Chapter #1 EEE8013 EEE3001. Linear Controller Design and State Space Analysis

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1 Chaper EEE83 EEE3 Chaper # EEE83 EEE3 Linear Conroller Design and Sae Space Analysis Ordinary Differenial Equaions.... Inroducion.... Firs Order ODEs Second Order ODEs General Maerial Roos are real and unequal Roos are Complex (and hence unequal) Roos are real and equal Tuorial Exercise I... 9 Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk /9

2 Chaper EEE83 EEE3 Ordinary Differenial Equaions. Inroducion To undersand he propeies (dynamics) of a sysem, we can model (represen) i using differenial equaions (DEs). The response/behaviour of he sysem is found by solving he DEs. In our cases, he DE is an Ordinary DE (ODE), i.e. no a paial derivaive. The main purpose of his Chaper is o learn how o solve firs and second order ODEs in he ime domain. This will serve as a building block o model and sudy more complicaed sysems. Our ulimae goal is o conrol he sysem when i does no show a saisfacory behaviour. Effecively, his will be done by modifying he ODE.. Firs Order ODEs The general form of a firs order ODE is: d f x, () where x, Analyical soluion: Explici formula for x() (a soluion which can be found using various mehods) which saisfies f ( x, ) d The proper noaion is x() and no x bu we drop he brackes in order o simplify he presenaion. Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk /9

3 Chaper EEE83 EEE3 Example.: Prove ha x e and xe are soluions of 3x d. de 3x 3e 3e 3e d d de 3x 3e 3e 3e d d Obviously here are infinie soluions o an ODE and for ha reason he found soluion is called he General Soluion of he ODE. Firs order Iniial Value Problem : f ( x, ), x x An iniial value problem is an ODE wih an iniial condiion, hence we do no find he general soluion bu he Specific Soluion ha passes hrough x a =. d Analyical soluion: Explici formula for x() which saisfies f ( x, ) and d passes hrough x when. Example.: Prove ha x e is a soluion, while x of 3 xx, d Boh expressions ( x e and x e ) saisfy he 3xbu a = d x e x x e x e is no a soluion Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 3/9

4 Chaper EEE83 EEE3 For ha reason some books use a differen symbol for he specific soluion:,, x. You mus be clear abou he difference beween an ODE and he soluion o an IVP! From now on we will jus sudy IVP unless oherwise explicily menioned. Linear Firs Order ODEs A linear s order ODE is given by: x' b x c, a a ax' bx c, a Non auonomous Auonomous () wih a,b,c and a. In engineering books he mos common form of () is (since a ): x' k x u (3) wih k,u Noe: We say ha u is he inpu o our sysem ha is represened by (3) The soluion of (3) (using he inegraing facor) is given by: k k k x e x e e u d Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 4/9

5 Chaper EEE83 EEE3 k The erm e x is called ransien response, while comes k k e e u d from he inpu signal u. If we assume ha u is consan: k k k x e x e e ud x e k x u e k k Hence: lim x u u / k, k k, k Thus we say ha if k> he sysem is sable (and he soluion converges exponenially a u/k) while if k< he sysem is unsable (and he soluion diverges exponenially o, ). Example.3: u= and k= & 5, x =.5 Transien Toal.5 Transien Toal Inpu componen Inpu componen Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 5/9

6 Chaper EEE83 EEE3 Example.4: u= and k=- & 5, x = Toal Transien Inpu componen.5 x Toal Transien Inpu componen.5.5 Example.5: u= and k=5, x = & Toal Transien Inpu componen 4 3 Toal Transien Inpu componen Example.6: u=- & and k=5, x =.5 Transien Inpu componen Toal Toal Transien Inpu componen.5.5 Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 6/9

7 Chaper EEE83 EEE3 Commens: In real sysems we canno have a sae (say he speed of a mass-spring sysem) ha becomes infinie, obviously he sysem will be desroyed when x ges o a high value. For he dynamics (seling ime, sabiliy ) of he sysem we should only focus on he homogenous ODE: x' k x 3. Second Order ODEs 3. General Maerial A second order ODE has as a general form: d x f x', x, d (4) A linear nd order ODE is given by: x'' A x' B x u, Non auonomous x'' Ax' Bx u, Auonomous (5) And again we focus on auonomous homogeneous sysems: x'' A x' B x (6) Again we define as an analyical soluion of (6) an expression ha saisfies i. Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 7/9

