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MECH 529
Dynamic System Modeling
Presentation Part 3
Dr. Clarence W. de Silva, P.Eng.
Professor of Mechanical Engineering
The University of British Columbia
e-mail: [email protected]
http://www.sites.mech.ubc.ca/~ial
C.W. de Silva
Graphical Interpretation
Analytical Linearization (state-
space and input-output models)
Linearization using Experimental
Data
Illustrative Examples
Plan
Nonlinear Analytical Models:
State-space Models
Input-Output Models
Model linearization
Real systems are nonlinear
They are represented by nonlinear analytical models
Common techniques (e.g., response analysis, frequency
domain analysis, eigenvalue problem analysis, simulation,
control) use linear models
Method: Linearize each nonlinear term (1st order Taylor
series approximation)
Note: A nonlinear term may be a function of more than one
independent variable.
Graphical Illustration
Nonlinear function f(x) of a single independent variable x
Linearization About an Operating Point Steps:
1. Express each nonlinear term at an operating point (or, reference condition)
2. Select (small) increments of the independent variables from the operating
condition
3. Express Taylor series expansion of nonlinear term about operating point
4. Retain O(1) terms
Note: We need the first-derivatives (local slopes) at operating condition
Example: A nonlinear function f(x,y) of two independent variables x and y
( , ) ( , )( , ) ( , ) with ,
or
( , ) ( , )ˆ ˆ ˆ ˆ( , ) ( , )
o o o oo o o o
f x y f x yf x y f x y x y x x x y y y
x y
f x y f x yf x x y y f x y x y
x y
Note: ( ) or ( )o denotes operating condition
ˆ( ) or δ( ) denotes a small increment about that condition.
Nonlinear State-Space Models
General nonlinear, time-variant, nth-order system
n first-order nonlinear differential equations (generally coupled
and time variant):
trrrqqqf
dt
dqmn ,,...,,,,...,, 21211
1
dq
dtf q q q r r r t
dq
dtf q q q r r r t
n m
nn n m
22 1 2 1 2
1 2 1 2
, , . . . , , , , . . . , ,
, , . . . , , , , . . . , ,
State Vector q Tnqqq ,...,, 21
Input Vector r r r rm
T
1 2, ,. . .,
Vector Notation: ( , , )tq f q r
State Model Linearization Equilibrium states 0q
Note: At equilibrium (operating point) response (sate vector)
remains steady
Equilibrium states q : Solving n equations f(q, r, t) = 0 for steady
input r
Slightly Excite from Equilibrium State:
If response builds up and deviates further unstable
equilibrium state
If returns to original operating point stable equilibrium
state
If it remains at new state neutral equilibrium state
(a) (b) (c)
(a) Stable (b) Neutral (c) Unstable
State Model Linearization (Cont’d)
Linearize about equilibrium point ( , )q r for small variations q and r:
( , , ) ( , , ) t t
f f
q q r q q r rq r
or x Ax Bu
Note: Higher-order terms are negligible for small q and r
Incremental state vector q x x x xn
T
1 2, ,. . .,
Incremental input vector r = u 1 2, ,. . .,T
mu u u
Linear system matrix A(t) =( , , )t
fq r
q
Input gain (distribution) matrix B(t) = ( , , ) t
fq r
r
Note 1: For a stationary (i.e., constant-parameter) system, for time
period of interest, A and B are constant
Note 2: A linear model is acceptable when the resulting error is small
(typically when response variations about operating point are small
compared to operating range).
Linearization About a Steady Operating Point
Results
( )( ) ( )
df xf f x x f x x
dx or
( )ˆ ˆdf x
f xdx
Note: ( ) = operating point; ˆ( ) = increment from operating point
Linear Term: f ax ˆ ˆf ax
Also, ˆ
ˆdx
x xdt
;
2
2
ˆˆ
d xx x
dt ; etc.
