Lecture Note
Models for Dynamic Systems and Systems Similarity
-
University:
California State University, Northridge -
Course:
ME 584 | Modeling and Simulation of Dynamic Systems Academic year:
2010
-
Views:
216
Pages:
44
Author:
Harrison Montoya
Models for Dynamic Systems
and Systems Similarity
ME584 Agenda
• Formulation of Models for Engineering
Systems
• Review of Solution of Differential
Equations
• Engineering Systems Similarity
• Simulation with Matlab
• Active learning: Pair-share questions,
Concept questions Chap2
3
Formulation of Models for
Engineering Systems Conservation Laws of
Engineering Systems
• Linear momentum: translational
momentum is equal to applied impulse
(rigid body an or fluid)
F
net
d
mv
dt
• Angular momentum: angular impulse is
equal to time rate of change of angular
momentum
d
T
net
dt
J Conservation Laws of
Engineering Systems
• Conservation of Charge/Kirchhoff’s law:
sum of all currents at a node is equal to
rate of storing charge
dQ
de
inode dt C dt
• Conservation of Mass: net mass flowrate
at a location equal to time rate of change
of mass that that location
m
net
d
V V V
dt Conservation Laws of
Engineering Systems
• Conservation of Energy
– Energy in a system is constant
– Reduces to Bernoulli equation: energy is not
exchanged or stored Review of Solution of
Differential Equations Input Signals
Step input
Ramp input Input Signals
Pulse input
Sinusoidal input Classical Differential Equations
dx
1 Order : a1 a0 x b0u
dt
System is stable if a1 0 and a0 0.
st
bo
d
a1
Let G , D ,
ao
dt
ao
[ D 1]x Gu(t ) Example Classical Differential Equations
• Use Table E.3 to find the solution, and find
steady state value. Is this system stable?
6 x 2 x 2u; u 0, xo 1
Solution :
a1 6, ao 2, bo 2, G
bo 2
a
6
1, 1 3
ao 2
ao 2
a1 0 and ao 0 System is stable. Table E.3 x xo e t / e t / 3
Steady state value : 2 methods
1) xss x(t ) e / 3 0
2) 6 sX ( s ) 2 X ( s ) 2U ( s )
xss lim s 0 s
X ( s)
0
U (s)
2
1
X (s)
U ( s ) 6 s 2 3s 1 Classical Differential Equations:
Pair-Share Exercise
• Use Table E.3 to find solution, and find steady state value for
this system. Is this system stable?
10 x 2 x 10u; u uo 1 (step input ), xo 0
Solution :
bo 10
a1 10
bo 10, a1 10, ao 2, G
5,
5
ao 2
ao 2
x xo e t / Guo [t (1 e t / )] 5[t 5(1 et / 5 )]
X ( s)
10
1
; U (s) ;
U ( s ) 10s 2
s
X (s)
lim s 0 X ( s ) .
xss lim s 0 s
U ( s)
System is unstable. Classical Differential Equations
2
d
x
dx
nd
2 Order : a2 2 a1 a0 x b0u
dt
dt
System is stable if a2 0, a1 0, and a0 0.
bo 1
a2 2 a1
d
Let G , 2 ,
, and D ,
ao n ao n a o
dt
1 : Underdamped case (Oscilliati on in response)
1 : Overdamped case ( No oscilliati on in response)
1 : Critically damped case ( No oscilliati on in response) Classical Differential Equations Classical Differential Equations
• Example: Using Table E.4 to find solution for this
system. Is this system stable?
