10. State Space Methods

Transcription

1 . Sae Space Mehods. Inroducion Sae space modelling was briefly inroduced in chaper. Here more coverage is provided of sae space mehods before some of heir uses in conrol sysem design are covered in he nex chaper. A sae space model, or represenaion, as given in equaion (.6), is denoed by he wo equaions x Ax Bu (.) y Cx Du (.) where equaions (.) and (.) are respecively he sae equaion and oupu equaion. The represenaion can be used for boh single-inpu single-oupu sysems (SISO) and mulipleinpu muliple-oupu sysems (MIMO). For he MIMO represenaion A, B, C and D will all be marices. If he sae dimension is n and here are r inpus and m oupus hen A, B, C and D will be marices of order, n x n, n x r, m x n and m x r, respecively. For SISO sysems B will be an n x column vecor, ofen denoed by b, C a x n row vecor, ofen denoed by c T, and D a scalar ofen denoed by d. Here he capial leer noaion will be used, even hough only SISO sysems are considered, and B, C, and D will have he aforemenioned dimensions. As menioned in chaper he choice of saes is no unique and his will be considered furher in secion.3. Firs, however, obaining a soluion of he sae equaion is discussed in he nex secion.. Soluion of he Sae Equaion Obaining he ime domain soluion o he sae equaion is analogous o he classical approach used o solve he simple firs order equaion x ax u (.3) The procedure in his case is o ake u =, iniially, and o assume a soluion for x( of e a x() where x() is he iniial value of x(. Differeniaing his expression gives a x ( ae x() ax( so ha he assumed soluion is valid. Now if he inpu u is considered his is assumed o yield a soluion of he form x( = e a f(, which on differeniaing gives a a x ( ae f ( e f (. Thus he differenial equaion is saisfied if ae a a a f ( e f ( ae f ( u(, giving f a ( [ e ] u(, which has he soluion

2 f a ( [ e ] u( ) d soluion can be wrien a a, giving x( e [ e ] u( ) d, where is a dummy variable. This x a( ) ( e u( ) d so ha he complee soluion for x( consiss of he sum of he wo soluions, known as he complimenary funcion (or iniial condiion response) and paricular inegral (or forced response), respecively and is a a( ) x( e x() e u( ) d (.4) For equaion (.) x is an n vecor and A an n x n marix no a scalar a and o obain he A complimenary funcion one assumes x( e x(). e A is now a funcion of a marix, which is defined by an infinie power series in exacly he same way as he scalar expression, so ha 3 3 e A I A A /! A / 3!... (.5)

3 where I is he n x n ideniy marix. Term by erm differeniaion of equaion (.5) shows ha he derivaive of e A is Ae A A and ha x( e x() saisfies he differenial equaion wih u =. e A is ofen denoed by ( and is known as he sae ransiion marix. Using he same approach as for he scalar case o ge he forced response he oal soluion is found o be x( ( x() ( ( ) Bu( ) d (.6) A I is easily shown ha he sae ransiion marix ( ) e has he propery ha ( ) ( ( ) so ha equaion (.6) can be wrien alernaively as x( ( x() ( ) Bu( ) d (.7) This ime domain soluion of equaion (.) is useful bu mos engineers prefer o make use of he Laplace ransform approach. Taking he Laplace ransform of equaion (.) gives sx ( s) x() AX ( s) BU ( s) (.8) which on rearranging as X(s) is an n vecor and A a n x n marix gives X ( s) ( si A) x() ( si A) BU ( s) (.9) Taking he inverse Laplace ransform of his and comparing wih equaion (.7) indicaes ha [ ( ] ( s) ( si A) (.) Also aking he Laplace ransform of he oupu equaion (.) and subsiuing for X(s) gives Y ( s) C( si A) x() [ C( si A) B D] U ( s) (.) so ha he ransfer funcion, G(s), beween he inpu u and oupu y is Y ( s) / U ( s) G( s) C( si A) B D C( s) B D (.) This will, of course, be he same independen of he choice of he saes. 3

