ENGI 3424 3 Laplace Transforms Page 3.01 3. Laplace …ggeorge/3424/handout/H3Laplace.pdf · y t g t h t x y x dx . ... ENGI 3424 3.02 – Laplace Transforms of Derivatives Page 3.07

ENGI 3424 3 – Laplace Transforms Page 3.01

3. Laplace Transforms

In some situations, a difficult problem can be transformed into an easier problem, whose

solution can be transformed back into the solution of the original problem. For example,

an integrating factor can sometimes be found to transform a non-exact first order first

degree ordinary differential equation into an exact ODE [Section 1.7]:

One type of first order ODE [not considered in this course] is the Bernoulli ODE, uy Py Ry , which, upon a transformation of the dependent variable, becomes a

linear ODE:

An initial value problem is an ordinary differential equation together with sufficient

initial conditions to determine all of the arbitrary constants of integration. A Laplace

transform will convert some initial value problems into much easier algebra problems.

The solution of the original problem is then the inverse Laplace transform of the solution

to the algebra problem.

Uses of Laplace transforms include:

1) the solution of some ordinary differential equations

2) the solution of some integro-differential equations, such as

0

ty t g t h t x y x dx .

ENGI 3424 3 – Laplace Transforms Page 3.02

Contents:

3.01 Definition, Linearity, Laplace Transforms of Polynomial Functions

3.02 Laplace Transforms of Derivatives

3.03 Review of Complex Numbers

3.04 First Shift Theorem, Transform of Exponential, Cosine and Sine Functions

3.05 Applications to Initial Value Problems

3.06 Laplace Transform of an Integral

3.07 Heaviside Function, Second Shift Theorem; Example for RC Circuit

3.08 Dirac Delta Function, Example for Mass-Spring System

3.09 Laplace Transform of Periodic Functions; Square and Sawtooth Waves

3.10 Derivative of a Laplace Transform

3.11 Convolution; Integro-Differential Equations; Circuit Example

Note: Sections 3.09, 3.10 and the latter part of 3.11 will be parts of this course only if

time permits.

ENGI 3424 3.01 – Definition; Polynomial Functions Page 3.03

3.01 Definition, Linearity, Laplace Transforms of Polynomial Functions

If f t is

defined on 0t ,

piece-wise continuous on 0t

(that is, only a finite number of finite discontinuities) and

of exponential order

[ tf t k e for all 0t and for some positive constants k and ],

then

0

limm

m

stF s e f t dt

exists for all 0s and

the Laplace transform of f t is

0

stF s e f t dt f t

L

Example 3.01.1

Find 1L .


Example 3.01.2

Find tL .

Linearity Property of Laplace Transforms

a f t b g tL (where a, b are constants):


Example 3.01.3

Find 2 3t L .

Example 3.01.4

Find ntL .


Example 3.01.5

Find 4 32 3t t L .

ENGI 3424 3.02 – Laplace Transforms of Derivatives Page 3.07

3.02 Laplace Transforms of Derivatives

0

stf t e f t dt

L

Continuing this pattern, we can deduce the Laplace transform for any higher derivative of

f t :

or

This result allows us to find the Laplace transform of an entire initial value problem.

ENGI 3424 3.02 – Laplace Transforms of Derivatives Page 3.08

Example 3.02.1

Find the Laplace transform of the initial value problem

2

3

26 6 , 0 0 , 0 1

d yy t y y

dt

and hence find the complete solution.

Let Y s y t L , then

6y y L

36t L

Therefore the Laplace transform of the initial value problem is

which is an algebra problem for Y s , (the Laplace transform of the complete solution).

Solving the algebra problem:

ENGI 3424 3.03 – Review of Complex Numbers Page 3.09

3.03 Review of Complex Numbers

Before proceeding further with Laplace transforms, some familiarity with complex

numbers is required. A very brief review of complex numbers is provided here.

