01 01 Functions

Math Transform Methods

Dr. Padraig Kirwan

Department of Computing, Mathematics and PhysicsWaterford Institute of Technology

EELEC C Year 2ELECT D Year 2

14th January 2013

1

Outline of Module

Syllabus

1 Use Functions in the solution of engineering problems.

2 Ordinary Differential Equations.

3 Laplace Transforms.

4 Z -Transforms.

Assessment

70% Final Exam

30% Continuous Assessment

Final Exam

Exam 2 Hours. Do all 3 questions.Detail

1 Functions & ODEs

2 Laplace Transforms

3 Z -Transforms2

Functions–Polynomials Lecture 1

A quantity whose value can change is known as a variable.

Functions are used to describe the rules which define the ways in whichsuch a change can occur.

We are usually interested in a variable (output) whose behaviourdepends on that of another variable (input).

A function is a rule which operates on an input and produces an output

Definition 1:

A function is a rule which operates on an input and produces a singleoutput from that input

input input

function

rule

3

Functions–Polynomials Lecture 1

The function with the rule ‘double the input’ is shown below.

t 2t

f

double

the input

This function is usually written as

f (t) = 2t

A very useful pictorial representation of afunction is its graph.

To plot the graph we first draw a pair of axes- a vertical axis and a horizontal axis.

These are drawn at right–angles to eachother and intersect at the origin.

Each pair of input and output values can berepresented on a graph by a single point.

t

y y = 2t

4

Functions–Polynomials Lecture 1

Definition 2:

An expression of the form

antn + an−1t

n−1 + · · ·+ a1t + a0

where n is a positive integer is known as a polynomial of degree n. Thevalues a0, . . . , an are constants and are called the coefficients.

Graph of a Polynomial

The graph of a polynomial function P(t)

is made up of all points (t, y) wherey = P(t).

is continuous, i.e. it has no breaks.

smooth, i.e. it has no sharp turns.

t

f (t)

turning points

If deg(P(t)) = n then the graph of P(t) has at most n − 1 turning points.5

Functions–Polynomials Lecture 1

Note

t

f (t)

f (t) f (t − c)c

The graph of f (t − c) is thegraph of f (t) moved right c units.

t

f (t)

f (t)f (t + c)c

The graph of f (t + c) is thegraph of f (t) moved left c units.

t

f (t)

f (t)f (t) + c

c

The graph of f (t) + c is thegraph of f (t) moved up c units.

t

f (t)

f (t)

f (t)− c

c

The graph of f (t)− c is thegraph of f (t) moved down c units.

6

Functions–Polynomials Lecture 1

Linear Functions

The simplest example of a polynomial is the linear function

f (t) = mt + c

This is often written as y = mt + cwhere the slope of the line is

m =∆y

∆t=

y2 − y1

t2 − t1

(0, c) = y − intercept

Frequently, these functions are writ-ten in the standard form

at + by = ct

f (t)

t1 t2

y1

y2

∆t

∆y

7

Functions–Polynomials Lecture 1

Example 3:

(a) Sketch the graph of the line given by the linear functions

y = 3t + 4

(b) Determine the equation of the line through the points (1, 2) and (3, 0).

t

f (t)

(1, 2)

(3, 0)

1 2 3

1

2

3

8

Functions–Polynomials Lecture 1

Quadratic Functions (Parabolas)

A quadratic function is a polynomial function of degree2, i.e. it is of the form

f (t) = at2 + bt + c where a 6= 0

The graph of a quadratic function is called a parabola.

Standard Form

To determine the turning point of a parabola it is usefulto express the quadratic function in the standard form

f (t) = a(t − h)2 + k

The turning point is at (h, k)

If a > 0 then the parabola is concave upwards

If a < 0 then the parabola is concave downwards

a > 0

(h, k)

a < 0

(h, k)

9

Functions–Polynomials Lecture 1

Example 4:

Consider the quadratic function

f (t) = 2t2 − 16t + 5

(a) Express f (t) in standard form.

(b) Determine the turning point of f (t).

(c) Determine where the graph of f (t) cuts the vertical axis.

