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I. MATHEMATICAL MODELS
Desired
Performance ControllerPlant /
ProcessComparison
Sensor
Output
Feedback Loop
InputIII.
II.
I.Errorsignals
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I.A Model Building Revised September, 2004
The process of model building
Real LifeSystem
PhysicalModel
Set of DifferentialEquations relating
Input/output of
plant/process
x
Up/down
movementRotational
movement
Comprised ofbasic elements
relating relevantinputs and outputs
I.A Model Building
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Governing Equations
For BasicElements
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I.A.1 Mechanical SystemUse Newtons Law
maF
..
IExample: Spring-Mass-damper system
)()()()( tFtkytydt
dbty
dt
dM
2
2
)()( ty
dt
dtv Define velocity:
)()()()( tFdttvktbvtvdt
dM
t
0
Input: force F ; Output: positiony
(As given in Table 2.2)
Springconst.k
M y
F
Dampingcoeff. b
Equilibriumposition
Equation forDamper elementEquation for
Mass element
Equation forSpring element
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I.A.2 Electric Circuit
Use Kirchoffs Laws
KCL:Sums of currents leaving and entering a node are equal
KVL: Sum of voltage around a closed path is zero
Example:LRCcircuit
)()()(
)(1
0tItV
dt
dC
R
tVdttV
L
t
I(t)RL C
VoltageV(t)
Using KCL (and Table 2.2):
Input: CurrentI(t)Output: Voltage V(t)
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)()()()( tVdttiC
tRitidt
dL
t
0
1
Velocity v(t)Force F(t)MassM
Friction b
Spring constantk
Current i(t)Voltage V(t)L InductanceR Resistance
1/C Inverse of capacitance
Force-Voltage analogy
V(t)
RL
C
)()()()( tFdttvktbvtvdt
dMt
0
Consider anotherLRCcircuit Voltage driven
i(t)
Compare mechanical system example:
+
-
Using KVL:
Input: Voltage V(t)Output: Current i(t)
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One can build mechanical, electrical, fluid, or thermalanalog to simulate performance of a system One may thus use easier-to-build analog to test out
performance of a system first before actual construction
Early example: Moniac a hydraulic systemto study US money flow by Bill Philips atLondon School of Economics, 1949
Summary of Model Building:Plant/process + Objective of study
Differential equations of
basic elements from Table 2.2
Appropriate physicalmodel comprisingof ideal basic element
Set of differentialequation relatingrelevant inputs andoutputs
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A Linearprocess:
Given: input is u1, output is y1input is u2, output is y2
Then, if input is u1+u2, output is y1+y2
I.A.4 Linear, Time-Invariant (LTI) process
A Time-invariantprocess:
input-output relation is not
dependent on time
Note: most processes can reasonably
be assumed as Time-invariant
Linear differential equation
with Time-invariantcoefficients
Scope of MAE3050:
Process describable by
Linear differential equation
Process describable by differentialequation of constant coefficients
Linear differential equationwith Time-varying coefficients
Nonlinear differential equationwith Time-invariant coefficients
Note: most processes are actuallynot linear (i.e., nonlinear) by nature
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)()(cos)()(
)()()(
)()()(
)()()()(
2
tutyCtydt
dAt
tutCytydt
dA
tutCytydt
dA
tutCytydt
dAt
Describing different equation for S:
S is Linear, Time-varying system
S is Linear, Time-invariant system
S is Nonlinear, Time-invariant system
S is Nonlinear, Time-varying system
Sinput u outputy
Given a system S:
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I.B Linearization
Example 2.1: Pendulum oscillator model
Idea: Taylor expansion of function
Can have a linear approximation of (X) about a certain Describing differential equation (X) is nonlinear (is it?)
PendulumTorque Input Output
2ML
T
L
g sin
..
L
Mg
Inputtorque
(X)
Nonlinear system may be studied through linearization ofits nonlinear model
ooperating point
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Recall: Taylor Series expansion of functionf(x) about point
.....)(!
)(.....)(')()()( )(
o
nn
oooo xf
n
xxxfxxxfxf
ox
)(')()( ooo xfxxxf
Range of validity of linear approximation depends on function
f(x) and operating point ox
)(xf
)( oxf
ox xOperating point
Linear approximation of )(xf
about operating point ox
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- From 1st order terms of (XX),
2MLL
go
)cos(
..
