+=

ss

BAsBA

Equating terms,

0: =+ BAs

132:1 =+− BA

These equations can be solved to give,

5

1−=A5

1=B

15


( )( )

−+

+−=

−+−−−

2

1

5

1

3

1

5

1

23

1 111

sL

sL

ssL

By inspection of the table of Laplace transforms the inverse Laplace transform is,

( )( )tt ee

ssL 231

5

1

5

1

23

1 +−=

−+−−

Worked Example

Find

( )

++−

9

122

1

ss

sL

Solution

The first thing to do is represent the Laplace transform using partial fractions,

( )( ) ( )

( )9

99

99

122

2322

2222 ++++++=

++++=

++

ss

DsCssBsAs

s

DCs

s

B

s

A

ss

s

( ) ( )( )9

9922

23

++++++=

ss

BAssDBsCA

Equating terms,

0:3 =+CAs

0:2 =+ DBs

19: =As

19:1 =B


9

1=A

9

1=B

9

1−=C

9

1−=D

( )

+−

+−

+

=

++ −−−−−

9

1

9

1

99

11

9

11

9

1

9

12

12

12

1122

1

sL

s

sL

sL

sL

ss

sL

16


The first three terms are easy to evaluate using the table of Laplace transforms,

tts

sL

sL

sL 3cos

9

1

9

1

9

1

99

11

9

11

9

12

12

11 −+=

+−

+

−−−

The fourth term requires a little more work,

ts

Ls

Ls

L 3sin27

1

9

3

27

1

9

1

93

3

9

1

9

12

12

12

1 =

+=

+×=

+−−−

the key point being multiplying the rational function by 3 and dividing by 3 to give a Laplace transform that has the same form as one of the ‘standard’ transforms.

This is a common operation in the business of finding inverse Laplace transforms

Putting this all together,

( ) tttss

sL 3sin

27

13cos

9

1

9

1

9

1

9

122

1 −−+=

++−

3.3.2 Finding inverses using the 1st shift theorem

The first shift theorem applied to the inverse Laplace transform,

( ){ } ( )tfeasFL at=−−1

Sometimes the alternative notation is simpler to apply,

( )[ ]{ } ( )tfesFL atass =−→

−1

We will only do one example in this section, however you will have ample opportunity to see this technique throughout the rest of these notes.

Worked Example

Evaluate

( )

+−

21

2

1

sL

17


Solution

( )t

ss

tes

Ls

L 2

22

12

1 1

2

1 −

+→

−− =

=

+

3.3.3 Completing the square

Sometimes a quotient of the form,

( )( )sq

sp

cannot be simplified using partial fractions. For example consider the quotient,

136

22 ++ ss

The quadratic term,

01362 =++ ss

has complex roots so cannot be factorised and therefore partial fractions does not offer a way forward in the next worked example. The alternative approach is to complete the square, see Appendix A2 if you have forgotten how to do this.

Worked Example

Evaluate

++−

136

22

1

ssL

Solution

As stated above the quotient cannot be represented as the sum of two partial fractions as the quadratic

01362 =++ ss

has no real roots. The answer is to complete the square.

( ) 43913)3(136 222 ++=−++=++ ssss

So,

18


( )

++=

++−−

43

2

136

22

12

1

sL

ssL

tes

L t

ss

2sin4

2 3

32

1 −

+→

− =

+=

Worked Example

Evaluate

+++−

52

72

1

ss

sL

Solution

Completing the square on the denominator,

( ) 4115)1(52 222 ++=−++=++ ssss

So,

( )

+++=

+++ −−

41

7

52

72

12

1

s

sL

ss

sL

( )

++++= −

41

612

1

s

sL

++

+=

++=

+→

−

+→

−

+→

−

12

1

12

1

12

1

4

6

44

6

ssssss sL

s

sL

s

sL

The first term is ready for inversion,

tes

sL t

ss

2cos4 1

21 −

+→

− =

+

The second term requires a little more manipulation to see the inverse.

tes

Ls

L t

ssss

2sin34

23

4

6

12

1

12

1 −

+→

−

+→

− =

+=

+

Putting the two terms together,

19


tetess

sL tt 2sin32cos

52

72

1 −−− +=

+++

Let us consider one more example, applying the same ideas to bed this in.

Worked Example

Find the inverse Laplace transform of

106

122 ++

−ss

s

Solution

Again we complete the square on the quadratic term,

( ) 13910)3(106 222 ++=−++=++ ssss

( )( )

( )

++−+=

++

−=

++− −−−

13

732

13

12

106

122

12

12

1

s

sL

s

sL

ss

sL

( ) ( )

++

−

++

+= −−

13

17

13

32

21

21

sL

s

sL

+−

+=

+→

−

+→

−

32

1

32

1

1

17

12

ssss sL

s

sL

tete tt sin7cos2 33 −− −=

This process at least initially has to be done step by step in a slow and methodical way otherwise silly mistakes will creep into the solution.

20


3.4 Solution of differential equations 1

This to some degree is the end point of this block on Laplace transforms as we are studying Laplace transforms as another technique for solving differential equations.

3.4.1 Laplace transforms of derivatives

Consider the Laplace transform of the derivative of a function,

∫∞ −=

0dte

dt

df

dt

dfL st

Applying integration by parts,

∫0

∞ dfdt

e−st dt=∫ 0

∞u

dvdt

dt=uv−∫0

∞v

dudt

dt

steu −=

dtdt

dfdv =

dtsedu st−−=

fv =

( )[ ] ( )∫∫∞ −∞−∞ − +=

000dtetfstfedte

dt

df ststst

( )( ) ( ){ }tfsLfdtedt

df st +−=∫∞ − 00

0

Laplace transformation of the first derivative of a function f(t).

( ) ( )0fsFsdt

dfL −=

Similarly the Laplace transformation of a functions second derivative can be found by integration by parts,

Laplace transformation of the second derivative of a function f(t).