Linearization Steps
1.Set all ()d
dt terms to zero, and solve the resulting algebraic
equations Steady operating conditions
2.Determine derivatives of all nonlinear terms wrt
independent variables, at operating point
3.For each term in original nonlinear equations, write
increment from operating point
Practical Considerations
Dev
ice
Ou
tpu
t
Dev
ice
Ou
tpu
t
Dev
ice
Ou
tpu
t
Saturation
Level
Device Input
Linear
Range
Saturation
Level
Device
Input
d -d Device Input
(a) (b) (c)
(a) Saturation; (b) Dead band (backlash); (c) Hysteresis
(d) Coulomb friction
Device Input
(Applied Force)
Frictional
(Resistance)
Force
Dynamic
Frictional Force
(d)
What are practical examples of
these nonlinearities?
Illustrative Examples
Example 1
Induction Motor-Pump Combination in Spray Painting System of Automobile Assembly Plant
Light gear with efficiency and speed ratio 1:r; flexible shaft torsional stiffness kp
Moments inertia of motor rotor and pump impeller: Jm and Jp
Dissipation in motor and bearings: viscous damping of damping constant bm
Pump load (paint load plus dissipation): viscous damping constant bp
Magnetic torque Tm of induction motor 0 0 0
2 20
( )
( )
mm
m
T qT
q
m = motor speed; T0 depends quadratically on phase voltage supplied to motor;
0 line frequency of the AC supply; q > 1.0
Related Questions
(a) Comment about the model accuracy
(b) State variables: motor speed m , pump-shaft torque Tp, pump speed p . Derive three
state equations nonlinear state model. What are the inputs to the system?
(c) What do parameters 0 and T0 represent with regard to motor behavior? Determine
m
m
T
,
0
mT
T
and
0
mT
. Verify that first is -ve and other two are +ve. Note: Under normal
operating conditions, we have 00 m (i.e., slip)
(d) Consider steady-state operating point: steady motor speed m . Obtain expressions for
p , pT , and 0T (at operating point), in terms of m and 0
(e) Let m
m
Tb
, 1
0
mT
T
, and 2
0
mT
at operating point.
Voltage control: Vary T0; Frequency control: Vary 0 .
Linearize state model about operating point and express it in terms of incremental
variables ˆm , ˆ
pT , ˆp , 0T , 0 . Let (incremental) output variables be incremental pump
speed ˆp and incremental angle of twist of the pump shaft. Obtain linear state space
model A, B, C and D.
(f) For frequency control ( 0T = 0) obtain input-output differential equation relating ˆp and
0 . Show that if 0 is suddenly changed by step 0ˆ then
3
3
ˆpd
dt
will instantaneously
change by step 2
0ˆ
p
m p
rk
J J
, but no lower derivatives of ˆ
p will change.
Solution (a) Backlash and inertia of gear transmission neglected (not accurate in general). Gear efficiency
is assumed constant but usually varies with speed
There is some flexibility in the shaft (coupling) between gear and drive motor
Energy dissipation (in pump load, bearings) is lumped into a single linear viscous-damping
element (In practice nonlinear and distributed)
(b) Motor speed m
m
d
dt
; Load (pump) speed
p
p
d
dt
; m = motor rotation;
p = pump rotation
Power = torque speed; Tg = reaction torque on motor shaft from gear; r = gear ratio mr =
output speed of gear Gear efficiency Output Power
Input Power
p m
g m
T r
T
g p
rT T
(i)
Constitutive Equations (State-Space Shell):
Newton’s 2nd
law for motor: m m m g m mJ T T b (ii)’
Newton’s 2nd
law for pump: p p p p pJ T b (iii)
Hooke’s law for flexible shaft: ( )p p m pT k r (iv)
State Equations: State vector = [ ]Tm p pT
Subst. (i) into (ii)’: m m m m m p
rJ T b T
(ii)
Input 0 0 0
2 20
( )
( )
mm
m
T qT
q
(viii) Nonlinear model; Note: Strictly, two inputs: 0 (= speed of
rotating magnetic field line frequency); T0 (phase voltage)2
