Tutorial 12

Ganesh

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1 Control Engineering CE447 Tutorial 12 Prepared by A/Prof Victor Sreeram School of Electronic and Computer Engineering University of Western Australia School of Electrical, Electronic and Computer Engineering. Control Engineering (CE) 447 Tutorial 12 1. A helicopter near hover can be described by the equations x -0.02 -1.4 9.8 x 9.8 1 1 x = -0.01 -0.4 0.0 x + 6.3 u 2 2 x 0.0 1.0 0.0 x 0.0 3 3 where x = horizontal velocity, x = pitch rate, x = pitch angle, 1 2 3 u = rotor tilt angle. (a) Find the open loop poles (b) Show that a state feedback law to move the poles to s = -2, s = -1– j is k = 0.0628 0.4706 0.9949 [ ] 2. Consider the system defined by x = Ax + Bu where 0 1 0 0 A = 0 0 1 , B = 0 , C = 1 2 3 [ ] -2 -4 -6 1 By using the state feedback control u = - Kx, is desired to have the closed-loop poles at s = - 2 – j4, s = - 6. Determine the state feedback gain matrix K. 3. Consider the system defined by x = Ax + Bu y = Cx where 0 1 0 0 A = 0 0 1 , B = 0 , c = 1 0 0 [ ] -6 -11 -6 1 2 Control Engineering CE447 ...

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• Publié par: Urne
• Nombre de lectures: 106
• Langue: English

Extrait

Control Engineering CE447 Tutorial 12 1 Prepared by A/Prof Victor Sreeram School of Electronic and Computer Engineering University of Western Australia     School of Electrical, Electronic and Computer Engineering. Control Engineering (CE) 447 Tutorial 12 1.   A helicopter near hover can be described by the equations   x 1  − 0.02 − 1.4 9.8 x 1 9.8                 x   2   =    − 0.01 − 0.4 0.0         x 2    +    6.3      u   x  3      0.0 1.0 0.0    x 3     0.0      where x 1 = horizontal velocity, x 2 = pitch rate, x 3 = pitch angle,      u  rotor tilt angle. (a) Find the open loop poles (b) Show that a state    =  feedback law to move the poles to s = -2, s =   − 1 ± j  is        k = [ 0.0628 0.4706 0.9949 ]     2. Consider the system defined by                     x  = Ax + Bu    where  0 1 0 0    = 0 0 1 , 0 , = [ ]   A      − 2 − 4 − 6      B =      1      C  1 2 3  By using the state feedback control u = − Kx , is desired to have the  closed-loop poles at s = − 2 ± j 4, s =   − 6. Determine the state feedback  gain matrix K .     3. Consider the system defined by x  = Ax + Bu                                              y = Cx    where  0 1 0 0  = 0 0 1 , 0 , = [ ] A      − 6 − 11 − 6      B =      1      c  1 0 0

Control Engineering CE447 Tutorial 12 2 Prepared by A/Prof Victor Sreeram School of Electronic and Computer Engineering University of Western Australia      It is desired to design a full-order state observer. Determine the observer  Gain matrix L by direct method. Assume that the desired eigenvalues of  the observers are: s   =   − 2 ± j 3.464, − 5 .   4. Antenna azimuth tracking system can be described by the following state-  space equations                           x = Ax + u   B                           y Cx      =  where            0 − 1 0 =           A           − 1 − 1    , B   =      1    , c = [ 1 0 ]   Design a controller/observer compensator by placing controller poles at s   =   − 3, − 4

  5.   Consider the undamped harmonic oscillator                 x 1 = x 2    2                x 2 = −ω 0 x 1 + u   a)   Using the observation of velocity, y = x 2, design an observer/state feedback compensator to control the position x 1 . Place the state- feedback controller poles at s   =   −ω 0 ± j ω 0 and both observer poles at s   =   −ω 0 . b)   Write the state-space equations for the full state observer and draw  also its block diagram. c)    Assuming that the state x 1 can be measured, derive the state-space  equations for a reduced order observer (from first principles) to  estimate the state x 2 . Draw also the block diagram of the reduced  order observer.    

