University:
California State University, Northridge
Course:
ME 584 | Modeling and Simulation of Dynamic Systems
Academic year:

2010
Views:

379

Pages:

29
Author:

Cullen Newton

1 Linearization and Review of Stability Linear 2 Lecture Objectives and Activities • • • • Review of stability Importance of linearization Linearization of nonlinear systems Active learning activities – Pair-share problems Linear 3 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. Linear 4 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 Linear 5 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 Linear 6 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 Linear 7 Stability Analysis with SS - Example Linear 8 Importance of linearization • Dynamics Analysis • Control and estimation systems design Linear 9 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  Linear 10 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 Linear 11 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 Linear 12 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 . Linear 13 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  Linear 14 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  Linear 15 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 Linear 16 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 Linear x x u u 0 17 Linear 18 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 Linear 19 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 Linear 20 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 Linear 21 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  Linear 22 Step 3 1  0  0  x*   x *  u*    20 0 .005 Linear 23 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 Linear 24 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 Linear 25 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  Linear 26 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 Linear 27 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 ) Linear 28 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 Linear 29 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

Linearization and Review of Stability

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