03 01 Laplace Transforms Slides Handout

Maths Transform MethodsTopic 00 : TOPIC NAME

Lectures 1326 : Laplace Transforms

Dr Kieran MurphyThis module and thesenotes were developedby Dr Pardaig Kirwanwith only minor modi-fications on my part.

Credits:

Department of Computing and Mathematics,Waterford Institute of Technology.

([email protected])

Autumn Semester, 2014

OutlineDefinition and properties of the Laplace transforms.

Convert a function in time domain (t) to a function in frequency domain (s), and back.

Solving constant linear, constant coefficient ODEs using Laplace transforms.

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Introduction to Laplace Transforms

Introduction to Laplace Transforms Lecture 13

Laplace transforms provide a method for representing and analyzing linearsystems using algebraic methods.The main idea behind the Laplace Transform is that we can solve an equation(or a system of equations) containing differential and integral terms bytransforming the equations in t-space to one in s-space.This makes the problem much easier to solve.If needed we can find the inverse Laplace transform which gives us thesolution back in t-space.

This can be illustrated as follows:

.

.

Differential EquationProblems

Algebra Problems

Solution of AlgebraProblems

Solution of DifferentialEquation

Laplace Transform

Very Easy

Inverse Laplace Transform

Difficult

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Pierre Simone de Laplace (1749-1827)Laplace was a French mathematician, astronomer and physicist who applied theNewtonian theory of gravitation to the solar system (an important problem of hisday). He played a leading role in the development of the metric system. NapoleanBonaparte was his student for a year.

Definition 1The Laplace transform of a function f (t) is denoted by L[f (t)] and is defined asfollows:

L[f (t)] =

0f (t)estdt

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Note:When the limits are inserted the result will become a function of s.

L[f (t)] =

0f (t)estdt = F(s)

We use lower-case f for the function in t while we use upper-case (capital) Ffor the function in s.

Example 2Use the definition above to determine the Laplace transform of

f (t) = a (constant)

Hence determine(a) L[3] (b) L[7] (c) L[2]

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Example 3Use the definition above to determine the Laplace transform of

f (t) = eat

Hence determine(a) L[e2t] (b) L[e7t] (c) L[e3t]

Class ExerciseUse the definition of the Laplace transform to determine

L[e8t]

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Example 4Use the definition of the Laplace transform to determine(a) L[e2t + e3t](b) L[7e3t]

The linearity property of the Laplace TransformIf f (t) and g(t) are one-sided functions and c is a constant then

L [f (t) + g(t)] = L [f (t)]+ L [g(t)] = F(s) + G(s)L [c f (t)] = c L [f (t)] = c F(s)

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We recall Eulers formula

ejat = cos(at) + j sin(at)

Example 5Determine the Laplace transform of

f (t) = ejat

Hence, determine the Laplace transform of(a) f (t) = cos(at)(b) f (t) = sin(at)

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Example 6Determine the Laplace transform of

f (t) =e3t + e3t

2

In a similar manner we find the following.

Other Transforms

L[sinh(at)] = as2 a2

L[cosh(at)] = ss2 a2

Class ExerciseDetermine the laplace transform of f (t) = sinh(7t).

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Example 7Determine the Laplace transform of f (t) = 4t3

In General

L[tn] = n!sn+1

Example 8Determine the Laplace transform of(a) f (t) = t2

(b) f (t) = 7t6

(c) f (t) = 3t2 + 6t 4

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Summaryf (t) F(s) = L[f (t)]

aas

eat1

s acos(at)

ss2 + a2

sin(at)a

s2 + a2

cosh(at)s

s2 a2

sinh(at)a

s2 a2

tnn!sn+1

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Class ExercisesDetermine the Laplace transforms of the following.(a) sin(7t)(b) e2t 3e4t(c) cosh(7t) + sinh(7t)(d) 3t6 + 4t2 6

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Shift Theorems

Shift Theorems Lecture 14

Multiplication by eat (the first shift theorem)If L[f (t)] = F(s) then

L[eatf (t)] = F(s+ a)

Example 9Determine the Laplace transforms of(a) sin(2t)(b) e3t sin(2t)

Example 10Determine the Laplace transforms of(a) t2

(b) e5tt2

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Shift Theorems


The following result is similar to the first shift theorem except that, in this case, it isthe time-variable that is shifted not the s-variable.