(d) Determine where the graph of f (t) cuts the horizontal axis.

(e) Sketch the graph of f (t).

Class Exercise

Sketch the graphs of the following functions. Determine where they cut thehorizontal and vertical axes.

(a) f (t) = 3t − 2 (b) f (t) = 3t2 + 6t − 8

10

Higher–order and Rational Polynomials Lecture 2

The Remainder Theorem

Let P(t) be a polynomial of degree n.

If P(k) = 0 then P(t) has a factor t − k ,

i.e.P(t)

t − kis a polynomial of degree n − 1.

We refer to k as a zero of P(t).

t

f (t)

zeros

Example 5:

Show that t − 2 and t + 3 are factors of the poly-nomial

f (t) = 2t4 + 7t3 − 4t2 − 27t − 18

Determine the remaining factors of f (t).

t

f (t)

11

Higher–order and Rational Polynomials Lecture 2

Example 6:

Determine a polynomial func-tion with zeros at −2,−1, 1, 2.

t

f (t)

−2 −1 1 2

12

Higher–order and Rational Polynomials Lecture 2

Definition 7:

An expression which consists of the ratio of two polynomials is known as arational polynomial, i.e. it is in the form

P(t)

Q(t)

where both P(t),Q(t) are polynomials.

If deg(P(t)) = m and deg(Q(t)) = nthen we assume that m ≤ n.

The zeros of the denominator, Q(t), areknown as the poles of the rationalpolynomial.

At these points the rational polynomial isundefined.

poles

13

Higher–order and Rational Polynomials Lecture 2

Example 8:

Determine a rational polynomial that

has poles at −2, 3± 5j , 1 and

has zeros at −3, 2.

Example 9:

Consider the rational polynomial

f (t) =2t2 + t

t + 1

(a) Determine the poles of f (t).

(b) Determine the slant asymptote of f (t).

14

Differentiation Lecture 3

Definition 10:

Let y = f (t) be a real valued function of t. The derivative of y = f (t) withrespect to t is given by

dy

dt= lim

∆t→0

∆y

∆t

This is a measure of the “rate of change” of f (t) with respect to t.

A

B

∆t

∆y

slope= ∆y∆t

Average rate of change

between A and B

A

slope=dydt

Instantaneous rate of change

at A15

Differentiation Lecture 3

The table of derivatives

y = f (t) dydt = f ′(t)

k , any constant 0t 1t2 2tt3 3t2

tn, any constant n ntn−1

et et

ekt kekt

ln(t) 1t

cos(t) − sin(t)cos(kt) −k sin(kt)sin(t) cos(t)sin(kt) k cos(kt)tan(t) sec2(t)tan(kt) k sec2(kt)

16

Differentiation Lecture 3

The table of derivatives

y = f (t) dydt = f ′(t)

cosec(t) = 1sin(t) −cosec(t) cot(t)

sec(t) = 1cos(t) sec(t) tan(t)

cot(t) = 1tan(t) −cosec2(t)

sin−1(t) 1√1−t2

cos−1(t) −1√1−t2

tan−1(t) 11+t2

cosh(t) sinh(t)sinh(t) cosh(t)tanh(t) sech2(t)sech(t) −sech(t) tanh(t)cosech(t) −cosech(t) coth(t)coth(t) −cosech(t)

17

Differentiation Lecture 3

We have the following three tools to help us in differentiating morecomplicated functions.

The Rules of Differentiation

The Product RuleIf y = uv then

dy

dt= u

dv

dt+ v

du

dt

The Quotient RuleIf y =

u

vthen

dy

dt=

v dudt − u dv

dt

v2

The Chain RuleIf y = f (u) and u = g(t) then

dy

dt=

dy

du· dudt

Example 11:

Differentiate each of the followingfunctions.