4 .3 A Sae Transformaion Obviously here mus be an algebraic relaionship beween differen possible choices of sae variables. Le his relaionship be x Tz (.3) where x is he original choice in equaions (.) and (.) and z is he new choice. Subsiuing his relaionship in equaion (.) givestz ATz Bu which can be wrien z T ATz T Bu (.4) Also subsiuing in he oupu equaion (.) gives y CTz Du (.5) Thus under he sae ransformaion of equaion (.3) a differen sae space represenaion ( T AT, T B, CT, D) is obained. If he new A marix is denoed by easy o show ha A and A z have he following properies A z T AT hen i is (i) (ii) (iii) The same eigenvalues The same deerminan The same race (Sum of elemens on he main diagonal) There are some specific forms of he A marix which are ofen commonly used in conrol engineering and no unsurprisingly hese relae o how one migh consider obaining a sae space represenaion for a ransfer funcion, he opic of he nex secion..4 Sae Represenaions of Transfer Funcions This opic was inroduced in secion.3 where he conrollable canonical form for a differenial equaion was considered. Here his and some oher forms will be considered by making use of block diagrams where every sae will be an inegraor oupu. To develop some represenaions consider he ransfer funcion Y ( s) U ( s) G( s) s 3 s 3s 4 7s 4s 8 (.6) 4

5 .4. Conrollable Canonical Form. As seen from equaion (.) he firs n- sae variables are inegrals of he nex sae, ha is x( j ) x jdx, or as shown in he equaion by x ( j ) x j, for j = o n. Thus he block diagram o represen his is n inegraors in series. The inpu o he firs inegraor is x n and is value is given by x n a x a x a x... u, he las row of he marix represenaion of 3 equaion (.). The numeraor erms are provided by feeding forward from he saes o give he required oupu. Thus, for our simple example, his can be shown in he block diagram of Figure., done in SIMULINK, where since he ransfer funcion is hird order n = 3, here are hree inegraors, blocks wih ransfer funcions /s, in series. Feedback from he saes, where he inegraor oupus from lef o righ are he saes x 3, x, and x, respecively, is by he coefficiens -8, -4 and -7. (negaive and in he reverse order of he ransfer funcion denominaor). The numeraor coefficiens provide feedforward from he saes, wih he s erm from x 3. 5

6 Figure. Conrollable Canonical Form Diagram for he Example. The marices for he sae represenaion are A, B and 4 3 C. MATLAB has a companion form, which for any sae space sysem G=ss(A,B,C,D), will be reurned on yping sys=canon(g, companion ). The companion represenaion sys will have an A marix which is he ranspose of he above A marix and a B = ( ) T..4. Observable Canonical Form The observable canonical form is relaed o he conrollable form by he following relaionships, A o = A c T, B o = C c T and C o = B c T. The subscrips c and o relae o he conrollable and observable form marices respecively and T denoes he ranspose. I is lef o he ineresed reader o develop a block diagram similar o Figure. for his form..4.3 Diagonal (or Modal) Form. If he impulse response of G(s) is required hen is evaluaion by inverse Laplace ransforms requires a parial fracion expansion of G(s). 6

7 / 3 4 / 3 This is G ( s), s s s 4 which is simply a parallel connecion of hree firs order ransfer funcions. The firs order ransfer funcion K/(s + a) can be modelled wih one inegraor as shown in Figure.. If he oupu of he inegraor is denoed by he sae variable x hen is sae and oupu equaions are x ax bu, y cx. Figure. Diagram of Model for One Sae Variable. Noe ha here is no unique value for b and c as all ha is required is ha heir produc should equal K. Thus a sae represenaion for G(s) has A, 4 B and C / 3 4 / 3 where we have chosen o ake all he B values as uniy. This form of he A marix is known as a diagonal form and can always be found if he denominaor of he ransfer funcion has real roos. To keep he marix real for complex roos, say ± j, he corresponding rows around he diagonal are replaced by he marix. For example if G ( s), he s( s s ) conrollable canonical form, wih he marices subscriped wih c, is A c, c C. B and c 7

