1. 1 Time Response. CHAPTER 3

2. 2 3.2 Poles and Zeros and System Response. Figure 3.1: (a) System showing input and output; (b) Pole-zero plot of the system; (c) Evolution of a system response.

3. 3  The transfer function,  For a unit step of; 1/s,  Its response, A, B and B’ are constant. For K=1 and  =1/a then, A, B and B’ are constant. For K=1 and  =1/a then,  Time constant (1/a), is defined as the time for e -at to decay to 37%of its initial value or the time it takes for step response to reach 63% of its final value. 3.3 First Order System.

4. 4  Step response, Cont’d… Figure 3.2: (a) First Order Response to a Unit Step.

5. 5  The pole of the transfer function is at –a, the farther the pole from the imaginary axis, the faster the transient response.  Rise time (T r ), the time the response to go from the 0.1 to 0.9 of its final value. T r =2.2/a.  Settling time (T s ), time range when the response to reach and stay within 2% of its final value. Let c(t) = 0.98 then the T s.=4/a. Cont’d…

6. 6 3.4 Second Order System.  The transfer function,  For Impulse response,  Where,  Standard Form, Where K is the dc gain,  is the damping ratio,  n is the undammped natural frequency. Where

7. 7  Example of 2 nd order system responses. Figure 3.3: Second Order System, pole plots and Step Response. Cont’d…

8. 8 General 2 nd Order System.  Natural Frequency (  n ),  Damping Ratio (  ), Example 3.1: Find the Natural Frequency (  n ) and Damping Ratio (  ), Solution:  n = 6 and  =0.35. Cont’d…

9. 9  From previous page, Figure 3.4: Left; Plot for an underdamped 2 nd Order System. Right; Step Response for 2 nd Order System Damping Cases. Cont’d…

10. 10  From previous page, Figure 3.5: Second Order Response as a Function of Damping Ratio. Cont’d…

11. 11  The transfer function,  Poles at the origin from the unit step and two real poles from the system.  Constant force response and force response. 3.4.1 Over Damped Response.

12. 12 Example 3.1: Over Damped Response. Find the step response of the system. Solution: Expand the partial fraction. Take the inverse Laplace Transform.

13. 13  Under Damped transfer function,  When 0 <  < 1 The transfer function is,  The Pole position is, 3.4.2 Under Damped Response.

14. 14  From previous page, Figure 3.6: Second Order Response as a Function of Damping Ratio. Cont’d…

15. 15  Performance Measures. Peak Time Overshoot Settling Time Figure 3.7: (Top) The 2nd Order Underdamped Response Specification. (bottom) Percent overshoot versus damping factor Cont’d…

16. 16  Performance Measures.  Poles position Figure 3.8: Lines of constant peak time, Tp, settling time, Ts, and percent overshoot, %OS Note: Ts2 < Ts1 ; Tp2 < Tp1; %OS1 < %OS2 Cont’d…

17. 17  Pole Placement. Figure 3.9: Step responses of second-order underdamped systems as poles move: (a) with constant real part; (b) with constant imaginary part; (c) with constant damping ratio Cont’d…

18. 18  The transfer function, 3.4.3 Critically Damped.

19. 19 Example 3.3: Critically Damped Response. Find the step response of the system. Solution: Expand the partial fraction.

20. 20 Dominant Pole.  The formula that describing %OS, t s, t p were derived only for system with two complex poles and no zeros.  A system with more than two poles or zeros can be approximated as a second order system that has just two complex dominant poles.

21. 21 Dominant Pole. Figure 3.11: Component responses of a three-pole system: (a) pole plot; (b) component responses: non-dominant pole is near dominant second-order pair (Case I), far from the pair (Case II), and at infinity (Case III).Cont’d… Cannot be represented as second order system Will approach second order system

22. 22Cont’d… Effect of adding a zero to a two-pole system  The closer is the zero to dominant poles, the greater its effect on transient response.  As the zero move away from dominant poles, the response approaches that of the two pole system. Starting at poles 1±j2.828, then consecutively add zeros at -3, -5, -10.

23. 23 (i) Stable system.  Natural response approaches zero.  Poles in LHP. (ii) Unstable system.  Natural response grows.  Poles in RHP. (iii) Marginally stable system.  Natural response neither grows/approaches zero.  Poles on j  axis. 3.5 Stability. Figure 3.12: Closed-loop poles and response: a. stable system; b. unstable system

24. 24 3.6 Routh-Hurwitz Stability Criteria. What is Routh-Hurwitz Criterion (RHC)?  Through the RHC method we can tell how many close- loop system poles are in the left half plane, in the right half- plane and on the j  -axis. We can find the number of poles in each section of the s-plane, but cannot find their coordinate.  The number of roots of the polynomial that are in the right half-plane is equal to the number of changes in the first column.  The RHC method requires two steps; (1) Generate the data table called Routh table. (2) Interpret the Routh table to tell number of close loop system poles in the left half plane, in the right half-plane and on the j  -axis.