1 y = sin(t)et

2 y =ln(t)

t4

3 y = cos(

2πt +π

4

)

18

Differentiation Lecture 3

Increasing Function

f (t) is increasing when f ′(t) > 0,i.e. a positive slope.

t

f (t)

increasing

Decreasing Function

f (t) is decreasing when f ′(t) < 0,i.e. a negative slope.

t

f (t)

decreasing

19

Differentiation Lecture 3

Stationary Points

If f ′(c) = 0 then c is called a critical value of f (t) and (c , f (c)) is called astationary point, i.e. a zero slope.

t

f (t)

stationary points

20

Differentiation Lecture 3

First Derivative Test

If a function f (t) has a local maximum or a local minimum at t = t0 then

f ′(t0) = 0

Example 12:

Determine the location of the stationary points of

f (t) = t2e−t

21

Differentiation Lecture 3

Second Derivative Test

If f ′(t0) = 0 then

f (t) has a local maximum at t0 if f ′′(t0) < 0,

f (t) has a local minimum at t0 if f ′′(t0) > 0,

Example 13:

Determine the nature of the stationary points of

f (t) = t2e−t

22

Differentiation Lecture 3

Point of Inflection

The graph of f (t) is concave-up at points where f ′′(t) > 0.

The graph of f (t) is concave-down at points where f ′′(t) < 0.

If f ′(t0) = f ′′(t0) = 0 then the graph of f (t) has a point of inflection att = t0.

Example 14:

Determine the point(s) of inflection of

f (t) = t2e−t

Hence determine the intervals where the graph is respectively concave-up orconcave-down.

23

Integration Lecture 4

Definition 15:

Let f (t) be a real-valued function of t. The (indefinite) integral of f (t)with respect to t is a function g(t) such that

dg(t)

dt= f (t)

We denote this by ∫f (t)dt = g(t)

24

Integration Lecture 4

Definition 16:

The (definite) integral of f (t) between t = aand t = b is the area under the curve f (t)between the limits and is given by∫ b

af (t)dt = g(b)− g(a)

∫ 3/2

0x2dx

x

f (x)

1 1 12

2

1

22 1

4

x2

25

Integration Lecture 4

f (t)∫f (t)dt

tntn+1

n + 1

1

tln(t)

1

tn1

(1− n)tn−1

√t

2

3

√t3

cos(ωt + φ)sin(ωt + φ)

ω

sin(ωt + φ)− cos(ωt + φ)

ω

1

a2 + t2

1

atan−1(

t

a)

26

Integration Lecture 4

When comparing the behaviour of a function f (t) with that of otherfunctions over a range a ≤ t ≤ b we can use the following measures.

Average (mean)

fav =

∫ ba f (t)dt

b − a

Root Mean Square

frms =

√∫ ba f (t)2dt

b − a

Example 17:

Determine fav and frms forf (t) = t2 − 4

where t ∈ [0, 4].27

Piecewise Polynomial Functions Lecture 5

Many signals are made up of different sections that are represented bydifferent polynomials.

Typically we might have the graph of a function as below and we needto express it in mathematical terms as follows.

t

y

0 1 2 3 4 5 6

1

2

3

f (t) =

0.2t if 0 < t < 11.8t − 3.6 if 1 < t < 22(t − 3)2 if 2 < t < 41 if 4 < t < 53t − 15 if t > 5

Deconstructing such graphs requires us to know about standardpiecewise functions.

28

Piecewise Polynomial Functions Lecture 5

The Hat Function

This function can be described as fol-lows.

hat(t) =

0 if t < −1t + 1 if − 1 < t < 0−t + 1 if 0 < t < 10 if t > 1

t

hat(t)

−2 −1 0 1 2

1

The Saw Function

This function can be described as fol-lows.

saw(t) =

0 if t < 0t if 0 < t < 10 if t > 1

t

saw(t)

−2 −1 0 1 2

1

29

Piecewise Polynomial Functions Lecture 5

The Rectangular Pulse Function

This function can be described as fol-lows.