8 A diagonal form is A.5.866, B and C.5774 This is he one given by MATLAB if he insrucion csys=canon(g, modal ) is used, where csys is he new sae space represenaion in he chosen canonical form modal. A special case is when G(s) has a repeaed roo, for example ifg ( s), which has a sae space s( s ) represenaion of A J, B J and C J. This can be seen from he parial fracion expansion of G(s), which isg ( s). The numeraor s s ( s ) coefficiens are in C, he hree roos, -, - remain on he diagonal bu he off-diagonal uni erm in A and he zero in B are due o he fac ha he las erm of he parial fracion expansion has as inpu he oupu from he second sae, no he inpu u. This form of A marix is known as a Jordan form and due o he numerical mehods used canno be found wih MATLAB..4.4 Guillemin Form Anoher simple way of obaining a sae space represenaion of a ransfer funcion is o make repeaed use of he sae represenaion of Figure.3 for he one-zero one-pole ransfer funcion ( ds c) /( s a) which has a sae space represenaion (a,,e-ad,d) for he single sae x. Figure.3 Diagram for One Sae Variable for Guillemin Form. 8

9 Thus for a ransfer funcion wih real poles and zeros in facored form given by G( s) ( s )( s ) / s( s 3)( s 5) one can spli i ino one of several possible series (cascade) s s combinaions such as ( )( )( ) and hen use he represenaion of Figure.3 for each s s 3 s 5 ransfer funcion o consiue he overall model as in Figure.4, which has, assuming he oupus of he inegraors from lef o righ are x 3, x and x, respecively, he equaions x 3 u x 3x x3 x 5x x x he derivaives gives he sae represenaion and y x x. Subsiuing appropriaely for 5 A 3, B and C 3 9

10 where he A marix is upper riangular. Figure.4. Diagram for Guillemin Form Sae Represenaion..5. Sae Transformaions beween Differen Forms Given a sae space represenaion hen one can evaluae he corresponding ransfer funcion and use his o obain a differen sae space represenaion. In some cases, however, i is more convenien if one can obain he specific sae ransformaion, T, discussed in secion.3, ha will do his direcly. I can be shown ha any A marix can be ransformed o a diagonal form by is own eigenvecor marix. Eigenvecors only define direcions, however, so ha such a marix is no unique wih a scalar muliplier being allowed on any column vecor i of T. The eigenvecor, i, corresponding o a paricular eigenvalue, s i, of a marix A is found from ( s I A) for i = n. For example, he A marix A has a characerisic equaion of 5 ( s )( s 5), giving s 3s ( s )( s ), so has eigenvalues of - and -. The corresponding eigenvecors are obained from ( I A ) and ( I A), a 3a yielding T and T a T 3a b 4b where a and b are any consans. Thus aking b hen he ransformaion T - AT will yield he diagonal 4b marix A whaever he choice of a and b. However, if he ransformaion was applied o a sae space represenaion (A,B,C,D) he resuling (A,B,C,D), would have differen resuls for B and C dependen on he choice of a and b. When he given A marix is in conrollable canonical form hen i can be shown ha he column eigenvecors i of T are given by i s i s i... s ( n) T i, which is known as a Vandermonde marix. i i

11 .5. Transforming o Conrollable Canonical Form If i is required o find he conrollable canonical form of a sae space represenaion (A,B,C,D) hen his can be achieved by a unique ransformaion as no only is he A c marix of a specific form bu so also is he vecor B c. From he wo equaions T AT A c and TB c B i can be shown ha he column vecors i of T are given, for i = n, by n B, A a n, A an, A3 a3n, ec. Here he a i, i =..(n-), are he coefficiens of he characerisic equaion of A, which of course form he las row of A c. Some algebraic manipulaions on hese equaions show ha he ransformaion marix T can be wrien as T B AB A B... A n B a n a a. n n a n. a a (.7).6 Evaluaion of he Sae Transiion Marix There are several ways o evaluae he sae ransiion marix ( e oulined below. A and some of hese are.6. Direc Expansion This is edious and involves calculaing powers of A, subsiuing hem in equaion (.5) and finding he exponenial series which give each erm in he summed marix expression..6. The Inverse Laplace Transform This involves finding ( from he inverse of equaion (.), ha is ( = (si-a) -. This is sraighforward bu very laborious for calculaing he required marix inversion, excep for low order marices, A. One hen has o find he inverse Laplace ransforms of he individual marix erms which are funcions of s.