This yields relation between o and operating point
- From 0
th
order terms of (XX),
gML
T
ML
T
L
goo
02
0 )sin()sin(
This gives the linear approximation of the pendulumdifferential equation relating torque variation (t)and about operating point
- Equation (X) becomes
2
0
ML
T
L
goo
)cos()sin(
..
(XX)
)(t
(XX1)
(XX0)
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)()( sFtf
I.C Laplace Transform
0dttfetfF(s)
st )()]([L
L
Powerful method in analyzing/solving LTI differentialequations
I.C.1 Definition
F(s) complex variable function of complex variable s is such that
0000)( tfortf)(tf
Function of t
Function ofcomplex variable s
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Example:
s
s
e
s
e
dtetutfF
t
st
t
st
st
1
0
0
)()]([L(s)
Q: Laplace transform F(s) exists only for real s>0?
A: Analytic Extension theorem extends existence ofF(s) towhole s-plane except at some singular point(s), if any.
0atfunctionStep ttuf(t) )(
becomes)(sF
Complex planelocations where
u(t)
1
t
(This term is 0 with real s > 0 )
Same for all Laplace Tansforms:F(s) exists in part of s-plane existence in
whole s-plane except of a few singular points!
In this case, singular point at s=0
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I.C.2 Properties of Laplace Transform
(a) Superposition of functions )(and)( 21 tftf
)()()]()([ 2121 sFsFtftf L
(b) Time Delay (Function shift to right by duration )0
)()]([ sFetf s L
L )( sF)(tf
Lt
)( tft
)( sFe s
Given
Then
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(c) Time Scaling (Expansion/contraction of time axis)
01 aFatfas
a ,)()]([L
Given
Then
L )( sF)(tf
tot
L
t
)(atf
ato
ato
)(1as
aF
a >1Contraction
of time axis
a < 1Extension
of time axis
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(d) Modulation by Exponential factor
)( tfe at
L)( sF
)(tf
Lt
t
Given
Then
e at
)()]([ asFtfe at L
)( asF
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(e) Differentiation of a function
)()()]([ 0fssFtfdtdL
(f) Integration of a function
s
sFt
df)(
0])([ L
)()]([ sFstf ndt
dn
n
L
- For function and derivatives all starting at zero
)0()0()0()()]([)1(
21
n
ffsfssFstf nnn
ndt
ndL
- Generalization Value ofat )( tfdt
d
0t
)(11
tfnn
dt
d
Value ofat 0t
)( tfValue ofat 0t
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dtff
dftftftf
t
t
)()(
)()()()(
20
1
20
121
Laplace Transform of selected function in Table 2.3
Laplace Transform of complicated functions maybeobtained from Laplace Transforms of simpler functionsusing Transform properties
(g) Convolution Theorem
where convolution of functionf1(t) andf2(t) defined as:
)()()]()([ 2121 sFsFtftf L
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Then z1
, z2
, ., zm
are the m zeros of the F(s), and
p1, p2, ., pn are the npoles of the F(s)
))...()()(())...(()(
n
m
pspspspszszsCsF
3211
I.C.3 Laplace Transform Theorems
Poles and zeros of F(s): factorize F(s) to obtain
If some of the pi are the same Repeated pole case
Repeated zero case not of concern
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Final Value Theorem: If all poles of sF(s) inside lefthalf plane (LHP) of s-plane, then exists and
Example:
Initial Value Theorem: Given any F(s),
Example:
)(lim)(lim ssFftfst 0
)(lim)( ssFfs
0
)( f
imply
existenceof
)(f
(Later!)?)()(
)()(
f
ss
ssF
102
332
)( f
?)()()( ys
ssY 102
?)()(
)(
0102
ys
ssY
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I.C.4 Inverse Laplace Transform
Partial Fraction Expansion and term-by-term Inversion
)()( sFtf
)4)(2(
1)(
ss
G s
4
1
2
1
2
1
2
1
ss
ttt eeg
42)(
2
1
2
1
L-1 L-1
L-1
Partial FractionExpansion
Term-by-termLaplace Inversion
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(All poles different)
))...()((
))...(()(
21
1
n
m
pspsps
zszsksF
- Write
-
)(...
