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Linearization and Review of Stability ME584 Fall 2010
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Lecture Objectives and Activities • • • •
Review of stability Importance of linearization Linearization of nonlinear systems Active learning activities – Pair-share problems
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Definition of Stability A stable system is a system with a bounded response to a bounded input
Response to a displacement/initial condition will produce either a decreasing, neural, or increasing response.
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Stability Analysis – 1st Order ODE dx a0 x b0u dt Characteri stic equation : a1 a0 0 a0 / a1
1st Order : a1
System is stable if 0, unstable if 0 Example : 6 x 2 x 2u; u 0, xo 1 Characteri stic equation : 6 2 0 3 x xo e t / 3 Example : 6 x 2 x 2u; u 0, xo 1 Characteri stic equation : 6 2 0 3 x xoet / 3
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Stability Analysis – 2nd Order ODE 2 d x dx 2nd Order : a2 2 a1 a0 x b0u dt dt Characteri stic Equation : 1 a 2 a1 Let 2 2 , , ao ao n
n
2 2 2 0 n
n
1, 2 j 1 2 n
n
System is unstable if 1 and / or 2 0 Stable response 1 : Underdampe d (Oscilliati on) 1 : Overdamped ( No oscilliation) 1 : Critically damped ( No oscilliation) Relative stability : degree of stability
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Stability Analysis – State Space (SS) State space format x Ax Let x ket , substitute
ke Ake or x Ax (I A) x 0 t
t
Non trivial solution if det I A 0
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Stability Analysis with SS - Example
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Importance of linearization • Dynamics Analysis • Control and estimation systems design
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Importance of linearization • Is nature linear or nonlinear? physical systems? – Many physical systems behave linearly within some range of variable, but become nonlinear as variables increase without limit – Possible to linearize nonlinear systems
• Example: Pendulum
mg sin( ) m K L For small , sin( )
m K (mg / L) • Tractable analysis with linear model
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Importance of linearization • Is this systemstable? m K (mg / l ) • State space model : state variables x [ ] x Ax Bu 1 0 0 where A ; B 1; u g / l K / m • Control u Gx Where G is the control matrix, x Ax Bu Ax B (Gx) ( A BG ) x Choose G to achieve desired performance •Estimatio n u Gxˆ where xˆ is estimate of x
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Linear Approximation dx x f ( x, u ) dt xx x* u u u* x : equilibrium value of x about which linearizat ion is taken also called steady state value/nominal value u : equilibrium value of u about which linearizat ion is taken x* : small perturbati on or variation of x u* : small perturbati on or variation of u To solve for x and u , set f ( x , u ) 0 x and u can also be provided from testing
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Taylor’s Expansion x x x* dx dx * f 0 f (x ,u ) dt dt x
f x x x * u u u
dx * Ax * Bu * dt where f A x
x x u u
f and B u
x x u u
A and B are called Jacobian matrices
x x u u
u * H .O.T .
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Linearization Procedure
Step 1. If x and u are not specified , set x 0 to solve for x and u Define x1 , x2 ,...,xn and u1 , u2 ,...,um , and form f1 , f 2 ,..., f n Step 2. Solve for A and B. Assume A is (n n) and B is (n m)
A
f x
x x u u
f1 x x ,u 1 f 2 x ,u x1 f n x ,u x1
f1 x ,u x2 f 2 x ,u x2 f n x ,u x2
Step 3. dx * Form Ax * Bu * dt
f1 x ,u xn f 2 ... f x ,u and B xn u f n ... x ,u xn ...
x x u u
f1 u 1 f 2 u1 f n u1
x ,u
x ,u
x ,u
f1 x ,u u2 f 2 x ,u u2 f n x ,u u2
f1 um f 2 ... um f n ... um ...
