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math guide - 34.9

34.2FUNCTIONS

34.2.1Discrete and Continuous Probability Distributions

Binomial

P( m) =

n

 

t

q

n t

q = 1 – p

q, p [ 0, 1]

t p

 

 

 

t

m

 

 

 

 

 

 

 

Poisson

 

 

 

λteλ

 

 

 

 

P( m)

=

 

 

λ > 0

 

------------

 

 

 

 

 

t

m

t!

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Hypergeometric

 

r

 

s

 

 

 

 

 

 

 

 

 

 

P( m)

=

t n t

 

 

-----------------------

 

 

 

 

 

 

r + s

 

 

 

 

t

m

 

n

 

 

 

 

Normal

 

 

 

 

 

 

 

 

 

 

P( x)

=

1

 

x

e

t2

dt

 

 

----------

 

 

 

 

 

 

2

π

 

 

 

 

 

 

Figure 34.9 Distribution functions

34.2.2Basic Polynomials

The quadratic equation appears in almost every engineering discipline, therefore is of great importance.

ax2 + bx + c = 0 = a( x r1) ( x r2)

r

, r

2

= --------------------------------------b ±

b2 – 4ac

 

1

 

 

2a

Figure 34.10 Quadratic equation

math guide - 34.10

• Cubic equations also appear on a regular basic, and as a result should also be considered.

x3 + ax2 + bx + c = 0 = ( x r1) ( x r2) ( x r3)

First, calculate,

 

 

 

Q =

3-----------------b a2

R =

9--------------------------------------ab – 27c – 2a3

S = 3 R + Q3 + R2 T = 3 R Q3 + R2

 

9

 

54

 

Then the roots,

a r1 = S + T --3

Figure 34.11

r2 =

S------------+ T

a--

+ j--------3

( S T)

r3 =

S------------+ T

a--

j--------3

( S T)

 

2

3

2

 

 

2

3

2

 

Cubic equations

• On a few occasions a quartic equation will also have to be solved. This can be done by first reducing the equation to a quadratic,

x4 + ax3 + bx2 + cx + d = 0 = ( x r1) ( x r2) ( x r3) ( x r4)

First, solve the equation below to get a real root (call it ‘y’),

y3 by2 + ( ac – 4d) y + ( 4bd c2 a2d) = 0

Next, find the roots of the 2 equations below,

r1, r2

2

a +

a2 – 4b + 4y

y +

y2

– 4d

= z + -------------------------------------------

 

z +

------------------------------

2

 

2

= 0

 

 

 

 

 

 

 

r3, r4

2

a

a2 – 4b + 4y

y

y2

– 4d

= z + ------------------------------------------

 

z +

-----------------------------

2

 

2

= 0

 

 

 

 

 

 

 

Figure 34.12 Quartic equations

math guide - 34.11

34.2.3Partial Fractions

The next is a flowchart for partial fraction expansions.

start with a function that has a polynomial numerator and denominator

is the order of the

numerator >= yes denominator?

no

Find roots of the denominator and break the equation into partial fraction form with unknown values

use long division to reduce the order of the numerator

 

OR

 

use limits technique.

 

use algebra technique

If there are higher order

 

 

roots (repeated terms)

 

 

then derivatives will be

 

 

 

 

required to find solutions

 

 

 

 

 

Done

Figure 34.13 The methodolgy for solving partial fractions

• The partial fraction expansion for,

math guide - 34.12

x( s)

=

1

 

 

A

+

B

+

 

 

C

 

 

 

 

 

 

 

 

 

 

2( s + 1)

 

= ----

--

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

s

 

s2

 

 

s

 

 

s + 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

( s + 1)

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C =

lim

 

 

 

 

 

= 1

 

 

 

 

 

 

 

 

 

 

s –1

 

 

 

 

 

s2( s + 1)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A =

lim

 

s2

 

1

 

 

 

=

lim

 

1

 

 

= 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

s 0

 

 

 

s2( s + 1)

 

 

 

 

s 0

 

s + 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B = lim

 

d

 

s

2

1

 

 

 

 

 

 

 

 

=

lim

 

 

d

1

 

= lim [ –( s + 1)

–2

] = –1

 

 

 

 

 

 

 

 

 

 

 

 

-----

 

---------------------

 

 

 

 

 

 

 

 

 

 

----

-----------

 

 

 

s 0

 

ds

 

 

s2( s + 1)

 

 

 

 

 

 

s 0

 

 

ds

s + 1

 

s 0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 34.14 A partial fraction example

• Consider the example below where the order of the numerator is larger than the denominator.

math guide - 34.13

( ) 5s3 + 3s2 + 8s + 6 x s = -------------------------------------------

s2 + 4

This cannot be solved using partial fractions because the numerator is 3rd order and the denominator is only 2nd order. Therefore long division can be used to reduce the order of the equation.

5s + 3

s2 + 4 5s3 + 3s2 + 8s + 6 5s3 + 20s

3s2 – 12s + 6

3s2 + 12

– 12s – 6

This can now be used to write a new function that has a reduced portion that can be solved with partial fractions.

x( s) = 5s + 3 +

---------------------12s – 6

solve

---------------------12s – 6

=

-------------A

+

------------B

 

s

2

+ 4

 

s

2

+ 4

 

s + 2j

 

s – 2j

 

 

 

 

 

 

 

 

Figure 34.15 Solving partial fractions when the numerator order is greater than the denominator

• When the order of the denominator terms is greater than 1 it requires an expanded partial fraction form, as shown below.

F( s) =

 

5

 

 

 

 

 

 

 

 

s-----------------------2( s + 1) 3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5

=

A

+

B

+

C

+

D

+

E

-----------------------

----

--s

------------------

------------------

(---------------s + 1)

s2( s + 1) 3

s2

 

 

( s + 1) 3

 

( s + 1) 2

 

Figure 34.16 Partial fractions with repeated roots

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