Induction Motor Characteristic Curve
Solution (Cont’d)
(c) From (viii): When 0m Tm = T0 T0 = starting torque of motor
From (viii): When Tm = 0 0m 0 = no-load speed = synchronous speed
Note: Under no-load conditions, no slip
Actual speed of induction motor = speed 0 of rotating magnetic field
Differentiate (viii) wrt 0T , 0 , and m : 0 012 2
0 0
( )
( )
m m
m
T q
T q
(at “O”) (ix)
2 2 2 20 0 0 0 0 0 0 0 0
22 2 2 2 2 20 0 0
[( )(2 ) ( )2 ] [( ) ( 1) ]
( ) ( )
m m m m m m
m m
T T q q q T q q
q q
(at “O”) (x)
2 2 2 20 0 0 0 0 0 0 0
2 2 2 2 2 20 0
[( )( 1) ( )( 2 )] [( 1) ( ) ]
( ) ( )
m m m m m
m m m
T T q q T q qb
q q
(at “O”) (xi)
Note: 1 > 0 ; 2 > 0 ; b > 0
(d) Steady-state operating point rates of changes of state variables = 0
Set 0m p pT in (ii) – (iv) 0 m m m p
rT b T
; 0 ( )p m pk r ; 0 p p pT b
p mr (xii); p p mT b r (xiii)
2 0 0 0
2 20
( ) (from (viii))
( )
mm m m p m
m
T qT b r b
q
2 2 20
0
0 0
( )( )
( )
m m p m
m
b r b qT
q
(xiv)
Note: For the steady operating point, steady inputs 0 0 and T are known. We have 3
equations (xii), (xiii), and (xiv) for the 3 unknowns , , and .p m pT
Solution (Cont’d) (e) Take increments of state equations (v) – (vii):
1 0 2 0ˆ ˆˆ ˆ ˆ ˆ
m m m m p m
rJ b T b T
(xv); ˆ ˆ ˆ( )p p m pT k r (xvi); ˆˆ ˆ
p p p p pJ T b (xvii)
Note: 0 0 1 0 2 0
0 0
ˆ ˆ ˆˆ ˆ ˆ ˆm m mm m m
m
T T TT T b T
T
Linearized State Equations: (xv)-(xvii)
State vector ˆˆ ˆT
m p pT
x ; Input vector 0 0ˆ ˆ
T
T
u ; Output vector ˆˆT
p p pT k
y
A =
( ) ( ) 0
0
0 1
m m m
p p
p p p
b b J r J
rk k
J b J
; B =
1 2
0 0
0 0
m mJ J
; C = 0 0 1
0 1 0pk
; D = 0
(f) For frequency control, 0ˆ 0T
Substitute (xvi) into (xv) to eliminate ˆm ; Substitute (xvii) into the result to eliminate
ˆpT
Input-Output Differential Equation: 3 2 2 2
2 03 2
ˆ ˆ ˆˆ ˆ[ ( )] [ ( ) ( )] ( )
p p p p p
m p m p p m p m p m p m p p
d d r J d r bJ J J b J b b k J b b b k b b rk
dtdt dt
(xix)
3rd
order differential equation; system is 3rd
order; state-space model is also 3rd
order.
Solution (Cont’d)
Observation From (xix):
When 0 is changed by “finite” step 0ˆ RHS of (xix) will be finite
If as a result,
2
2
ˆpd
dt
or lower derivatives change by a finite step
3
3
ˆpd
dt
should
change by an infinite value (Note: derivative of a step = impulse)
LHS will be infinite
Contradicts RHS of (xix) is finite
2
2
ˆpd
dt
,
ˆpd
dt
, and ˆ
p will not change instantaneously
Note: These three may be considered state variables (which, by definition, can’t
change instantaneously, without infinite input)
Only
3
3
ˆpd
dt
will change instantaneously by a finite value due to finite step change
of 0
From (xix): Resulting change of
3
3
ˆpd
dt
is
2
0ˆ
p
m p
rk
J J
• Mechanical damping constant bm comes from bearing
friction and other mechanical sources
• Electrical damping constant b comes from the
electromagnetic interactions in the motor
• The two occur together (and should be treated
together, in model analysis, simulation, design,
control, etc.).
E.g., whether the response is underdamped or
overdamped depends on the sum bm +b and not the
individual components
Electro-mechanical coupling
Electro-Mechanical Coupling
1. Electro-mechanical coupling in damping
2. Mechanical damping and electrical damping should be treated
together. System behavior depends on the combined effect a
case for integrated analysis, simulation, design, control, etc.)
3. System nonlinearity can come from nonlinear coupling of
inputs and state variables
4. A state variable cannot change instantaneously
5. When an input changes instantaneously, the highest derivative
of the response changes with it (instantaneously); and the
lower derivatives don’t change.
