EE/ME/AE324: Dynamical Systems - Clarkson...
Transcript of EE/ME/AE324: Dynamical Systems - Clarkson...
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EE/ME/AE324:Dynamical Systems
Chapter 3: Standard Forms for System Models
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State‐Variables• A common way to model systems is using state‐variables:y y g
0
independent variables that completely describe the systemresponse for all t t given knowledge of the state variables≥ 0
0 0
response for all t t given knowledge of the state variablesat t (initial conditions) and the system inputs for all t t
≥
≥
• 1 1
1 2 m 1 2
Suppose a system has state variables , ,..., , inputs u , u ,..., u , and outputs , ,..., ;
n
p
q q qy y y
1 11 1 12 2 1 11 1 12 2 1
a general state equation for a linear system is:... ...n n m mq a q a q a q b u b u b u= + + + + + + +
q2 21 1 22 2 2 21 1 22 2 2... ...n n m ma q a q a q b u b u b u= + + + + + + +
1 1 2 2 1 1 2 2... ...n n n nn n n n nm mq a q a q a q b u b u b u= + + + + + + +
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State‐Variables(Space) Equations• This set of -first order differential state equationsn q
can be written in matrix form as:q = Aq +Buwhere
q u
q Aq Bu
1 1
2 2, ,
q uq u
q = u =
n mq u b b b 11 12 1
21 22 2 and
n
n
a a aa a a = =
A B
11 12 1
21 22 2
m
m
b b bb b b
1 2n n nna a a 1 2n n nmb b b
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State‐Variables(Space) Equations• Similarly, the system outputs can be expressedpSimilarly, the system outputs can be expressed
algebraically as:p
y = Cq +Duwhere
yq u
y Cq +Du
11 1
22 2, ,
yq uyq u
q = u = y =
pn m yq u
d d d 11 12 1
21 22 2 and
n
n
c c cc c c =
C
11 12 1
21 22 2
m
m
d d dd d d =
D
1 2
and
p p pnc c c
C
1 2p p pmd d d
D
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State‐Variables(Space) Equations• The complete state variable model can now be expressed as:The complete state variable model can now be expressed as:q = Aq +Buy = Cq +Duwhere
n m p∈ ∈ ∈
y Cq Du
q u yR R R
• The choice of system states is not unique meaning
, , , , and n n n m p n p m× × × ×
∈ ∈ ∈
∈ ∈ ∈ ∈
q u yA B C D
R R RR R R R
• The choice of system states is not unique, meaning different A, B, C, D can describe the same system
• Mechanical systems typically require two states per• Mechanical systems typically require two states per inertial element, e.g., and x x
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Steps to Create State‐Variable Eqns. • Use free‐body diagrams to develop system EoM
• Manipulate EoM into state‐space form, i.e., first‐order diff. state eqns. = fcns. of states + inputs
• Express system outputs in terms of the desired p y psystem states and specified inputs
• Ex 3 1:Ex 3.1:
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Ex. 3.1: Mass‐Spring‐Damper System
• EoM: ( )Mx Bx Kx f t+ + =EoM:
• Desired outputs are spring force, and velocity and acceleration of the mass
( )aMx Bx Kx f t+ + =
and acceleration of the mass
• Generally, you can choose system states as d l f hposition and velocity of each system mass, e.g.:
1 2q qx x
1 2
2 TBD from EoM
q qx q x
= = ⇒ = =
q q
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Ex. 3.1: Mass‐Spring‐Damper System• Manipulating the EoM we obtain:Manipulating the EoM, we obtain:
1 ( )aK Bx x x f tM M M
= − − +
( )
1 ( )
afM M MK B f t
• Substituting the RHS of the above term into the
1 2 ( )aq q f tM M M
= − − +
Substituting the RHS of the above term into the state eqn. yields:
q 2
1 ( )
qx
K B f
= =
q1 2 ( )aq q f tx
M M M − − +
q
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Ex. 3.1: Mass‐Spring‐Damper System• Assuming the applied force f (t) as the input e gAssuming the applied force fa(t) as the input, e.g.,