)()()(
2
2
1
1
n
n
ps
C
ps
C
ps
CsF
- Determine C1,,Cn as
ipsii sFpsC )()( i = 1,,n
Partial Fraction Expansion: Distinct Pole Case
nppp 21
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Partial Fraction Expansion: Repeated Pole Case
))...(()(
))...(()(
11
1
nr
r
m
pspsps
zszsksF
-
nr ppppp ...... 1111
- Write
)(...
)()(...
)()()(
n
n
r
r
r
r
ps
C
ps
C
ps
C
ps
C
ps
CsF
1
1
121
2
1
1
(first rpoles equal)
Cr+1,,Cn of non-repeatedpoles determined by formula
of distinct pole case
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1)]()[(
!
11 ps
r
i
i
ir sFpsds
d
iC
- Determine C1,,Cras
, i=0,,r-1
1
1
211
212
ps
ps
sFpsds
dC
sFpsC
)]()[(
)]()[(
* Special case: r =2
))...(()(
))...(()(
12
1
1
nr
m
pspsps
zszsksF
)(...
)()()()(
1
12
1
2
1
1
n
n
r
r
ps
C
ps
C
ps
C
ps
CsF
1)]()[(
!
1 212 psi
i
i sFpsds
d
iC
i=0,1
r-1
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* Special case:
1
1
1
)]()[(!2
1
)]()[(
)]()[(
312
2
1
312
313
ps
ps
ps
sFpsds
dC
sFpsds
dC
sFpsC
r =3
))...(()(
))...(()(
13
1
1
nr
m
pspsps
zszsksF
)(...
)()()()()(
1
13
1
32
1
2
1
1
n
n
r
r
ps
C
ps
C
ps
C
ps
C
ps
CsF
1)]()[(
!
1 313 psi
i
i sFpsds
d
iC
i=0,1,2r-1
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- Distinct Real Pole:
- Multiple Real Pole:
- Distinct Complex Pole:
ptCe
ps
C
1
L
ptr
r et
rC
ps
C 1
)!1(
1
)(
1
L
)cos(||21
*
*
CteCps
C
ps
C t
L
where jpeCC Cj ,||
Term-by-Term Laplace Transform Inversion
- Multiple Complex Pole: Complicated!
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Example 2.2 Application of Laplace Transform to solvingLTI differential equation
)()()()( trtytyty 34
)0()0( ,yy Initial conditions :
Apply Laplace transform :
))(())(()()(
)())(())()()(( )(
312
314
23400 02
sssssssY
ssYyssYysysYs
Excitation on RHS:r(t)= 2 (meaning 2u(t) )
Step function
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Taking Inverse Laplace Transform
Response due to InitialCondition
(zero if =0,=0)
)())(()( tetetetety 331
323
223
Response due toexcitation r(t)
(zero if r(t)=0)
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I.C.5 Pole Locations and Time Response
Partial Fraction Expansion yields
......)(*
*
1
1
ps
C
ps
C
ps
CsY
is real for real pole; and complex for complex poleiC
-- From pole locationpi of Y(s): more information on time behavior ofy(t)
Information abouty(t) from Y(s)?
-- From FVT and IVT:
information on final and initial values ofy(t)
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Inverse Laplace Transform Each pole contributing
a term toy(t) Contribution from a single real pole:
)(
)()(1
satpole
tyinesYin
s
t
Single real pole location and its contribution toy(t):
te
te
t
e
te te
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-- Graphic illustration: Location (of complex polepair) and time contribution
dj
dj
y(t)
t
y(t)
t
te
y(t)
te
Faster
decayenvelop
Fastergrowthenvelop
IncreasingFrequency
Growth Envelop
d
Increasing
Frequency
FrequencyDecay Envelop
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-Representation forp in LHP(b) ),( n
-- Conversion to/from CartesianRepresentation
-- Correspondingly,y(t)becomes sinusoidal signal with
tne
22
22
2
or1
d
dn
nd
n
,
,
cos
Frequency:
Decay Envelop:
21 n
21 nn jp
or
Distance between pole and origin
Cosine of pole angular positionrelative to ve real axis
:
:
n
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Fixed n , varying 10:
-- Pole location and time response in (n, ) representation
0
1
7070 .
