illustrations the transfer function of linear systems

Illustrations

The Transfer Function of Linear Systems

V1 s( ) R1

Cs

I s( ) Z1 s( ) R

Z2 s( )1

CsV2 s( )1

Cs

I s( )

V2 s( )

V1 s( )

1

Cs

R1

Cs

Z2 s( )

Z1 s( ) Z2 s( )

Illustrations


Example 2.2

2ty t( )d

d

24

ty t( )d

d 3 y t( ) 2 r t( )

Initial Conditions: Y 0( ) 1ty 0( )d

d0 r t( ) 1

The Laplace transform yields:

s2 Y s( ) s y 0( ) 4 s Y s( ) y 0( )( ) 3Y s( ) 2 R s( )

Since R(s)=1/s and y(0)=1, we obtain:

Y s( )s 4( )

s2 4s 3 2

s s2 4s 3

The partial fraction expansion yields:

Y s( )

3

2

s 1( )

12

s 3( )

1s 1( )

1

3

s 3( )

2

3

s

Therefore the transient response is:

y t( )3

2e t

1

2e 3 t

1 e t 1

3e 3 t

2

3

The steady-state response is:

ty t( )lim

2

3

Illustrations


i

o

i

o

i

o

Z

Z

V

V

R

R

IR

IR

V

V

1

2

1

2

Illustrations


Illustrations

)/)(/1(/1/1

1

/1/1

/10

0

/1

0

00

/1

0

121121

1

21

1

1

22

2

2

1

21

1

12

2

1

1

1

vvRRRsCr

v

LssCR

R

v

v

Ls

v

sC

v

R

vv

R

vv

R

v

sC

vr

Illustrations

R

XFind

))((

1

R

X

))(()(

)(

)(

)(

)(

2

1122

12

1

2

1122

12

121212

211

12

112112

1112112

21

2

1

1

12

2

2

1

22

22112

222112

22

2

2

sbsbsbsMX

X

sbsbsbsMX

XXsbXsbsbsMXR

XsMRsXsXbsXb

xMRxxbxb

dt

xdMF

sb

sbksM

X

X

XsMsXsXbkX

xMxxbkxdt

xdMF

Illustrations


Illustrations

Kf if

Tm K1 Kf if t( ) ia t( )

field controled motor - Lapalce Transform

Tm s( ) K1 Kf Ia If s( )

Vf s( ) Rf Lf s If s( )

Tm s( ) TL s( ) Td s( )

TL s( ) J s2 s( ) b s s( )

rearranging equations

TL s( ) Tm s( ) Td s( )

Tm s( ) Km If s( )

If s( )Vf s( )

Rf Lf s


Td s( ) 0

s( )

Vf s( )

Km

s J s b( ) Lf s Rf

Ia = constant, controlling using if

2

)(

JsBsTT

JBTTF

JsBwJswBwTT

wJBwTT

wJBwTT

wJJJT

dm

dm

dm

dm

dm

BsJs

k

sLRTI

Tm

v

I

vm

ffmff

f

f

2

1*

1*

1**

Illustrations


Illustrations


V 2 s( )

V 1 s( )

1RCs

V 2 s( )

V 1 s( )RCs

Illustrations


V 2 s( )

V 1 s( )

R 2 R 1 C s 1 R 1

V 2 s( )

V 1 s( )

R 1 C 1 s 1 R 2 C 2 s 1 R 1 C 2 s

Illustrations


s( )

V f s( )

K m

s J s b( ) L f s R f

s( )

V a s( )

K m

s R a L a s J s b( ) K b K m

Illustrations


Vo s( )

Vc s( )

K

Rc Rq

s c 1 s q 1

c

Lc

Rc

q

Lq

Rq

For the unloaded case:

id 0 c q

0.05s c 0.5s

V12 Vq V34 Vd

s( )

Vc s( )

Km

s s 1

J

b m( )

m = slope of linearized torque-speed curve (normally negative)

Illustrations

The Transfer Function of Linear SystemsY s( )

X s( )

K

s Ms B( )