Translation of the t-variable (the second shift theorem)If L(f (t)) = F(s) then

L(f (t a)) = easF(s)

L1(easF(s)) = f (t a)

Example 11Determine the Laplace transform of(a) t2

(b) (t 3)2

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Shift Theorems


Class ExercisesDetermine the Laplace transforms of

(a) f (t) = e2t cosh(4t)(b) g(t) = 3t2e4t

(c) h(t) = sin(t + 3)(d) k(t) = 6et3

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Multiplication and Division by t

Multiplication and Division by t Lecture 15

Multiplication by tIf L[f (t)] = F(s) then

L[tf (t)] = d[F(s)]ds

Example 12Determine the Laplace transforms of(a) sin(2t)(b) t sin(2t)

Example 13Determine the Laplace transforms of(a) e2t

(b) 3te2t

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Division by tIf L[f (t)] = F(s) then

L[f (t)t

]=

s

F(s)ds

Note:

This rule is only applicable if the limit of[f (t)t

]exists as t 0. But to test this,

when we put t = 0 we get division by zero. So we use lHopitals rule. This statesthat

limxa

f (x)g(x)

= limxa

f (x)g(x)

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Example 14(a) Determine the Laplace transform of f (t) = sin(2t).

(b) Determine whether the limit limt0

sin(2t)t

exists or not.

(c) Determine the Laplace transform of f (t) =sin(2t)

t.

Class ExercisesDetermine the Laplace transform of the following

(a) f (t) = 3t cos(2t)

(b) f (t) =1 cos(2t)

t

(c) f (t) = t cosh(4t)

(d) f (t) =e3t 1

t

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Inverse Laplace Transforms

Inverse Laplace Transforms Lecture 16

Definition 15If L[f (t)] = F(s) then the inverse Laplace Transform is

f (t) = L1[F(s)]

f (t) F(s)

L

L1

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Recall that the Laplace transform of a function f (t) is defined by an integralfrom t = 0 to t = as follows:

L[f (t)] =

0f (t)estdt

Hence the value of a function f (t) for negative t is immaterial to the Laplacetransform.This is reflected in the inverse Laplace transform by the use of the Heavisidestep function.Consequently we can write

L1[

7s2 + 49

]= sin(7t)U(t)

L1[

6s3

]= 3t2U(t)

The importance of U(t) will be seen later, however we note this by means ofthe following table.

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Summaryf (t) F(s) = L[f (t)]

aU(t)as

eatU(t)1

s acos(at)U(t)

ss2 + a2

sin(at)U(t)a

s2 + a2

cosh(at)U(t)s

s2 a2

sinh(at)U(t)a

s2 a2

tnU(t)n!sn+1

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The summary table can be used backwards. For example

L[sin(at)U(t)] = as2 + a2

implies that L1[

as2 + a2

]= sin(at)U(t)

sin(at)U(t)a

s2 + a2

L

L1

Example 16Determine

(a) L1[

7s2 + 49

](b) L1

[6s3

]21 of 40



Because the Laplace transform is a linear operator it follows that the inverseLaplace transform is also linear.

The Linearity Property of the Inverse Laplace TransformIf c is a constant then

L1(F(s) + G(s)) = L1(F(s)) + L1(G(s))L1(cF(s)) = cL1(F(s))

Example 17

(a) L1[

9ss2 + 16

](b) L1

[1

s+ 2+

2s 3

]

Class Exercise

(a) L1[

18s2

](b) L1

[3

s+ 6+

4s 7

]22 of 40

Partial Fractions

Partial Fractions Lecture 17

Example 18(a) Show that

1s+ 2

+2

s 3 =3s+ 1

s2 s 6

(b) Determine

L1[

3s+ 1s2 s 6

]This is not one of the standard transforms.