Note: These lower derivatives may be taken as state variables.
Main Concepts Illustrated in Example 1
Example 2
A Wood Cutting Machine
Jm = axial moment of inertia of motor rotor
bm = equivalent viscous damping constant of motor bearings
k = torsional stiffness of flexible shaft; Jc = axial moment of inertia of cutter blade
bc = equivalent viscous damping constant of cutter bearings; Tm = magnetic torque of motor
m = motor speed; Tk = torque transmitted through flexible shaft; c = cutter speed
TL = load torque on cutter from workpiece (wood).
Note: In comparison with flexible shaft, coupling unit is rigid; shafts are light
Cutting load L c cT c
Constant parameter c depends on depth of cut and material properties of workpiece
Related Questions
(a) Use Tm = input; TL = output; T
m k cT = state vector. Develop a
complete (nonlinear) state model. What is the system (model) order?
(b) From state model, obtain a single input-output differential equation, with
Tm = input and c = output.
(c) Steady operating conditions: Input and state variables are all constant.
Express operating point values m , kT , c , and LT in terms of mT and
model parameters only. Consider cases: mT > 0 and mT < 0.
(d) Consider incremental change ˆmT in motor torque and corresponding
changes ˆm , ˆ
kT , ˆc , ˆ
LT . Determine a linear state model (A, B, C, D) using
x = ˆˆ ˆ[ ]T
m k cT as state vector, u = ˆmT
as input and y = ˆLT
as output.
(e) In incremental model, if twist angle of flexible shaft ( m c ) is output
determine a suitable state model. What is the system order then?
(f) In the incremental model, if c is the output, how should the state model
in Part (a) be modified? Then what is the system order?
Solution
System Free-Body Diagram
(a) Constitutive equations for the three elements State Equations:
m m m m m kJ T b T (i); ( )k m cT k (ii); c c k c c c cJ T b c (iii)
State vector = T
m k cT ; Input vector = mT ; Output vector = LT = [ ]c cc
3rd order system (3 state equations)
(b) Note: sgnc c c 2( ) ( sgn ) 2 sgn 2c c c c c c c c c
d d
dt dt (iv)
22
2( ) 2 2 sgn( )c c c c c c
d
dt (v) Note: Since sgn( ) 1 for 0; 1 for 0;c c c it is a constant
Substitute (ii) in (i), to eliminate m : [ ] [ ]m k c m m k c kJ T k T b T k T
Substitute (iii) in this equation, to eliminate Tk (use properties (iv) and (v)):
2k c c c c c cT J b c ; 22 2 sgn( )k c c c c c c c cT J b c c
22 2 sgn( ) 2m mc c c c c c c c m c m c c c c c c m c c c c c c c
J bJ b c c J T J b c b J b c
k k
3 2 22
3 2 2( ) 2 ( ) 2 2 sgn( )( ) ( )c c c c c c
m c m c c m m c m c m c m c m c m c c c c m
d d d d d dJ J J b J b cJ J k J k b b b c cJ k b b kc kT
dt dt dtdt dt dt
Jc
Tk c
L c cT c
c cb
Solution (Cont’d) (c) Operating Point: 0 m m m kT b T (vi); 0 ( )m ck (vii); 0 k c c c cT b c (viii)
m c ; ( )k c c cT b c ; ( )m m c c cT b b c Signs of m , c , and mT are identical
Case 1: mT > 0 c > 0
Eliminate kT using (vi) and (viii): 20 m m m c c cT b b c
2 ( ) 0c m c c mc b b T
2( ) ( ) 4
2
m c m c mc
b b b b cT
c
Positive root:
2( ) 4 ( )
2 2
m c m m cc m
b b cT b b
c c
From (vi):
2( ) 4 ( )
2 2
m c m m ck m m
b b cT b bT T b
c c
From (viii):
2( ) 4 ( )( )
2 2
m c m m cL c c m m c
b b cT b bT c T b b
c c
Case 2: mT < 0 c < 0 (vi)-(viii): 20 m m c c c cT b b c
2 ( ) 0c m c c mc b b T
2( ) ( ) 4
2
m c m c mc
b b b b cT
c
; Note: mT < 0. Use the negative root
2( ) 4( )
2 2
m c mm cc
b b cTb b
c c
. The rest follows as in Case 1.