KA kx

kp

B bA2

kp

kxx

gd

dkp

Pgd

dg g x P( ) flow

A = area of piston

Gear Ratio = n = N1/N2

N2 L N1 m

L n m

L n m

Illustrations


V2 s( )

V1 s( )

R2

R

R2

R1 R2

R2

R

max

V2 s( ) ks 1 s( ) 2 s( ) V2 s( ) ks error s( )

ks

Vbattery

max

Illustrations


V2 s( ) Kt s( ) Kt s s( )

Kt constant

V2 s( )

V1 s( )

ka

s 1

Ro = output resistanceCo = output capacitance

Ro Co 1s

and is often negligible for controller amplifier

Illustrations


T s( )

q s( )

1

Ct s Q S1

R

T To Te = temperature difference due to thermal process

Ct = thermal capacitance= fluid flow rate = constant= specific heat of water= thermal resistance of insulation= rate of heat flow of heating element

QSRt

q s( )

xo t( ) y t( ) xin t( )

Xo s( )

Xin s( )

s2

s2 b

M

sk

M

For low frequency oscillations, where n

Xo j Xin j

2

k

M

Illustrations


x r

converts radial motion to linear motion

Illustrations

Block Diagram Models

Illustrations


Original Diagram Equivalent Diagram

Original Diagram Equivalent Diagram

Illustrations


GH

G

R

Y

YGYHR

1

)(

Illustrations


Example 2.7

431322432131

4321

43)1)2)2)4/()1)3(((((

GGHGGHGGGGH

GGGG

R

Y

YGGYHGHGYGYHR

..)21(1( LiLjLkiLjLLL

PD

R

Y

Illustrations

Block Diagram Models Example 2.7

Illustrations

Signal-Flow Graph Models

For complex systems, the block diagram method can become difficult to complete. By using the signal-flow graph model, the reduction procedure (used in the block diagram method) is not necessary to determine the relationship between system variables.

Illustrations


Y1 s( ) G11 s( ) R1 s( ) G12 s( ) R2 s( )

Y2 s( ) G21 s( ) R1 s( ) G22 s( ) R2 s( )

Illustrations


a11 x1 a12 x2 r1 x1

a21 x1 a22 x2 r2 x2

Illustrations


Example 2.8

Y s( )

R s( )

G 1 G 2 G 3 G 4 1 L 3 L 4 G 5 G 6 G 7 G 8 1 L 1 L 2

1 L 1 L 2 L 3 L 4 L 1 L 3 L 1 L 4 L 2 L 3 L 2 L 4

..)21(1( LiLjLkiLjLLL

PD

R

Y

Illustrations


Example 2.10

db

ab

dd

aa

TGGks

GV

GGks

GG

TT

VV

21

2

21

21

1

1

1

1

Illustrations

Illustrations


Y s( )

R s( )

P1 P2 2 P3

P1 G1 G2 G3 G4 G5 G6 P2 G1 G2 G7 G6 P3 G1 G2 G3 G4 G8

1 L1 L2 L3 L4 L5 L6 L7 L8 L5 L7 L5 L4 L3 L4

1 3 1 2 1 L5 1 G4 H4

Illustrations

Design Examples

Illustrations

Speed control of an electric traction motor.

Design Examples

Illustrations

Design Examples

Illustrations

The Simulation of Systems Using MATLAB

Illustrations

error


Sys1 = sysh2 / sysg4

Illustrations


Illustrations

error


Num4=[0.1];

Illustrations


Illustrations

Sequential Design Example: Disk Drive Read System

Illustrations

=

Sequential Design Example: Disk Drive Read System

Illustrations

P2.11

Illustrations

1

L c s R c

Vc

Ic

K1

1

L q s R q

Vq

K2

K3-Vb

+Vd

Km

Id

1

L d L a s R d R a

Tm

1

J s b

1

s

P2.11

Illustrations

Illustrations

http://www.jhu.edu/%7Esignals/sensitivity/index.htm

Illustrations

http://www.jhu.edu/%7Esignals/

illustrations the transfer function of linear systems

Documents

matlab slide

illustrations design

simulation of systems

illustrations speed

complex systems

block diagram method

reduction procedure

electric traction motor