The problem is how did we get1

s+ 2+

2s 3 from

3s+ 1s2 s 6 .

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Partial Fractions


Partial FractionsTo determine constants A and B such that

cs+ d(s a)(s b) =

As a +

Bs b

we multiply both sides by (s a)(s b) to leave us with

cs+ d = A(s b) + B(s a)

By letting s = a and then s = b we obtain values for A and B respectively.

Example 19Determine

L1[

4s+ 1s2 s 12

]

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Partial Fractions


Example 20Determine

L1[

s+ 142s2 + 7s+ 3

]

Class ExercisesDetermine

(a) L1[

8s 6s2 2s

](b) L1

[s

3s2 + 14s+ 8

]

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Completing the Square

Completing the Square Lecture 18

Previously we noted that the Laplace transform of a function multiplied by eat isgiven by

L(eatf (t)) = F(s+ a) where L(f (t)) = F(s)Reversing this we get that

L1 (F(s+ a)) = eatf (t) where L(f (t)) = F(s)

Example 21Determine the inverse Laplace transform of

3(s 1)2 + 9

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If we expand the denominator in the previous example then we can rewrite it as

3(s 1)2 + 9 =

3s2 2s+ 10

The process of writing s2 2s+ 10 as (s 1)2 + 9 is called completing thesquare.If the denominator of the function of s is an irreducible quadratic thencompleting the square and the first shift theorem allows us to determine theinverse Laplace transform of it.

Example 22Complete the square in the denominator, and determine the inverse Laplacetransform of

(a)s 3

s2 6s+ 13 (b)6s

s2 + 8s+ 17

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Class ExerciseComplete the square in the denominator, and determine the inverse Laplacetransform of

(a)6

s2 + 4s+ 13(b)

2ss2 + 10s+ 29

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Solving ODEs using Laplace Transforms1

Solving ODEs using Laplace Transforms1 Lecture 19

Method:This involves 4 distinct stages:(1) Determine the Laplace transform of the differential equation.(2) Insert the given initial conditions.(3) Rearrange the equation algebraically to give the transform of the solution.(4) Determine the inverse transform to obtain the solution of the differential

equation.

This can be modelled as follows:

.

.

Differential EquationProblems

Algebra Problems

Solution of AlgebraProblems

Solution of DifferentialEquation

Laplace Transform

Very Easy

Inverse Laplace Transform

Difficult

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Example 23If L[x(t)] = X(s) then show that

L[dx(t)dt

]= sX(s) x(0)

Transforms of derivatives:If L[x(t)] = X(s) then

L[dx(t)dt

]= sX(s) x(0)

L[d2x(t)dt2

]= s2X(s) sx(0) x(0)

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Example 24Solve the following differential equation using Laplace transforms.

dy(t)dt

+ 2y(t) = 10e3t

given that y = 6 when t = 0.

Class ExerciseSolve the following differential equation using Laplace transforms.

di(t)dt 4i(t) = 10

given that i = 0 when t = 0.

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The following example uses partial fractions where the denominator is aproduct of distinct linear factors such as as+ b. These result in partialfractions of the form

Aas+ b

Inverting the Laplace transform will be straightforward from the table oftransforms.


d2x(t)dt2

+ 3dx(t)dt

+ 2x(t) = 0

given that x(t) = 6 anddx(t)dt

= 3 when t = 0.

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The following example uses partial fractions where the denominator contains aproduct of repeated linear factors such as (as+ b)2. These results in twopartial fractions of the form

Aas+ b

+B

(as+ b)2

Inverting the Laplace transform will require the firstshift theorem and thetable of transforms.


d2x(t)dt2

+ 6dx(t)dt

+ 9x(t) = 0


= 1 when t = 0.