Control Engineering CE447 Tutorial 12 Prepared by A/Prof Victor Sreeram School of Electronic and Computer Engineering University of Western Australia     Tutorial Solutions   Question 1:  

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The desired closed-loop characteristic polynomial                   α ( s ) = ( s + 2 ) ( s + 1 − j ) ( s + 1 − j )     = ( s + 2 ) s 2 + 2 s + 2 = s 3 + 4 s 2 + 6 s + 4                     +   s 0 0         − 0.02 − 1.4 9.3       9.8             sI − A bk  =  0 s 0 − − 0.01 − 0.4 0 + 6.3 [ k 1 k 2 k 3 ]       0 0 s    0 1.0 0      0      s 0.02 1.4 9.8 9.8 k 1 9.8 k 2 9.8 k 3  = + 0.01 s + 0.4 − 0    +    6.3 k 1 6.3 k 2 6.3 k 3        0 − 1.0 s     0 0 0      where k = [ 0.0628 0.4706 0.9949 ]             sI − A + bk  = s 3 + 4 s 2 + 6 s + 4 (verify  this matrix using matlab).     Question 2:                                      x = Ax + Bu       where  0 1 0   0  =                         A       − 0 − 0 − 1     , B   =       0       2 4 6 1  

Control Engineering CE447 Tutorial 12 Prepared by A/Prof Victor Sreeram School of Electronic and Computer Engineering University of Western Australia     Note A , b is controllable. ∴ Pole placement is possible. Desired closed-loop system poles = − 2 ± j 4, − 6.      The desired characteristic equation for the closed loop is               α ( s ) = ( s + 2 + j 4 ) ( s + 2 − j 4 ) ( s + 6 )                           = s 2 + 4 s + 20 ( s + 6 ) = s 3 + 10 s 2 + 44 s + 120    Let the desired feedback gain matrix                                     k =   [ k 1 k 2 k 3 ]      The closed-loop characteristic polynomial is given by    s 0 0   0 1 0  0    0 0 s       − 2 − 4 − 6        1      k k k            sI − A + bk  =  0 s 0  −  0 0 1  +  0 [ 1 2 3 ]          −  =  2  0 s k 1 4 + s 1 k 2 s + 6 − 0 1 + 3          + k             sI − A + bk  = s 3 + ( 6 + k 3 ) s 2 + ( 4 + k 2 ) s + ( 2 + k 1 )            Equating sI − A + bk  = α ( s )    we get    6 + k 3 = 10, 4 + k 2 = 44 and 2 + k 1 = 120  

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Control Engineering CE447 Tutorial 12 5 Prepared by A/Prof Victor Sreeram School of Electronic and Computer Engineering University of Western Australia      Solving we get          k 1 =4, k 2 =40, k 3 =118   =        k   [ 4 40 118 ]     Question 3:   c c A       1 0 0                   = cA 2    =   0 0 1 0 0 1            Since rank [  ] = 3, c , A is observable, Therefore full state observer design is possible. The desired characteristic polynomial of the observer                      α ( s ) = ( s + 2 − j 3.464 ) ( s + 2 + j 3.464 ) ( s + 5 )      = s 3 + 9 s 2 + 36 s + 80 (1)   The characteristic polynomial of the observer    s     l sI A LC s 0 1 + l 1 0 0           − +  =     0 0 0 0 s 0 0     −         −     6 0 0   −         1  1 1   −   6 0         l 3 2 1      [ ]      s + l 1 − 1 0    − 1     = l 2   s l 3 + 6 11 s + 6            sI − A + LC  = s 3 + ( l 1 + 6 ) s 2 + ( 6 l 1 + l 2 + 11 ) s + ( 11 l 1 + 6 l 2 + l 3 + 6 )  (2)