25. 25  The Close-Loop Transfer function.  Initial layout for the Routh-Hurwitz Table.  Completed Routh Table. Cont’d…

26. 26 Example 3.4: Routh-Hurwitz. Make a Routh table from the system shown below. Solution: Find the equivalent close loop system. Figure (b) above. Interpretation: There are two sign changes in the first column. 1  -72  103 The system is unstable, two poles exist in the right half plane.

27. 27 Example 3.5: Routh-Hurwitz.  The number of RHP poles = The number of SIGN CHANGES of COL 1  TWO sign changes:  RHP Poles =2

28. 28 Example 3.6: Routh-Hurwitz. Solution:  Two sign changes: 2 RHP (UNSTABLE)  Poles: 2 LHP and 2 RHP

29. 29 Example 3.7: Routh-Hurwitz.  Assume  is small POSITIVE : TWO sign changes  Poles: 2 RHP, 3 LHP Solution:

30. 30 Example 3.8: Routh-Hurwitz. Solution:  Assume  is small positive: Two sign changes  Poles: 2 RHP, 3 LHP

31. 31 Example 3.9: Routh-Hurwitz. Solution:  NO sign changes: No RHP (STABLE)  Row of ZEROS indicate existence of complex poles & Symmetric Equations  Poles: 1 LHP and 4 on jw axis

32. 32  When a purely even or odd polynomial is a factor of the original polynomial.  Even polynomial only have roots that are symmetrical about the origin.  The symmetry can occur under 3 conditions: 1.The roots are symmetrical and real. 2.The roots are symmetrical and imaginary. 3.The roots are quadrantal. Entire row is zero 2 2 1 1 3 3 jj 

33. 33 Example 3.10: Routh-Hurwitz. Solution:  2 sign changes: 2 RHP (symmetric)  Poles: 2 RHP, 4 LHP and 2 on j  axis

34. 34  Main use is to determine the position of the poles, which in turns can determine the stability of the response. Use of Routh Hurwitz Criteria Example A closed-loop transfer function is given by Determine the range for K for the system to be always stable and its oscillating frequency before it becomes unstable.

35. 35 Solution:  Characteristic equation is:  Expand the equation  Form the Routh’s array s4s4 13K s3s3 32 s2s2 K s1s1 s0s0 K

36. 36 Solution:  For no sign change  Referring to row 4  which gives,  and row 5,  Hence its range,  Oscillating frequency,

37. 37 Steady state  From the diagram  Consider  And  Use the final value theorem and define steady state error, e ss that is given by Y(s)Y(s)R(s)R(s)E(s)E(s) +- B(s)

38. 38 Unit step  Unit step input,  From  Steady state error,  We define step error coefficient,  Thus, the steady state error is  By knowing the type of open-loop transfer function,  we can know step error coefficient and thus the steady state error

39. 39  For open-loop transfer function of type 0:  For open-loop transfer function of type 1:  For open-loop transfer function of type 2: Unit step

40. 40 Example: A first order plant with time constant of 9 sec and dc gain of 5 is negatively feedback with unity gain, determine the steady state error for a unit step input and the final value of the output. Solution: The block diagram of the system is As we are looking for a steady state error for a step input, we need to know, Knowing the open-loop transfer function, then And steady state error of, Its final value is, Unit step Y(s)Y(s) R(s)R(s) + -

41. 41 Unit Ramp, while its Laplace form is As in the above section, we know that Thus, its steady state error is Define ramp error coefficient, Which the steady state error as Just like for the unit step input we can conclude the steady state error for a unit ramp through the type of the open-loop transfer function of the system. For open-loop transfer function of type 0: For open-loop transfer function of type 1: For open-loop transfer function of type 2:

42. 42 Example: A missile positioning system is shown. (i) Find its closed-loop transfer function (ii) Determine its undamped natural frequency and its damping ratio if (iii) Determine the steady state error, if the input is a unit ramp. (iv) Cadangkan satu kaedah bagi menghapuskan ralat keadaan mantap untuk (iii). Compensator DC motor + - Unit Ramp

43. 43 Solution: (a) By Mason rule, the closed-loop transfer function is (b) If, Comparing with a standard second order transfer function Unit Ramp

44. 44 Comparing Thus undamped natural frequency rad.s -1 and damping ratio of (c) To determine the ramp error coefficient, we must obtain its open-loop transfer function As it is a type 1, the system will have a finite ramp error coefficient, putting Hence steady state error of Unit Ramp

45. 45 Unit Parabola Its time function,while its Laplace,thus its steady state error is Define parabolic error coefficient, Similarly we can determine its steady state error by knowing the type of the open-loop transfer function For open-loop transfer function of type 0: For open-loop transfer function of type 1: For open-loop transfer function of type 2:

46. 46 In summary we can make a table of the steady state error for the above input Unit stepUnit rampUnit parabolic Type 0  Type 10  Type 200