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(a)d2x(t)dt2

+ 8dx(t)dt

+ 10x(t) = 0, x(0) = 1, x(0) = 0

(b) 4dy(t)dt

+ 10y(t) = 10 cos(2t), y(0) = 1, y(0) = 0

(c)d2x(t)dt2

+ 10dx(t)dt

+ 25x(t) = 10, x(0) = 0, x(0) = 0

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The following example uses partial fractions where the denominator contains aquadratic factor which cannot be factored such as as2 + bs+ c. This results ina partial fraction of the form

As+ Bas2 + bs+ c

Inverting the Laplace transform will require completing the square, thefirstshift theorem and the table of transforms.


d2x(t)dt2

+ 4dx(t)dt

+ 9x(t) = sin(4t)


= 0 when t = 0.

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Example 28Use Laplace transforms to solve

d2x(t)dt2

+ x(t) = 2t x(0) = 0, x(0) = 5


d2x(t)dt2

dx(t)dt 2x(t) = e2t


= 1 when t = 0.

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Electrical Circuits

Electrical Circuits Lecture 22

Example 29Consider the following electrical circuit.

10 0.004F 0.1H

i(t)20V

Determine expressions for1 the charge on the capacitor after t seconds,2 the current in the circuit after t seconds.

Assume that there is no initial charge on the capacitor.37 of 40

Electrical Circuits



CF LH

cos(t) i(t)~


Assume that there is no initial charge on the capacitor.38 of 40

Electrical Circuits


q(t) =cos(t) cos( 1

LCt)

L2 C 6=1LC

t

q(t)

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Electrical Circuits


q(t) =t sin(t) cos(t)

22 =

1LC

t

q(t)

This is the result of Resonance where the applied voltage has the frequency asthe systems natural frequency.The derivation of the above solution is much easier using the ConvolutionTheorem (next section).

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The Convolution Theorem

The Convolution Theorem Lecture 23

In this Section we introduce the convolution of two functions f (t), g(t) whichwe denote by (f g)(t).The convolution is an important construct because of the convolution theoremwhich allows us to find the inverse Laplace transform of a product of twotransformed functions:

L1(F(s)G(s)) = (f g)(t)

Definition 31If f (t) and g(t) are one-sided functions then their convolution is defined by:

(f g)(t) = t

0f (t x)g(x)dx

The rationale for this peculiar definition when we look at its discrete version,i.e. the convolution of sequences.

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This is an odd looking definition but it turns out to have considerable use bothin Laplace transform theory and in the modelling of linear engineeringsystems.One should note that the variable of integration is x.As far as the integration process is concerned the t-variable is (temporarily)regarded as a constant.

Example 32Let f (t) = t and g(t) = t2. Determine(a) (f g)(t)(b) (g f )(t)

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The above result is true in general

Commutativity Property of Convolution

(f g)(t) = (g f )(t)The convolution of f (t) with g(t) is the same as the convolution of g(t) with f (t).

Example 33Determine the Laplace transforms of(a) f (t) = t(b) g(t) = t2

(c) (f g)(t)

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The Convolution TheoremIf L(f (t)) = F(s) and L(g(t)) = G(s) then it can be shown that

L((f g)(t)) = F(s)G(s)

L1(F(s)G(s)) = (f g)(t)

The second version will be particularly useful.

Example 34Use the convolution theorem to determine the inverse Laplace transform of

3s(s2 + 9)(s2 + 4)

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Class ExerciseUse partial fractions to determine the inverse Laplace transform of

7(s 1)(s+ 4)

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The Convolution TheoremApplication

The Convolution TheoremApplication Lecture 24


0.001F 0.1H

20 cos(100t) i(t)~


Assume that there is no initial charge on the capacitor.

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The Convolution TheoremApplication

The Convolution TheoremApplication Lecture 24

Class ExerciseConsider the following electrical circuit.

0.001F 0.1H

20 cos(25t) i(t)~


Assume that there is no initial charge on the capacitor.

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The Impulse Function

The Impulse Function Lecture 25

There is often a need for considering the effect on a system (modelled by adifferential equation) by a forcing function which acts for a very short timeinterval.

