page 62

Kp = 0.1

Kd = 0.02

Kp = 0.2

5.3.3 Back-Shift Transform Table

•The general application of the table is,

1.Develop the differential equation.

2.Look for the equation in the table. Sometimes the equation will have to be rearranged to match the form in the table. If it is not in the table, derive the transfer function the long way.

3.Select the appropriate transfer from the right column, and substitute in values and variables.

page 63

• A few of the transforms are given below,

Time Domain

Back-Shift Domain

y = Kx( t – D)

D

y( B)

INT

---

=

KB

T

----------

x( B)

d

Kx

y( B)

=

KBT

---- y =

----------

-----------

dt

x( B)

1 – B

T

τ

d

y + y =

Kx

–-τ-

----

y( B)

KB 1 – e

dt

=

----------

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

x( B)

T

–--

1 – Be τ

d 2

y =

Kx

y( B)

=

KT2B( 1 + B)

----

----------

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

dt

x( B)

2( 1 – B)

2

d

2

d

y( B)

K( b1B + b2B2)

----------

=

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

τ

----

----

dt

y + dt

y = Kx

x( B)

1 – a1B – a2B2

T

T

–--

–--

a1 = 1 + e

τ

a2 = –e τ

T

b

= T – τ

–--

1

1 – e

τ

T

T

–--

–--

b2 = – Te

τ

+ τ

1 – e

τ

page 64

Time Domain

Back-Shift Domain

1

d 2

2ζ

d

y + y = Kx

y( B)

K

2

2

y +

( b1B + b2B )

-----

----

-----

----

ω

n

dt

ω n

dt

----------

=

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

AND/OR

x( B)

1 – a1B – a2B

2

( τ

1τ 2)

d

2

( τ 1

+ τ

2)

d

y + y = Kx

y +

----

----

dt

dt

Case 1:

ζ

>

1

T

T

–

T

–

T

–----

–----

----

----

a1 = e

τ 1 + e

τ 2

a2 = –e

τ 1

τ 2

T

T

T

–

T

–----

–----

–----

----

e

τ 1

e τ

2

–

----

–

----

e τ 2

e

τ 1

--------

– --------

--------

– --------

τ 2

τ 1

T

T

τ 2

τ 1

b

= 1

b

= e

τ 1

τ 2

1

+ ------------------------

2

+ ------------------------

1

1

1

1

----

– ----

----

– ----

τ 1

τ 2

τ 1

τ 2

Case 2:

ζ

= 1

a1 = 2e–Tω n

a2 = –e–2Tω n

b1 = 1 – e–Tω n – Tω ne–Tω n

b2 = e–Tω n ( e–Tω n + Tω n – 1)

Case 3:

ζ

<

1

ω

d

= ω n

1 – ζ

2

a

1

= 2e–Tζω

n cos ( Tω

)

a

2

= –e–2ζ

Tω

n

d

b1

= 1 –

ζω

ne–Tζω n

d)

– e

–Tζω

n

cos ( Tω

d)

------------------------ sin ( Tω

ω

d

–Tζω n –Tζω n

ζω

n

b2

= e

e

+

---------

sin ( Tω

d)

– cos ( Tω

d)

ω

d

• Try finding the transfer function for the system below,

page 65

N

Ns

---

------

Ks = 2m

Kd = 0.5 m

M=1kg

y

F

5.3.3.1 - A Summary of Differential Equation Solutions

•A quick look at how differential equations relate to difference equations will be useful.

•First order homogenous difference equation,

Equation form,

yn – ayn – 1 = 0

General solution form,

1 –n yn = C --a

• second order homogenous difference equation,

page 66

General equation form,

yn – a1yn – 1 – a2yn – 2 = 0

First we must examine the roots,

r1, r2

=

a1 ± a12 + 4a2

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

–2a2

Case 1 - both roots real, and unequal,

yn = C1r1–n + C2r2–n

Case 2 - both roots real, and equal,

y

=

C

–a1

–n

–a1 –n

n

1

--------

+ C

2

n --------

2a

2a

2

2

Case 3 - complex roots

α

2

β

2

θ

β

2

2

r , r

=

α

± jβ

r

=

+

= atan

---

C = C1

+ C2

2

α

1

yn

=

r

–n( C1 sin ( θ

n)

+ C2 cos ( θ n) )

= C

r

–n sin ( θ n + φ )

• Higher order homogenous difference equations,

Given the general form,

yn – yn – 1 + a2yn – 2 – …

– amyn – m = 0

Find the roots for an m-order polynomial,

r1, r2, … , rm

Equation roots

yn = C1r1–n + C2r2–n + …

+ Cmrm–n

• nonhomogeneous equations