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# State space analysis, eign values and eign vectors

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State space analysis concept, state space model to transfer function model in first and second companion forms jordan canonical forms, Concept of eign values eign vector and its physical meaning,characteristic equation derivation is presented from the control system subject area.

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### State space analysis, eign values and eign vectors

1. 1. By: Shilpa Mishra M.E., NITTTR Chandigarh . STATE SPACE ANALYSIS Conversion of T.F. models to canonical state variable models and concept of Eignvalues & Eignvectors.
2. 2. Contents_______________ 1. Introduction 2. Need of realization of transfer function into state variable models 3. Realization of transfer function into a state space model or mathematical model in following possible representation: a. First companion form (controllable form) b. Second companion form (observable form) c. Jordan canonical form 4. Eigenvalues and Eigen vectors 08/05/142 Shilpa Mishra ME IC 122509
3. 3. Plant Mathematical Model : Differential equation Linear, time invariant Frequency Domain Technique Time Domain Technique Two approaches for analysis and design of control system: 1.Classical Technique or Frequency Domain Technique.(T.F. + graphical plots like root locus, bode etc.) 2.Modern Technique or Time Domain Technique (State variable approach). 08/05/143 Shilpa Mishra ME IC 122509
4. 4. Transfer Function form Need of conversion of transfer function form into state space form: 1.A transfer function can be easily fitted to the determined experimental data in best possible manner. In state variable we have so many design techniques available for system. Hence in order to apply these techniques T.F. must be realized into state variable model. 08/05/144 Shilpa Mishra ME IC 122509
5. 5. 08/05/145 Shilpa Mishra ME IC 122509
6. 6. BuAxx += DuCxy += x x y u A B C D = state vector = derivative of the state vector with respect to time = output vector = input or control vector = system matrix = input matrix = output matrix = Feed forward matrix State equation output equation General State Space form of Physical System 08/05/146 Shilpa Mishra ME IC 122509
7. 7. Deriving State Space Model from Transfer Function Model  The process of converting transfer function to state space form is NOT unique. There are various “realizations” possible.  All realizations are “equivalent” (i.e. properties do not change). However, one representation may have some advantages over others for a particular task.  Possible representations: 1. First companion form 2. Second companion form 3. Jordan canonical form 08/05/147 Shilpa Mishra ME IC 122509
8. 8. 1. First Companion Form (SISO System) If LTI SISO system is described by transfer function of the form; Decomposition of transfer function: . 012 23 3 01 2 2 )( )( )( )( asasasa bsbsb sR sC sU sY +++ ++ == 08/05/148 Shilpa Mishra ME IC 122509
9. 9. ( ) ( ) ( ) ( )sXbsbsbsCsY 101 2 2 ++== ( ) 10 1 12 1 2 2 xb dt dx b dt xd bty ++= )2........(..........322110)( xbxbxbty ++= ( ) ( ) ( ) ( )sXasasasasRsU 101 2 2 3 3 +++== ( ) )( )()()( 10 1 12 1 2 23 1 3 3 txa dt tdx a dt txd a dt txd atu +++= 43322110)( xaxaxaxatu +++= I. II. )1...().........(32211043 tuxaxaxaxa +−−−= 21 xx = 32 xx =1)( xtx = 43 xx =& Select state variables like : 08/05/149 Shilpa Mishra ME IC 122509
10. 10. 21 xx = 32 xx =1)( xtx = from equation (1) & (2) and state equation, block diagram realization in first companion form of TF will be  43 xx = 08/05/1410 Shilpa Mishra ME IC 122509