u = fa(t), we can express the state eqn. as:
q 2
1 ( )
qK B f
=
q1 2 ( )aq q f t
M M M − − +
q
0 1 01K B
+ 1K B
M M M
= + − −
q uM M M
= Aq+Bu
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Ex. 3.1: Mass‐Spring‐Damper SystemW th d i d t t• We can now express the desired outputs as:
1Kf Kx Kq= =
2
1x q
B K=
2 1
1 B Kx u q qM M M = − −
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Ex. 3.1: Mass‐Spring‐Damper System[ ]If th t t b ittTf• [ ]If , , , the output eqn. can be written:Kf x x=y
0 0K
0 1 01K B
= +
y q u1K B
M M M − −
=Cq+Du
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Why Model Using State Eqns.?• The use of state eqns. allows the application of
more sophisticated system analysis and design techniques, e.g., “modern” state‐space control
• Assume all parameters in the MSD system of Ex. 3.1 were unity and the applied force was a unit step, then the state space representation is:
1 0 00 1 0
0 1 0
, , 0 1 , 01 1 1
1 1 1
= = = = − − − −
A B C D1 1 1
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Why Model Using State Eqns.?• This model can be entered directly into Simulink• This model can be entered directly into Simulink
to simulate the system response, as shown:
Click to open scope window
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Spring force (mass position)
Mass velocity
Mass acceleration
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Ex. 3.2
•
1 1 1 1 2 1 2 2 1
EoM:( ) ( )M x K x B x x K x x+ = − + −1 1 1 1 2 1 2 2 1
2 2 2 2 1 2 1
( ) ( )( ) ( ) ( )a
M x K x B x x K x xM x K x x B x x f t
+ ++ − + − =
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Ex. 3.2• Input is f (t) output is tensile spring force of K2 andInput is fa(t), output is tensile spring force of K2 and
total momentum of the masses mT
• Choose state variables as shown:x x x x• 1 2 1 2
1 1 1 2
Choose state variables , , , , as shown:x x x x
x q x q
1 2 1 TBD from EoMx q x = ⇒ =
q = q =2 3 2 4
2 4 2 TBD from EoM
x q x qx q x
q q
2 4 2 TBD from EoMx q x
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Ex. 3.2• EoM:
1 1 1 1 2 1 2 2 1
EoM:
( ) ( )M x K x B x x K x x+ = − + −
2 2 2 2 1 2 1( ) ( ) ( )aM x K x x B x x f t+ − + − =
From the EoM, we have:
K K KB B 1 2 21 1 1 2 2
1 1 1 1
K K KB B
M M M Mx x x x x
+= − − + +
2 22 1 1 2 2
1( )a
K KB B
M M M M Mx x x x x f t= + − − +
2 2 2 2 2M M M M M
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Ex. 3.2• Substituting the EoM into the state eqn. yields:
0 1 0 0 1 2 2
1 1 1 1
K K KB B
M M M M
+− −
q = q0 0 0 1
K KB B
q = q
2 2
2 2 2 2
K KB B
M M M M− −
0
+
u = Aq + Bu
2
1
M
+
u = Aq + Bu
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Ex. 3.2• The desired outputs can now be expressed as:
2 2 2 1 2 1 2 3
p p( )Kf K x x K q K q= − = − +
1 1 2 2 1 2 2 4
0 0 0T
K K
m M x M x M q M q−
= + = +
2 2
1 2 0
0 0 0
0 0
K K
M M⇒
+ y = u = Cq + Duq
• The complete state‐space system is given by
A B C DA, B, C, D
• See Matlab file EX3_2_Main.m and Simulinkfile EX3_1.mdl for simulation on class web site
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Spring force (relative position)
Momentum of masses
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Repeat Ex. 3.2 Using New States• Repeat Ex 3 2 using the following states:
1 2 1 1
Repeat Ex. 3.2 using the following states:, , , R Rx x x x x x= −
2 1 2 1 R Rx x x x x xx x x x x x x x x
⇒ = − ⇒ = −⇒ = + ⇒ = + ⇒ = +2 1 2 1 2 1R R Rx x x x x x x x x⇒ = + ⇒ = + ⇒ = +
( )f
2 1( )RM x x+
( )af t2 RK x
RBx
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Repeat Ex. 3.2 Using New States• Substituing the new state info. into the prior EoM:
1 1 1 1 2
( ) ( )R RM x K x Bx K x
M x x K x Bx f t
+ = +
+ + +2 1 2
1 2
( ) ( )R R R aM x x K x Bx f t
