1. This Week’s Objectives Establish Dynamic Models of System to be Controlled –Second Order Systems Obtain Solutions using LaPlace Transforms Create Simulink Model and Generate Simulated Results

2. Modeling of a Spring-Mass- Damper System (Modeling of a Second Order System)

3. let’s start with a one car system –suppose we know: m 1 mass of the car k 1, L 01 spring constant and free length cdamping coefficient m1m1 k 1,L 01 c f(t) x1x1

4. one car system –suppose at start (initial condition): x 1 = L 01 spring is applying zero force = 0car is at rest f(0) = 0there is no applied force m1m1 k 1,L 01 c f(t) x1x1

5. one car system –now we want (final condition): x 1 = L 01 + 3 cm we want to move there ‘as fast as possible’ we don’t want to overshoot ‘much’ we want the system to settle down ‘quick’ m1m1 k 1,L 01 c f(t) x1x1 We’re going to need to properly define these terms.

6. one car system –we have a desired final position –we can measure the position of the car –we can set the value of f, the applied force, to any value we want and constantly adjust it m1m1 k 1,L 01 c f(t) x1x1 We’ll pick a value for f that is related to the difference between the current measured position and the desired final position.

7. m1m1 k 1,L 01 c f(t) x1x1 In Controls Terminology: BLOCK DIAGRAM

8. problem statement –given: m 1 mass of the car k 1, L 01 spring constant and free length cdamping coefficient x 1 (0)initial position at t=0 (0)initial velocity at t=0 x d desired position –generate values for the input force such that the car reaches the desired position within 3 sec the car will not overshoot by more than 0.25 cm the car will settle within  0.1 cm within 5 sec m1m1 k 1,L 01 c f(t) x1x1

9. how to start? let’s first write the equation of motion assume the current position x 1 >L 01 and that 1 is positive m1m1 k 1,L 01 c f(t) x1x1 m1m1 c k 1 (x 1 - L 01 ) f(t)

10. how do you solve for x 1 (t) if you know f(t) as well as x 1 (0) and 1 (0)? m1m1 k 1,L 01 c f(t) x1x1

11. we must solve a second order differential equation we will write the equation in the Laplace domain Laplace transformations substitute easily solved algebraic equations for differential equations m1m1 k 1,L 01 c f(t) x1x1

12. Review of Laplace transforms if f(t) is some function of time, then the Laplace transformation of f(t) can be written as change from the time domain to the 's' domain

13. Example: Let f(t) = 1 for t > 0. Find F(s).

14. Example: Let f(t) = e at for t > 0 where ‘a’ is a constant. Find F(s).

15. The LaPlace transform is a linear operation. For any functions f(t) and g(t) and for any constants, a and b, L{a f(t) + b g(t)} = a L {f(t)} + b L {g(t)}

16. Example: Let f(t) = cosh(at) = (e at + e -at )/2 for t > 0. Find F(s).

17. LaPlace transform of the derivative of f(t). L(f '(t)) = sL (f) - f(0) L (f "(t)) = s 2 L (f) - sf(0) - f'(0) L (f (n) ) = s n L (f) - s n-1 f(0) - s n-2 f '(0) -... - f (n-1) (0)

18. LaPlace transform of the integral of f(t).

19. Laplace transforms have been calculated for a large variety of functions. Sampling of these are listed in Table 2.3 of the text –Modern Control Systems, 9 th ed., R.C. Dorf and R.H. Bishop, Prentice Hall, 2001.

21. Inverse transformations are often of the form where G(s) and H(s) are polynomials in s This ratio must be re-written in terms of its partial fraction expansion. Review the techniques for partial fraction expansion when the polynomial H(s) has repeated roots or complex roots.

22. Example: Leta≠b. Find f(t) first perform partial fraction reduction Multiplying the left and right side of the above equation by (s-a)(s-b) gives

23. 1= A 1 (s-b) + A 2 (s-a) Can pick two values for s and then solve for A 1 and A 2 from the two equations in two unknowns. When s=a, 1= A 1 (a-b). When s=b, 1=A 2 (b-a). Thus

24. substituting A 1 and A 2 into F(s) gives Using the linearity of the Laplace transform,

25. from previous example END OF REVIEW

26. car 1 car 2 car 3 Back to our problem.