Note: In view of the rotational symmetry of the problem, Case 2 can be derived from the
results of Case 1 (i.e., in the expression for c in case 1, change the sign of mT and then change
the sign of the entire result.
Solution (Cont’d) (d) Linearize:
ˆm mJ =
ˆmT ˆˆ
m m kb T
ˆkT = ˆ( mk ˆ )c
ˆc cJ = ˆ ˆ
k c cT b ˆ2 c cc Note: 2( ) ( sgn ) 2 sgn 2c c c c c c c
c c
d d
d d
with x = ˆ[ m ˆ ˆ ]Tk cT and u = ˆmT
A =
1 0
0
0 1 ( 2 )
m m m
c c c c
b J J
k k
J b c J
; B =
1
0
0
mJ
and y = ˆLT
= ˆ2 c cc C = 0 0 2 cc and D = [0]
(e) k
m c
Ty
k
Exactly the same state equations plus this new output equation System order = 3
(f) cy
Here, c cannot be expressed as an algebraic equation of the 3 previous state variables.
A new state variable c has to be defined (Note: Not a natural state variable for Jc; Presence of
rigid-body mode) Additional state equation: cc
d
dt
System order = 4
1. Analysis of nonlinearities that involve the signum
function (sgn(.)).
2. Proper choice of state variables.
3. The natural state variable for a rigid-body mode is
velocity (not position). When this is violated,
inconsistency in system order may result.
Main Concepts Illustrated in Example 2
Example 3
A Simplified Model of an Elevator
J = moment of inertia of cable pulley; r = radius of pulley
K = stiffness of cable; m = mass of car and occupants
Related Questions (a) Which system parameters are variable? Explain.
(b) Damping torque ( )dT at pulley bearing is nonlinear in angular speed of pulley.
State vector [ ]Tf vx
f = tension force in cable; v = velocity of car (positive upwards)
Input vector [ ]TmTu
Tm = torque applied by motor to pulley (positive in direction shown in Figure)
Output vector y = [v]
Obtain a complete, nonlinear, state-space model for the system.
(c) With Tm = input; v = output, convert state model into a nonlinear I/O differential
equation model. What is the system order?
(d) Obtain an equation whose solution steady-state operating speed v of elevator car.
(e) Lienarize the I/O differential-equation model for small changes ˆmT of input and v of
output, about an operating point.
Note: mT = steady-state operating-point torque of motor (assumed known).
Hint: Denote ddT
d by b().
(f) Linearize the state model and determine model matrices A, B, C, and D.
Obtain linear I/O differential equation from this state-space model
Verify that it is identical to the result in Part (e)
Solution
(a) r is a variable due to winding/unwinding of cable around pulley; m varies with
occupancy.
(b) State Equations: Apply Newton’s 2nd law to two inertia elements and Hooke’s law
to spring element: ( )m dJ T rf T (i); ( )f k r v (ii); mv f mg (iii)
Output y =v
(c) Eliminate f by substituting (iii) into (i) and (ii):
( ) ( )m dJ T r m v g T (iv); ( )mv k r v (v)
From (v): 1
( )m
v vr k
1
( )m
v vr k
Substitute these into (iv), to eliminate : 1
( ) ( ) ( ( ))m d
J m mv v T r m v g T v v
r k r k (vi)
3rd order model (Highest derivative in (vi) is 3rd order).
(d) At Steady State: 0v 0 and 0v v as well. Substitute into (vi)
Steady-state Equation: ( ) 0m d
vT r mg T
r Solve steady-state operating speed v
Note 1: mT = steady-state value of input Tm.
