Linearization and Review of Stability

Linear 1 Linearization and Review of Stability ME584 Fall 2010 Linear 2 Lecture Objectives and Activities • • • • Review of stability Importanc...
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Linear

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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  ket , 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 xx  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   x1   f    k g   x  sin x 1  f 2   x2   m 2 L 

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Step 1  x1  Set x     0,  x2  x1  x2  0  x2  0 k g x2  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

0

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Step 3 dx *  x1 *  0   g dt  x2 *   L x1*  x2 *

1   x * k  1     x2 * m

g k x2 *   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, mx  u  mg  k1 x  k2 x 3

u

where , m  200kg g 10 m / s 2 k1  1000 N / m k2  1000 N / m

3

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Step 1 Let x1  x and x2  x x   2   f1   x1   f      u k1 k2 3   f 2   x2   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   x1   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 x1  0 and x2  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     x2   I  I   mgL u   x2  (sin x1 ) x1  I I  1

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When Arm is Hit x  x , so  2

 1

 mgL u x1  (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

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