K K Bx x x x
+ + + =
⇒ = − + + 1 1
1 1 1
1
R Rx x x xM M M
K B
⇒ = − + + 2
1
2 2 2
1( )R R R a
K Bx x x x f t
M M M= − − − +
11 2 2
1 1 2 1 2 2
1 ( )R R a
K K K B B
M M M M M Mx x x f t+ += − − + +
1 1 2 1 2 2
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Repeat Ex. 3.2 Using New States• Substituting into the new state eqn. yields:
1 1 1 2x q x q
1 2 1
3 4
TBD from EoM
R R
x q xx q x q
⇒ =
q = = q =3 4
4 TBD from EoMR R
R R
x q x qx q x
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Repeat Ex. 3.2 Using New States
1x
q =
1 2
1
K K Bx x x
x
− + +
1
1 1 1
R Rx x xM M M
x
+ +
1 2 21
( )
R
K K K B Bf
x
1 2 2
1
1 1 2 1 2 2
( )R R a
x x x f tM M M M M M
− + − + +
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Repeat Ex. 3.2 Using New States• Substituting into the new state eqn. yields:
0 1 0 0
q =
1 2
0 1 0 0
0
00
K K B
1 2
1 1 1
0 00
M M M−
+
q u
1 2 2
0 0 0 11
K K K B B
q
1 2 2
2
1 1 2 1 2
0 MM M M M M
− + − +
= Aq+Bu
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Repeat Ex. 3.2 Using New States•The desired outputs can now be expressed as:
2 2 2 1 2 2 3
The desired outputs can now be expressed as:( )K Rf K x x K x K q= − = =
1 1 2 2 1 1 2 1( )( ) ( )
T Rm M x M x M x M x xM q M q q M M q M q
= + = + += + + = + +
2
1 2 2 2 4 1 2 2 2 4
0 0 0 0
( ) ( )K
M q M q q M M q M q
⇒
= + + = + +
+ y = u = Cq + Duq
St t t i i b A B C D
1 2 2 00 ( 0)M M M⇒ + + y u Cq + Duq
• State‐space system is given by A, B, C, D
• See Matlab file EX3_2b_Main.m and Simulink file EX3_1.mdl for simulation on class web site
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Spring force (relative position)
Momentum of masses
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Ex. 3.4• The following system is similar to Ex.3.2, except forThe following system is similar to Ex.3.2, except for
the addition of grounded spring K3
• What is the effect of this additional element on theWhat is the effect of this additional element on the state‐space representation of the system?
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Ex. 3.4• The new EoM yields the following state eqn.:
0 1 0 0 1 2 2
1 1 1 1
K K KB B
M M M M
+− −
q = q0 0 0 1
KK B BK+
q = q
322
2 2 2 2
KK B B
M M M M
K− −
+ 0
+
u = Aq + Bu
2
1
M
+
u = Aq + Bu
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Ex. 3.7• Find the state‐space representation of the systemFind the state space representation of the system
when the input is fa(t) and the output is x2
• Assume EQ when 0 and 0 (as drawn);x x x x> >• 1 2 1 2
2 2 2 2
Assume EQ when 0, and 0 (as drawn);this implies that: stretched by , stretched by ,
x x x xK x B x
= = > >
1 1 1 1 2 stretched by and stretched by B x K x x−
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Ex. 3.7
E M i t f fi t d d d d diff•1 1 1 1 1 1 2
EoM consist of a first order and a second order diff. eqn: ( ) ( )aM x B x K x x f t+ + − =
•2 2 2 2 1 1 2( )B x K x K x x+ = −
1 1 2EoM can be rewritten in terms of the states and :x x x1 1 2
1 1 11 1 1 2
EoM can be rewritten in terms of the states , and : 1 ( )a
x x xK B Kx x x x f tM M M M
= − − + +
1 1 1 2
1 1 2
aM M M M
K K Kx x x
+= − 2 1 2
2 2
x x xB B
= −
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Ex. 3.7Manipulating the system into matrix form yields:•
1 1
Manipulating the system into matrix form yields:x q
1 2
2 3
x qx q
=
q =
2q
1 1 11 2 3
21
11
( )a
K B Kf t
M M M Mq q
qxx q− − + +
⇒ =
q =
1 1 21 3
2
M M M M
K K Kx
q q+−
1 3
2 2B Bq q
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Ex. 3.7
0 1 0 0
1K B K
1 1 1 1K B K
M M M M− −
⇒
q = q + u
1 1 2
2 2
00
K K K
B B
+−
= Aq +Bu
Si th t t i it•[ ] [ ]
2Since the output is , we can write:0 0 1 0
x=y = q + u Cq +Du
• How does model change if B2 is removed?