x ,u x ,u x ,u
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Example mg sin( ) m K 0 L let x1 , x2 , u 0 (no control input ) x2 f1 x1 f k g x sin x 1 f 2 x2 m 2 L
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Step 1 x1 Set x 0, x2 x1 x2 0 x2 0 k g x2 x2 sin x1 0 m L x1 0, ,2 For this example, let us consider x1 0 x1 0 x x2 0
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Step 2 f A x f1 x1 f1 x2 f 2 x1 f 2 x2
x ,u
x ,u
x ,u
x ,u
x x u u
f1 x 1 f 2 x1
x ,u
x ,u
f1 x2 x2 f 2 x ,u x2
x2 f1 k x g sin x 1 f 2 m 2 L
( x2 ) x ,u 0 x1 ( x2 ) x ,u 1 x2 g k g cos x1 ( x2 sin x1 ) x ,u L x1 m L
k g k ( x2 sin x1 ) x ,u m x2 m L
x1 0
g L
Step 2 (continued) f B u
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x x u u
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Step 3 dx * x1 * 0 g dt x2 * L x1* x2 *
1 x * k 1 x2 * m
g k x2 * x1 * x2 * L m g k The same as L m
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Pair-Share Exercise Linearize the system about the poin t where the mass compresses the spring by1m and the applied force u 0, mx u mg k1 x k2 x 3
u
where , m 200kg g 10 m / s 2 k1 1000 N / m k2 1000 N / m
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Step 1 Let x1 x and x2 x x 2 f1 x1 f u k1 k2 3 f 2 x2 m g m x1 m x1 x1 1 x x2 0 u 0
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Step 2 f1 x1 A f 2 x1 f1 u B f 2 u
x ,u
x ,u
0 x ,u k1 3k2 2 f 2 m m x1 x2 x ,u f1 x2
0 x ,u .005 x ,u
1 0 1 0 20 0
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Step 3 1 0 0 x* x * u* 20 0 .005
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Pair-Share Example A simple robot arm is modeled as
I T mgL cos where I is moment of inertia of arm, m is the mass, and T is the torque that the motor supplies. We want the motor to hold the arm at five angles:
e 0o , 45o , 90o , 135o , 180o ,225o Find the torque required and determine what will happen if something hits the arm and slightly alters its position?
m L T
arm θ
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Step 1 .
Let x1 , x2 , u T x2 f1 x1 f u mgL cos x f x 1 2 2 I I x1 e x x2 0 To find the torque u at x , set x x and x1 0 and x2 0 u mgL cos x1 x1 0 o ,
u mgL
x1 45 o ,
u 0.707mgL
x1 90 o ,
u 0
x1 135 o ,
u 0.707mgL
x1 180o ,
u mgL
x1 225o ,
u 0.707mgL
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Step 2 f1 x 1 A f 2 x1 f1 u B f 2 u
x ,u
x ,u
0 1 x ,u mgL sin x1 0 f 2 I x2 x ,u f1 x2
0 x ,u 1 I x ,u
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Step 3 0 0 1 x x 1 mgL 1 x sin x1 u 0 x2 x2 I I mgL u x2 (sin x1 ) x1 I I 1
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When Arm is Hit x x , so 2
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mgL u x1 (sin x1 ) x1 I I Characteri stic equation mgL 2 (sin x1 ) 0 I has roots
mgL 1, 2 sin x1 I
For x1 0o , 1, 2 0,0
(neutrally stable )
x1 45o , 1, 2
.707 mgL I
(unstable )
x1 90o , 1, 2
mgL I
(unstable )
.707 mgL I x1 180o , 1, 2 0,0 x1 135o , 1, 2
x1 225o , 1, 2 j
.707 mgL I
(unstable ) (neutrally stable ) (undamped / neutrally stable )
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Lecture Recap • Many nonlinear systems behave linearly with small perturbation • Linearization procedure – Establish equilibrium – Solve for A and B
• Analysis is tractable with linear models • Next lecture: Stability analysis and simulation with Matlab
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References • Woods, R. L., and Lawrence, K., Modeling and Simulation of Dynamic Systems, Prentice Hall, 1997. • Palm, W. J., Modeling, Analysis, and Control of Dynamic Systems • Close, C. M., Frederick, D. H., Newell, J. C., Modeling and Analysis of Dynamic Systems, Third Edition, Wiley, 2002