27. equation of motion take Laplace transform of left and right side of equation m1m1 k 1,L 01 c f(t) x1x1

28. m1m1 k 1,L 01 c f(t) x1x1

29. m1m1 k 1,L 01 c f(t) x1x1

30. m1m1 k 1,L 01 c f(t) x1x1 suppose m 1 = 2 kg = 2 N sec 2 /m k 1 = 3 N/cm = 300 N/m L 01 = 6 cm = 0.06 m c = 2.5 N sec/cm = 250 N sec/m x 1 (0) = 4 cm = 0.04 m (0) = -2 cm/sec = -0.02 m/sec f(t) is a step input of 8 N starting at t=0

31. m1m1 k 1,L 01 c f(t) x1x1 are units consistent? seems like s has units of 1/sec and X(s) has units of m sec

33. x 1 (t) = 0.08667 + 0.00667 e -62.5 t [-7cosh(61.29 t) – 7.187sinh(61.29 t)] how do we check the results ? check boundary conditions check steady state final position - final value theorem:

35. How to model the system in Simulink? m1m1 k 1,L 01 c f(t) x1x1

36. m1m1 k 1,L 01 c f(t) x1x1 suppose m 1 = 2 kg = 2 N sec 2 /m k 1 = 3 N/cm = 300 N/m L 01 = 6 cm = 0.06 m c = 2.5 N sec/cm = 250 N sec/m x 1 (0) = 4 cm = 0.04 m (0) = -2 cm/sec = -0.02 m/sec f(t) is a step input of 8 N starting at t=0

37. let’s model this system using Simulink governing differential equation for our case, f(t) = 0, (0) = -0.02 m/sec, x 1 (0) = 0.04 m thus m1m1 k 1,L 01 c f(t) x1x1

40. blocks can be grouped into subsystems

41. x

42. m1m1 k 1,L 01 c f(t) x1x1

44. d1 = d2 = 15 cm d3 = 80 cm m1 = 2.5 kg m2 = 5.5 kg c1 = 8 N sec/m c2 = 10 N sec/m k1 = 3 N/cm ; L01 = 6 cm k2 = 5 N/cm ; L02 = 8 cm k3 = 4 N/cm ; L03 = 10 cm Cars 1 and 2 start at rest at their static equilibrium position. A constant force f 1 = 20N and f 2 = -5 N. Obtain the motion response of the two cars, i.e. obtain x 1 (t) and x 2 (t). x1x1 x2x2 d1d1 f1f1 f2f2 d2d2 d3d3 c2c2 c1c1 m2m2 m1m1 k 2, L 02 k 1, L 01 k 3, L 03

45. Determine the position of the two cars at static equilibrium when no forces are applied. x1x1 x2x2 d1d1 m1m1 k 2, L 02 k 1, L 01 Car 1 will be in equilibrium whenk 1 (x 1 – L 01 ) = k 2 [x 2 – (x 1 + d 1 + L 02 )] x1x1 x2x2 d1d1 d2d2 d3d3 m2m2 k 2, L 02 k 3, L 03 Car 2 will be in equilibrium when k 2 [x 2 – (x 1 + d 1 + L 02 )] = k 3 [d 3 – (x 2 + d 2 + L 03 )] Substituting the given parameters and solving for x 1 and x 2 gives x 1_equil = 17.06 cm x 2 _ equil = 46.70 cm

46. Now write equations of motion of the system. x1x1 m1m1 k 2 [x 2 – (x 1 + d 1 + L 02 )] k 1 (x 1 – L 01 ) f1f1 x1x1 x2x2 d1d1 d2d2 d3d3 m2m2 k 3 [d 3 – (x 2 + d 2 + L 03 )] k 2 [x 2 – (x 1 + d 1 + L 02 )] f2f2

49. time, sec position, meters car 2 car 1

50. This Week’s Objectives Establish Dynamic Models of System to be Controlled –Second Order Systems Obtain Solutions using LaPlace Transforms Create Simulink Model and Generate Simulated Results