( ) ( ) ( )0022

2

fsfsFsdt

fdL ′−−=

21


It has to be said that in the analysis above it is assumed that the function f(t) and its derivatives are sufficiently smooth for the integrals to exist.

Note the Laplace transforms of derivatives are included in the table of Laplace transforms.

3.4.2 Solution of first order differential equations

Now we have the Laplace transforms of derivatives and we can find the inverse Laplace transform we can now solve differential equations without having to integrate or differentiate anything!

Consider the following initial value problem,

teydt

dy 322 =−

(3.6)( ) 20 =y

Take the Laplace transform of both sides of the differential equation,

{ }teLydt

dyL 322 =

−

{ } { }teLyLdt

dyL 322 =−

( ) ( ) ( )3

220

−=−−

ssYysYs (3.7)

where Y(s) denotes the Laplace transform of the function, y(t).

We can now reorganise the above equation (3.7) to make the Laplace transform of y(t) the subject of the equation.

( ) ( ) 23

22 +

−=−

ssYs

( ) ( )( ) 2

2

32

2

−+

−−=

ssssY

( ) ( )( ) ( ) ( ) ( ) 3

2

32

42

32

322

−=

−−−=

−−−+=

sss

s

ss

ssY

Now all we have to do is find the inverse Laplace transform. In this case the inverse Laplace transform can be found directly from the table of Laplace transforms,

22


( ){ } ( ) tes

LtysYL 311 23

2 =

−== −−

This is the solution to the initial value problem (3.6).

We have solved our first differential equation using the Laplace transform.

If you review the example given above you will see the Laplace transform method for solving differential equations can be separated into three steps,

The Laplace Transform Method

Step 1

Take the Laplace Transform of the given differential equation

Step 2

Make the transformed variable (Y(s) above) the subject of the transformed equation.

Step 3

Apply the inverse Laplace transform to find y(t).

As you can see from the example given above the initial value is included in the solution without having to introduce a constant A and then use the initial value to find the value of A.

The Laplace transform method of solving differential equations comes into its own when considering second order differential equations.

3.4.3 Solution of 2nd order differential equations

The nonhomogeneous second order constant coefficient differential equation reads,

( )tfcydt

dyb

dt

yda =++

2

2

Note using Laplace transforms we can solve this equation without the need to separate the solution into the complementary function (solution to the homogeneous problem (f(t)=0) and a particular integral to extend the solution to the nonhomogeneous differential equation. The solution of second order differential equations follows a similar line to the solution of first order differential equations using the Laplace transform method.

23


Example

Solve the second order differential equation,

0652

2

=++ ydt

dy

dt

yd

Subject to the initial values,

( ) ( ) 00,10 ==dt

dyy

Solution

Apply the Laplace transform to both sides of the differential equation,

{ } 0652

2

=+

+

yLdt

dyL

dt

ydL

( ) ( ) ( ) ( ) ( )( ) ( ) 0605002 =+−+−− sYyssYdt

dysysYs

Substitute the initial values into the transformed equation,

( ) ( )( ) ( ) 06152 =+−+− sYssYssYs

Reorganise the transformed equation such that Y(s) is the subject of the equation,

( ) ( ) 5652 +=++ ssYss

( )65

52 ++

+=ss

ssY

We can now take the inverse Laplace transform of both sides,

( ){ }

+++= −−

65

52

11

ss

sLsYL (3.8)

( ){ } ( )tysYL =−1 (3.9)

Considering the right hand side of (3.8),

( )( ) 3232

5

65

52 +

++

=++

+=++

+s

B

s

A

ss

s

ss

s

24


( ) ( )( )( )

( )( )( )32

23

32

23

+++++=

+++++=

ss

BAsBA

ss

sBsA

Equating terms

1: =+ BAs

523:1 =+ BA

These can be solved to give,

2,3 −== BA

32

12

1 12

13

3

2

2

3

65

5

+→+→

−−

−

=

+−

+=

+++

ssss ssssL

ss

sL

tt ee 32 23 −− −=

Putting the above together with (3.9),

( ) tt eety 32 23 −− −=

Let us do another worked example that includes a nonhomogeneous degeneracy!

Worked Example


teydt

dy

dt

yd 22

2

65 −=++


( ) ( ) 10,10 −==dt

dyy

Solution


{ } { }teLyLdt

dyL

dt

ydL 2

2

2

65 −=+

+

( ) ( ) ( ) ( ) ( )( ) ( )2

1605002

+=+−+−−

ssYyssY

dt

dysysYs


25


( ) ( )( ) ( )2

161512

+=+−++−

ssYssYssYs


( ) ( ) 42

1652 ++

+=++ s

ssYss

( ) ( ) ( ) ( ) ( )32

4

32

1

65

42

1

22 ++++

++=

++

+++=

ss

s

ssss

sssY


( ){ } ( ) ( ) ( )( )

++++

++= −−

32

4

32

12

11

ss

s

ssLsYL (3.10)

( ){ } ( )tysYL =−1

Considering the right hand side of the above equation we have two sets of partial fractions to calculate,

( ) ( ) ( ) 32232

122 +

++

++

=++ s

C

s

B

s

A

ss

( )( ) ( ) ( )( ) ( )32

23322

2

++++++++=

ss

sCsBssA(3.11)

This can be solved in the same way as the first worked example for 2nd order differential equations.

Alternatively you can initially set s=-2 and equating the LHS and RHS,

( ) 1321 =⇒+−= BB

Setting s=-3,

( ) 1231 2 =⇒+−= CC

A can now be found by equating the s2 terms on the RHS of (3.11),

10 −=⇒=+ ACA

Similarly (the details have been omitted),

26


( )( ) 3

1

2

2

32

4

+−

+=

+++

ssss

s

Putting the partial fractions together,

( ) ( ) ( ) ( ) ( ) 3

1

2

2

3

1

2

1

2

1

32

4

32

122 +

−+

++

++

++

−=++

++++ sssssss

s

ss

( ) 22

1

2

1

++

+=

ss

We can now take the inverse Laplace transform of the RHS of (3.10).

