optimal and multivariable control

1. OPTIMAL and MULTIVARIABLE
CONTROLS
LECTURE # 11
THE VARIATIONAL APPROACH TO
OPTIMAL CONTROL PROBLEMS

2. Necessary Conditions for Optimal
Control
• Problem
• Assumptions
– Admissible states and controls are not bounded.
– Initial state x(t0)=x0 and initial time t0 are specified.
– x is an nx1 vector and u is mx1 vector.

3. Necessary Conditions for Optimal
Control
• Previously, the following problem was considered
• In control problems the trajectory is determined by the
control u
– We wish to consider functionals of n + m functions
– Only m are independent known as controls
– Function must satisfy ‘n’ different. equation constraints given in 5.1-1
• The difference between cost function given in 5.1-2 and
previously considered functional is of is term involving final
state and time i.e.,
• Assuming h is differential, it can be rewritten as

4. • The performance measure can now be expressed as

5. • Define
• To determine variation of Ja, the variations
are introduced.

7. Necessary Conditions for Optimal Control
• Consider the terms involving h

8. Necessary Conditions for Optimal Control
• In the integral term we have

11. Boundary Conditions
• To determine boundary conditions 5.1-18
needs to be modified accordingly.
• In all cases it is assumed that we have the n
equations
• The following boundary problems are
discussed
– Problems with fixed final time.
– Problems with free final time.

12. Problems with Fixed Final Time
• Different scenarios might occur such as
• CASE 1 FINAL STATE SPECIFIED

14. Example 5.1-1

16. Example 5.1-1 (a)

17. Example 5.1-1 (a)

18. Problems with Fixed Final Time
• CASE 2 FINAL STATE FREE

19. Example 5.1-1 (b)

20. Example 5.1-1 (b)

22. Problems with Fixed Final Time
• CASE 3 FINAL STATE LYING ON A SURFACE DEFINED
BY m(x(t))=0
• Example
– Suppose that the final state
of a second order system is
required to lie on a circle
– The admissible changes in x(tf)
are tangent to the circle at point

23. CASE 3 FINAL STATE LYING ON A SURFACE
DEFINED BY m(x(t))=0
• The tangent line is normal to the gradient vector

24. CASE 3 FINAL STATE LYING ON A
SURFACE DEFINED BY m(x(t))=0
5.1-18

25. CASE 3 FINAL STATE LYING ON A
SURFACE DEFINED BY m(x(t))=0
• Generalized Case
• Each component of m represents a hyper-surface in the n-dimensional
state space.
• Thus, the final state lies on the intersection of the k hyper-surfaces.
• is tangent/ normal to the each of the hyper-surface at the point

28. Example 5.1-1 (c)

29. Example 5.1-1 (c)

31. Problems with Free Final Time
• If final time is free several situations might occur
1. Final state fixed
2. Final state free
3. x(tf) lying on a moving point
4. Final state lying on a surface defined by
m(x(t))=0
5. Final state lying on a moving surface defined by
m(x(t))=0

32. Case 1 Final State Fixed with Free Final Time

33. Case 2 Final State Free with Free Final Time

34. Case 3 x(tf) lying on a moving point
with Free Final Time

35. Table 5-1

36. Case 4 Final state lying on a surface
defined by m(x(t))=0

37. Case 4 Final state lying on a surface
defined by m(x(t))=0

38. Case 4 Final state lying on a surface
defined by m(x(t))=0

39. Case 4 Final state lying on a surface
defined by m(x(t))=0

40. Generalized Case
• Each component of m describes a hyper surface in the n-
dimensional state space.
• The final state lies at the intersection of the hyper
surfaces defined by m.
• ᵟxf is (to first order) tangent to each of the hyper surfaces
at the point x*(tf,tf)

41. Generalized Case
• Thus, ᵟxf is normal to each of the gradient vectors
which are assumed to be linearly independent
• The (2n+k+1) equations are