x x x u; u 0, xo 1, xo 0
Solution :
a2 1, a1 1, a0 1, System is stable. bo 1,
G
bo
1
2 a1
a
1, 2 2 1 n 1,
1 .5,
n ao
n ao
ao
1 : Underdamped case
d
n
1 2 0.866
x t
x(t ) xo e
sin( d t ) o e n sin( d t )
cos( d t )
1 2
d
.5
.5t
e cos(0.866t )
sin(0.866t )
0.866
t
n State Space Description
• State variables describe present configuration of
a system and can be used to determine future
response, given inputs and dynamic equations
• Number of state variables
– Should be as small as possible, but a minimum
number of state variables must be selected as
components of the state vector (sufficient to describe
completely the state of the system)
– Usually equal to order of system’s differential
equation or number of energy-storage elements
• Choice of state variables is not unique State Space Representation
x Ax Bu
y Cx Du
x state vect or y output vector
x time deriv ative of t he state v ector
u input vect or
A system matrix
B input matr ix
C output mat rix
D feedforwar d matrix State Space Representation
• Example: Express x 2 x 3x f in State
Space format
x 2 x 3 x f
z x
z x
x z 2 x 3 x f 2 z 3 x f
1 x 0
x 0
z 3 2 z 1 f
• Is this system stable? Yes State Space Representation:
Pair-Share Exercise
• Example: Express in state space format
x 2x 3x 4 x f
z x
y z x
y z x 2 x 3 x 4 x f
2 y 3 z 4 x f
1
0 x 0
x 0
z 0
z 0 f
0
1
y 4 3 2 y 1
• Is this system stable? Yes State Space Representation
• Example: develop a state space model with output vL
• Ldi/dt+Ri+(1/c)∫idt=v(t)
• i(t)=dq/dt, i’=q”
• Lq”+Rq’+(1/c)q=v(t)
• To make two 1st-order equations
– dq/dt=i
– di/dt=-(1/LC)q-R/Li+v(t)/L
• Output: vL=Ldi/dt State Space Representation:
Concept Question
• Which variables can be chosen to describe the state of the system?
A.
B.
C.
D.
E.
F.
G.
Only i
Only q
i and q
Only VR
Only VC
L, R, or C
Other possibilities
• How many state variables are sufficient to describe the state of the
system?
• Some possible pairs
– i and q
– VR and VC
– …… State Space Representation
Choose q and i :
x Ax Bu
dq
di
R
1
1
i
q i v(t )
dt
dt
L
L
Lc
1
0
dq / dt
R
x
,
A 1
Lc
di / dt
L
0
q
x ,
B 1 , u v(t)
L
i
output equations
y Cx Du
1
y vL (t ), C - R , D 1
c
State Space Representation:
Pair-Share Exercise
Form state space model with vR and vC
•
•
•
•
•
•
•
•
vL=Ldi/dt=(L/R) dvR/dt
As vL+vR+vC=v(t)
(L/R) dvR/dt+vR+vC= v(t)
dvR/dt=-(R/L) vR-(R/L) vC+(R/L)v(t)
vC=(1/c) ∫(vR/R)dt
Differentiate
dvC/dt=(1/RC)vR
New state equations State Space
Graphical Representation
• q,i were selected
as state variables
• Graphical
representation
of state space
and a state vector
q
i
Function of time t State Space Representation
• Example: RLC Electrical circuit
Output iR(t)
dvC
1
1
vC iL
dt
RC
C
1
1
diL
vC v(t )
dt
L
L
1
output i R vC
R
1
1
vC RC C vC 0
1 v(t )
i 1
L
0 iL L
L
1
vC
iR
0
R
iL State Space Solution
• Solution for state space equation
x Ax Bu
sX ( s ) X (0) AX ( s ) BU ( s )
( sI A) X ( s ) X (0) BU ( s )
X ( s ) ( sI A) 1 X (0) ( sI A) 1 BU ( s )
t
x(t ) (t ) x(0) (t )]Bu ( )d
0
where (t ) is Laplace transform of ( sI A) 1 ,
With (t ), x(0), and u ( ) known, x(t ) can be solved or evaluated numericall y.
(t ) : state transition matrix State Space Solution
• Example
1
1
0
x
x, x(0) 0 and u 0.
2
3
1
1 0 0
1
s 3 1
1
1
( sI A) s
2 3 ( s 1)( s 2) 2 s
0
1
s3
1
( s 1)( s 2) ( s 1)( s 2)
2
s
( s 1)( s 2) ( s 1)( s 2)
Take inverse Laplace transform,
( 2e t e 2 t )
(e t e 2 t )
(t )
t
2t
t
2t
(
2
2
)
(
2
)
e
e
e
e
2e t e 2 t
x(t ) (t ) x(0)
t
2t
2
e
2
e
State Space Solution:
Pair-Share exercise
• Find solution for this system
0 2
1
x
x, x(0) 1 and u 0.
