Lecture 4: Time Response Reference: Nise Chapter 4, Sections 4.1 through 4.8

1. Lecture 4: Time ResponseReference: Nise Chapter 4, Sections 4.1 through 4.8 ESE 441 Control Systems I William J. Murphy, D.Sc.

2. Deducing Time Response via Parametric Form of Transfer Functions • Effect of Poles & Zeros on Time Response • First-Order Systems: Gain/Time-Constant Format • Second-Order Systems: Damping Ratio/Natural Frequency Format • Higher-Order Systems: • Effect of Additional Zeros • Effect of Additional Poles

3. Poles, Zeros & Time Response • Poles: The poles of a transfer function are those values of s for which the function is undefined (becomes infinite). • Zeros: The zeros of a transfer function are those values of s for which the function is zero. N.B.: A transfer function with n poles and m zeros is said to have (n-m) zeros at s equals infinity.

4. Example of the Effect of Pole/Zero Locations 1. Input pole generates form of forced response. 2. Transfer function pole generates form of natural response. 3. Real axis pole generates exponential character of natural response. 4. Location of zero affects the amplitude (magnitude of residues) of each term of the response.

5. First-Order System Format The response of a first-order system can be approximately deduced from its gain and time constant.

6. Impulse Response & Step Response of First-Order System The step response is shown on the following slide. Note that the steady-state value of the step response is K. Thus, K is interpreted as the system’s gain. For convenience, it is common to plot the step response with a gain of unity (K=1).

7. Response of a Unity-Gain, First-Order System to a Step Input

8. Experimental Determination of Gain & Time Constant Input = 1u(t) Gain = 0.72/1=0.72 Time Constant = 0.13

9. Second-Order System Format The response of the system can be approximately deduced from its gain, natural frequency, and damping ratio.

10. Step Response of Second-Order System with Various Damping Ratios

11. Pole Plot of Underdamped Second-Order System

12. S-Plane for Various Damping Ratios Note: The overdamped case can be considered to be two first-order systems. One with a slow time constant and one with a fast time constant.

13. Step Response for Various Damping Ratios

14. Second-Order Underdamped Response Specifications 1. Peak Time, Tp: Time to reach first, or maximum, peak. 2. Percent Overshoot, %OS: Amount response overshoots steady-state value, expressed as a percentage. 3. Settling Time, Ts: Same as for first-order system. 4. Rise Time, Tr: Same as for first-order system.

15. Peak Time & Settling Time

16. Percent Overshoot

17. Rise Time See footnote on p. 196 for approximate analytic expression of normalized rise time as a function of damping ratio.

18. MATLAB’s “sgrid” & “grid” Functions MATLAB command window: sgrid, grid ‘sgrid’ plots lines of constant damping ratio & natural frequency. “grid” plots lines of constant damped natural frequency & (damping ratio)x(natural frequency) sgrid(Z,Wn) plots constant damping and frequency lines for the damping ratios in the vector Z and the natural frequencies in the vector Wn.

19. Effect of Pole Locations a. constant real part;b. constant imaginary part;c. constant damping ratio

20. Example 4.6, p. 200

21. MATLAB Function: “timespecs” function [peak_time,settling_time,percent_overshoot,rise_time]=timespecs(zeta,wn) % Computes time response specifications from damping ratio and natural frequency wd=wn*sqrt(1-zeta^2); peak_time=pi/wd; settling_time=4/zeta/wn; percent_overshoot=exp(-zeta*pi/sqrt(1-zeta^2))*100; rise_time=(1.76*zeta^3-0.417*zeta^2+1.039*zeta+1)/wn;

22. echo % Nise Problem 4.49 GcoverK=tf(1,conv([1 10],[1 4 10])); GoverK=zpk(GcoverK) wn=sqrt(10),zeta=4/2/wn % Part a. [Tp,Ts,OS,Tr]=timespecs(zeta,wn),pause % Part b. ss(GcoverK),pause % Part c. use ltiview % Nise Problem 50 PoverU=tf(0.0763,[1 1.15,0.28]) zpk(PoverU) damp(PoverU) GAIN=dcgain(PoverU) step(PoverU) Percent_isoflorane_for_complete_paralysis=1/GAIN echo off

23. Effect of Additional Poles Consider the impulse response of a system with two real poles: How much smaller (faster) must the second time constant be for it to be negligible? Let t equal the slow time constant and the slow time constant be k times the fast time constant. Then,

24. Effect of Additional Poles (continued) Examine the term exp(1-k). How large must k be for this term to be negligible? K exp (1-k) If the fast time constant is 4 to 6 time faster than the slow one, the effect of the fast time constant is only 5 to 1%. Nise (p.203) assumes that the exponential decay due to an additional pole is negligible after 5 time constants, I.e., if the pole is 5 time farther to the left of the dominant (slower) poles. However, he warns on p.205 that this is an approximation that should always be checked by simulation in a final design.

25. Extension to Higher-Order Systems a. pole plot;b. componentresponses: nondominant pole is neardominant second-order pair (Case I), far from the pair (Case II), andat infinity (Case III)

26. Chalk Talk: Effect of Zeros

27. Chalk Talk: Zeros in Proximity of Poles • Effect on Pole Dominance • Pole-Zero Cancellation - A WARNING! • Design Technique • Notch Filters

28. Demo of MATLAB’s Time Response Capabilities • Commands ‘dcgain’ & ‘damp’ • Commands ‘impulse’ & ‘step’ • How to deal with higher-order polynomials • LTI GUI command, ‘ltiview’ • Example transfer function

29. Problem 4.48 P(s)/V(s) = (s+0.5)/(s+2)(s+5)

30. Problem 4.49 Wo/Vi = K/(s+10)(s^2+4s+10) Figure P4.11Vacuum robot liftstwo bags of salt

31. MATLAB M-File echo % Nise Problem 4.49 GcoverK=tf(1,conv([1 10],[1 4 10])); GoverK=zpk(GcoverK) wn=sqrt(10),zeta=4/2/wn % Part a. [Tp,Ts,OS,Tr]=timespecs(zeta,wn),pause % Part b. ss(GcoverK),pause % Part c. use ltiview % Nise Problem 50 PoverU=tf(0.0763,[1 1.15,0.28]) zpk(PoverU) damp(PoverU) GAIN=dcgain(PoverU) step(PoverU) Percent_isoflorane_for_complete_paralysis=1/GAIN echo off See M-File “TimeResponse.m”