1. Autumn 2008 EEE8013 Revision lecture 1 Ordinary Differential Equations

2. Autumn 2008 Model: Ordinary Differential Equations (ODE): Modeling Dynamics: Properties of the system, we have to solve/study the ODE.

3. Autumn 2008 First order ODEs: First order systems: Study approaches Analytic: Explicit formula for x(t) (a solution – separate variables, integrating factor) which satisfies INFINITE curves (for all Initial Conditions (ICs)).

4. Autumn 2008 First order linear equations First order linear equations - (linear in x and x’) General form: Numerical Solution: k=5, u=0.5

5. Autumn 2008 Analytic solution: Step input

6. Autumn 2008 Response to a sinusoidal input

7. Autumn 2008 Response to a sinusoidal input

8. Autumn 2008 Second order ODEs Second order ODEs: u=0 => Homogeneous ODE So I am expecting 2 arbitrary constants Let’s try a

9. Autumn 2008 Overdamped system Roots are real and unequal 0123456 -0.5 0 0.5 1 1.5 Overall solution x 2 x 1

10. Autumn 2008 Critically damped system Roots are real and equal A=2, B=1, x(0)=1, x’(0)= 0 => c 1 =c 2 =1

11. Autumn 2008 Underdamped system Roots are complexUnderdamped system r=a+bj r=a+bj Theorem: If x is a complex solution to a real ODE then Re(x) and Im(x) are the real solutions of the ODE:

12. Autumn 2008 Underdamped system, example A=1, B=1, x(0)=1, x’(0)=0 => c1=1, c2=1/sqrt(3)

13. Autumn 2008 Undamped Undamped system A=0, B=1, x(0)=1, x’(0)=0 =>c1=1, c2=0:

14. Autumn 2008 Stability In all previous cases if the real part is positive then the solution will diverge to infinity and the ODE (and hence the system) is called unstable.

15. Autumn 2008 Natural frequency, damping frequency, damping factor is the damping factor andis the natural frequency of the system. => Overdamped system implies that Case 1:

16. Autumn 2008 Natural frequency, damping frequency, damping factor => Critically damped system implies that => Underdamped systems implies called damped frequency or pseudo-frequency No damping the frequency of the oscillations = natural frequency Case 2: Case 3: Case 4:

17. Autumn 2008 Summary

18. Stability revised If then cases 1-3 are the same but with

19. Autumn 2008 NonHomogeneous (NH) differential equations  u=0 => Homogeneous => x 1 & x 2.  Assume a particular solution of the nonhomogeneous ODE: xp If u(t)=R=cosnt => Then all the solutions of the NHODE are So we have all the previous cases for under/over/un/critically damped systems plus a constant R/B. If complementary solution is stable then the particular solution is called steady state.

20. Autumn 2008 Example x(0)=1, x’(0)=0 => c1=-1, c2=-1/sqrt(3)

21. Autumn 2008 State Space Very difficult to be studied => so we use computers Computers are better with 1st order ODE 1 n th => n 1 st Powerful tools from the linear algebra Use sensors: Output = x =>

22. Autumn 2008 State Space

23. Autumn 2008 Block Diagram X is an n x 1 state vector U is an q x 1 input vector Y is an p x 1 output vector A is an n x n state matrix B is an n x q input matrix C is an p x n output matrix D is an p x m feed forward matrix (usually zero)

24. Autumn 2008 State space rules The state vector describes the system => Gives its state => The state of a system is a complete summary of the system at a particular point in time. If the current state of the system and the future input signals are known then it is possible to define the future states and outputs of the system. The choice of the state space variables is free as long as some rules are followed: 1.They must be linearly independent. 2.They must specify completely the dynamic behaviour of the system. 3.Finally they must not be input of the system.

25. Autumn 2008 State space The system’s states can be written in a vector form as:  A standard orthogonal basis (since they are linear independent)  for an n-dimensional vector space called state space. Matlab definition

26. Autumn 2008 Solution

27. Solution II How can we solve that ??? Assume is a parameter => A homogeneous linear system (This last equation is the characteristic equation of the system, why???).

28. Autumn 2008 Solution III I assume that a 2 =1 so a 1 =2 I assume that a 1 =2 so a 2 =-2 Matlab example

29. Autumn 2008 General Solution The roots of this equation are called eigenvalues negative eigenvalues => stable positive eigenvalues => unstable repeated eigenvalues => eigenvectors are not linearly independent. Complex eigenvalues => conjugate and the eigenvector will be complex =>solution will consists of sines, cosines and exponential terms

30. Autumn 2008 Properties of general solution If we start exactly on one eigenvector then the solution will remain on that forever. Hence if I have some stable and some unstable eigenvalues it is still possible (in theory) for the solution to converge to zero if we start exactly on a stable eigenvector. Determines the nature of the time response (stable, fast..) Determines the extend in which each state contributes to Determines the extend in which the IC excites the To find the eigenvalues and eigenvectors use the command eig()

31. Autumn 2008 Example

32. Example

33. Example

34. State Transition Matrix Until now the use of vector ODEs was not very helpful. We still have special cases => No special cases are needed then Use the command expm (not exp)! X(0)=[1 2] X(5) =? 4 ways to calculate it!!!

35. Autumn 2008 State Transition Matrix

36. Autumn 2008 State Transition Matrix

37. Autumn 2008 Solution

38. SS => TF??? TF Response to ICs

39. Autumn 2008 SS => TF??? CE of the TF The TF is a matrix

40. Autumn 2008 Example: Find the TF of SS => TF???

41. Autumn 2008 Basic properties of state space State space transformations State space representations are not unique Same input/output properties, => same eigenvalues T is an invertible matrix Z is the new state vector

42. Autumn 2008 Do these two systems have the TF? Matlab example Basic properties of state space

43. Autumn 2008 Observability - Controllability Notice the structure of A and C Observability - Controllability

44. Autumn 2008 There is a pole zero cancellation pzmap(ss_model)Matlab verification The cancellation is due to C=[3 0]. We can influence x 2 through U but we cannot observe how it behaves and hence there is no way to feedback that signal to a controller!!! Observability - Controllability

45. Autumn 2008 In this case we can see how both states behave but we can not change U in any way so that we can influence x 2 due to the form of B. Unobservable & uncontrollable Minimal realisation. Difficult task if the system is nonlinear!!!! Observability - Controllability

46. Autumn 2008 Check the rank >> rank(obsv(A,C)) >> rank(ctrb(A,B)) Observability - Controllability