Note 2: Same result is obtained from state equations (i)-(iii) under steady-state
conditions: 0 ( ); 0 ( ); 0m dT rf T k r v f mg
Convert into a single equation by eliminating and f
Solution (Cont’d)
(e)Linearize (vi): ˆ ˆˆ ˆ ˆ( ) ( )m
J mv v T r mv b
r k (vii) where
( )( ) mdT
bd
From (v): ˆˆ ˆ( )mv k r v (viii)
Substitute (viii) into (vii), to eliminate : 1ˆˆ ˆ ˆ ˆ ˆ( ) ( ) ( )m
J m mv v T r mv b v v
r k r k
( ) ( ) ˆˆ ˆ ˆ ˆ( ) m
J m b m J bv v rm v v T
r k rk r r
(f) Linearize (i)-(iii): ˆˆˆ ˆ( )mJ T rf b (ix); ˆ ˆ ˆ( )f k r v (x);
ˆˆmv f (xi)
Output ˆy v ; Input ˆmu T ; State vector x = ˆˆ ˆ[ ]Tf v
A =
( )0
0
10 0
b r
J J
rk k
m
; B=
1
0
0
J
; C = 0 0 1 ; D = 0
Substitute (xi) into (ix) and (x), to eliminate f : ˆˆ ˆˆ ( )mJ T rmv b ; ˆˆ ˆ( )mv k r v
Eliminate same result as before, for the input/output equation
1. Linearize a state-space model and then convert it into
a linear input-output model
2. Convert a nonlinear state-space model into a
(nonlinear) input-output model and linearize
3. Results of 1 and 2 are the same (as long as same
forms of nonlinearity are used and the same
linearization procedures are used)
Main Concepts Illustrated in Example 3
Example 4
Coordinate System for the Spacecraft Problem.
Horizontal on Earth’s
Surface
Vertical
Direction
Y
X
f(t)
Spacecraft
Mass m
(X, Y)
Example (Cont’d) A rocket-propelled spacecraft of mass m is fired vertically up (in the Y-
direction) from earth’s surface (see Figure).
Vertical distance of spacecraft centroid (altitude) = y
(measured from earth’s surface)
Upward thrust force of rocket = f(t)
Gravitational pull on spacecraft =
2R
mgR y
g = acceleration due to gravity at earth’s surface
R = “average” radius of earth ( 6370 km).
Magnitude of aerodynamic drag force resisting motion of spacecraft
2 y rky e
Note: Exponential term represents loss of air density at higher elevations,
k and r are positive and constant parameters; dy
ydt
.
Related Questions
(a) Derive the input-output differential equation for the system.
f = input; y = output.
(b) Spacecraft accelerates to altitude yo and maintains a constant speed vo
(still moving in the same vertical (Y) direction)
Determine an expression for rocket force needed for this
constant-speed motion.
Express answer using yo, vo, time t, and system parameters m, g, R, r, k.
Show that this force decreases as the spacecraft ascends.
(c) Linearize the I-O model (Part (a)) about steady operating condition
(part (b)), for small variations y and y in position and speed
of spacecraft, due to a force disturbance ˆ ( )f t .
Related Questions (Cont’d)
(d) Uing y and y as state variables and y as the output, derive a complete
(nonlinear) state-space model for vertical dynamics of the spacecraft.
(e) Linearize state-space model (d) about steady conditions (b) for small
variations y and y in position and speed of spacecraft, due to force
disturbance ˆ ( )f t .
(f) From linear state model (e) derive the linear input-output model and
show that the result is identical to what you obtained in Part (c).
Solution
(a) Newton’s 2nd law in Y-direction:
2
( )y rRmy mg k y ye f t
R y
(i)
Note: Gravity ~ nonlinear spring y can serve as its state variable
(b) At constant speed vo: oy v (ii); 0o
dy v
dt (iii)
Integrate (ii) and use IC oy y at t = 0.
Position under steady conditions: o oy v t y (iv)
Substitute (ii)-(iv) in (i):
( )
20 ( )
1
o ov t y ro o s
o o
mgk v v e f t
v t y
R
where ( )sf t = force of rocket at constant speed vo.
vo > 0
( )2
2( )
1
o ov t y rs o
o o
mgf t kv e
v t y
R
Note: This expression decreases as t increases, reaching zero in the limit
(1st term: quadratically; 2nd
term: exponentially).
Solution (Cont’d) (c) Derivatives needed for linearization (O(1) Taylor series terms):
2 3
1 2 1; ; 2
1 1
y r y r y r y ry ydy ye e y ye y e
dy R y r yy y
R R
Linearized I-O equation: 3
2 1 ˆˆ ˆ ˆ ˆ2 ( )
1
y r y rmg kmy y y ye y k y e y f t
R ry
R
(v)
(Note: Steady terms cancel out because they satisfy the steady operating condition)
Note: y is given by (iv) and is time-varying.