8 Chaper EEE83 EEE3 Example.7: Given x'' x' 3x prove ha soluions: e '' e ' 3 e 9e 6e 3e e '' e ' 3 e e e 3e x e and x e are wo Assume ha you have soluions for a nd order ODE x and x (we will see laer how o ge hese wo soluions), hen: x A x B x x A x B x obviously I can muliply hese wo equaions wih arbirary consans: Cx C A x CB x Cx CA x CB x and now I can add hem and collec similar erms: C x C x '' A C x C x ' B C x C x Common Term Common Term Common Term which means ha C x C x is also a soluion of he ODE. (i.e. he linear combinaion of x and x ) Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 8/9

9 Chaper EEE83 EEE3 Example.8: Given x'' x' 3x prove ha e e '' e e ' 3 e e e e e e e e e e 6e 4e 3e 6e 9e 6e 3e e 4e 6e x e e is a soluion: Now, he quesion is, if we have x and x, can ALL oher soluions of he ODE, be expressed as a linear combinaion of x and x? So assume a hird soluion : A B '' ' Now, he quesion can be wrien as, can we find consans C and C such as: Cx Cx ' C x ' C x ' This equaion can be seen as a by sysem wih unknowns C and C as: C x x x' x' C ' From linear algebra his sysem of equaions has a unique soluion if: ' ' x x x x x x ' x x ' Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 9/9

10 Chaper EEE83 EEE3 Noe: The marix W x, x of he ODE. ' ' x x x x is called he Wronskian We also know from linear algebra ha he deerminan is no zero if: ' ' x x C x x So if he wo soluions x and x are linear independen (LI) hen ANY oher soluion can be described by he linear combinaion of x and x. So now we have o look for wo LI soluions for he nd order ODE. Example.9: Prove ha wo soluions of x'' x' 3x, x e are linear independen. x e and x x e e e e W x, x W x' x' 3e e 3e e W e e e e e e e Example.: Prove ha wo soluions of x'' x' 3x, x e are NOT linear independen. 3 x e and From he Polish mahemaician Józef Maria Hoëne-Wroński Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk /9

11 Chaper EEE83 EEE3 x x e e, x' x' 3e 6e W x x e e W e e 3e 6e Example.: For he ODE x'' x' 3x prove ha he soluion 3 x e e canno be wrien as any combinaion of x e and x e x C x C x e e C e C e C C e From his expression we have ha C C (and hence we have he erm e ) bu here is no erm e for e. Bu how can we find wo LI soluions? For homogeneous s order ODEs wih k u= he soluion was: order ODEs: x e C so we will ry a similar approach for nd x '' Ax' Bx, assume 3 x e => x' re & x' ' r e => x '' Ax' Bx r e Are Be r Ar B (7) This is called he Characerisic Equaion (CE) and we have o check is roos: A r A 4B, hese are he Characerisic values or Eigenvalues. 3 Noice ha we do NOT know wha is he value of r. Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk /9

12 Chaper EEE83 EEE3 3. Roos are real and unequal If A 4B he sysem is called Overdamped and he wo roos are r and r wih r r, r, r soluions as: ' '. Then r r x x e e x x re r e x e and r x e are wo linear independen r r e re e re hence he general soluion is r Cx Cx Ce Ce (8) x If r and r < hen x and he sysem is sable. If r or r > hen x and he sysem is unsable. Example.: The CE of x'' x' 3x is r r 3 which means 43 r 5 ha he wo roos are: r, r 6 5 x e e and hence he LI soluions are 6 x e e 5 6 This means ha he general soluion is x Ce Ce and hence he ODE is sable. The Wronskian is x x e e 6 5 e x x e e e e e e 5 6 ' ' 5 6 If he iniial condiion is x, x' hen: C C C 6 x 6e 5e 5C 6C C Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk /9