090
0
Double real pole at
ns
nj
nj
y(t)
t
y(t)
t
Complex pole pair
Complex pole pair on j-axis
y(t)
t
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Example: Spring-Mass-Damper system
000 )(,)( yyy o
)()()()( tFtkytydt
dbty
dt
dM
2
2
k
M y
F
Wall friction, b
- With F(t)=0,
M
k
M
bnn 22 ,
oy
MksMbs
Mbs
sY
2)(
- Two independent design parameters b/Mand k/M toyieldand n
o
nn
n yss
ssY
22 2
2
)(
No externalexcitation!
kM
b
2
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where , and hence
*)(
*
)(*))((
)(
ps
C
ps
Cy
psps
ssY o
n
2
- Underdamped case: , i.e.,10
901212
122
2
CyCyj
jC oo ;
21 nn jp
cos
)sin()(
tey
ty nto n 2
21
1
kMb 20
representation!),( n
Y(s) has two complex poles:
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22
2
)()()()(
nno
n
n
s
B
s
Ay
s
ssY
- Critically damped case: or1 kMb 2
on
o
yB
yA
tno
netyty )()( 1
Response reaches its final value at the shortest timewithout oscillation
Y(s) has two overlapping real poles:
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)()())((
)(
2121
2
ps
B
ps
A
psps
yssY on
- Overdamped case: or1
oo yByA12
112
12
2
2
2
,
1221 nnpp ,
otptp yeety
21
121
121 2
2
2
2
)(
kMb 2
Y(s) has two distinct (non-overlapping) real poles:
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- Example of responses:
http://www.efunda.com/formulae/vibrations/sdof_free_damped.cfm
- For the case F(t)=0, , see000 vyyy o )(,)(
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Design Concept #1:
To properly place the poles of Y(s) so that the resultingy(t) yields satisfactory performance
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Example 2.15: Design of an accelerometer
- Idea: constant Force Fo on massMs
Wanty as measurement of acceleration:s
o MFay
x = case position
Fo
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ydt
dbkyxy
dt
dM )(
2
2
os Fxdt
dM
2
2
s
o
MFy
Mky
Mby
- Absolute movement of mass inside case:x+y
- Force acting on mass:
- Combining
- Apply Laplace Transform:
M
ks
M
bss
MF
M
ks
M
bs
yMbysy
sY so
22
000 )()()()(
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- Conduct PFE,
)()()()(
))(())((
)()()(
)(
2121
2121
000
ss
D
ss
C
s
E
ss
B
ss
A
sssss
MF
ssss
yMbysy
sY so
- Inverse Laplace transform,
Particularly,,
s
o
M
F
k
ME
tsts
s
otsts DeCeM
F
k
MBeAety 2121
)(
Due to initial value
M
ks
M
bs
221 and sswhere are roots of
Due to excitation
- ConstantsA,B, C,D andEcan be determined as in PFE
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- Acceleration measurement:
* As t large,
and
* Hence, plot verse tyield a as steady state value!
21 and ss
021 tsts ee ,
s
oss
MF
kMyty
)(
021 tsts ee ,
Fo/Ms=a is acceleration!
ssy acceleration!
)(tyM
k
* With b, k, M +ve, inside LHP21 and ss
Can also obtain from FVT
- The design issues
* How fasty(t) becomesyss depends on how fast
* Design b/Mand k/Msuch that are farther in theLHP so that and hence y(t)yss fast enough
* However, faster decay expensive components trade-off
021 tsts ee ,
- Specific numerical design in Example 2 15
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- Specific numerical design in Example 2.15
* What are ? How long to reach a reasonable acceleration measurement? Can we redesign to get as a faster measurement?
21 and ss
- Question: what if the applied force F, hence a, is time varying?
2.46Fig.3,2,3,2)0(,1)0( aM
F
M
k
M
b
yy s
o
)(tyM
k
2.46
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I.D. Transfer Function
I.D.1 Definition
Given
r(t) Plant y(t)
Transfer Function = L.T. of outputL.T. of input )(
)(sR
sY
(with zero initial conditions)
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u(t)
Laplace Transform converts differential equation description
to transfer function description (zero initial conditions)
Plant y(t)
cbsasd
2
)()()()( tudtcytybtya
U(s) Y(s)
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I.D.2 Block Diagram and Transformation Rules
Pictorical representation of system in termsof its components
- Tedious to obtain Transfer Function)()(0
sVsV
i
- Divide system into three components of BridgeCircuit
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)()()()()( sGsGsG
sVsV
i
o321
Complicated system?
