1. It is the time response of a system to an input that sets the criteria for our control systems. Many quantitative criteria have been defined to characterise the time response of a system. The time constant and system gain of a first order system are useful in its analysis, but other criteria describe the time response more accurately to an engineer. A first order system may be written as: or where K=1 and T = 1/a → and for a unit step → using Laplace tables we obtain the standard response ©2000, John Wiley & Sons, Inc. Nise/Control Systems Engineering, 3/e First Order Time Response

2. First-order system response to a unit step Rise Time, T r Rise time is defined as the time for the response to go from 0.1 to 0.9 of its final value: Settling Time, T s Settling time is defined as the time for the response to reach, and stay within, 2% of its final value (2% is standard) ©2000, John Wiley & Sons, Inc. Nise/Control Systems Engineering, 3/e First Order Time Response ©2000, John Wiley & Sons, Inc. Nise/Control Systems Engineering, 3/e First Order Time Response

3. ©2000, John Wiley & Sons, Inc. Nise/Control Systems Engineering, 3/e Second Order Time Response Second-order step response components generated by complex poles Consider the characteristic equation of to give roots The real part generates: The complex part generates:

4. Damping Ratio Consider a second order system: with unity in front of the s 2 term and k such that note that the constants a`, b`, c` are not equivalent to a, b Natural Frequency: When there is no damping in the system (a = 0 in this case) We obtain with poles We know that as a complex number. So the frequency of oscillation, which is termed the natural frequency Exponential Decay Frequency: Considering an underdamped system The real part of, is, where is termed the exponential decay frequency Damping ratio: The damping ratio is defined to be = Exponential decay frequency = = Natural frequency

5. Damping Ratio Consider a second order system: Written in terms of damping ratio and natural frequency: Consider the roots of the characteristic equation:

6. Second-order underdamped responses for damping ratio values ©2000, John Wiley & Sons, Inc. Nise/Control Systems Engineering, 3/e Second Order Time Response

7. Second-order underdamped response specifications Can not use equations for first order system! Still would like to consider: Rise time, Settling time,but also Time to peak, Percentage overshoot (%OS), Steady-state error, ©2000, John Wiley & Sons, Inc. Nise/Control Systems Engineering, 3/e Second Order Time Response

8. Time to Peak Determine the time to the first peak (zero slope): Differentiate the time response to a step input and set to zero: [ this is straightforward thanks to Laplace!] → But it is easy to differentiate this equation: the differential is Now we complete the square to utilise the Laplace tables: We now need to find something familiar in the table:

9. Time to Peak Determine the time to the first peak (zero slope): Differentiate the time response to a step input and set to zero: [ this is straightforward thanks to Laplace!] We can return to the time domain: We can set this to zero and note that sin -1 0 = nπ: → We now no the time to all n peaks, but only need n=1:

10. Similarly: Rise time, calculated by iterating c(t), see final value theorem for c final Settling time, Time to peak, [note c max = c(t p )] Percentage overshoot (%OS), [note %OS is a function of damping ratio only ] Steady-state error, is the difference between the input and output for a prescribed test input as Second Order Time Response

11. Normalized rise time vs. damping ratio for a second-order underdamped response ©2000, John Wiley & Sons, Inc. Nise/Control Systems Engineering, 3/e Second Order Rise Time

12. 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 ©2000, John Wiley & Sons, Inc. Nise/Control Systems Engineering, 3/e Second Order Time Response

13. Controllers Another form of controller is the P+I controller Its transfer function can be written as Here C = I and T = P/I. P and I can be chosen so the 1+sT term (controller zero) cancels Plant pole. Suppose Plant If apply P+I to this Plant, and make T = T 2, then So Note, the I term means that the steady state value is 1.

14. Controllers P+D controller Make 1+s T cancel Plant pole. If Plant is and P+D is applied, then so PID controller - can cancel two lags in a plant: then In all these examples, by careful arrangement, systems is first/second order. Cancellation may not give best response, but analysis of systems is easier!

15. Controllers Suppose the system is one with controller C and plant G. We want Rise time Fast Settling timewithin two seconds Time to peak< 0.5 Percentage overshoot (%OS)< 90% Steady-state error< 0.1 But have: I O G C

16. Gain Controller Try a gain controller K p. Rise time Fast Settling timewithin two seconds Time to peak< 0.5 Percentage overshoot (%OS)< 90% Steady-state error< 0.1 But have: I O G C

17. Controllers Try a gain controller 1 +K i /s. We want Rise time Fast Settling timewithin two seconds Time to peak< 0.5 Percentage overshoot (%OS)< 90% Steady-state error< 0.1 But have: I O G C

18. Controllers Try a gain controller K p +K i /s. We want Rise time Fast Settling timewithin two seconds Time to peak< 0.5 Percentage overshoot (%OS)< 90% Steady-state error< 0.1 But have: I O G C

19. D Controller Try a gain controller K d s. We want Rise time Fast Settling timewithin two seconds Time to peak< 0.5 Percentage overshoot (%OS)< 90% Steady-state error< 0.1 But have: I O G C

20. PD Controller Try a gain controller 1 +K d s. We want Rise time Fast Settling timewithin two seconds Time to peak< 0.5 Percentage overshoot (%OS)< 90% Steady-state error< 0.1 But have: I O G C

21. PD Controllers Try a gain controller K p +K d s. We want Rise time Fast Settling timewithin two seconds Time to peak< 0.5 Percentage overshoot (%OS)< 90% Steady-state error< 0.1 But have: I O G C

22. Pole Cancellations Try a lead/lag controller (1 +K 1 s)/(1+K 2 s). We want Rise time Fast Settling timewithin two seconds Time to peak< 0.5 Percentage overshoot (%OS)< 90% Steady-state error< 0.1 But have: I O G C