( )tt

ssss

teessss

L 22

22

22

1 11

2

1

2

1 −−

+→+→

− +=

+

=

++

+

( ) ( )tety t += − 12

So we see that in principle solving homogeneous and nonhomogeneous second order differential equations can be treated in the same way using the Laplace transform method.

One last example before we move on.

Example


092

2

=+ ydt

yd


( ) ( ) 10,00 ==dt

dyy

Solution


{ } 092

2

=+

yLdt

ydL

( ) ( ) ( ) ( ) 09002 =+−− sYdt

dysysYs


27


( ) ( ) 0912 =+− sYsYs


( ) ( ) 192 =+ sYs

( )9

12 +

=s

sY


( ){ }

+= −−

9

12

11

sLsYL

( ){ } ( )tysYL =−1

Considering

ts

Ls

L 3sin3

1

9

3

3

1

9

12

12

1 =

+=

+−−

giving the solution,

( ) tty 3sin3

1=

28


3.4.3.1 An LCR circuit example

Before we move on let us consider one last worked example. This is an initial value problem you have seen before, but solved using techniques considered in EPS Maths 3.

The worked example is a model for an LCR circuit, see the figure below. An LCR circuit is one that includes a resistor, a capacitor and an inductor, connected in series with a voltage source e(t).

Figure 3.1 – An LCR circuit.

Before closing the switch at time t=0 the charge on the capacitor, q and the resulting current in the circuit are zero. Applying Kirchhoff’s second law to the circuit gives a second order nonhomogeneous differential equation for the charge on the capacitor,

( )teqCdt

dqR

dt

qdL =++ 1

2

2

( ) ( ) 000 ==dt

dqq

In the circuit given above the different components have the following values,

( ) VoltsteFaradCHenryLOhmsR 2010,1,160 4 ==== −

Solution

Substituting the values of the electrical properties into the differential equation,

2010160 42

2

=++ qdt

dq

dt

qd

Taking Laplace transforms of both sides of the differential equation,

t=0R L

C q(t)e(t) i(t) q

29


{ } { }2010160 42

2

LqLdt

dqL

dt

qdL =+

+

( ) ( ) ( ) ( ) ( )( ) ( )s

sQqssQdt

dqsqsQs

2010016000 42 =+−+−−

Substituting in the initial values,

( ) ( )s

sQss20

10160 42 =++

Making the Laplace transform of the solution of the differential equation the subject of the equation,

( ) ( )42 10160

20

++=

ssssQ

To take the inverse Laplace transform of the RHS it needs to be represented using partial fractions,

( )( )

( )42

242

4242 10160

10160

1016010160

20

++++++=

++++=

++ sss

CsBsssA

ss

CBs

s

A

sss

Equating terms,

BAs +=0:2

CAs +=1600:

A41020:1 =

These equations have the solution,

500

160,

500

1,

500

1 −=−== CBA

( )

+++−=

++ 4242 10160

1601

500

1

10160

20

ss

s

ssss

The quadratic term in the equation above has complex roots so the approach to take to find the inverse Laplace transform is to complete the square,

( ) ( ) ( )

++

−++

+−=

++

+−222222 6080

80

6080

801

500

1

6080

1601

500

1

ss

s

ss

s

s

The first two terms are now in a form that the inverse Laplace transform can be taken. The final term requires further manipulation,

30


( ) ( )

++

−++

+−=2222 6080

60

3

4

6080

801

500

1

ss

s

s

Therefore,

( )

+−

+−

=

++ +→+→

−

8022

802242

1

60

60

3

4

60

1

500

1

10160

20

ssss ss

s

ssssL

( )

−−=

++−−− tete

sssL tt 60sin

3

460cos1

500

1

10160

20 808042

1

So the solution reads,

( )

−−= −− tetetq tt 60sin

3

460cos1

500

1 8080

3.5 Solution of Differential Equations 2

So far we have solved differential equations using the Laplace transform method without the direct application of techniques taken from calculus. It has offered a number of advantages compared to other techniques. We will consider the application of the Laplace transform method to a class of differential equations that are not easily solved any other way.

3.5.1 Differential equations and Dirac functions

Laplace transforms can be used to solve problems involving an impulsive force or current, where the impulse is delivered over a time interval, (t0, t1)

( ) ( )∫= 1

0

t

tdttftI

I(t) is the total momentum input to the system. For an electrical circuit the equivalent property is the applied voltage, V(t).

Dirac delta function

Suppose the applied force is given by the function.

( )

+<<=

otherwise

ttttf0

100 ε

εε

The above function should be interpreted as a constant force (1/ε) applied over a time interval of length ε. By construction,

31


( ) ( ) 110

00=== ∫∫

+∞ εεε ε

t

tdtdttftI

Iε(t) is the area under the curve fε(t) and it is independent of ε. Taking the limit,

( ) ( )00 tttf −→→ δε ε

Where δ(t-t0) is called the Dirac delta function. This is a peculiar name for this construct as it does not have all of the properties of a function. For this reason it is a member of a class called generalised functions. The Dirac delta function is zero everywhere except at t=t0

where it is a singularity and therefore undefined. The important point is the Dirac delta function has the property,

( ) 10 0 =−∫∞

dtttδ

This is important as it represents an impulse of magnitude 1 acting over an infinitely short time interval. Another important property of the Dirac delta function is that,

( ) ( ) ( )00 0 tfdttftt =−∫∞δ

This means the Laplace transform of a Dirac delta function can be evaluated,

( ) sast edteat −∞ − =−∫0δ

This is called the sifting property of Dirac functions as it makes it possible to isolate a particular value of a function. To see this more clearly, consider,

( ){ } 10 == etL δ

and it thus follows that

{ } ( )tL δ=− 11

Therefore in principle any differential equation involving an impulse delivered over a very short time interval can be solved using Laplace transforms.

The last construct we need before we can start solving differential equations involving Dirac delta functions is the second shift theorem.