The set of real numbers is closed under addition, subtraction, multiplication and (with

the exception of zero) division. However, the square root of any negative number is

not real. This generates the set of imaginary numbers, which are all real multiples of the

imaginary number 1j . For example: 49 49 1 49 1 7 j .

Imaginary and real numbers can be combined by addition to form the set of complex

numbers, which can be represented on an Argand diagram:

Example 3.03.1

Two complex numbers are defined by 3 2z j and 1 3w j :

The real parts of these numbers are Re 3z and Re 1w

The imaginary parts of these numbers are Im 2z and Im 3w

The sum (or difference) of any pair of complex numbers is calculated by taking the sum

(or difference) of the real parts and imaginary parts separately:


The real and imaginary parts of a complex number z x y j behave on the Argand

diagram very like the Cartesian x and y components of a vector x

y

v :

Also like vectors in 2, complex numbers can be expressed in terms of their magnitude

and direction.

The modulus (or “magnitude” or “amplitude”) of z x y j

(which is also the length of the vector x

y

v ) is 2 2z x y j x y

The argument (or “phase”) of z x y j

(which is also the angle that the vector x

y

v makes with the positive x axis) is ,

where tany

x . When Re 0z x , add or subtract radians from the inverse

tangent in order to be in the correct quadrant (2nd

or 3rd

, not 4th

or 1st).

The principal argument is usually taken to be in the range or 0 2 .

In Example 3.03.1,


Multiplication and Division of Complex Numbers

Example 3.03.1 (continued)

3 2 1 3zw j j

It is generally true that

zw z w and arg arg argzw z w

Also, for division (where the divisor is not zero),

zz

w w and arg arg arg

zz w

w

The argument of the product or quotient may need to be adjusted by the addition or

subtraction of 2 in order to bring it back inside the range of the principal argument.


In the eighteenth century Euler extended the definition of the exponential function to

non-real exponents:

cos sinj

e j

It then follows that

cos sinj

z x j y r j r r e

,

where and argr z z .

Alternative rules for multiplication and division then follow immediately:

1 2 1 2 1 2 1 21 21 2 1 2

jj j j jz z re r e rr e e rr e

and

1 1 1 1

2

2 2 22

1 11 2

2 2, 0

j jj

j j

z re r ree r

z r rr e e

Exponents and roots of complex numbers are best found in exponential form.

Example 3.03.2

For the complex numbers 3 4 , 2 2z j w j , find 3 1, , and

zzw z

w z.



Definition: The complex conjugate of z x y j is *z x y j .

For any complex number z ,

22* 2 2 2zz x y j x y j x y j x y z

For any non-zero complex number, 1

z

Example 3.03.2(d) (again)


Example 3.03.3

Find all distinct cube roots of the real number 8.

The only real cube root of 8 is 2. However, there is a pair of non-real roots also.

Method 1

Let z be a cube root of 8, then 3 8z

Method 2

3 3

1/30 2 0 2 2 /3

8 2 , 2 2j n j n j n

z e n z e e

2 22cos 2 sin

3 3

n nz j

0 2cos0 2 sin0 2n z j

2 21 2cos 2 sin 1 3

3 3n z j j

2 21 2cos 2 sin 1 3

3 3n z j j

and these are the only three distinct values.

n = ..., –7, –4, 2, 5, 8, ... return the same value for z as n = –1.

n = ..., –6, –3, 3, 6, 9, ... return the same value for z as n = 0.

n = ..., –5, –2, 4, 7, 10, ... return the same value for z as n = +1.



On the Argand diagram the roots are equally spaced around a circle

of radius 2 centred on 0.

A similar graphical approach can be used to find the n distinct nth roots of any non-zero

complex number.


Example 3.03.4

On the Argand diagram sketch the locations of the five distinct fifth roots of the complex

number w. You may assume that the radius of the guide circle is 5 w .