1
3
1
1 0 0 2
s 3 2
1
( sI A) 1 s
1 3
1
s
0
1
(
s
1
)(
s
2
)
Take inverse Laplace transform,
(2e t e 2t ) (2e t 2e 2t )
(t ) t 2t
t
2t
e
e
e
e
(
)
(
2
)
e 2 t
x(t ) (t ) x(0) 2t
e Engineering Systems
Similarity System Similarity
• Models in different engineering systems
have same or similar differential equations
• Similar basis elements
– Resistance
– Capacitance
– Inductance
• Dynamic behaviors have different time
constants, influencing modeling choices
and assumptions System Similarity Effort and Flow Variables
• Variables used for differential equations are
either effort or flow variables
– Effort variables: express effort placed on a
component, e.g., force, voltage
– Flow variables: express flow or time rate of change of
a system variable, e.g., velocity, current
• Impedance: is any transfer function between a
flow variable as input and effort variable as
output
– Effort = impedance x flow
– Impedance can be static (resistance) or dynamic
(capacitance/inductance) General Model Structure for
Mechanical System General Model Structure for
Electrical System Simulation with Matlab 1st Order System: Differential
Equation and Transfer Function
Free body diagram: Sliding block with friction
•
•
•
•
•
•
f(t) - m v' - b v = 0
L{ v(t) } = V(s), L{ v’(t) } = sV(s) – v(0), L{ f(t) } = F(s)
Assume v(0) = 0,
m[sV(s)] + b[V(s)] – F(s) = 0
Rearranging we get: [ms + b]V(s) = F(s)
The transfer function is: V(s)/F(s) = 1/[ms+b] 1st Order System: TF in Matlab
• Let m = 20,000kg, b = 500kg/s
• Matlab commands:
>> m = 20000;
>> b = 500;
>> num = [ 1 ];
>> den = [ m b ];
>> first_tf = tf(num, den)
Transfer function:
1
------------20000 s + 500 1st Order System: Step Response
>> u = 1;
>> step(u * first_tf)
>> u = 20000;
>> step(u * first_tf) 1st Order System: Impulse
Response
>> u = 1;
>> impulse(u * first_tf)
>> u = 20000;
>> impulse(u * first_tf) 1st Order System: State Space
x' = Ax + Bu
y = Cx + Du
>> [A, B, C, D] = tf2ss(num,den)
A=
-0.0250
B=
1
C=
5.0000e-005
D=
0
>> first_ss = ss(first_tf)
a=
x1
x1
-0.025
x1
u1
0.007813
y1
x1
0.0064
b=
c=
d=
u1
y1
0 2nd Order System: Differential
Equation and Transfer Function
bq’
• Sum of all torques
• kq + bq' + Iq" = 0
• L{ q(t) } = Q(s)
• L{ q’(t) } = sQ(s) – q(0)
• L{ q"(t) } = s2Q(s) – sq(0) – q'(0)
Iq”
kq
• k[Q(s)] + b[sQ(s)] + I[s2Q(s) - w(0)] = 0
• Or [Is2 + bs + k]Q(s) = I*w(0)
• Q(s) = [ 1/[Is2 + bs + k] ] * I*w(0)
• The transfer function is:
• G(s) = 1/[Is2 + bs + k]
a rotational mass inertia I,
fiction oefficient b, spring
coefficient k 2nd Order System:
I = 1,500,000kgm2
b = 200,000Nm/s
k = 4,000,000Nm
>> I = 1500000;
>> b = 200000;
>> k = 4000000;
>> num = [ 1 ];
>> den = [ I b k ];
>> second_tf = tf(num, den)
1
-----------------------------1.5e006 s^2 + 200000 s + 4e006
• >> u = 0.002 * 1.5e6;
>> impulse(u * second_tf) Matlab Simulation Exercise
• Find the transfer function, state space
model, and simulate the impulse and step
response for this system
x 2 x 3 x u
x (0) 0
x(0) 0
Models for Dynamic Systems and Systems Similarity
of 44
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