32


The Second Shift Theorem

If

( ) ( )

<≥−

=at

atatgtf

0

then

( ) ( )sGesF sa−=

where( ) ( ){ }tgLsG =

This is pretty dry stuff. It will become a little clearer when you see an application of the second shift theorem.

Worked Example


( )2

3 1

sesF s−=

Solution

By inspection of the second shift theorem in this example,

3=aand

( )2

1

ssG =

hence,

( ){ } ts

LsGL =

= −−

211 1

then the 2nd shift theorem implies,

<≥−

=

−−

30

3312

31

t

tt

seL s

One more worked example to see how the second shift theorem can be used.

33


Worked Example


( )7

12

+= −

sesF s

Solution

( )7

1

+=

ssG

hence,

( ){ } t

ss

ess

LsGL 7

7

11 1

7

1 −

+→

−− =

=

+=

Applying the 2nd shift theorem,

( )

<≥

=

+

−−−−

20

2

7

1 2721

t

te

seL

ts

We now have all of the tools in place to solve problems involving instantaneous impulses.

Consider a freely vibrating system without damping. The dynamics of the system is governed by a differential equation of the form,

022

2

=+ ydt

yd ω

See section 1.3.1 in the first semester of material. If the system is subjected to an instantaneous impulse of magnitude, a at a time, t=b, the second order differential equation is modified to,

( )btaydt

yd −=+ δω 22

2

Worked Example

Use Laplace transforms to solve the initial value problem,

( )342

2

−=+ tydt

yd δ

( ) ( ) 00,10 ==dt

dyy

34


Solution

We follow the same solution strategy as previous examples looking at solving differential equations using the Laplace transform method.

Step 1

Take the Laplace transform of the differential equation,

{ } ( ){ }342

2

−=+

tLyLdt

ydL δ

( ) ( ) ( ) ( ) sesYdt

dysysYs 32 400 −=+−−

(See the table of Laplace transforms for the Laplace transform of a Dirac function)

Step 2

Substitute the initial values into the equation,

( ) ( ) sesYssYs 32 4 −=+−

Step 3

Make Y(s) the subject of the equation,

( ) ( ) sesYs s +=+ −32 4

( )44

122

3

++

+= −

s

s

sesY s

Step 4

Find the inverse Laplace transform for Y(s).

( ){ }

++

+= −−−−

44

12

12

311

s

sL

seLsYL s

ts

sL 2cos

421 =

+−

To find the inverse Laplace transform,

+−−

4

12

31

seL s

35


we use the 2nd shift theorem. Consider,

ts

Ls

L 2sin2

1

4

2

2

1

4

12

12

1 =

+=

+−−

Therefore the 2nd shift theorem says,

( )( )

<

≥−=

+−−

30

332sin2

1

4

12

31

t

tts

eL s

Putting these results together gives the solution,

( ) ( )

≥−+

<=

362sin2

12cos

32cos

ttt

ttty

See the figure below for a plot of y(t).

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 2 4 6 8 10

x

y

Figure 3.2 – The displacement of a vibrating body subjected to an impulse (t=3.)

Let’s look at one last example before we move on.

36


Example

Use Laplace transforms to solve the initial value problem,

( )72862

2

−=++ tydt

dy

dt

yd δ

(3.12)

( ) ( ) 60,00 ==dt

dyy

Solution

We follow the same solution strategy as in the previous example.

Step 1

Take the Laplace transform of the differential equation,

{ } ( ){ }72862

2

−=+

+

tLyLdt

dyL

dt

ydL δ

( ) ( ) ( ) ( ) ( )( ) ( ) sesYyssYdt

dysysYs 72 280600 −=+−+−−

Step 2

Substitute the initial values into the equation,

( ) ( ) ( ) sesYssYsYs 72 2866 −=++−

Step 3

Make Y(s) the subject of the equation,

( ) ( ) 686 72 +=++ − sesYss

( )86

6

86

222

7

+++

++= −

ssssesY s

Step 4

Find the inverse Laplace transform for Y(s). This is a long process in this example but provided you take your time you will see the approach is the same as the previous example.

( ){ }

+++

++= −−−−

86

6

86

22

12

711

ssL

sseLsYL s

37


To find the inverses we have to represent the quotients using partial fractions as,

( )( )42862 ++=++ ssss

So,

( )( )( ) ( )

( ) ( )42

24

4242

2

+++++=

++

+=

++ ss

sBsA

s

B

s

A

ss

Equating terms,

BA

BAs

242:1

0:

+=

+=

This has solution,

1,1 −== BA

It follows that,

4286

2 77

27

+−

+=

++

−−−

s

e

s

e

sse

sss

and (steps omitted in the derivation)

4

3

2

3

86

62 +

−+

=++ ssss

42

112

1 13

13

4

13

2

13

86

6

+→+→

−−−

−

=

+−

+=

++ ssss sssL

sL

ssL

tt ee 42 33 −− −=

+−

+=

++−−−−−−

4

1

2

1

86

2 71712

71

seL

seL

sseL sss

t

ss

ess

L 2

2

1 1

2

1 −

+→

− =

=

+

t

ss

ess

L 4

4

1 1

4

1 −

+→

− =

=

+

38


So by the second shift theorem,

( )

<≥

=

+

−−−−

70

7

2

1 7271

t

te

seL

ts

Similarly

( )

<≥

=

+

−−−−

70

7

4

1 7471

t

te

seL

ts

Putting this altogether, (i.e. the results in the boxes above) we get

( ) ( ) ( )

≥−+−<−

=−−−−−−

−−

733

733747242

42

teeee

teety

tttt

tt

See the figure below for a graphical solution.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 2 4 6 8 10

x

y

Figure 3.3 – Graphical solution of the initial value problem, (3.12).

