State Space: Observer Design Lecture 11

Transcription

1 State Space: Oberver Deign Lecture Advanced Control Sytem Dr Eyad Radwan Dr Eyad Radwan/ACS/ State Space-L

2 Controller deign relie upon acce to the tate variable for feedback through adjutable gain. Thi acce can be provided by hardware. Sometime it i impractical to ue thi hardware for reaon of cot, accuracy, or availability (i.e. ome of the tate variable may not be available at all, or it i too cotly to meaure). It i poible to etimate the tate that are not available or eay to meaure. Etimated tate, rather than actual tate, are then fed to the controller. An oberver, or an etimator, i ued to calculate tate variable that are not acceible from the plant. Dr Eyad Radwan/ACS/ State Space-L0 2

3 An oberver cheme i hown in Figure -. r=0 u Plant Plant output y Controller Oberver xˆ Etimated tate, The plant ha the following tate equation: Figure - Etimated output ŷ x y Ax Cx Bu (.) Dr Eyad Radwan/ACS/ State Space-L0 3

4 The oberver ha the following equation: xˆ Axˆ Bu yˆ Cxˆ Subtracting eq.(.2) from (.) give: x x = A(x x) y y = C(x x) (.2) (.3) From eq.(.3) the dynamic of the difference between the actual and etimated tate i unforced, and if the plant i table, thi difference, due to difference in initial tate vector, approache zero. However, the peed of convergence between the actual tate and the etimated tate i the ame a the tranient repone of the plant ince the characteritic equation for eq.(.3) i the ame a that for eq.(.). Dr Eyad Radwan/ACS/ State Space-L0 4

5 To increae the peed of convergence between the actual and etimated tate we ue feedback, hown conceptually in Figure -2a and in more detail in Figure -2b. The error between the output of the plant and the oberver i fed back to the derivative of the oberver' tate. r=0 u Plant Oberver Plant output y u B ò A xˆ C ŷ - Etimated error output y Controller xˆ Etimated tate, Figure -2a Etimated output ŷ Dr Eyad Radwan/ACS/ State Space-L0 L To controller Figure -2b 5

6 The ytem correct to drive thi error to zero. With feedback we can deign a deired tranient repone into the oberver, that i much quicker than that of the plant or controlled cloed-loop ytem. In deigning an oberver, it i the oberver canonical form that yield the eay olution for the oberver gain. Figure -3a how an example of a third-order plant repreented in oberver canonical form. In Figure -3b, the plant i configured a an oberver with the addition of feedback, a previouly decribed Dr Eyad Radwan/ACS/ State Space-L0 6

7 b 2 b 3 R() b yˆ ( ) ˆx 3 ˆx 2 -a 2 ˆx -a -a 0 Figure -3a R() b b 3 b 2 y ˆ( ) - y() ˆx 3 ˆx 2 -a 2 ˆx -a l -a 0 l 3 l 2 Figure -3a Dr Eyad Radwan/ACS/ State Space-L0 7

8 The deign of the oberver conit of evaluating the contant vector, L, o that the tranient repone of the oberver i fater than the repone of the controlled loop in order to yield a rapidly updated etimate of the tate vector. finding the tate equation for the error between the actual tate vector and the etimated tate vector, (x x): x = Ax Bu L y y (.4) y = Cx and from the tate equation (.): x = Ax Bu y = Cx Dr Eyad Radwan/ACS/ State Space-L0 8

9 Subtracting eq.(.4) from (.), yield: x x = A x x L y y y y = C x x x x = A x x L C x x x x = (A L C) x x Or e x = A L C e x (.5) Where e x = x x The characteritic equation reulting from eq.(.5) can be olved to meet the deign requirement by deigning the value of L to yield tability and tranient repone that i Dr Eyad Radwan/ACS/ State Space-L0 9

10 fater than the controlled cloed-loop repone. For an n-th order ytem repreented in oberver canonical form; a n 0 0 l a n l 2: A LC = 0 0 a 0 0 l n a o 0 0 l n (a n l ) 0 0 (a n 2 l 2 ) = (.6) (a l n ) 0 0 (a o l n ) 0 0 Dr Eyad Radwan/ACS/ State Space-L0 0

11 The characteritic equation of (.6) can be written a: det(a LC) =0 n a n l n a n 2 l 2 n 2 a l n a 0 l n = 0 Equating (.7) with deired deign equation: (.7) n d n n d n 2 n 2 d d 0 = 0 The oberver gain can be found a: l i = d n i a i i = 0,,2, n (.8) Dr Eyad Radwan/ACS/ State Space-L0

12 Example. Deign an oberver for the plant ( 4) G = ( )( 2)( 5) which i repreented in oberver canonical form. The oberver will repond 0 time fater than the controlled loop deigned in example 0.2 ( 2 n n 2 5 ). Solution: G = ( 4) ( )( 2)( 5) = ( 4) Dr Eyad Radwan/ACS/ State Space-L0 2

13 repreenting the ytem in oberver canonical form a in Figure.4, R() 4 yˆ ( ) ˆx 3 ˆx 2-8 ˆx -7-0 Figure.4 Or in tate pace form: x = Ax Bu, y = Cx 8 0 x = x 0 4 u Dr Eyad Radwan/ACS/ State Space-L0 3

14 Writing the oberver error a in eq.(.5): (8 l ) 0 e x = A L C e x = (7 l 2 ) 0 (0 l 3 ) 0 0 Now writing the characteritic polynomial: det I A LC = 0 = 3 8 l 2 7 l 2 0 l 3 = 0 e x Comparing the above equation with the deired deign equation to give tranient 0 time fater than: 2 2 n 2 n Dr Eyad Radwan/ACS/ State Space-L0 4

15 The pole of the oberver hould be at: -0±j20 and the third pole i elected 0 time the real part of the dominant pole thu it i placed at -00. hence the deired polynomial equation: = Thi give the oberver gain a: L=2, l2=2483, and l3=49,990. the SFG of the oberver: R() 4 ˆx 3 ˆx 2-8 ˆx y ˆ( ) - y() Dr Eyad Radwan/ACS/ State Space-L0 5