Π(t) =

0 if t < −11 if − 1 < t < 10 if t > 1

t

Π(t)

−2 −1 0 1 2

1

The Absolute Value Function

This function can be described as fol-lows.

|t| =

{t if t ≥ 0−t if t < 0

(= abs(t)) t

|t|

−2 −1 0 1 2

1

2

30

Piecewise Polynomial Functions Lecture 5

The Signum Function

This function can be described as fol-lows.

sgn(t) =

{−1 if t < 01 if t > 0

t

sgn(t)

−2 −1 0 1 2

−1

1

The Ramp Function

This function can be described as fol-lows.

r(t) =

{t if t ≥ 00 if t < 0 t

r(t)

−2 −1 0 1 2

1

2

31

Piecewise Polynomial Functions Lecture 5

Example 18:

Sketch the graphs of

(a) |t − 3|(b) 2|t|(c) |3t − 6|

Example 19:

Differentiate|t|

32

Heaviside Step Function Lecture 6

Often referred to as the Unit Step Function or just the StepFunction, the Heaviside function was invented by the English electricalengineer Oliver Heaviside.

It is used in the mathematics of control theory to represent a signal thatswitches on at a specified time and stays on indefinitely.

Definition 20:

The Heaviside step function is given by

U(t) =

{0 if t < 01 if t ≥ 0 t

U(t)

−2 −1 0 1 2

1

This is the building block for piecewise polynomial graphs. It picks outwhere the graph is non-zero.

33

Heaviside Step Function Lecture 6

Consider a function that maintains a zero value for all values of t up toc and a unit value for t = c and all values of t > c .

This can be expressed as

f (t) =

{0 if t < c1 if t ≥ c

= U(t − c)

where U(t − c) is a single symbol in which c indicates the value of t atwhich the function changes from a value of 0 to a value of 1.

The graph of f (t) is the graph of U(t) translated c units to the right.

t

U(t − c)

c

1

34

Heaviside Step Function Lecture 6

Example 21:

Consider the graph of f (t) = t2. If we want to “switch” on this function atsay t = 2 then all values of f (t) for t < 2 would have to be zeroed, i.e.

f (t) = t2U(t − 2) =

{0 if t < 2t2 if t ≥ 2

Sketch the graphs of

(a) f (t) = t2

(b) f (t) = t2U(t − 2)

Note

In the above example we normally deal only with positive values of t.

35

Heaviside Step Function Lecture 6

We may want to switch a signal off as well.

This is done by using the Heaviside step function as a rectangular pulse,as follows.

Consider the graph of U(t − a)− U(t − b) where a < b.

t U(t − a) U(t − b) U(t − a)− U(t − b)

t < a 0 0 0

a < t < b 1 0 1

t > b 1 1 0

t

U(t − a)− U(t − b)

a b

1

36

Heaviside Step Function Lecture 6

Example 22:

Sketch the graph of

g(t) =

0 if t < 2t2 if 2 ≤ t ≤ 50 if t > 5

and express it in terms of the Heaviside Step Function.

Example 23:

Express

(a) |t|(b) sgn(t)

(c) r(t)

in terms of the Heaviside step function.

37

Heaviside Step Function Lecture 6

Example 24:

Derive an expression based on the Heaviside step function for the signal h(t)whose graph is shown below.

t−2 −1 1 2 3

−2

−1

1

Hence evaluate the definite integrals

(a)

∫ ∞−∞

h(t)dt

(b)

∫ ∞−∞

(h(t))2dt

38

Heaviside Step Function Lecture 6

The step function has a useful property: multiplying an ordinaryfunction f (t) by the step function U(t) changes it into a one-sided orcausal function.

Example 25:

Skecth the graphs of the following functions.

(a) 3 sin(t)

(b) 3 sin(t)U(t)

The step function can be used to introduce a delay into the function.

Example 26:

Skecth the graphs of the following functions.

(a) 3 sin(t − π)

(b) 3 sin(t − π)U(t − π)

39

The Exponential & Logarithmic Functions Lecture 7

For any quantity Q(t) that varies over time such that

dQ(t)

dt∝ Q(t)

then Q(t) will be in the form of an exponential function.

Note:

The following are examples of quantities that vary in this way.

(a) The rate of decay of a radioactive substance is proportional to theamount present.

(b) The rate of growth of a bacteria is proportional to the amount present.

(c) The rate at which the temperature of a dead person falls is proportionalto the difference between the corpse and the room temperature.

40

The Exponential & Logarithmic Functions Lecture 7

Definition 27:

We define the natural exponential function to be

exp(t) = 1 + t +t2

2!+

t3

3!+

t4

4!+ · · · =

∞∑k=0

tk

k!