- Then
Vi
Use Block Diagram transformation rules
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T-1
T-2
T-3
T-4
T-5
T-6
Same relationsbetweenX1,X2
andX3
Example 2.7
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p
T-4
T-1
T-6
T-6
T-6
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T 6
)(sGY(s)
U(s) )(sGY(s)
U(s)
Useful fact about pickoff point
Summarizing example of all things taught (so far)
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A water tank with input u and water levely
The differential equation:
(for simplicity, letA=1 m2
)
Laplace transform leads to transfer function representation
If u is a step function (constant water flowing in)
Summarizing example of all things taught (so far)
udt
dyu
dt
dyA
AreaA
y (m)
u (m3/sec)
s
1)(sU )(sY
ttys
sYs
sU )()()(2
11
t (sec)
y (m)
2
2
)0)0((assuming y
D i d f l l 2 ?
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Desired performance: water levely to stay aty=2 m?
Open loop control: knowing that water levely=2 at t=2,just stop water flow at t=2
Closed loop control: measurey, compare with the desiredperformance to determine the input u(t)=r(t)-y(t)
t (sec)
y (m)
2
2
t (sec)
u (m3/sec)
2
1
s
1)(sU)(sY)(sR
+_
s
sR2
)(Desired performance:water level =2 Feedback actualy to compare
with desired performance
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Closed loop transfer function using Transformation Rule #6:
With desired performance r=2,
Moreover, input u can be determine as
tetu
ssYsRsU
2)(
)1(
2)()()(
)1(2)(
)1(
2)(
2)(
tety
sssY
ssR
)()(
)()()( sRs
sYsRGH
GsY
1
1
1
t (sec)
u (m3/sec)
4
2
t (sec)
y (m)
4
2
Indeed,y(t)
stayed at y=2
I E Modeling of DC Motor
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I.E. Modeling of DC Motor
Mathematical model: Newtons Law; Kirchoffs Laws; and
laws describing the mechanical/electrical interactions:
- Magnetic field induced by current if:
if
Electromechanical System interacting mechanical andelectrical components
if
if
i i i i fi ld
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- Induced voltage on wire moving with velocity v in :
e vPolarity of e induced current to yield
force/torque against velocityv
Induced force F
ia
Wire movingwith velocity v
+
_
- Force on wire carrying currentia
in magnetic field :
F ia
Induced voltage e
E l 2 5 DC M t~
Vf
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Field circuit:- current ifto generate magnetic field
Armature circuit:- current ia flowing inside magnetic field
- TorqueTm on rotor:
- Induced back voltage e on armature circuit:
Example 2.5: DC Motor
fiKKve 32
ffiK
affam iiKKiKT 11
Rf
Lf
if
Magnetic field
Field
circuitArmature
circuit
+
-e
K1,Kf,K2,K3 someproportional
constants
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- Rotor equation (with friction term):
- With ia
constant: Torque
)()()( tTtbtJ L
- Possible existence of disturbance torque total torque on rotor:
dmL TTT
- Field circuit equation:
)()()( tVtiRtidt
dL fffff
afm iKKK 1
fmaffm
iKiiKKT 1
Case #1: Field current if
-controlled motor
- Armature current ia kept constant (can use currentgenerator instead of armature circuit)
Mechanical Component:i ( i h f i i )
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Rotor equation (with friction term)
Or, inserting the variable (t) asangular speed
Electrical component:
Field circuit equation
)()()( tTtbtJ L
Possible presence of
disturbance torquedmL TTT
)()()( tVtiRtidt
dL fffff
Torque applied on rotordue to current if :
Coupling electrical and
mechanical components
fmm iKT
)()( )()()( tt
tTtbtJL
- Block diagram transform
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- With
)()(
1)())((
)( sTbJss
sVbJsRsLs
Ks dfff
m
,,and,0 ff
fLd R
L
b
JT
)())((
)( sVsss
bRK
s fLf
f
m
11
fL - Typically,
)()(
)( sVss
bRKs f
L
f
m