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Ex. 3.7 with B2 Removed
• As shown is Chapter 2, the springs are in series and can be combined to form a simple MSD s stem ithcan be combined to form a simple MSD system with
1 2 1dK K KK 1 2 1
2 11 2 1 2
and eqK x xK K K K
= = + + • System can be expressed in terms of either x1 or x2
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Ex. 3.9• Here x2(t) is an input• Here x2(t) is an input,
so the motion of M2 is knownknown
• As shown is Chapter 2, the EoM is:the EoM is:
1 1 1 1 2 2 2 2( ) ( ) ( )eqM x Bx K x B x t K x t u t+ + = +• This be can treated like a simple MSD system, or
alternately we rewrite the system as:
1 1 2 2 1 1 2 2 ( ) ( )eqM x B x t K x Bx K x t
KB KB
− = − − +
2 21 2 1 1 2
1 1 1 1
( ) ( )eqKB KBx x t x x x tM M M M
⇒ − = − − +
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Ex. 3.91Defining the system states as and where:Ax x1Defining the system states as and where:
( ) ( )
Ax x
B Bx x x t x x x t
= − ⇒ = + 1 2 1 21 1
( ) ( )A Ax x x t x x x tM M
B
= ⇒ = +
1 2
1
( )ABx x x tM
⇒ = −
21 1 2 ( )eqK KBx x x t
= − − + 1 1 21 1 1
( )M M M
K B
2K B
11 1
eqK BxM M
= − −
2
221 1
( )AK Bx x tM M
+ −
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Ex. 3.91 1
This yields the state eqn.: x q 1 1
2A
x qx q
= ⇒
q =
211
( )ABx x tMx
+
11
22
1 22 ( )A eqA
x K KB Bx x x tM M M M
= − − + −
q =
1 221 1 1 1
0 1
AM M M M
B
1
2
0 1
eq
BM
K B
= + q
=
u Aq +Bu2
21 1
1
qK BM MM
− − − 21M
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Input‐Output Eqns.• For a system with scalar input u(t) and output y(t)For a system with scalar input u(t) and output y(t),
the system input‐output eqn. has the form:( ) ( )n mb b b
• The derivatives can be expressed using the p
( ) ( )2 1 0 1 0
n mn ma y a y a y a y b u b u b u+ + + + = + + +
• The derivatives can be expressed using the p‐operator with the equation factored as shown:
n mb b b
( ) ( )1 0 1 0
n mn m
n m
a p y a py a y b p u b pu b u
a p a p a y b p b p b u
+ + + = + + +
+ + + = + + +
• This is a useful first step towards developing f f i b d d l l d
( ) ( )1 0 1 0n ma p a p a y b p b p b u+ + + = + + +
transfer function based system models, explored further in Chapter 8, e.g., replace “p” by “s”
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Ex. 3.2 Revisited
• EoM:
1 1 1 1 2 1 2 2 1
2 2 2 1 2 2 1
( ) ( )( ) ( ) ( )a
M x K x B x x K x xM x B x x K x x f t
+ = − + −+ − + − =
• Find the input‐output eqn. assuming input fa(t) and output x1
2 2 2 1 2 2 1( ) ( ) ( )af
and output x1
( ) ( )21 1 2 1 2 2M p Bp K K x Bp K x+ + + = +
( ) ( )22 2 2 2 1 ( )aM p Bp K x Bp K x f t+ + − + =
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Ex. 3.2 Revisited2Eliminate from the eqns. as shown:x
( )2
21 1 2
Eliminate from the eqns. as shown:x
M p Bp K Kx x
+ + +=
( )( )( )
2 12
2 2
x xBp K
=+
( )( )( )( ) ( )
2 22 2 1 1 2 1
( )
M p Bp K M p Bp K K x
Bp K Bp K x Bp K f t
⇒ + + + + +
− + + = +( )( ) ( )2 2 1 2
4 32 1 1 2 1 1
( )
( )aBp K Bp K x Bp K f t
M M p x M M Bp x
+ + = +
⇒ + +2
2 1 2 1 2 1 1 1( ( ) )( ) ( )
M K K M K p x BK pxK K B f t K f t
+ + + ++ +2 1 1 2( ) ( )a aK K x Bpf t K f t+ = +
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Ex. 3.2 RevisitedConverting from p notation back to diff eq formConverting from p-notation back to diff. eq. formyields the input-output eqn.:
( )4 (3)2 1 1 2 1 1( )
( ( ) )M M x M M BxM K K M K BK K K
⇒ + ++ + + + +2 1 2 1 2 1 1 1 2 1 1
2
( ( ) )
( ) ( )a a
M K K M K x BK x K K x
Bf t K f t
+ + + + +
= + 2( ) ( )a af f
System1x( )af t
System
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Questions?