Example 28:

Evaluate

(a) exp(0)

(b) exp(1)

(c) exp(2)

(d) exp(3.5)

Example 29:

Differentiate exp(t) with respect to t.

41

The Exponential & Logarithmic Functions Lecture 7

Notation

It is usual to writee = exp(1) ≈ 2.71828 . . .

This is called Euler’s number or the natural base. Consequently

exp(t) = exp(t · 1) = [exp(1)]t = et

We will generally use this notation

et = exp(t)

Properties

(a)det

dt= et

(b)deat

dt= aeat

(c) e0 = 1

(d) eaeb = ea+b

(e)ea

eb= ea−b

(f) [ea]n = ena

42

The Exponential & Logarithmic Functions Lecture 7

t

y

−2 −1 0 1 2

5

10

15

ete−t

The graph of e−t is simply a reflection of the graph of et in the vertical axis.

43

The Exponential & Logarithmic Functions Lecture 7

Definition 30:

We define the Logarithmic function to base e by

y = ln(t) if t = ey

The graph of ln(t) is obtained by reflecting the graph of y = et in the linethrough the origin at an angle of 45◦.

.

.

1 2−1

1

−1

et

ln(t)

t44

The Exponential & Logarithmic Functions Lecture 7

Note1 ln(t) is not defined for t ≤ 0.

2 ln(t) has a vertical asymptote (infinite discontinuity) at t = 0.

3 ln(t) is increasing in value, albeit at a decreasing rate.

Properties of ln(t)

(a) ln(1) = 0(b) ln(ab) = ln(a) + ln(b)

(c) ln(ab

)= ln(a)− ln(b)

(d) ln(an) = n ln(a)

(e)d(ln(t))

dt=

1

t

(f) ln(et) = t

(g) e ln(t) = t

45

The Sinusoidal Functions Lecture 8

The study of angles in triangles(trigonometry) usually gives rise todefinitions of sin(A), cos(A) andtan(A) as ratios of angles in aright-angled triangle.

sinα

cosα

tanα =sinα

cosα

−1 −12

1

−1

−12

12

1

This approach is very limited and for most engineering problems wedefine these functions in an equivalent manner.

Definition 31:

cos(t) = 1− t2

2!+

t4

4!− t6

6!+ · · · =

∞∑n=0

(−1)nt2n

(2n)!=

Adjacent

Hypotenuse

sin(t) = t − t3

3!+

t5

5!− t7

7!+ · · · =

∞∑n=0

(−1)nt2n+1

(2n + 1)!=

Opposite

Hypotenuse

46

The Sinusoidal Functions Lecture 8

t−π 0 π 2π 3π 4π

−1

1

sin(t)

cos(t)

Properties

1 Period: cos(t) and sin(t) both have period T = 2π, i.e.

cos(t) = cos(t + 2π) and sin(t) = sin(t + 2π)

2 Amplitude cos(t) and sin(t) both have amplitudes of 1, i.e. theirgraphs reach a maximum of 1 and a minimum of −1.

47

The Sinusoidal Functions Lecture 8

General Properties

The general sinusoidal function y = A cos(ωt + α) has a graph the sameshape as that of y = cos(t).

(a) The period of a sinusoid indicates how long it takes the graph to go

through one full cycle and is given by T =2π

ω.

(b) The frequency of a sinusoid indicates how many cycles of the graph

happen in one second and is given by f =ω

2π. Clearly we have that

f = 1T .

(c) The amplitude of a sinusoid is the maximum value that the graph ofthe function attains and it is denoted by A.

(d) The phase angle, α, of a sinusoid y = A cos(ωt + α) dictates howmuch the graph of y = A cos(ωt) has been shifted left or right asappropriate. Since

sin(ωt) = cos(ωt − π

2)

we have that the sine wave has the same shape as the cosine wave justmoved to the right by π

2 .

Example 32:

Sketch the graphs of

(a) y = cos(t) and y = sin(t).

(b) y = cos(2t) and y = sin(2t).

(c) y = cos(2πt) and y = sin(2πt).

