Download Modeling Transfer Function

Transfer Function
LTI System
an
dc
d n 1c
d nc
a

 a 1  a 0 c
n 1
n
n 1
dt
dt
dt
Modeling
 bm
G  s 
M. Sami Fadali
Professor of Electrical Engineering
University of Nevada
dr
d m1r
d mr
b

b1  b0 r
m 1
m
m 1
dt
dt
dt
C  s
b s m  bm1 s m1 b1 s  b0
 m n
R s zero IC a n s  a n 1 s n 1  a 1 s  a 0
Example
1
Response of LTI System
• Convolution Theorem:
Example: Transfer Function of Point Mass
= impulse response
L
f ( t )  m x
L
x
F ( s)  m s 2 X ( s)
G ( s) 
L
• Impulse Input
2
1
X ( s)

F ( s) m s 2
f
m
F(s)  1 s
• The transfer function and the impulse response
are Laplace transform pairs.
3
1 1
t2
X (s)  2   x(t ) 
2m
ms s
4
Analytical Modeling
Relations
• Physical systems store and dissipate energy.
• Lumped idealized elements: represent energy
dissipation and energy storage separately.
• Physical elements may approximate the
behavior of the idealized elements.
• Physical elements are not lumped. They
involve both energy dissipation and energy
storage.
• Constitutive (elemental) Relations
– Govern the behavior of the idealized elements.
– Hold only approximately for physical elements.
• Connective Relations
– Govern connections of elements.
– Often derived from conservation laws.
5
Electrical Systems
•
•
•
•
6
Constitutive Relations for R, L, C
Table 2.3
Ideal R: energy dissipation.
Ideal L: magnetic energy storage.
Ideal C: electrostatic energy storage.
Connective Relations: Kirchhoff’s Laws
Impedance Admittance
– Current Law: conservation of charge
– Voltage Law: conservation of energy
7
8
Cramer’s Rule
Mesh Analysis
 a11 a12   x1   b1 
 
a



 21 a22   x2  b2 
1. Replace passive elements with Z(s).
2. Define clockwise current for each mesh.
3. Write KVL in matrix form with source
voltages that drive clockwise current
positive.
4.Use Cramer’s rule to solve for the transfer
function.
• Provided that a solution exists
b1
x1 
b1
a21 a22
a11 a12
 a11a22 a12a21
a21 a22
Example: Mesh Analysis
10
Cramer’s Rule
Z 1  1  sC Vi
Z 1  L1 s  R1  R2
R2 L1 s  R1 
L1 s  R1  R2
a11
b2 a22
a
b
, x2  21 2
a11 a12
a11 a12
a21 a22
9

a12
+
i
I 2 ( s) 
C

Z(s)I(s)  Vi (s)
1 sC
Z1 1 sC
  I1 (s)  Vi (s)
 1 sC R  L s 1 sC I (s)   0 


 2  
3
2
G(s) 

Vo (s)  R3 I 2 (s)
11
 1  sC
0
i
 1  sC
Z 1  1  sC
 1  sC
R3  L2 s  1  sC
Vo R3 I 2 ( s )

Vi
Vi
R3 sC 
Z1  1 sC R3  L2 s  1 sC   1 sC 2
12
Example: Node Analysis
Node Analysis
1.Replace passive elements with Y(s).
2. For each node, define a node voltage relative
to a reference node.
3. Write KCL in matrix form with source
currents that drive current into a node
positive.
4.Solve for the transfer function using
Cramer’s rule.
R1
Change voltage
source to
current source.
Vs/Rs
L1
Rs
L2
R3
R2
13
14
Node Equations
 Y(s)V(s)  I(s)
G 1/ R,Vo Vc

