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Plan of the Lecture Review: coordinate transformations; conversion of any controllable system to CCF. Todays topic: pole placement by (full) state feedback. Plan of the Lecture Review: coordinate transformations; conversion of any


  1. Plan of the Lecture ◮ Review: coordinate transformations; conversion of any controllable system to CCF. ◮ Today’s topic: pole placement by (full) state feedback.

  2. Plan of the Lecture ◮ Review: coordinate transformations; conversion of any controllable system to CCF. ◮ Today’s topic: pole placement by (full) state feedback. Goal: learn how to assign arbitrary closed-loop poles of a controllable system ˙ x = Ax + Bu by means of state feedback u = − Kx .

  3. Plan of the Lecture ◮ Review: coordinate transformations; conversion of any controllable system to CCF. ◮ Today’s topic: pole placement by (full) state feedback. Goal: learn how to assign arbitrary closed-loop poles of a controllable system ˙ x = Ax + Bu by means of state feedback u = − Kx . Reading: FPE, Chapter 7

  4. State-Space Realizations x = Ax + Bu ˙ u y y = Cx ↓ G ( s ) = C ( Is − A ) − 1 B

  5. State-Space Realizations x = Ax + Bu ˙ u y y = Cx ↓ G ( s ) = C ( Is − A ) − 1 B Open-loop poles are the eigenvalues of A : det( Is − A ) = 0

  6. State-Space Realizations x = Ax + Bu ˙ u y y = Cx ↓ G ( s ) = C ( Is − A ) − 1 B Open-loop poles are the eigenvalues of A : det( Is − A ) = 0 Then we add a controller to move the poles to desired locations: + KD ( s ) G ( s ) R Y −

  7. Goal: Pole Placement by State Feedback Consider a single-input system in state-space form: x = Ax + Bu ˙ u y y = Cx Today, our goal is to establish the following fact: If the above system is controllable , then we can assign arbitrary closed-loop poles by means of a state feedback law   x 1 x 2   � �   u = − Kx = − k 1 k 2 . . . k n .  .  .   x n = − ( k 1 x 1 + . . . + k n x n ) , where K is a 1 × n matrix of feedback gains.

  8. Review: Controllability Consider a single-input system ( u ∈ R ): x ∈ R n x = Ax + Bu, ˙ y = Cx The Controllability Matrix is defined as � � B | AB | A 2 B | . . . | A n − 1 B C ( A, B ) = We say that the above system is controllable if its controllability matrix C ( A, B ) is invertible .

  9. Review: Controllability Consider a single-input system ( u ∈ R ): x ∈ R n x = Ax + Bu, ˙ y = Cx The Controllability Matrix is defined as � � B | AB | A 2 B | . . . | A n − 1 B C ( A, B ) = We say that the above system is controllable if its controllability matrix C ( A, B ) is invertible . ◮ As we will see today, if the system is controllable, then we may assign arbitrary closed-loop poles by state feedback of the form u = − Kx .

  10. Review: Controllability Consider a single-input system ( u ∈ R ): x ∈ R n x = Ax + Bu, ˙ y = Cx The Controllability Matrix is defined as � � B | AB | A 2 B | . . . | A n − 1 B C ( A, B ) = We say that the above system is controllable if its controllability matrix C ( A, B ) is invertible . ◮ As we will see today, if the system is controllable, then we may assign arbitrary closed-loop poles by state feedback of the form u = − Kx . ◮ Whether or not the system is controllable depends on its state-space realization.

  11. Controller Canonical Form A single-input state-space model x = Ax + Bu, ˙ y = Cx is said to be in Controller Canonical Form (CCF) is the matrices A, B are of the form     0 1 0 . . . 0 0 0 0 0 1 0 0 0 . . .         . . . . . . ...     . . . . . . A = , B = . . . . . .         0 0 0 0 1 0 . . .     ∗ ∗ ∗ . . . ∗ ∗ 1 A system in CCF is always controllable !! (The proof of this for n > 2 uses the Jordan canonical form, we will not worry about this.)

  12. Coordinate Transformations

  13. Coordinate Transformations ◮ We will see that state feedback design is particularly easy when the system is in CCF.

  14. Coordinate Transformations ◮ We will see that state feedback design is particularly easy when the system is in CCF. ◮ Hence, we need a way of constructing a CCF state-space realization of a given controllable system.

  15. Coordinate Transformations ◮ We will see that state feedback design is particularly easy when the system is in CCF. ◮ Hence, we need a way of constructing a CCF state-space realization of a given controllable system. ◮ We will do this by suitably changing the coordinate system for the state vector.

  16. Coordinate Transformations and State-Space Models T x = ¯ x + ¯ ˙ x = Ax + Bu ˙ − − − − → ¯ A ¯ Bu y = ¯ y = Cx C ¯ x where ¯ ¯ ¯ A = TAT − 1 , C = CT − 1 B = TB,

  17. Coordinate Transformations and State-Space Models T x = ¯ x + ¯ ˙ x = Ax + Bu ˙ − − − − → ¯ A ¯ Bu y = ¯ y = Cx C ¯ x where ¯ ¯ ¯ A = TAT − 1 , C = CT − 1 B = TB, ◮ The transfer function does not change.