1
Electric response
much faster thanmechanical response
3rdorder differential equationrelating input and outputsimplified to 2ndorder
- Block diagrams
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** After neglecting the electrical response by omitting f
)(sPosition
1
fR
Electrical component Mechanical component
Coupling Presence of disturbance torque
- Block diagrams
** Full dynamics
Figure 2.19
Electrical component Mechanical component
Coupling Presence of disturbance torque
)(s
Position
Case #2: Armature current (ia ) controlled motor
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- Rotor equation (with friction term):
- With if constant:
- Armature circuit equation:
Case #2: Armature current (ia ) controlled motor
)()()( tTtbtJ L
- Possible existence of disturbance torque total torque on rotor:
dmL TTT
)()()()( tKtVtiRti
dt
dL baaaaa
Back emf term on armaturecircuit introduces another
source of coupling between
electrical and mechanical
components
ffm iKKK 1
amaffm iKiiKKT 1
- Field current if kept constant constant magnetic field (or use permanent magnet)
fiKe 3
fb iKK 3
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- Block diagram: full dynamics
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- After transformation rules
)(
))((
)()(
))(()( sT
KKbJsRsLs
RsLsV
KKbJsRsLs
Ks
dbmaa
aa
abmaa
m
)(
))((
)( sV
bR
KKsss
bRKs a
a
bmLa
a
m
11
- Typically, ,,,a
aaLd R
Lb
JT withand0
Figure 2.20
)(sVa)(s
Position
L - Typically,Again, 3rdorder differentialequation simplified to 2ndorder
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s1)(s
Speed
)(s
Position
aL Typically,
)()(
)()(
)(
)( sVss
KKbRK
sV
bR
KKss
bRK
s abma
m
a
a
bmL
a
m
11 1
)( bma
a
KKbR
JR
1where
- Block diagram: neglecting electrical response (omitting )a
equation simplified to 2 order
Electrical
Coupling: current-
induced torque Mechanical
Coupling: angular speed-induced back emf
Presence of disturbance torque
Design Example: Disk Drive Read System y(t)= t)
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- Output = from DC Motor (Flexture flexibility ignored)
- Armature-current controlled motor with zero back emf:
- Controller = gain amplification transfer function=Ka- Perfect measurement sensor transfer functionH(s)=1
Kb=0
Figure 2.76 (a)
- Disturbance torque=0
Head
Flexture
Rotor
Magnets
Arm
2.75
r(t)
Armature-current controlledDC Motor/Read Arm:G(s)
Figure 2.76 (b)
with added details
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Va(s)
- Parameters:
)20)(1000(
5000
))(()(
sssbJsRsLs
KsG
aa
m
Kb=0
Ka
H(s)=1
0R(s)
- Armature-current controlled DC Motor Transfer Functionwith Kb=0:
a
Y(s)
= s)
-G(s) in terms of :L
aL and
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Figure 2.742.78
)105.0)(1001.0(
25.0
)1)(1(
)(
ssssss
bRK
sGLa
a
m
msbJmsRL
La
aa 50,1
aL - Reduced model with
)().(
.)(
20
5
1050
250
ssss
sG
- Closed loop transfer function usingreduced model:
a
a
Kss
K
sR
sY
520
52 )(
)(
40aK
- Closed loop with full model yield nearly the same response
(Full model)
% textbook design example (Reduced
model)
Ka=40;
- Matlab program to generate plot:
The case in Fig 2 78
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Ka=40;
num=5*Ka;den=[1 20 5*Ka];
t=[0:0.002:0.7];
y=step(num,den,t);
plot(t,y)
grid
xlabel('Time (s)')
ylabel('y(t) (rad)')
Title('Figure 2.74')
hold
% Do case for ka=20
Ka=20;
num=5*Ka;
den=[1 20 5*Ka];
t=[0:0.002:0.7];
y=step(num,den,t);
plot(t,y,'--')
% Do case for ka=60
Ka=60;
num=5*Ka;
den=[1 20 5*Ka];t=[0:0.002:0.7];
y=step(num,den,t);
plot(t,y,'-.')
% Label the cases
gtext('Ka=20')
gtext('Ka=40')
gtext('Ka=60')
Exercise: generating plot for full model?
The case in Fig. 2.78
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Third Reading Assignment:
PP. 256-276, PP. 295-301 of Textbook