11. 11. Again from equation (1) & (2) complete state model will be ; )3).....(( 3 /1 0 0 3 2 1 210 100 010 3 1 3 2 1 , )( 3 1 3 3 2 2 3 1 1 3 0 43 )(32211043 tu a x x x aaa a x x x or tu a x a a x a a x a a xx tuxaxaxaxa                                     + −−− = +−−−== +−−−=     A B 08/05/1411 Shilpa Mishra ME IC 122509
12. 12. Equation (3)&(4) combining together gives the complete realization of the given transfer function. Matrix A has coefficients of the denominator of the TF preceded by minus sign in its bottom row and rest of the matrix is zero except for the superdiagonol terms which are all unity. In matrix theory matrix with this structure is said to be in companion form therefore this realization is called first companion form of realizing a TF. [ ] )4.......( 3 2 1 210 )( , 322110)(           = ++= x x x bbbty or xbxbxbty C 08/05/1412 Shilpa Mishra ME IC 122509
13. 13. Example :TF to State Space (constant term in numerator) rcccc 2424269 =+++  cx =1 cx =2 cx =3 1. Inverse Laplace 2. Select state variables 21 xx = 32 xx = rxxxx 2492624 3213 +−−−= 1xcy == ( ) ( ) ( )sD sN sG = ( )sN ( )sD numerator denominator 08/05/1413 Shilpa Mishra ME IC 122509
14. 14. r 24 0 0 x x x 92624 100 010 x x x 3 2 1 3 2 1           +                     −−− =              [ ]           = 3 2 1 001 x x x y 08/05/1414 Shilpa Mishra ME IC 122509
15. 15. Example: 08/05/1415 Shilpa Mishra ME IC 122509
16. 16. 2. Second Companion Form (SISO System) In second companion form coefficient of the denominator of the transfer function appear in one of the column of the A matrix. This form can be obtained by the following steps: Let the transfer function is, n nnn n nnn asasas bsbsbsb sU sY sH ++++ ++++ == −− −− ...... ........ )( )( )( 2 2 1 1 2 2 1 10 )5)].......(()([ 1 ....)]()([ 1 )()( 0)]()([...)]()([)]()([, )()....()().....( 110 11 1 0 1 10 1 1 sYasUb s sYasUb s sUbsY sUbsYasUbsYassubsYsor sUbsbsbsYasas nnn nn nn n nn n nn −++−+= =−++−+− +++=+++ − −− On dividing by and solving for Y(s);n s 08/05/1416 Shilpa Mishra ME IC 122509
17. 17. By equation (5) block diagram can be drawn as follows which is called second companion form of realization:- 08/05/1417 Shilpa Mishra ME IC 122509
18. 18. To get state variable model, output of each integrator is identified as state variables starting at the left and preceding to the right. The corresponding differential equations are, ub n xy u n bub n x n ax u n bub n x n axx ubub n xa n x n x ubub n xa n x n x 0 isequationoutputtheand ) 0 ( 1 1 ) 0 ( 112 2 ) 0 ( 221 1 ) 0 ( 11 += ++−= − ++ − −= ++− − = − ++− − =      08/05/1418 Shilpa Mishra ME IC 122509
19. 19. Now from here state and output equations organized in matrix form are given below: BuAxx += DuCxy += [ ] 0 011 011 0 1 2 1 ;1000 B; 100 010 001 000 bDC bab bab bab a a a a A nn nn n n n ==           − − − =                 − − − − = −−− −       08/05/1419 Shilpa Mishra ME IC 122509
20. 20. •In this form of realizing a TF the poles of the transfer function form a string along the main diagonal of the matrix A. •In Jordan canonical form state space model will be like:- 3. Jordan canonical form (Non-repeated roots) 08/05/1420 Shilpa Mishra ME IC 122509
21. 21. n nnn n nnn asasas dsdsdsd sU sY sH ++++ ++++ == −− −− ...... ........ )( )( )( 2 2 1 1 2 2 1 10 Let general transfer function; )( )( )( )( 1 21 1 ssx dt tdx L xx dt tdx xtx if =      == =  08/05/1421 Shilpa Mishra ME IC 122509
22. 22. urxλx . . . urxλx urxλx nnnn += += +=    2222 1111 nxxxudty ++++= ......)( 210 & output equation; With the help of above equations jordan canonical state model can be obtained as; State equation: 08/05/1422 Shilpa Mishra ME IC 122509