39


3.5.2 Nonlinear differential equations

Although Laplace transforms provide a good way of solving initial value problems without resorting to calculus based techniques they are restricted to linear differential equations with constant coefficients. This is differential equations of the form,

( )tfcydt

dyb

dt

yda =++

2

2

where a, b and c are constants. In this section we want to extend the range of application to nonlinear equations of the form,

( )tfydt

dyytc

dt

dy

dt

dyytb

dt

yd

dt

dyyta =

+

+

,,,,,,

2

2

For example

( ) 01 =−− yydt

dy

is a nonlinear first order differential equation. This is nonlinear as

1−= yc

There is no easy way of solving the above differential equation using Laplace transforms.

In its place we solve a linear differential equation that approximates the original nonlinear differential equation.

Approximating a nonlinear differential equation by a linear equation is called linearization. The derivation of an approximate linear differential equation uses a 1st order Taylor Series.

Recall that a Taylor series of a function f(y) centred on y0 is given by,

( ) ( ) ( ) ( ) ( ) ( ) +′′−+′−+= 02

0000 2

1yfyyyfyyyfyf

So for example consider the function,

( ) ( ) 21 yyyyyf −=−= (3.13)

( ) yyf 21−=′

Then a linear approximation to (3.13) centred on y0=2 is,

( ) ( ) ( ) ( )222 fyfyf ′−+≈

40


( ) ( ) yyyf 34)3(22 −=−−+−≈ (3.14)

See the Figure below to compare the function and it’s linear approximation.

-14

-12

-10

-8

-6

-4

-2

0

2

4

6

0 1 2 3 4

y

fun

cti

on

y(1-y)

4-3y

Figure 3.4 – Comparison of a function with its linear approximation centred on y0=2.

We see that near the centre of the Taylor series expansion, y0=2, the approximation is very accurate and accuracy deteriates as you move away from the centre. Let us see how this impacts on the solution of an initial value problem,

Consider the initial value problem,

( ) 01 =−− yydt

dy

(3.15)( ) 20 =y

This is a separable differential equation (some revision) so can be solved by separating the dependent and independent variables,

( ) dtdyyy

=−1

1

and integrating both sides of the equation.

( ) ∫∫ =−

dtdyyy 1

1

CtdtRHS +== ∫41


( ) ∫ ∫∫ −−=

−= dy

ydy

ydy

yyLHS

1

11

1

1 “From partial fractions”

( )

−

=−−=1

ln1lnlny

yyy

“I only need one integration constant”So

Cty

y +=

−1

ln

Rearranging the result such that y is the subject of the equation.

tCty

y

Aeey

ye ==

−= +

−

11

ln

tt

t

eA

A

Ae

Aey −−

=−

=1

the integration constant, A can be found using the initial value,

21

2 =⇒−

= AA

A

Giving the analytical solution

tey −−

=2

2

Quite a lot of work to produce an analytical solution to compare with the solution of the linear approximation to the linear problem,

Worked Example

Find an approximate solution to the initial value problem given below using a linear approximation to the function (3.13).

( ) 01 =−− yydt

dy

( ) 20 =y

42


Solution

The linear approximation to the function has already been calculated, (3.14) giving the linear initial value problem,

034 =+− ydt

dy

( ) 20 =y

This is now in a form that the Laplace transform method can be applied to.

Step 1

Applying the Laplace transform

{ } { } 034 =+−

yLLdt

dyL

( ) ( ) ( ) 034

0 =+−− sYs

yssY

Step 2

Substitute the initial value and make Y(s) the subject of the equation,

( ) ( )s

s

ssYs

42423

+=+=+

( ) ( )3

42

++=

ss

ssY

Step 3

Find the inverse Laplace transform using partial fractions,

( )( )

( )3

3

33

42

+++=

++=

++

ss

BssA

s

B

s

A

ss

s

Equating terms in the numerator gives A and B,

3

2,

3

4 == BA

( ) 3

111 1

3

2

3

4

3

1

3

21

3

4

3

42

+→

−−−

+=

++

=

++

ssssL

sL

ss

sL

43


te 3

3

2

3

4 −+=

Giving the result,

( ) tety 3

3

2

3

4 −+=

The figure below shows a comparison between the exact analytical solution and the approximate solution.

1

1.2

1.4

1.6

1.8

2

2.2

0 0.2 0.4 0.6 0.8 1

t

y

Exact Solution

Approximate Solution

Figure 3.5 – Comparison of the exact and approximate solution to (3.15).

From the figure above the two solutions are in close agreement for t<0.3 and the difference between the two curves grows with time. This is consistent with Figure 3.? Where as y decreases the linear approximation to the function (3.13) becomes less and less accurate.

This is not the first example of the use of linear approximation to produce a linear differential equation that can be solved that approximates a nonlinear differential equation. Recall last semester in the discussion of the oscillations of an undamped pendulum, the application of Newton’s second law (F=ma) gave the model equation,

θθsin

2

2

mgdt

dml −= (3.16)

Where θ is the angular displacement from the vertical, m is the pendulum mass, l is the length of the string and g is gravity. See Section 1.2.2.1 in the printed notes from last semester for more details. This is a second order nonlinear differential equation. Last semester the model was converted to an approximate linear model by noting that for small angular displacement,

θθ ≈sin

44


Giving the linear model after a little simplification,

02

2

=+ θθl

g

dt

d(3.17)

which can be solved the conventional way or by using Laplace transforms. Now we see that the argument could run slightly differently! The linear approximation to sin θ centred on the origin is given by,

( ) ( ) ( ) ( )000 θθθθθ fff ′−+≈

( ) ( ) θθ =−+≈ 0cos00sinyf

giving the result.

The figure below shows a comparison of the analytical solution of (3.17) and a numerical solution of (3.16) calculated using MATLAB, showing in this case that introducing the linear approximation produces an error in the phase that grows progressively with time..

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15 20

t

θ


Exact Solution

Figure 3.6 – Comparison of the nonlinear and linear undamped oscillator model.

45


Exercise

Solve the initial value problem

02

2

=+ θθl

g

dt

d

( ) 00,2.00 === twhendt

dθπθ

using the Laplace transform method.