Control Engineering CE447 Tutorial 12 Prepared by A/Prof Victor Sreeram School of Electronic and Computer Engineering University of Western Australia      Equating (1) and (2) we have                     l 1 + 6 = 9 ∴   l 1 = 3    6 l 1 + l 2 + 11 = 36 ∴   l 2 = 7    11 l 1 + 6 l 2 + l 3 + 6 = 80 ∴   l 3 = -1                         ∴     L =         3 7      − 1     Question 4:         The desired characteristic polynomial            α c ( s ) = ( s + 2 ) ( s + 3 ) = s 2 + 5 s + 6 (1)              sI − A + bk  =   s 0 s 0      0 1    1 1   +   1 0     [ k 1 k 2 ]    −    − −         s − 1                                       =   1 + k 1 s + 1 + k 2               sI A bk  = s 2 + ( 1 + k 2 ) s + ( 1 + k 1 )  (2) − +    Equating (1) and (2) we have                  1 + k 2  = 5 ∴   k 2 = 3  

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Control Engineering CE447 Tutorial 12 Prepared by A/Prof Victor Sreeram School of Electronic and Computer Engineering University of Western Australia      1 + k 1 = 6 ∴   k 1 = 5                                        K = [ 5 4 ]             Observer design.                  α o ( s ) = ( s + 3 ) ( s + 4 ) = s 2 + 7 s + 12 (1)               sI − A + LC  =    0 s 0 s    −      −   1 0  −   1 1    +    l l 2 1      [ 1 0 ]        s + l 1 − 1                                        = 1 + l 2 s + 1             sI − A + LC  = s 2 + ( 1 + l 1 ) s + ( 1 + l 1 + l 2 )  (2)    Equating (1) and (2) we have   =  1 + l 1  7 ∴   l 1 = 3  1 + l 1 + l 2                     ∴   l 2 = 5   6                          ∴    L = 5                Alternate solution                       x  = Ax + Bu , y = cx             =   0 − 1   , =     0  , = [ 1 ] A  − 1 − 1  B  1   c 0

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Control Engineering CE447 Tutorial 12 Prepared by A/Prof Victor Sreeram School of Electronic and Computer Engineering University of Western Australia                   = [ B AB ]  =  1 0 − 1 1        rank [  ] = 2. ∴    A , b is controllable.

                        = c c A   =   0 1 1 0           rank [  ] = 2. ∴    c , A is observable.     Open-loop characteristic polynomial                ( )  = s − 1    a s = sI − A   1 s + 1                            2 2     = s + s + 1 = s + a 1 s + a 2                        ∴      a 1 = 1, a 2 = 1         Controller design: determination of Gain k      =                W   0 1 a 1 1     =   1 1      0 1     The desired controller polynomial                   α ( s ) = ( s + 2 ) ( s + 3 )  = s 2 + 5 s + 6   2 =                              s +α 1 s +α 2    

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Control Engineering CE447 Tutorial 12 Prepared by A/Prof Victor Sreeram School of Electronic and Computer Engineering University of Western Australia                        ∴      α 1 = 5, α 2 = 6        Using Bass Guras formula             K =   (α 1 − a 1 )( α 2 − a 2 )     W − 1 e − 1                   − −                     ( 5 − 1 ) ( 6 − 1 )    1 1   1    0 − 1    1   = 0 1 1 1                             = [ 4 5 ]   0 1   −  1 1         1 1 0 1                      = [ 4 1 ]      1 1 0 1     = [ 5 4 ]      The desired observer polynomial             α ( s ) = ( s + 3 ) ( s + 4 )  = s 2 + 7 s + 12    = s 2 +α 1 s +α 2                   ∴      α 1 = 7, α 2 = 12    Using Bass Guras formula         1 1  1                l = ( W  ) − α α − − a    2 a 2                 =       a 1 1 1 0      c c A       − 1    ( ( 1 7 2 − − 1 1 ) )     =       a 1 1 1 0         c c A       − 1    1 6 1      

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