0oy y v Steady state : 2
3
2 1 ˆˆ ˆ ˆ ˆ2 ( )
1
y r y ro o
mg kmy y v e y kv e y f t
R ry
R
(v)
Note: Unstable system.
(d) State vector 1 2
T Tx x y y x
(i) State equations: 1 2x x (vi)
1
2 2 22
1
1( )
1
x rg kx x x e f t
m mx
R
(vii)
1 1 y xOutput equation :
Solution (Cont’d) (e) To linearize, use derivatives (local lopes) as before:
1 1 1 12 2
2 2 2 2 22 31 1 21 1
1 2 1; 2
1 1
x r x r x r x rx xdx x e e x x e x e
dx R x r xx x
R R
at steady operating (constant speed) conditions: 1 o ox v t y and 2 0ox v
Linearized state-space model: 1 2ˆ ˆx x (viii)
1 1( )2
2 1 1 23
1
2 1 ˆˆ ˆ ˆ ˆ2 ( )
1
x r x ro o
g k kx x v e x v e x f t
mr m mxR
R
(ix)
Output equation: 1 1ˆ ˆy x (x)
(f) Substitute (viii) in (ix) 1 1( )2
1 1 1 13
1
2 1 ˆˆ ˆ ˆ ˆ2 ( )
1
x r x ro o
mg k kx x v e x v e x f t
mr m mxR
R
(Identical to (v), since 1ˆx y )
1. Handling of gravitational pull as a nonlinear
spring two state variables: velocity for inertia and
position (which gives spring force) for nonlinear
spring. Compare with simple pendulum (needs two
state variables: velocity and position corresponding to
the nonlinear spring of gravitational pull) and free-
falling mass under gravity (needs only one state
variable: velocity; gravity appears as a constant
external force, not a nonlinear spring).
2. Time-varying operating state.
3. Linearization concepts, as illustrated in Example 3
Main Concepts Illustrated in Example 4
Graphical Solution to
Linearization:
Use of Experimental Operating
Curves
1. Linearization is done by finding the local slopes
(derivatives with respect to the independent
variables) of experimental characteristic curves
2. As a matter of experimental feasibility, the curves are
determined by varying one variable at a time
(keeping the other variables constant at some
operating condition)
3. On a two-axis coordinate frame, only two such
variables can be represented
Graphical Linearization Using Experimental Characteristic Curves
1. Keep one variable constant at some operating condition and
vary the other variable in steps
2. Plot the data curve
3. Repeat steps 1 and 2 by changing the operating condition of
the first variable by a small step. Obtain a series of
characteristic curves in this manner
4. Linearization: For a specific operating condition, find the
slope of the curve passing through it (i.e., with the first
variable kept constant); Also find the vertical increment of two
adjacent curves enclosing the operating point (i.e., with the
second variable is kept constant)
5. Slopes with respect to the two independent variables are
obtained from the two values in Step 4.
Graphical Linearization (Two-variable Case)
Graphical Linearization (Two-variable Case)
Response
y
Variable
y
x2+x2
b x2
1 O
Slope at O wrt x1: 2
1 constantx
yb
x
Slope at O wrt x2: 1
2 2 constantx
y yk
x x
2 1
1 2 1 2
1 2
1 2
ˆ ˆ ˆor
x x
y yy x x b x k x
x x
y bx kx
Linearized model of motor.
x1
1 2
1 2
1 2
( , )y y x x
y yy x x
x x
Note: Partial derivatives
(Only one variable changes)
DC Motor Example (Control Circuit and Mechanical Loading)
(a) Equivalent circuit of a dc motor (separately excited); (b) Armature mechanical loading diagram
DC Motor Operating Curves
Torque-Speed Curves: (a) Shunt-Wound; (b)Series-Wound; (c) Compound-Wound; (d) General Case
Note: In each curve, excitation voltage vc is maintained constant
Linear Model for Motor Control
Figure (d): One curve is at control voltage vc and the at vc+vc.
Tangent drawn at a selected point (operating point O).
