Control Systems, Lecture04

Transcription

1 Control Systems, Lecture04 İbrahim Beklan Küçükdemiral Yıldız Teknik Üniversitesi / 53

2 Transfer Functions The output response of a system is the sum of two responses: the forced response and the natural response. Many techniques, such as solving a differential equation or taking the inverse Laplace transform, enable us to evaluate this output response but these techniques are laborious and time-consuming. The concept of poles and zeros, fundamental to the analysis and design of control systems, simplifies the evaluation of a system s response. 2 / 53

3 Poles and Zeros Poles The poles of a transfer function are the values of the Laplace transform variable, s, that cause the transfer function to become infinite any roots of the denominator of the transfer function Zeros The zeros of a transfer function are the values of the Laplace transform variable, s, that cause the transfer function to become zero any roots of the numerator of the transfer function 3 / 53

4 Example 4 / 53

5 Conclusions A pole of the input function generates the form of the forced response (that is, the pole at the origin generated a step function at the output). A pole of the transfer function generates the form of the natural response (that is, the pole at 5 generated e 5t ). A pole on the real axis generates an exponential response of the form e αt, where α is the pole location on the real axis. Thus, the farther to the left a pole is on the negative real axis, the faster the exponential transient response will decay to zero. 5 / 53

6 Example Given the system of Figure, write the output, c(t), in general terms. By inspection, each system pole generates an exponen- tial as part of the natural response. The input s pole generates the forced response. Thus, K2 K3 K4 K C (s) = s s +2 s +4 s +5 Taking the inverse Laplace transform, we get c(t) = [K1 + K2 e 2t + K3 e 4t + K4 e 5t ]1(t) 6 / 53

7 First Order Systems A first-order system without zeros can be described by the transfer function If the input is a unit step, where R(s) = 1/s, the Laplace transform of the step response is a C (s) = R(s)G (s) = s(s + a) 7 / 53 c(t) = [1 e at ]1(t)

8 First Order Systems: Significance of a When t = 1/a c(1/a) = / 53

9 Rise Time, Tr Rise time is defined as the time for the waveform to go from 0.1 to 0.9 of its final value. 2.2 Tr = a Settling Time, Ts Settling time is defined as the time for the response to reach, and stay within, 2% of its final value Ts = 4 Time Constant = 9 / 53 4 a

10 2nd order Systems Compared to the simplicity of a first-order system, a second-order system exhibits a wide range of responses that must be analyzed and described. Lets consider a typical 2nd order system G (s) = b s 2 + as + b By assigning appropriate values to parameters a and b, we can show all possible second-order transient responses. The unit step response then can be found using C (s) = R(s)G (s), where R(s) = 1/s, followed by a partial-fraction expansion and the inverse Laplace transform. 10 / 53

11 Different responses of a 2nd order System 11 / 53

12 Different responses of a 2nd order System 12 / 53

13 2nd order Systems Overdamped responses Poles: Two real at σ1 and σ2 Natural response: Two exponentials with time constants equal to the reciprocal of the pole locations, or c(t) = K1 e σ1 t + K2 e σ2 t Underdamped responses Poles: are complex and at σd ± jωd Natural response: Damped sinusoid with an exponential envelope whose time constant is equal to the reciprocal of the pole s real part. The radian frequency of the sinusoid, the damped frequency of oscillation, is equal to the imaginary part of the poles, or c(t) = Ae σd t cos(ωd t φ) 13 / 53

14 2nd order Systems Critically damped responses Poles: Two real poles at σ1 Natural response: One term is an exponential whose time constant is equal to the reciprocal of the pole location. Another term is the product of time, t, and an exponential with time constant equal to the reciprocal of the pole location, or c(t) = K1 e σ1 t + K2 te σ2 t Undamped responses Poles: Two imaginary at ±jω1 Natural response: Undamped sinusoid with radian frequency equal to the imaginary part of the poles, or c(t) = A cos(ω1 t φ) 14 / 53

15 4 different responses 15 / 53

16 Exercise 16 / 53

17 General 2nd Order System Natural Frequency, ωn The natural frequency of a second-order system is the frequency of oscillation of the system without damping. For example, the frequency of oscillation of a series RLC circuit with the resistance shorted would be the natural frequency. Damping Ratio ζ ζ= 1 Natural Period (seconds) Exponential decay frequency = Natural frequenc y(rad/sec) 2π Exponential Time Constant 17 / 53

18 Consider a 2nd order system G (s) = s2 b + as + b Without damping, poles be on the jω axis, and the response would be an undamped sinusoid. For the poles to be purely imaginary, a = 0: G (s) = s2 b +b By definition, the natural frequency, ωn, is the frequency of oscillation of this system. ωn = b Hence b = ωn2 18 / 53