For example, how does the current in a circuit behave if the voltage is switchedon and then very shortly afterwards switched off?

How does a cantilevered beam vibrate if it is hit with a hammer (providing aforce which acts over a very short time interval)?

Both of these engineering systems can be modelled by a differential equation.

There are many ways the kick or impulse to the system can be modelled.

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The three most important input signals in engineering applications areStep functions - U(t)Oscillation input - R cos(t + )Impulse - very large for a very short time.

Let us define

ga(t) =

{1

2a a < t < a0 otherwise

.

.

a a

12a

t

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Clearly the area under the graph is equal to 1, i.e.

ga(t)dt = 1

As we let a 0 the graph of ga(t) gets taller and narrower, but always with an areaof 1.

Definition 36The impulse function is given by

(t) = lima0

ga(t)

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Properties1 (t) = 0 when t 6= 02

(t) dt = 1

3

(t)f (t) dt = f (0)

4

(t c)f (t) dt = f (c)

Consequently the Laplace transform of the impulse function is

L((t)) =

0est(t)dt = es0 = e0 = 1

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It is also worth noting thatdU(t)dt

= (t)

Example 37Determine the solution of

d2y(t)dt2

+ 2dy(t)dt

+ 10y(t) = (t 5)

subject to y(0) = y(0) = 0.

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Transfer Functions

Transfer Functions Lecture 26

In this section we introduce the concept of a transfer function and then use thisto obtain a Laplace transform model of a linear engineering system.

We shall also see how to obtain the impulse response of a linear system andhence to construct the general response by use of the convolution theorem.

Linear engineering systems are those that can be modelled by lineardifferential equations.

We shall only consider those systems that can be modelled by constantcoefficient ordinary differential equations.

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Transfer Functions


Consider a system modelled by the second order differential equation.

ad2ydt2

+ bdydt

+ cy = f (t)

in which a, b, c are given constants and f (t) is a given function.

In this context f (t) is often called the input signal or forcing function and thesolution y(t) is often called the output signal.

We shall assume that the initial conditions are zero (in this casey(0) = 0, y(0) = 0).

Now, taking the Laplace transform of the differential equation, gives:

(as2 + bs+ c)Y(s) = F(s)

where we have designated L(y(t)) = Y(s) and L(f (t)) = F(s).

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Transfer Functions


Definition 38We define the transfer function of a system to be the ratio of the Laplacetransform of the output signal to the Laplace transform of the input signal withthe initial conditions as zero.The transfer function (a function of s), is denoted by H(s).In this case

H(s) Y(s)F(s)

=1

as2 + bs+ c

Now, in the special case in which the input signal is the delta function,f (t) = (t), we have F(s) = 1 and so, H(s) = Y(s).

We call the solution to the differential equation in this special case the unitimpulse response function and denote it by h(t)U(t).

We include the step function U(t) to emphasize its one-sidedness.

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Transfer Functions


Soh(t)U(t) = L1(H(s)) when f (t) = (t)

Now, keeping this in mind and returning to the general case in which the inputsignal f (t) is not necessarily the impulse function (t), we have:

Y(s) = H(s)F(s)

The solution for the output signal is, as usual, obtained by taking the inverseLaplace transform:

y(t) = L1(Y(s))= L1(H(s)F(s))= (h f )(t)

using the convolution theorem.

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Transfer Functions


Linear System SolutionThe solution to a linear system, modelled by a constant coefficient ordinarydifferential equation, is given by the convolution of the unit impulse responsefunction h(t)U(t) with the input function f (t).

Example 39Determine the impulse response function h(t) to a linear engineering systemmodelled by the differential equation:

d2ydt2

+ 9y = cos(3t) y(0) = 0, y(0) = 0

and hence solve the system.

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Introduction to Laplace TransformsShift TheoremsMultiplication and Division by tInverse Laplace TransformsPartial FractionsCompleting the SquareSolving ODEs using Laplace Transforms1Solving ODEs using Laplace Transforms2Solving ODEs using Laplace Transforms3Electrical Circuits

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