  18. Coordinate Transformations and State-Space Models T x = ¯ x + ¯ ˙ x = Ax + Bu ˙ − − − − → ¯ A ¯ Bu y = ¯ y = Cx C ¯ x where ¯ ¯ ¯ A = TAT − 1 , C = CT − 1 B = TB, ◮ The transfer function does not change. ◮ The controllability matrix is transformed: C ( ¯ A, ¯ B ) = T C ( A, B ) .

  19. Coordinate Transformations and State-Space Models T x = ¯ x + ¯ ˙ x = Ax + Bu ˙ − − − − → ¯ A ¯ Bu y = ¯ y = Cx C ¯ x where ¯ ¯ ¯ A = TAT − 1 , C = CT − 1 B = TB, ◮ The transfer function does not change. ◮ The controllability matrix is transformed: C ( ¯ A, ¯ B ) = T C ( A, B ) . ◮ The transformed system is controllable if and only if the original one is.

  20. Coordinate Transformations and State-Space Models T x = ¯ x + ¯ ˙ x = Ax + Bu ˙ − − − − → ¯ A ¯ Bu y = ¯ y = Cx C ¯ x where ¯ ¯ ¯ A = TAT − 1 , C = CT − 1 B = TB, ◮ The transfer function does not change. ◮ The controllability matrix is transformed: C ( ¯ A, ¯ B ) = T C ( A, B ) . ◮ The transformed system is controllable if and only if the original one is. ◮ If the original system is controllable, then B ) [ C ( A, B )] − 1 . T = C ( ¯ A, ¯

  21. Coordinate Transformations and State-Space Models T x = ¯ x + ¯ ˙ x = Ax + Bu ˙ − − − − → ¯ A ¯ Bu y = ¯ y = Cx C ¯ x where ¯ ¯ ¯ A = TAT − 1 , C = CT − 1 B = TB, ◮ The transfer function does not change. ◮ The controllability matrix is transformed: C ( ¯ A, ¯ B ) = T C ( A, B ) . ◮ The transformed system is controllable if and only if the original one is. ◮ If the original system is controllable, then B ) [ C ( A, B )] − 1 . T = C ( ¯ A, ¯ This gives us a way of systematically passing to CCF.

  22. Example: Converting a Controllable System to CCF � − 15 � � 1 � 8 A = , B = ( C is immaterial) − 15 7 1

  23. Example: Converting a Controllable System to CCF � − 15 � � 1 � 8 A = , B = ( C is immaterial) − 15 7 1 Step 1: check for controllability. � 1 � − 7 C = det C = − 1 – controllable 1 − 8

  24. Example: Converting a Controllable System to CCF � − 15 � � 1 � 8 A = , B = ( C is immaterial) − 15 7 1 Step 1: check for controllability. � 1 � − 7 C = det C = − 1 – controllable 1 − 8 Step 2: Determine desired C ( ¯ A, ¯ B ). � 0 � 1 C ( ¯ A, ¯ B ) = [ ¯ B | ¯ A ¯ B ] = 1 − 8

  25. Example: Converting a Controllable System to CCF � − 15 � � 1 � 8 A = , B = ( C is immaterial) − 15 7 1 Step 1: check for controllability. � 1 � − 7 C = det C = − 1 – controllable 1 − 8 Step 2: Determine desired C ( ¯ A, ¯ B ). � 0 � 1 C ( ¯ A, ¯ B ) = [ ¯ B | ¯ A ¯ B ] = 1 − 8 Step 3: Compute T . � 0 � � 8 � � 1 � 1 − 7 − 1 B ) · [ C ( A, B )] − 1 = T = C ( ¯ A, ¯ = 1 − 8 1 − 1 0 1

  26. Finally, Pole Placement via State Feedback Consider a state-space model x ∈ R n , u ∈ R x = Ax + Bu, ˙ y = x

  27. Finally, Pole Placement via State Feedback Consider a state-space model x ∈ R n , u ∈ R x = Ax + Bu, ˙ y = x Let’s introduce a state feedback law u = − Ky ≡ − Kx   x 1 x 2   � �   = −  = − ( k 1 x 1 + . . . + k n x n ) k 1 k 2 . . . k n .   . .  x n

  28. Finally, Pole Placement via State Feedback Consider a state-space model x ∈ R n , u ∈ R x = Ax + Bu, ˙ y = x Let’s introduce a state feedback law u = − Ky ≡ − Kx   x 1 x 2   � �   = −  = − ( k 1 x 1 + . . . + k n x n ) k 1 k 2 . . . k n .   . .  x n Closed-loop system: x = Ax − BKx = ( A − BK ) x ˙ y = x

  29. Pole Placement via State Feedback Let’s also add a reference input: + x = Ax + Bu ˙ u y r − y = x K

  30. Pole Placement via State Feedback Let’s also add a reference input: + x = Ax + Bu ˙ u y r − y = x K x = Ax + B ( − Kx + r ) = ( A − BK ) x + Br, ˙ y = x

  31. Pole Placement via State Feedback Let’s also add a reference input: + x = Ax + Bu ˙ u y r − y = x K x = Ax + B ( − Kx + r ) = ( A − BK ) x + Br, ˙ y = x Take the Laplace transform: sX ( s ) = ( A − BK ) X ( s ) + BR ( s ) , Y ( s ) = X ( s ) Y ( s ) = ( Is − A + BK ) − 1 B R ( s ) � �� � G

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