12 .6.3 Use of a Diagonal Transformaion If he marix T is a ransformaion which diagonalises he A marix o hen i can be shown ha ( e A Te T. Thus, once T and is inverse have been found his approach requires evaluaion of he produc of hree n x n marices..6.4 Use of he Cayley Hamilon Theorem This heorem saes ha a marix saisfies is own characerisic equaion. Thus, if he marix A n n has a characerisic equaion s) s a s... a s a, hen ( n n n ( A) A an A... a A a. This means ha A n can be calculaed in lower 3 powers of A and ha any infinie series of A, f ( A) ci c A c A c3 A... o., 3 n can be expressed as f ( A) I A A 3 A... n A for a marix of order n. Furher all eigenvalues of A mus also saisfy his equaion. Thus when he funcion of he marix A of order n is he exponenial is eigenvalues and he marix mus saisfy n n k e si A k si and k e k A (.8) k k

13 The firs equaion when used for all he eigenvalues provides n equaions which can be solved for he n coefficiens. Subsiuing hese in he second equaion enables he sae ransiion marix ( e A o be found..7 Conrollabiliy and Observabiliy Consider a sae space represenaion (A,B,C,) wih A, 3 4 B and C Then, in block diagram form his consiss of he four modes a -, -, -3, -4 which are conneced respecively o he inpu and oupu, oupu only, inpu only and o neiher inpu or oupu. The ransfer funcion from inpu o oupu is simply /(s+) as he - mode is he only one conneced o boh he inpu and oupu. Since he - mode is conneced o boh inpu and oupu i is said o be boh conrollable and observable. The - mode is said o be unconrollable and observable being conneced o he oupu only; he -3 mode conrollable and unobservable being conneced o he inpu only; and he -4 mode is said o be unconrollable and unobservable being conneced o neiher he inpu or he oupu. Given a sae space descripion i is desirable, as will be seen in he nex chaper, o know which modes are in he differen siuaions exemplified by he four above modes. A sysem is said o be conrollable if all he saes are conrollable, and observable if all he saes are observable. The formal definiions are given below. From he above example i is clear ha only hose modes which are conrollable and observable appear in he ransfer funcion beween inpu and oupu. Thus, if a sysem wih an n x n A marix is conrollable and observable he denominaor of is ransfer funcion will be of order n (i.e. i will have n poles)..7. Conrollabiliy A sysem is conrollable if here exiss an inpu u which ransfers he iniial sae x() o he zero sae x( = in a finie ime. Given any SISO sysem, A (n x n) and B (n x ) marices hen i can be shown ha he sysem will be conrollable if he (n x n) conrollabiliy n marix B AB A B... A B has rank n. I will be noiced ha his marix is he firs par of he ransformaion marix for T in equaion (.7) and, as a consequence, a sysem can only be pu ino conrollable canonical form if i is conrollable. Or, alernaively, a sysem which has a conrollable canonical form sae space represenaion is conrollable. 3

14 .7. Observabiliy A sysem is observable if he iniial sae x() can be uniquely deermined by observing he oupu over a finie ime. Given any SISO sysem, A (n x n) and C ( x n) marices hen i can be C CA shown ha he sysem will be conrollable if he (n x n) observabiliy marix CA :: n CA rank n. Again i can be shown ha a sysem can only be pu ino observable form if i is observable. has.8 Cascade Connecion In previous chapers on conrol sysem design significan aenion has been given o cascade compensaion and he effec on he open loop frequency response locus of adding a compensaor. If he compensaor and plan are given in sae space form hen i may be desirable o obain a sae space represenaion for heir cascade combinaion. Thus, le he compensaor G c (s) wih sae z, inpu e, and oupu u have he sae space represenaion (A,B,C,D ) and he plan G(s) wih sae x, have inpu u, and oupu c have he sae space represenaion (A,B,C,D ), hen z A z Be, u Cz De and A x B u, C x D u x c z A Wriing a combined sae vecor (z, x) T one can wrie x BC z x D C C D D e c z B e A x BD which gives a sae space represenaion (A,B,C,D) wih and A B A, BC A BD B, D C C C and D D D. (.9) 4