19 Now what is the term a? Assuming an underdamped system, the complex poles have a real part, σ, equal to a/2. The magnitude of this value is then the exponential decay frequency ζ= σ a/2 Exponential decay frequency = = Natural frequenc y(rad/sec) ωn ωn = a = 2ζωn Our general second-order transfer function finally looks like this: G (s) = 19 / 53 ωn2 s 2 + 2ζωn s + ωn2

20 Example Given the transfer function G (s) = s s + 36 Find ζ and ωn Solution ωn = 6, 20 / 53 2ζωn = 4.2 = ζ = 0.35

21 2nd Order System Pole Locations Now that we have defined z and vn, let us relate these quantities to the pole location. Solving for the poles of the transfer function G (s) = ωn2 s 2 + 2ζωn s + ωn2 yields s1,2 = ζωn ± ωn 21 / 53 p 1 ζ2

22 Pole Locations vs ζ 22 / 53

23 Example 23 / 53

24 Underdamped Second-Order Systems Now that we have generalized the second-order transfer function in terms of ζ and ωn, let us analyze the step response of an underdamped second-order system Not only will this response be found in terms of z and vn, but more specifications indigenous to the underdamped case will be defined. A detailed description of the underdamped response is necessary for both analysis and design 24 / 53

25 Underdamped Second-Order Systems... Let us begin by finding the step response for the general second-order system. Assume ζ < 1 and ωn2 K1 K2 s + K3 = + 2 s(s 2 + 2ζωn s + ωn2 ) s s + 2ζωn s + ωn2 p (s + ζωn ) + ζ 2 ωn 1 ζ ζ C (s) = 2 s (s + ζωn ) + ωn2 (1 ζ 2 ) C (s) = G (s)r(s) = Taking the Inverse Laplace Transform yields c(t) = 1 e ζωn t! p p ζ cos(ωn 1 ζ 2 t) + p sin(ωn 1 ζ 2 t) 1 ζ2 p p e ζωn t cos(ωn 1 ζ 2 t φ), φ = tan 1 (ζ/ 1 ζ 2 ) 1 =1 p 1 ζ2 25 / 53

26 c(ωn t) for different values of ζ 26 / 53

27 Performance Specifications for a 2nd Order System Rise Time, Tr The time required for the waveform to go from 0.1 of the final value to 0.9 of the final value. Peak Time, Tp The time required to reach the first, or maximum, peak. Percent overshoot, %OS The amount that the waveform overshoots the steady- state, or final, value at the peak time, expressed as a percentage of the steady-state value. Settling Time, Ts The time required for the transient s damped oscillations to reach and stay within ±2% of the steady-state value. 27 / 53

28 Performance Specifications for a 2nd Order System 28 / 53

29 Evaluation of Tp Tp is found by differentiating c(t) iand finding the first zero crossing after t = 0. L{c (t)} = sc (s) = = ωn2 s 2 + 2ζωn s + ωn2 ωn2 (s + ζωn )2 + ωn2 (1 ζ 2 ) Therefore c (t) = p ωn e ζωn t sin(ωn 1 ζ2 p c (t) = 0 = ωn 1 ζ 2 t = nπ Tp = 29 / 53 π p ωn 1 ζ 2 = ωn 1 ζ 2 ωn p 1 ζ2 (s + ζωn )2 + ωn2 (1 ζ 2 ) p 1 ζ 2 t) n = 0, 1, 2,...

30 Evaluation of %OS Percent Overshoot is defined as %OS = cmax cfinal 100 cfinal The term cmax is found by evaluating c(t) at the peak time, c(tp ).! ζ (ζπ/ 1 ζ 2 ) cmax = c(tp ) = 1 e cos π + p sin π) 1 ζ2 2 cmax = 1 + e (ζπ/ 1 ζ ) Since cfinal = 1, we have %OS = e (ζπ/ 30 / 53 1 ζ 2 ) 100

31 Evaluation of %OS For a given %OS, the required ζ can be calculated by ln(%os/100) ζ=q π 2 + ln2 (%OS/100) 31 / 53

32 Evaluation of Ts In order to find the settling time, we must find the time for which c(t) reaches and stays within 2% of the steady-state value, cfinal. Using our definition, the settling time is the time it takes for the amplitude of the decaying sinusoid to reach 0.02, or 1 e ζωn t p = ζ2 p ln( ζ 2 ) Ts = ζωn You can verify that the numerator varies from 3.91 to 4.74 as ζ varies from 0 to 0.9. Hence, a good approximation is Ts 32 / 53 4 ζωn

33 Evaluation of Tr A precise analytical relationship between rise time and damping ratio, ζ, cannot be found. 33 / 53

34 Evaluation of Tr But for 0.3 ζ 0.8, a good approximation is Tr 2.16ζ ωn For a fixed ζ, response is faster for a larger ωn and For a fixed ωn, response is faster for a smaller ζ 34 / 53