13 Chaper EEE83 EEE3 3.4 Roos are Complex (and hence no equal) If A 4B hen he sysem is called Underdamped and he wo roos are r a bj and r r a bj wih r r, r, r. Then and a bj x e e are wo linear independen soluions as x e e a bj e abj e a bj e a bj e e a bj e abj abj a bj e a bj e a a a a a bje a bje e a bj a bj e bj a bj a bj a bj a bj Hence he general soluion is x C x C x C e C e (9) bu remember ha C and C are complex now variables such as x. Example.3: The CE of x'' x' 5x is r r 5 which means 6 4 j r j ha he wo roos are: r, j r j j x e e and hence he LI soluions are j x e e j j This means ha he general soluion is x Ce Ce and hence he ODE is sable. The Wronskian is x x e j j j x' x' j e j e j j j j j e e j e e j e j e j j e 4je e j Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 3/9

14 Chaper If he iniial condiion is x C jc, x' hen: C 4 j C C j x j e j e 4 4 C j j j 4 EEE83 EEE3 An alernaive approach is no o use x & x bu a linear combinaion of hem: y e e, y e e e e e e Noe ha re re re re Using Euler s formula: abj a e e cosb jsinb and hence: a bj a bj a a y e e e cosb jsinb cosb jsinb e cosb a bj a bj a a y e e e cosb jsinb cosb jsinb je sinb As y and y are soluions so do y, y. So he general soluion when j we have complex roos is: a x e C cosb C sin b, C, C () Example.4: The CE of x'' x' 5x is r r 5 which means 6 4 j r j ha he wo roos are: r, j r j Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 4/9

15 Chaper EEE83 EEE3 x e cos and hence he LI soluions are x e sin This means ha he general soluion is x e C cos C sin and hence he ODE is sable. The Wronskian is x x e cos e sin e x ' x ' e cos e sin e sin e cos If he iniial condiion is x, x' hen: C C C C C.5 x e cos.5sin 3.3 Roos are real and equal If A 4Bhen he sysem is called Criically damped and he wo roos are r r r wih r. One soluion is x e bu how abou x? We can use x e and he general soluion: r x C x C x C e C e () r The Wronskian is: e r e r re e e r r r r r r r r r r r r e e e re e e e e e Example.5: The CE of x'' x' x is r he wo roos are: r, r r r which means ha Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 5/9

16 Chaper EEE83 EEE3 x e and hence he LI soluions are x e This means ha he general soluion is x Ce Ce and hence he ODE is sable. The Wronskian is e e e e e e If he iniial condiion is x C C C C C x e e, x' hen: No assessed maerial To see why x e is he nd soluion go o he ODE and place e '' Ae ' Bx e r Ar B x e : Since r is a double roo of he CE: a. So: e '' A e ' Bx e a r r Taking he ime derivaive w r: '' ' d e d e d e A B d e a r r dr dr dr dr r Ar B a r r for some consan And as we can change he sequence of he differeniaion: d e ar r '' ' d e d e d e A B dr dr dr dr By using simple calculus: Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 6/9

17 Chaper EEE83 EEE3 '' ' e Ae Be e ar r e ar r '' ' d e d a r r e A e Be ar r e dr dr '' ' By placing now where r=r : Which means ha e r e r re re e r r r And hence e A e Be e mus be a soluion of my ODE and: r r r r r r r r r e re e e re re e re e x e is my second soluion. Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 7/9

18 Chaper EEE83 EEE3 Roo Space jb jb a a Criical or overdamped underdamped Sable Unsable Name Oscillaions? Componens of soluion Overdamped No Two exponenials: Criically damped No k k e, e, k k, Two exponenials: k k e, e, k Underdamped Yes One exponenial and one cosine e k cos,, k Undamped Yes one cosine cos Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 8/9

19 Chaper EEE83 EEE3 4. Tuorial Exercise I. By using he general form of he analyic soluion ry o predic he response of he following sysems. Your answer mus describe he sysem as sable/unsable, convergen o zero/nonzero value. Crosscheck your answer by solving he DE: d 5 6x, x, x, x d 5 6x, x, x, x d 5 6x, x, x, x d 5 6x, x, x, x d 3, x, x, x. Find he soluion of x 6x 5x, x, x 3. Briefly describe how he soluion behaves for hese iniial condiions. Draw a skech of he response. 3. Find he soluion of x x 6x, x, x. Briefly describe how he soluion behaves for hese iniial condiions. Draw a skech of he response. 4. Find he soluion of x x.5x, x, x / 3. Briefly describe how he soluion behaves for hese iniial condiions. Draw a skech of he response. 5. Find he Wronskian marices of he soluions of Q-5. Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 9/9