23. 23. Jordan Canonical Form (Non-repeated roots): Block Diagram 08/05/1423 Shilpa Mishra ME IC 122509
24. 24. Jordan canonical form: Example (Non repeated roots) 08/05/1424 Shilpa Mishra ME IC 122509
25. 25. 08/05/1425 Shilpa Mishra ME IC 122509
26. 26. Jordan canonical form: Example (repeated roots): 08/05/1426 Shilpa Mishra ME IC 122509
27. 27. 08/05/1427 Shilpa Mishra ME IC 122509
28. 28. Eignvalues and Eignvector Definition : Given a linear transformation A, a non-zero vector x is defined to be an eigenvector of the transformation if it satisfies the eigenvalue equation for some scalar λ. In this situation, the scalar λ is called an eigenvalue of A corresponding to the eigenvector x.  It indicates that vector x has the property that its direction is not changed by the transformation A, but that it is only scaled by a factor of λ.  Only certain special vectors x are eigenvectors, and only certain special scalars λ are eigenvalues.  The eigenvector must be non-zero because the equation A0 = λ0 holds for every A and every λ. λxAx = 08/05/1428 Shilpa Mishra ME IC 122509
29. 29. A acts to stretch the vector x, not change its direction, so x is an eigenvector of A. The eigenvalue λ is simply the amount of "stretch" or "shrink" to which a vector is subjected when transformed by A. If λ = 1, the vector remains unchanged (unaffected by the transformation). A transformation I under which a vector x remains unchanged, Ix = x, is defined as identity transformation If λ = −1, the vector flips to the opposite direction; this is defined as reflection. 08/05/1429 Shilpa Mishra ME IC 122509
30. 30. Computation of eigenvalues & the characteristic equation  When a transformation is represented by a square matrix A, the eigenvalue equation can be expressed as; Ax = λx (A − λI)x = 0 As x must not be zero, this can be rearranged to; det(A − λI) = 0. which is defined to be the characteristic equation of the n × n matrix A.  Expansion of the determinant generates a polynomial of degree n in λ and may be written as; here λ1, λ2……λn are called eignvalues. For each eign value there exist eign vector x. 08/05/1430 Shilpa Mishra ME IC 122509
31. 31. Example 1. find the eigenvalues and eigenvectors of the matrix Solution So the eigenvalues are 3.421 and 0.3288. 08/05/1431 Shilpa Mishra ME IC 122509
32. 32. Let be the eigenvector corresponding to Hence .421x1 +1.5x2 = 0 .75x1 +2.671x2 = 0 08/05/1432 Shilpa Mishra ME IC 122509
33. 33. If then The eigenvector corresponding to then is The eigenvector corresponding to is Similarly, the eigenvector corresponding to is 08/05/1433 Shilpa Mishra ME IC 122509
34. 34. Example 2. Find the eigenvalues and eigenvectors of Solution The characteristic equation is given by The roots of the above equation are Note that there are eigenvalues that are repeated. Since there are only two distinct eigenvalues, there are only two eigenspaces. But, corresponding to λ = 0.5 there should be two eigenvectors that form a basis for the eigenspace 08/05/1434 Shilpa Mishra ME IC 122509
35. 35. To find the eigenspaces, let Given then For , Solving this system gives 08/05/1435 Shilpa Mishra ME IC 122509
36. 36. So So the vectors and form a basis for the eigenspace for the eigenvalue . For , Solving this system gives The eigenvector corresponding to is ; Hence the vector is a basis for the eigenspace for the eigenvalue of . 08/05/1436 Shilpa Mishra ME IC 122509
37. 37. Thanks for Attention 08/05/1437 Shilpa Mishra ME IC 122509