Magnitude b of slope (negative)
Damping Constant (including electrical and mechanical damping effects): constantc
m
m v
Tb
Note: Mechanical damping included in this b depends on mechanical damping present in the
measured torque during test (primarily bearing friction)
Draw a vertical line through operating point O Torque intercept mT between two curves
Note: Vertical line is a constant speed line
Voltage Gain: constantm
m mv
c c
T Tk
v v
ˆ ˆ ˆ or
c m
m mm m c m v c m m v c
m cv
T TT v b k v T b k v
v
Linearized model of motor.
Note: Torque needed to drive rotor inertia is not included in this equation
(because steady-state curves are used in determining parameters)
Inertia term is explicitly present in Mechanical Equation of motor rotor (See Figure (b)):
mm m L
dJ T T
dt
Linearized (incremental) model:
ˆ ˆ ˆ ˆˆ ˆmm m L m v c L
dJ T T b k v T
dt
Note: Tm and b include overall damping (both electrical and mechanical) of motor.
1. Keep one variable constant at some operating condition and
vary the other variable in steps
2. Repeat for an increment of the variable that was kept constant
3. Plot the successive operating curves obtained in this manner
4. Determine the local slopes at an operating point
5. The rest is the same as with analytical linearization
Linearization Steps
Steady-state
operating curves
Example ,T Q
tJ
tbk
kT
p
pJ
pb
Q
Fuel Rate
Gas Turbine Centrifugal Pump Water In
Water Out
0
20
60
80
100
120
140
160
200 400 600 800
40
456789
10
(rpm)
(N.m)T
(gal/s)Q
Speed
TorqueFuel Input
Rate
(Gas turbine operating a water pump)
2
g
Fuel input rate= (gal/s)
Moment of inertia of rotor= 0.1 k ×m
Mechanical damping constant : = 0.5 N×m/rad/s
Torional stiffness 20.0 N×m/rad
Torque in the shaft =
t
t
k
Q
J
b
k
T
For Gas Turbine :
For Shaft :
2
gMoment of inertia of rotor 0.05 k ×m
may include added mass due to fluid load
Mechanical damping constant of pump 3.0 N×m/rad/s
linear viscous;may include fluid load
p
t
J
b
For Centrifugal Pump :
Example (Cont’d) Related Questions:
(a) Determine a complete state-space model for the system.
Use: ( , )T Q = generated torque of the turbine;
[ , ,k pT ] = state variables; Q = input; p = output.
(b) Linearize the model about an operating point.
Incremented variables: , , , andk k p pT Q q
Operating point: 8 gal/sQ and 400 as the.
,T Q
tJ
tb
mT
kT
k
p
kT
pT p
pJ
Gas Turbine Rotor
Shaft
Pump Rotor
p pb
Example (Cont’d) State-space Model (Linearized)
Incremental state vector, [ , , ] ; [ ]T
k p q x u
State-space model,
x Ax Bu
y Cx Du
where,
/( ) / 1/ 0
0 ; 0
0 1/ / 0
(0 0 1); (0)
q tt t t
p p p
k Jb b J J
k k
J b J
A B
C D
Example (Cont’d) Local Slopes (from Experimental Curves) b = slope at operating point on the 8 gal/sQ curve
(130 80)(N×m) 501.326 N×m/rad/s
2 2(560 200) (rad/s) 360
60 60
qk = increment ofT over the change in Q from7 to 9 gal/s withkept constant at 400
rpm(110 92)(N×m)
9.0 N×m/gal/s(9 7)(gal/s)
With the given parameter values,
-1
2 -1
2 1
-1
3
2
1.326 0.518.26 s
0.1
110.0 (kg.m )
1 120.0 (kg.m )
0.05
3.060.0 s
0.05
9.0 (N.m/gal/s)90.0 gal/s
0.1 (kg.m )
t
t
t
p
p
p
q
t
b b
J
J
J
b
J
k
J
0
20
60
80
100
120
140
160
200 400 600 800
40
45678
910
(rpm)
(N.m)T
(gal/s)Q
Speed
TorqueFuel Input
Rate
Example (Cont’d) State-space Model (Linearized)
Incremental state vector, [ , , ] ; [ ]T
k p q x u
State-space model,
x Ax Bu
y Cx Du
where,
18.26 10.0 0 90.0
20.0 0 20.0 ; 0
0 20.0 60.0 0
(0 0 1); (0)
A B
C D