(d) y = cos(2πt + π) and y = sin(2πt + π).

48

The Sinusoidal Functions Lecture 8

Addition of Sinusoidal Functions of the Same Period

The following identities are noted on page 9 of the log tables.

cos(A + B) = cos(A) cos(B)− sin(A) sin(B)sin(A + B) = sin(A) cos(B) + cos(A) sin(B)

We can use these to simplify sums of sinusoidal functions.

Example 33:

Express x(t) = 3 cos(2πt) + 4 sin(2πt) in the form y = A cos(2πt + α).Sketch the graphs of

1 3 cos(2πt)

2 4 sin(2πt)

3 x(t)

Example 34:

Express x(t) = 5 cos(100πt) + 6 sin(100πt + π8 ) in the form

y = A cos(100πt + α).

49

The Sinusoidal Functions Lecture 8

Differentiation of Sinusoidal Functions1 Use the power series definitions to determine the derivatives of cos(t)

and sin(t).

2 Use the chain rule to determine the first derivatives of cos(ωt + α) andsin(ωt + α).

3 Determine the second derivatives of A cos(ωt) + B sin(ωt).

Consequently we have

tan(t) =sin(t)

cos(t)

sec(t) =1

cos(t)

cosec(t) =1

sin(t)

cotan(t) =1

tan(t)

sinc(t) =

{1 if t = 0sin(πt)πt if t 6= 0

50

The Hyperbolic Functions Lecture 9

Definition 35:

A function x(t) is said to be even if

x(−t) = x(t) for all t.

The graph of such a function is symmetric about the vertical axis.

Example 36:

1 t2

2 |t|3 cos(t)

4 Π(t)

5 hat(t)

6 sinc(t)

51

The Hyperbolic Functions Lecture 9

Definition 37:

A function x(t) is said to be odd if

x(−t) = −x(t) for all t.

The graph of such a function is preserved under reflection in the vertical axisfollowed by reflection in the horizontal axis.

Example 38:

1 t3

2 t5

3 sin(t)

4 sgn(t)

52

The Hyperbolic Functions Lecture 9

Examples of functions that are neither even nor odd include

et , ln(t), t2 + 2t − 1

Most functions are neither even nor odd, however it is possible to expresssuch functions as the sum of an even and odd function.

Example 39:

Express h(t) = Π(t − 2) as the sum of an odd function and an even function.

Definition 40:

We define the hyperbolic sine and hyperbolic cosine functions as follows;

sinh(t) =et − e−t

2

cosh(t) =et + e−t

2

53

The Hyperbolic Functions Lecture 9

Example 41:1 Show that cosh(t) is even and sinh(t) is odd.

2 Graph cosh(t) and sinh(t).

3 Show that et =cosh(t)+sinh(t).

Differentiating cosh(t) and sinh(t) twice will lead us to the solution ofanother standard type of differential equation.

Example 42:

Determine the first and second derivatives of

1 cosh(kt)

2 sinh(kt)

54

The Hyperbolic Functions Lecture 9

We now compare the graphs of

e−kt cos(ωt) and e−ktcosh(mt)

The function e−kt cos(ωt) inherits some of the features of cos(ωt).

Its graph cuts the t axis every 2πω .

Its maxima are 2πω apart.

Its minima are 2πω apart.

However, the exponential term causes the maxima to decrease as t increasesand the minima to increase as t increases. Consequently the function is notperiodic.

Example 43:

Sketch the graph ofe−0.2t cos(πt)

55

The Hyperbolic Functions Lecture 9

The function e−ktcosh(mt) on the other hand is simply the sum of twoexponentials.

e−ktcosh(mt) = e−kt[emt + e−mt

2

]=

e(m−k)t + e−(m+k)t

2

→ e(m−k)t

2as t increases

Example 44:

Sketch the graph ofe−0.2tcosh(2t)

56

The Hyperbolic Functions Lecture 9

Properties

1 cosh(A + B) =cosh(A)sinh(B)+sinh(A)cosh(B)

2 sinh(2A) = 2sinh(A)cosh(A)

3 cosh2(A)−sinh2(A) = 1.

57