Engineering Application – Outflow from a water tank.

Volume conservation applied to a water tank emptying gives the equation,

outVdt

dV −=

where the right hand side is the volume flow rate out of the tank. See the schematic figure below.

Figure 3.7 Gravity driven outflow from a water tank.

It is more convenient to work with the water height, h rather than the volume, V, where,

hAV cros=

h

Acros

outV

46


as if the outflow is due to gravity alone a model for the outflow is given by,

ghACV pipeDout ρ2=

There are a lot of symbols here that will not be defined as this is a mathematics module not a course in fluid mechanics, suffice it to say everything is a constant except h and V and these change with time, the remainder will soon be fixed to some convenient values so we can focus on the mathematics. Using the above the volume conservation equation becomes

ghACdt

dhA pipeDcros ρ2−=

Therefore given the initial height of the water, hmax the time the tank takes to empty can be estimated. To allow us to focus on the mathematics we group all of the constants together to give the initial value problem,

hCdt

dh −=

(3.18)( ) max0 hh =

where

gAA

CC pipe

cros

D ρ2=

Letting C=1 and hmax=1, the initial value problem can be solved (revision) to give the exact solution,

( ) 202

12

≤≤

−= tfor

tth (3.19)

Worked Example

Linearise the initial value problem (3.18) using a Taylor series of order one centred on h0=1 for the source term in (3.18) and solve using the Laplace transform method.

Solution (some steps have been omitted)

The first thing to do is derive the linear approximation to the source term. The general form of the truncated Taylor series is given below,

( ) ( ) ( ) ( )000 hfhhhfhf ′−+≈

Where in this case,

( ) hhf =

47


and

( )h

hf2

1=′

thus

( ) ( ) hhhf2

1

2

1

12

111 +=−+=

The linear approximation to the original initial value problem is then,

+−= h

dt

dh

2

1

2

1

( ) 10 =h

Applying the Laplace transform method to both sides and substituting in the initial value,

+−=− H

ssH

2

1

2

11

Make H the subject of the equation,

( )12

12

+−=

ss

sH

Using partial fractions

( )2

11

21

12

12

++−=

+−=

ssss

sH

Taking inverse Laplace transforms

( ) t

ss

es

th 2

1

2

121

121

−

+→

+−=

+−=

Giving the result,

( ) 12 2

1

−=− t

eth (3.20)

The figure below shows a comparison between the approximate ( 3.20) and exact solution (3.19) to the initial value problem (3.18).

48


0

0.2

0.4

0.6

0.8

1

1.2

0 0.5 1 1.5 2

t

h

Exact Solution


Figure 3.8 – Comparison of the exact and approximate solution to (3.18).

Exercise – Heating a slab up

The temperature of a slab being heated by an electric heater is given by the initial value problem,

( )44ambTTq

dt

dTC −−= α

( ) ambTT =0

where Tamb is the surrounding air temperature and has a value of 300K. C is the heat capacitance of the slab (a constant). q is the heat flux imposed by the heater (a constant) and α is called the coefficient of radiation (another constant).

Linearise the differential equation using a 1st order truncated Taylor Series centred on the initial temperature for the function,

44)( ambTTTf −=

Solve the resulting linear initial value problem using the Laplace transfer method given,

K

JC 000,1=

Wq 000,1=

49


48101

K

W−×=α

Exercise – Flow across a distillation column tray

The flow dynamics for a tray in a distillation column is governed by volume conservation,

outintray VVdt

dhA −=

where the outflow term is given by a correlation for flow over a weir of the form,

5.1ChVout =

Note C is a constant. See the figure below for the configuration of the distillation tray.

Figure 3.9 – Flow over a tray in a distillation column.

The values for the physical properties in this scenario are given convenient values rather than physically correct values.

h

inV

Atray

50


Given the initial value problem,

5.1232

1h

dt

dh −=

( ) 10 =h

Linearise the differential equation using a Taylor series centred on h0=1 and solve using the Laplace transform method.

3.6 Systems of differential equations

The extension of the Laplace transform method for solving differential equations is natural. Recall that for one differential equation the Laplace transform method is a three step process.

The Laplace Transform Method

Step 1

Take the Laplace Transform of the given differential equation

Step 2

Make the transformed variable (Y(s) above) the subject of the transformed equation.

Step 3

Apply the inverse Laplace transform to find y(t).

For a system of differential equations the second step is more complicated as it involves the solution of a system of (simultaneous) algebraic equations rather than one equation. By way of example consider the following initial value problem.

Example

Solve the following initial value problem using the Laplace transform method.

211 2xx

dt

dx +=

212 22 xx

dt

dx −=

( ) 201 =x , ( ) 102 =x

51


Solution

The first step is to take the Laplace transform of the differential equations.

{ } { }211 2 xLxL

dt

dxL +=

{ } { }212 22 xLxL

dt

dxL −=

( ) 2111 20 XXxsX +=−

( ) 2122 220 XXxsX −=−

Substituting the initial values into the equations and reorganising to put all of the unknowns on one side of the equations,

( ) 221 21 =−− XXs (3.21)

( ) 122 21 =++− XsX (3.22)

This is two equations in two unknowns, X1 and X2 which can be solved. From (3.21)

1

12 2

1 −+=

s

XX (3.23)

Substituting for X1 in (3.22),

( ) 121

14 2

2 =++−+− Xs

s

X

which can be rearranged,

1

3

1

41

1

412

1

42 −

+=−

+−=−

+=

++

−−

s

s

s

s

sXs

s

( )( )1

3

1

1242 −

+=

−−++−

s

sX

s

ss

( )( ) 2

1

32

3

6

322 −

=+−

+=−+

+=sss

s

ss

sX

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Having found X2 we can substitute into (3.23) to give X1.