Also note that cos sinj

e j

and

cos sin cos sinj

e j j

and

ENGI 3424 3.04 – First Shift Theorem, Cosine and Sine Page 3.17

3.04 First Shift Theorem, Transform of Exponential, Cosine and Sine Functions

ate f t L

The first shifting theorem then follows:

If , then atf t F s e f t F s a L L

Example 3.04.1

Find 3 5tt eL .

Example 3.04.2

Find ateL .


Example 3.04.3

Find cos tL .

Example 3.04.4

Find sin tL .

It then follows that

1

2 2cos

st

s

L and

1

2 2

1 sin t

s

L


Example 3.04.5

Find sin tL directly, from the definition of a Laplace transform.

Example 3.04.6

Find sin tL , using cos tL .


Example 3.04.7

Find sinate tL .

Using the first shift theorem, we have, immediately,

Similarly,

2 2cosat s a

e ts a

L

Example 3.04.8

Find tan tL .

Example 3.04.9

Find sinh atL .

Similarly, 2 2cosh

sat

s a

L


Summary so far:

Definition:

0

stf t e f t dt

L

Linearity:

, constantsa f t b g t a f t b g t a b L L L

Polynomial functions:

11

1

! 1

1 !

nn

nn

n tt

s ns

L L

First Shift Theorem:

atf t F s e f t F s a L L and 1 1 ates a

L

Trigonometric Functions:

1

2 22 2

1 sinsin

atat e t

e ts a s a

L L

1

2 22 2cos cosat ats a s a

e t e ts a s a

L L

Hyperbolic Functions:

1

2 22 2

1 sinhsinh

2

a b t a b tatat b e bt e e

e btb bs a b s a b

L L

1

2 22 2cosh cosh

2

a b t a b tat ats a s a e e

e bt e bts a b s a b

L L

Derivatives:

0f t s f t f L L

2 0 0f t s f t s f f L L

etc.

When trying to find the inverse Laplace transform of a rational function F s , one

usually attempts to break the function up into its partial fractions, with real linear and

quadratic denominators, so that one may read the inverses from the table on this page.

ENGI 3424 3.05 – Applications to Initial Value Problems Page 3.22

3.05 Applications to Initial Value Problems

Example 3.05.1

Solve the initial value problem

5 6 0 , 0 1, 0 0y y y y y

Let Y s y t L .

y t L

y t L

Taking the Laplace transform of the entire initial value problem:

5 6 0y y y L L

Check 1:



Check 2:

Solution by Chapter 2 method:

5 6 0y y y


Example 3.05.2

Solve the initial value problem

44 13 26 , 0 5, 0 29ty y y e y y

Let Y s y t L .

y t L

y t L

426 te L

Taking the Laplace transform of the entire initial value problem:

44 13 26 ty y y e L L

Recall the three standard inverse Laplace transforms:

1 1 ates a

L

1

2 2cosats a

e ts a

L

1

2 2sinate t

s a

L



We need to express Y s in partial fractions and to complete the square in the quadratic

denominator.


Example 3.05.2 (by a Chapter 2 method)

44 13 26 , 0 5, 0 29ty y y e y y

ENGI 3424 3.06 – Laplace Transform of an Integral Page 3.27

3.06 Laplace Transform of an Integral

0

Let , thent

f t g d

Example 3.06.1

Find the function f t whose Laplace transform is 2 2

1F s

s s

.

ENGI 3424 3.06 – Laplace Transform of an Integral Page 3.28

Example 3.06.1 (Alternative solution, using partial fractions)

Example 3.06.2

Find the function f t whose Laplace transform is 2 2 2

1F s

s s

.

Alternative method, using partial fractions:

ENGI 3424 3.07 – Heaviside Function, Second Shift Theorem Page 3.29

3.07 Heaviside Function, Second Shift Theorem; Example for RC Circuit

The Heaviside function H t a (also known as the unit step function u t a ) is

defined by

00

1

t aH t a a

t a

The Laplace transform of the

Heaviside function is:

F s H t a L

If

0,

g t t af t

h t t a

then

ON OFF

f t g t h t g t H t a

[The function h t is “switched on” and g t is “switched off” at time t = a.]