1
Gs  G1  sL
1

1



sL1

 G1


1

sL1
1
1
G2 

sL1 sL2
1

sL2
Va
Gs  G1 
R1
L1
Rs
R2
Vb
 1
 
 Vs 
Vo
Vs/Rs
L2
R3

 G1

 Va ( s ) Vs ( s )Gs 
1
 V ( s )    0 


 b  
sL2
(
)
0
V
s




1  c  
G1  G3 

sL2 
15
V
G ( s)  o 
Vs

1
sL1
1
sL1
 G1
Gs  G1 

1
sL1
 G1
1
sL1
1
sL1
1
1
G2 

sL1 sL2
1

sL2

1
sL1
1
1
G2 

sL1 sL2
1

sL2

Vs
Rs
0
0
 G1

1
sL2
G1  G3 
1
sL2
16
a) Ideal Spring
Translational Mechanical Systems
•
•
•
•
• Elastic energy (neglect plastic deformation)
• Linear element: force proportional to the
.
deformation
k
• No energy dissipation
f
• No mass
x
x
Ideal Damper b: energy dissipation.
Ideal Spring k: potential energy storage.
Pure Mass m: kinetic energy storage.
Connective Relations: Newton’s 2nd Law
1
17
b) Ideal Viscous Damper
2
18
c) Point Mass
f
• Energy dissipation
• No mass
x1
b
x2
•Perfectly rigid
• No elastic deformation
•No dissipation
• Linear element: force proportional to rate of
deformation
.
•Linear element
f
m
x
19
20
Constitutive Relations for Spring,
Mass, Damper
Ex. Mass-Spring-Damper
m x  f  f s  f d
 f  k x  b x
m x  b x  kx  f
X  s
1
G s 
 2
F s ms  bs  k
k
f(t)
m
b
x
21
Example 2.11
22
Transfer Function (Cramer’s Rule)
m1 
x 1  b1 x 1  b3  x 1  x 2   k 1 x 1  k 2  x 1  x 2   f
G s  
m2 
x 2  b2 x 2  b3  x 2  x 1   k 3 x 2  k 2  x 2  x 1   0
2
 1  m1s  b1  b3 s  k1  k 2 F
 
0
 b3 s  k 2 
F

2
m1s  b1  b3 s  k1  k 2
 b3 s  k 2 
2
m2 s  b2  b3 s  k 2  k3
 b3 s  k 2 
b3 s  k 2 
m1 s 2  b1  b3  s  k1  k 2
 X 1   F 



2
m2 s  b2  b3  s  k 2  k 3  X 2   0 
b3 s  k 2 


23
X2
F
b3 s  k 2
m1s  b1  b3 s  k1  k2 m2 s 2  b2  b3 s  k2  k3  b3 s  k2 2
2
24
3-D.O.F. Translational
Mechanical System
Rotational Mechanical Systems
Three equations of motion.
25
26
Example 2.19
Transfer Function (Cramer’s Rule)
J11  D11  K 1   2   
2
 1  J1s  D1s  K 
 
0
K

2
G ( s) 

2

J1s  D1s  K
K
J 2 s 2  D2 s  K
K
J 22  D22  K  2  1   0
J1s2  D1s  K
 1  
K

    0
2



K
J
s
D
s
K

 2  
2
2
K

  J1s 2  D1s  K J 2 s 2  D2 s  K   K 2

27
28
Example 2.20
Gears
N1
1 1
Assume:
1- No losses.
2- No inertia.
3- Perfectly rigid.
J11  D11  K 1   2   
J 22  D2 2  3   K  2  1   0
J 33  D2 3  2   D33  0
J1s  D1s  K
1  
K
0

   
2

K
J
s

D
s

K

D
s
2
2
2

2   0

0
D2 s
J3s2   D2  D3s3  0
2
N2
J
2 2
Single velocity at point of contact
Equal arc length
29
30
Energy Storage
Energy Balance
• Translation
• Assume no losses
• Mass
• Trade speed for torque
• Spring
N2 > N1 : output side slower but delivers
more torque
• Rotation
N2 < N1 : output side faster but delivers
less torque
• Spring
31
• Inertia
32
Equivalent Inertia
Energy Dissipation
N1
1
1
E  J 112  J 2 22
2
2
• Power dissipated
• Translation
2
J 1   2   N1 
  

J 2  1   N 2 
N 
J e  J1  J 2  1 
 N2 
• Rotation
1 , 1
2
N2
2
J
2 , 2
Equivalent to
inertia on
output side
 no. teeth of destination 
Je  J 

 no. teeth of source 
2
33
Effect of Loading Output Side on Input Side
34
Damper and Spring
N1
Damper
1 , 1
Inertia
J
N2
2 , 2
 2  J2
N

N2
 1  J  1 1 
N1
 N2 
N 

Je  1  J  1 


1
 N2 
2
2
2
Spring
1
2
 no. teeth of destinatio n 
 J

 no. teeth of source 
2
N 
 no. teeth of destinatio n 

De  1  D 1   D


1
 no. teeth of source 
 N2 
N 
 no. teeth of destinatio n 

Ke  1  K  1   K 

1
 no. teeth of source 
 N2 
1 , 1
2
2
2 , 2
35
36
Example (Special Case)
Example
Db
,m
N1
Jb
Jg
L
2
JL
K
DL
N2
J s
e
2
• Two equations of motion.
• Cannot simply add all rotational masses!
N 
 De s  K  2 ( s )  T2 ( s )   2 T1 ( s )
 N1 
37
38
Solution
Gear Train
 l   2   3    l 

     
1  1   2    l 1 
Redraw the schematic with
(i) added “e” for elements (and variables)
moved, and (ii) gears removed.
Dbe
e,me
Jbe

 N  N   N
  1  3    2l 3 
 N 2  N 4   N 2l  2 
 ne
L
2
Jge
1
N1
N2
N3
2
l-1
3

N4
l
l 1

 1 ne
JL
K
DL
J l  ne2 J1
39
Bl  ne2 B1
K l  ne2 K1
40
DC Motor
TFs of Electromechanical Systems
• Electrical Subsystem
– Varies with motor type
– armature (rotor) conductors current
– field (stator) conductors or permanent magnet
• Mechanical Subsystem
– Varies with load.
– Write equations of motion.
DC motor armature
(rotor)
DC motor.
National Instruments:
http://zone.ni.com/devzone/cda/ph/p/id/52
41
42
Torque Equation
• Magnetic flux
Field Control
Wb
• Changing
• Force
r
= conductor length