2

2

2

21

1

12

1

12

1

21 −=

−−+

−=

−

+−=

ss

s

sssX

Having solved the simultaneous equations taking the inverse Laplace transform gives the result,

{ } t

ss

esss

LXL 2

2

11

1 21

22

12

2

2 =

=

−=

−=

−→

−−

using the table of Laplace transforms and the 1st shift theorem.

Similarly

{ } tes

LXL 212

1

2

1 =

−= −−

Giving the result,

( ) tetx 21 2=

( ) tetx 22 =

What about 2nd order systems?

What about nonhomogeneous systems?

The method is exactly the same. A second order system example is given below, a nonhomogeneous first order system is left as an exercise.

Worked Example

Solve the following second order system using the Laplace transform method.

022

2

=−+ yxdt

xd

022

2

=+− yxdt

yd

( ) 40 =x , ( ) 20 =y

dt

dy

dt

dxt == 0

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Solution

Some of the details of the solution will be omitted as it is essentially the same steps as the last example.

Take the Laplace transform of the differential equations

( ) ( ) 02002 =−+

−− YX

dt

dxsxXs

( ) ( ) 02002 =+−

−− YX

dt

dysyYs

Substitute for the initial values and rearrange such that the “unknowns” are on one side

( ) sYXs 422 =−+ (3.24)

( ) sYsX 222 =++− (3.25)

From (3.24),

( ) sXsY 422 −+= (3.26)

Substituting for Y in (3.25) and rearranging such that X is the subject of the equation,

( )( )31

104

34

10422

3

24

3

+++=

+++=

ss

ss

ss

ssX

Substituting this into (3.26) and tidying up the equation gives,

( )( )31

8222

3

+++=ss

ssY

Representing X and Y using partial fractions,

( )( )( ) ( ) ( ) ( )

( )( )31

13

3131

10422

22

2222

3

+++++++=

+++

++=

+++=

ss

sDCssBAs

s

DCs

s

BAs

ss

ssX

Multiplying out the brackets on the numerator and equating terms gives the following equations.

4:3 =+CAs

0:2 =+ DBs

103: =+CAs

03:1 =+ DB

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01,0,3 ==== DandCBA

Thus

31

322 +

++

=s

s

s

sX

Similarly,

31

322 +

−+

=s

s

s

sY

Taking the inverse Laplace transforms

{ }

++

+= −−−

313

21

211

s

sL

s

sLXL

Both of these inverse Laplace transforms can be found using the table of Laplace transforms,

( ) tttx 3coscos3 +=

Similarly

( ) ttty 3coscos3 −=

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There are many examples of systems of differential equations that are formulated as problems in vibration and circuit simulation, as an example application a mathematical model for the simulation of a circuit is given below.

Application

A two loop circuit is shown schematically in Figure 3.10 shown below.

Figure 3.10 – A double loop circuit.

Applying Kirchhoff’s 1st law at node X gives,

21 iii +=

Applying Kirchhoff’s 2nd law to the two loops in turn gives the differential equations,

( ) ( ) ViRiidt

dLiiR =++++ 12211211

012232

2 =−+ iRiRdt

diL

( ) ( ) 000 21 == ii

Given

OhmsROhmsRR 20,10 231 ===

HenrysLL 521 ==

VoltsV 200=

R1

R2 R

3

L1 L

2

V(t)

i1

i2

i

X

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Find the currents i1 and i2 using the Laplace transform method.

Solution

Firstly substitute the numerical values for the constant terms into the differential equations,

( ) 20020510 121

21 =+

+++ i

dt

di

dt

diii

020105 122 =−+ ii

dt

di

Dividing through by common factors in the differential equations,

( ) 4042 121

21 =++++ idt

di

dt

diii

042 122 =−+ ii

dt

di

and taking Laplace transforms

( )( ) ( )( )s

IisIisIII40

40022 1221121 =+−+−++

( )( ) 0420 1222 =−+− IIisI

Substituting in the initial values and tidying up,

( ) ( )s

IsIs40

26 21 =+++ (3.27)

( ) 024 21 =++− IsI (3.28)

From (3.28)

2

4 12 +

=s

II (3.29)

Substituting for I2 in (3.27) and making I1 the subject of the equation,

( )10

401 +

=ss

I

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From (3.29)

( ) ( )102

160

2

4 12 ++

=+

=ssss

II (3.30)

Using partial fractions,

( )( )

( )10

10

1010

401 +

++=+

+=+

=ss

BssA

s

B

s

A

ssI

equating terms in the numerators

4410

40 −=−=== ABandA

Therefore

+−

=

10

14

141 ss

I

Taking inverse Laplace transforms and applying the first shift theorem to the second term,

( ) ( )t

ss

ess

ti 10

101 14

144

10

144 −

+→

−=

−=

+−=

Similarly representing (3.30) using partial fractions and applying the 1st shift theorem (details omitted)

( ) tt eeti 1022 2108 −− +−=

3.7 Summary

This block has been organised into three separate sets of theory. The first looks at finding the Laplace transforms of functions. This can be done by evaluating the improper integral definition of the Laplace transform,

( ){ } ( )∫∞ −=

0dttfetfL st

This approach is not advocated here as a better approach is to use a table of Laplace transforms together with the first shift theorem and the linear property of the Laplace operator.

Having mastered Laplace transforms we focused on finding inverse Laplace transforms. This is a more difficult problem as it requires a certain ability to recognise patterns. The different

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techniques for finding inverse Laplace transforms, such as the use of partial fractions and completing the square are all about representing the Laplace transform in a way that it can be matched up with functions in the table of Laplace transforms. The first shift theorem in reverse can also be used to find inverse Laplace transforms.

We looked at the application of Laplace transforms to the solution of differential equations. This involves three steps,

Step 1 – Apply the Laplace transform to the differential equation

Step 2 – Rearrange the transformed equation to be explicit in the Laplace transform of the solution, Y(s)

Step 3 – Take the inverse Laplace transform to find the solution of the differential equation.

The Laplace transform method of solution for differential equations has a number of advantages over standard methods.