Example 3.07.1

Express the function

2 2

4 2

x xf x

x

in a single line definition.

Answer: f x

The Second Shift Theorem

The result of shifting the graph y f t (defined only for t > 0) a units to the right, is

the graph y f t a H t a :

The Laplace transform of the shifted function of t is:

0

stH t a f t a e H t a f t a dt

L

H t a f t a L

The second shift theorem (for a > 0) is

asH t a f t a e f t L L


Another way to express the second shift theorem is:

1 1 asF s f t e F s H t a f t a L L

Example 3.07.2

Find 34 4H t t L .

34 4H t t L

Example 3.07.3

Find the function whose Laplace transform is 2

5

4

se

s

.


Example 3.07.4

In the RC circuit shown here, there is no

charge on the capacitor and no current

flowing at the time t = 0.

The input voltage Ein is a constant Eo

during the time 1 2t t t and is zero at all

other times.

Find the output voltage Eout for this circuit.

Initial conditions:

q(0) = i(0) = 0 Note:dq

idt

o 1 2

in0 otherwise

E t t tE

inE

Equating voltage drops around the circuit with inE :

Let Q s q t L . Taking the Laplace transform of this initial value problem:



Graph of the output voltage against time:


Example 3.07.5

Find the complete solution of the initial value problem

2

2

0 0 34 ; 0 0 0 .

3

td yy f t y y

t td t

Let Y s y t L



ENGI 3424 3.08 – Dirac Delta Function Page 3.36

3.08 Dirac Delta Function, Example for Mass-Spring System

1

Let

0 otherwise

a t af t

0

Area under Area of

graph rectangle

11f t dt

Also

f t

Define the Dirac delta function to be

0

limt a f t

then

f t L

t a L

Therefore for 0a

and

Also, for 0a , the total area under the graph is 1 (even in the limit as 0+), so

0

1t a dt


For g t any function that is continuous everywhere on 0, ,

0

g t t a dt

0

g t t a dt g a

(which is the sifting property of the Dirac delta function).

Example 3.08.1

A damped mass-spring system, (with damping constant = c = 3m and spring modulus =

k = 2m, where m is the mass), is at rest in its equilibrium position, until an impulse of

5 Ns is applied at time t = 4 s. Find the response y t .

One can anticipate some features of the response. The forcing function is zero until

4t . Thus there is nothing to disturb the system until that instant. We can deduce that

0y t during 4t . The mass-spring system is set into motion abruptly by the arrival

of the impulse at the instant 4t , but the forcing term is again zero thereafter. The

damping force will asymptotically restore the system to its equilibrium state. The only

question remaining is whether the system is under-damped or not (see the diagram on the

next page).



Anticipated response:

The auxiliary equation is 2 3 2 0 . The discriminant is

Solution, using Laplace transforms:

3 2 5 4y y y t

0 0 0y y

Let Y s y t L



From the first shift theorem: 1 1 ates a

L

The second shift theorem states:

1 1 asF s f t e F s f t a H t a L L

Therefore

or, equivalently,

ENGI 3424 3.09 – Periodic Functions Page 3.40

3.09 Laplace Transform of Periodic Functions; Square and Sawtooth Waves [only if time permits]

If the constant 0p and f t p f t for all 0t , then

f t is a periodic function of t, with period p.

Example of a periodic function (with one finite discontinuity in each period):

Define a new set of functions ng t , each of which captures only a single period of f t :

1

0 otherwisen

f t np t n pg t

Then 0n

nf t g t

0

Let n

stn nG s g t e g t dt

L

1 1

0 0n p n p

st stn

n p n pe g t dt e f t dt

Let F s f t L , then


Therefore, for a periodic function f t with fundamental period p,

0

1

1

pst

s pf t e f t dte

L

Example 3.09.1

Use the formula above to verify the Laplace transform of sinf t t .