= magnetic flux density
f
and fixing
• Back EMF (Faraday’s Law)
• Voltage induced in moving coil proportional to the
rate of cutting of lines of magnetic flux.
= armature (rotor) current
is only approximately constant through the use of
high resistance (inefficient)
Control
• Vary torque by changing
or
43
44
Armature Control
• Changing
and fixing
Schematic
Za
• Used in practice.
Za
ea
1
vb
• KVL
ea
N1
vb
D2
J2
J1, D1
2
N2
K
45
L
46
Mechanical Subsystem
Matrix Form
Equations of motion for rotational system
J11  D11  K e 1   L'     K t ia
J 2 eL'  D2 eL'  K e  L'  1   0
J 11  D11  K e 1   L'     K t ia
J 2 eL'  D2 eL'  K e  L'  1   0
, 1
D1
Kb s

J s2  D s  K
1
e
 1

 Ke
D2e
J2e
J1
L
0
 Ke
J 2 e s  D2 e s  K e
2
La s  Ra   1   Ea 
 K t   'L    0 
0   I a   0 
Ke
47
48
Evaluation of Motor Parameters
Transfer Function
Dynamometer measure speed & torque for constant
Dynamometer Test gives speed-torque curves
supplied by manufacturer
Assume
 '  N N 
G(s)  L  L 1 2
Ea
Ea
Kbs
Ea La s  Ra
1 2
 N1 N2   J1s  D1s  Ke 0  Kt
 Ea 
0
0
 Ke

Kb s
0
La s  Ra
J1s2  D1s  Ke
 Ke
 Kt
2
J2e s  D2es  Ke
0
 Ke

500
=stall torque
=no-load speed
Torque (N-m)
400
 N1 N 2 K e K t
ea= 100 V
300
200
100
La s  Ra J1s 2  D1s  K e J 2e s 2  D2e s  K e   K e2   K t K b sJ 2e s 2  D2e s  K e 
0
0
10
20
30
40
50
Speed (rad/s)
49
Solve for Parameters
50
Transfer Function
Ra
ea
 J e s  De  m ( s)  K t I a ( s)  K t
vb
• Stall torque:
G(s) 
T, m
• No-load speed:
 m (s)
K t Ra 

Ea ( s ) J e s  De   K t Ra K b
Da
JLe
Ja
DLe
51
Ea ( s )  K b  m ( s )
Ra
J e  J a  J Le
De  Da  DLe
52
Linearity
Nonlinearities
(i) Homogeneity
(ii) Additivity
• Affine
53
54
Equilibrium Point
Linearization
1st order approximation (in the vicinity of
• System at an equilibrium stays there unless perturbed.
)
• Set all derivatives equal to zero for equilibrium.
80
= value of forcing function at equilibrium
60
• Cancel constants
40
and
20
for small
0
0
2
4
6
8
55
56
Linearized Differential Equation
In the Vicinity of the Equilibrium
Special case: nonlinearity in output only
dc d c  c0  dc
d i c d i c


Similarly
 i , i  1,2,, n
dt i
dt
dt
dt
dt
n 1
d c
d c
dc
 a n 1
   a1
 f ( c )  f ( c 0 )  r  r0   r
n 1
n
dt
dt
dt
n
d n c
d n  1 c
d c
df

 a 1
 a 0 c  r , a 0 
a
n1
n
n1
dt
dc
dt
dt
c0
Linear: can Laplace transform to get the TF
1st order approximation
d n c
d n1c
dc df

a
   a1

c  r
n 1
n
n 1
dt dc c0
dt
dt
G ( s) 
C
1
 n
n1
R s  a n  1 s  a n  2 s n  2  a 1 s  a 0
57
58
Example: Pendulum
Procedure
Moment of inertia
1. Determine the equilibrium point(s).
2. Find the first order approximation of all
nonlinear functions.
3. Rewrite the system differential equation in
terms of perturbations canceling the
constants using Step 1.
59
Equation of Motion
• Linearize about  = 30o
• Equilibrium at  = 30o
sin( )
60
Linearization
Potentiometer
• Using Trigonometric Identity
sin 30 o     sin 30 o cos   cos30 o sin  
•
•
•
•
 1 2   3 2 
cos   1
sin    
• Using 1st order approximation formula
sin   sin30 o  
d sin 
d
30o
  sin30 o   cos30 o  
10 turns
1 turn = 2  rad
20 V
Pot Gain = 20/(10 X 2  )
= (1/ ) V/rad
J  B  mgl 1 2   3 2     0  
J  B  mgl  3 2   
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Fluid Systems
Linearized Model
h=Rq
Conservation of Mass
dCh
 Q  qin  Q  qo
dt
C  Area
C
qi
h
H
qo
dh
h
dh
 qin     h  Rqin
dt
dt
R
R
h( s )

qin ( s )  s  1
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