• The incorporation of initial values is simpler.• The solution technique requires no knowledge of calculus.• Certain problems involving near instantaneous disturbances to the modelled system

can be incorporated.

The last point above involves representing instantaneous disturbances using Dirac functions and solving the differential equation using the 2nd shift theorem.

The final part of the block looked at how this technique can be applied to nonlinear differential equations and systems of differential equations. Rather than solving a nonlinear differential equation the nonlinear term is replaced using a 1st order truncated Taylor Series centred on the initial value. This gives a linear differential equation that can be solved with Laplace transforms that is initially accurate and diverges from the exact solution for increasing time.

Linear systems of differential equations are relatively easy to solve using the Laplace transform method as you can follow the 3 step process given above. The second step for systems of differential equations includes the solution of simultaneous equations.

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Appendix A1- Partial Fractions (1st year notes)

In the same way you can add algebraic fractions together you can also separate them into combinations of algebraic fractions, such terms are called partial fractions.

Worked Example

Consider

23

1072 ++

+xx

x

Suppose we want to convert this into the sum of partial fractions.

Solution

Step 1. Factorise the denominator.

We have already used this as a worked example so we know

( )( )12

107

23

1072 ++

+=++

+xx

x

xx

x

Step 2. Assume a solution with unknown coefficients (A and B)

( )( ) 1212

107

++

+=

+++

x

B

x

A

xx

x

Step 3. Multiply both sides of the equation by the denominator (x+2)(x+1).

( )( )( )( )( )

( )( ) ( )( )1

12

2

12

12

10712

++++

+++=

+++++

x

Bxx

x

Axx

xx

xxx

and tidy up

( ) BxAxx )2(1107 +++=+

Step 4. Find A and B.

The above equation is true for all values of x. So let x be -1 and then -2.

1−=x

( ) ( )( ) ( ) BA )21(111017 +−++−=+−×

So 3=B

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2−=x

( ) ( )( ) ( ) BA )22(121027 +−++−=+−×

So 4=A

1

3

2

4

23

1072 +

++

=++

+xxxx

x

Where you have an algebraic fraction of the form,

( ) 21

2

+−

x

x

that is the denominator has a repeated root the above approach does not work you can try it if you like.

In this case you have to represent the algebraic fraction as the sum of two partial fractions in the form,

( ) ( ) 22 111

2

++

+=

+−

x

B

x

A

x

x

Multiply both sides by (x+1)2,

( ) ( )( )

( ) ( )( ) 2

22

2

2

1

1

1

1

1

21

+++

++=

+−+

x

Bx

x

Ax

x

xx

Tidy up,

( ) BAxx ++=− 12

You now equate terms to get two simultaneous equations in A and B to solve.

Ax =1:

BA +=−2:1

So

)(1 EasyA =

3−=B

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( ) ( ) 22 1

3

1

1

1

2

+−

+=

+−

xxx

x

You can also get algebraic fractions where the denominator includes quadratic terms (ax2+bx+c) that cannot be factorised into two linear factors as the roots are complex. This gives rise to a partial fraction of the form,

cbxax

BAx

+++

2

Where A and B are real constants.

Worked Example

Consider the algebraic fraction,

( )( )21

72 −++ xxx

x

Convert this into partial fractions.

Solution

Note the x2+x+1 has complex roots so cannot be factorised. The partial fraction representation is given below.

( )( ) 2121

722 −

+++

+=−++ x

C

xx

BAx

xxx

x

Now we have the partial fraction representation of the algebraic fraction we find the unknowns, A, B and C in the same way as the other examples.

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Step 1 Recombine the partial fractions over a common denominator

( ) ( ) ( )( )( )21

12

21 2

2

2 −+++++−+=

−+

+++

xxx

xxCxBAx

x

C

xx

BAx

Step 2 Multiply out the factors in the numerator, using F.O.I.L if required.

( ) ( ) ( )( )( ) ( )( )21

22

21

122

22

2

2

−+++++−+−=

−+++++−+

xxx

CCxCxBBxAxAx

xxx

xxCxBAx

Step 3 Tidy up

( )( )( ) ( )

( )( )21

22

21

72

2

2 −+++−++−++=

−++ xxx

CBxCBAxCA

xxx

x

Step 4 Equate terms

CAx +=0:2 (A1.1)

CBAx ++−= 27: (A1.2)

CB +−= 20:1 (A1.3)

Step 5 Solve system of equations to give A,B and C.

Three equations in three unknowns but the approach is the same as the other worked examples. From (A1.3),

AC −=

Substituting into the two remaining equations,

BAABA +−=⇒−+−= 3727 (A1.4)

AB −−= 20 (A1.5)

From (A1.5)

BA 2−=

Substituting for A in (A1.4) gives B.

( ) 177237 =⇒=⇒+−−= BBBB

Back substituting,

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22 −=−= BA

2=−= AC

Putting this all together gives the final result.

( )( ) 2

2

1

12

21

722 −

+++

+−=−++ xxx

x

xxx

x

The general rules for partial fractions are

Given a, b, c, l, m, n, p, q, r and s i.e. they are numbers then

( )( ) srx

B

qpx

A

srxqpx

nmx

++

+=

+++

( ) ( ) 22 qpx

B

qpx

A

qpx

nmx

++

+=

++

( )( ) qpx

C

cbxax

BAx

qpxcbxax

nmxlx

++

+++=

+++++

22

2

Multiply through by the denominator on the left hand side of the equation, tidy up and compare terms in the top line.

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Appendix A2 – Completing the square (1st year notes)

In any quadratic equation you can complete the square. Consider the quadratic function,

232 +−= xxy

You can complete the square by noting that

4

92

2

323

22 −+

−=+− xxx

This representation comes from noting that

42

222 bb

xbxx −

+=+

Where b is any number you like, such as -3. Once you have completed the square you can find the roots of the original quadratic although in the Laplace transform method that is not the point of the operation.

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Block 2 Laplace Transform

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Transcript of Block 2 Laplace Transform