Example 3.09.2 The Square Wave

The square wave is periodic, with fundamental period 2p a .

In the “zeroth” period 0 2t a ,

0

1 0

1 2

t af t g t

a t a

The Laplace transform F s of the square wave f t then follows.


Example 3.09.3 The Saw-tooth Wave

The saw-tooth wave is periodic, with fundamental period p a .

In the “zeroth” period 0 t a ,

0

btf t g t

a

The Laplace transform F s of the saw-tooth wave f t then follows.


Example 3.09.4

Find the function f t whose Laplace transform is 2

1tanh

2

asF s

s

(where a is a constant).

ENGI 3424 3.10 – Derivative of a Laplace Transform Page 3.45

3.10 Derivative of a Laplace Transform [only if time permits]

Example 3.10.1

Find cosF s t t L .

Method via an ODE:

Let cosf t t t

Then f t

and f t

Also 0f

0f

Taking the Laplace transform of this initial value problem,

0 0

st stdF s f t e f t dt F s e f t dt

ds

L

Using Leibnitz differentiation of an integral:

00

0 0 st stF s e f t dt e t f t dts

Therefore

d

f t t f tds

L L 1 1F s t F s L L

ENGI 3424 3.10 – Derivative of a Laplace Transform Page 3.46

Example 3.10.1 (again)

Find cosF s t t L .

Method via the derivative of a transform:

Let cosg t t

Example 3.10.2

Find sinF s t t L .

ENGI 3424 3.11 – Convolution Page 3.47

3.11 Convolution; Integro-Differential Equations; Circuit Example

If f t F sL

and g t G sL

then the convolution of f t and g t is denoted by f g t , is defined by

0

t

f g t f g t d

and the Laplace transform of the convolution of two functions is the product of the

separate Laplace transforms:

f g t F s G s L

An equivalent identity is

1 1 1F s G s F s G s L L L

Convolution is commutative:

0 0

t t

g f f t g d f g t d f g

Example 3.11.1

Find the inverse Laplace transform of 2 2 2

1.R s

s s



Note:

This inverse transform can also be found using partial fractions or the division of another

Laplace transform by s. Both of these alternatives are in example 3.06.2.


Example 3.11.2

Find

1

22 2

1.

s

L

The alternative methods that were available in example 3.11.1 are not available here.

The only obvious way to proceed is via convolution.

22 2

1Let R s

s


Some More Identities involving Convolution [only if time permits]

1 y


Transfer Function [only if time permits]

For a second order linear ordinary differential equation with constant coefficients, when

both initial conditions are zero, the complete solution can be expressed as a convolution.

y by cy r t and 0 0 0y y

2s bs c Y s R s

Y s Q s R s , where Q(s) is the transfer function:

2

1Q s

s bs c

The complete solution is just y t q t r t , where 1

2

1.q t

s bs c

L

Example 5.11.3

A unit impulse is delivered at time 1t to an harmonic system (with c/m = 4) that is in

equilibrium until that instant. Find the response y t .


Integro-Differential Equations [only if time permits]

Example 3.11.4

Find the solution y t of the integro-differential equation

2

01 9

ty t t t x y x dx


Example 3.11.5 [only if time permits]

Solve the system of simultaneous integral equations

1 1 2

2 2 1 2

0

0 0

4 12 1

2 2

t

t t

i i i d

i i d i i d

1 1 2 2Let , .I s i t I s i t L L

Taking the Laplace transform of the entire system of integral equations,



These are the currents in the circuit

with initial conditions 1 2

14

0 , 0 0i i .

END OF CHAPTER 3

Documents

ENGI 3424 3 Laplace Transforms Page 3.01 3. Laplace …ggeorge/3424/handout/H3Laplace.pdf · y t g t h t x y x dx . ... ENGI 3424 3.02 – Laplace Transforms of Derivatives Page 3.07