35 How the response changes by the locations of poles? Tp = ωn π 1 ζ 2 = Ts = π ωd 4 ζωn cos θ = ζ 35 / 53

36 How the response changes by the locations of poles? 36 / 53

37 How the response changes by the locations of poles? Tp is inversely proportional to the imaginary part of the pole. Since horizontal lines on the s-plane are lines of constant imaginary value, they are also lines of constant peak time. Settling time is inversely proportional to the real part of the pole. Since vertical lines on the s-plane are lines of constant real value, they are also lines of constant settling time. Finally, since ζ = cos θ, radial lines are lines of constant ζ. Since percent overshoot is only a function of ζ, radial lines are thus lines of constant percent overshoot, %OS 37 / 53

38 How Pole Locations Affect Peak Time? 38 / 53

39 How Pole Locations Affect Overshoot? 39 / 53

40 How Pole Locations Affect Overshoot? 40 / 53

41 EX: Given the pole locations, calculate ζ, ωn, Tp, %OS, and Ts 41 / 53 ζ = cos θ = cos[arctan(7/3)] = ωn = = Tp = π ωd = π 7 Ts = 4 σd = 4 3 = 0.449sec. 2 %OS = e ( ζπ/ 1 ζ ) 100 = 26% = 1.333sec.

42 Exercise Given the system shown in figure, find J and D to yield 20% overshoot and a settling time of 2 seconds for a unit step input T (t). 42 / 53

43 System Response with Additional Poles Now that we have analyzed systems with two poles, how does the addition of another pole affect the response? The formulas describing percent overshoot, settling time, and peak time were derived only for a system with two complex poles and no zeros. If a system has more than two poles or has zeros, we cannot use the formulas to calculate the performance specifications that we derived. Under certain conditions, a system with more than two poles or with zeros can be approximated as a second-order system that has just two complex dominant poles. 43 / 53

44 Consider a system having 3 poles: Consider a three-pole system with complex poles and a third pole on the real axis. Assume the poles are at p p1,2 = ζωn ± jωn 1 ζ 2, p3 = αr the step response of the system can be determined from a partial-fraction expansion. C (s) = A B(s + ζωn ) + C ωd D s s + αr (s + ζωn ) + ωd In time-domain: h i c(t) = A + e ζωn t (B cos ωd t + C sin ωd t) + De αr t 1(t) 44 / 53

45 System Responses for different locations of αr We shall consider 3 different scenarios: 45 / 53 For Case I, αr = αr1 and is not much larger than ζωn ; For Case II, αr = αr2 and is much larger than ζωn For Case III, αr =

46 System Responses for different locations of αr / 53

47 If αr is not much greater than ζωn (Case I), the real pole s transient response will not decay to insignificance at the peak time or settling time generated by the second-order pair. In this case, the exponential decay is significant, and the system cannot be represented as a second-order system. How much farther from the dominant poles does the third pole have to be for its effect on the second-order response to be negligible? The exponential decay is negligible after five time constants. Thus, if the real pole is five times farther to the left than the dominant poles, we assume that the system is represented by its dominant second-order pair of poles. 47 / 53

48 Example Consider the following 3 systems and lets evaluate the step responses s 2 + 4s T2 (s) = 2 (s + 10)(s + 4s ) T3 (s) = (s + 3)(s 2 + 4s ) T1 (s) = Step responses are c1 (t) = e 2t cos(4.532t 23.8 ) c2 (t) = e 10t 1.189e 2t cos(4.532t ) c3 (t) = e 3t 0.707e 2t cos(4.532t ) 48 / 53

49 Example / 53

50 System Response With Zeros Now that we have seen the effect of an additional pole, let us add a zero to the second-order system T (s) = (s + a) (s + b)(s + c) We add a real-axis zero to a two-pole system. The zero will be added first in the left half-plane and then in the right half-plane and its effects noted and analyzed. Starting with a two-pole system with poles at 1 ± j2.828, we consecutively add zeros at 3, 5, and 10. We can see that the closer the zero is to the dominant poles, the greater its effect on the transient response. As the zero moves away from the dominant poles, the response approaches that of the two-pole system. 50 / 53

51 System Response With Zeros 51 / 53

52 Some MATLAB Code deng=[1 2 9]; Ta=tf([1 3]*9/3,deng); Tb=tf([1 5]*9/5,deng); Tc=tf([1 10]*9/10,deng); T=tf(9,deng); step(t,ta,tb,tc) text(0.5,0.6, no zero ) text(0.4, 0.7, zero at-10 ) text(0.35, 0.8, zero at -5 ) text(0.3,0.9, zero at -3 ) 52 / 53

53 If zero moves to RHP we obtain a non-minimum-phase system 53 / 53