Lecture 7:Time Response Pole-Zero Maps Influence of Poles and Zeros Higher Order Systems and Pole Dominance Criterion

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1 Cleveland State University MCE441: Intr. Linear Control Lecture 7:Time Influence of Poles and Zeros Higher Order and Pole Criterion Prof. Richter 1 / 26

2 First-Order Specs: Step : Pole Real inputs contain randomness (for example, noise): unpredictable. Test inputs help to predict response quality under other inputs. step inputs are useful to simulate sudden changes (startup, jump loading, etc) sinusoidal inputs are used to test frequency response (more later) impulses are used to simulate shock conditions ramp inputs are used to simulate transitions between setpoints 2 / 26

3 First-Order Specs: Step : Pole z 2 Im G(s) = p 5 p 4 z 1 p2 p 1 z 1 p 4 p 5 (s z 1 )(s z 1 )(s z 2 ) (s p 5 )(s p 5 )(s p 2 ) 2 (s p 1 )(s p 4 )(s p 4 )(s p 5 )(s p 5 ) Re 3 / 26

4 First-Order First-Order Specs: Step : Pole First-order systems are described by the I/O differential equation or in TF form τċ+c = r(t) C(s) R(s) = 1 τs+1 The number τ is called the time constant of the system. The TF has one real pole at s = 1 τ. Step Applying the unit step input R(s) = 1 s gives the response c(t) = 1 e t τ 4 / 26

5 First-Order Specs: Step : Pole (Stable) first-order systems always track the step command (the error approaches zero asymptotically). The time it takes for the output to reach approx. 98% of its steady value is called settling time, also referred to as 2 % settling time. It equals 4 time constants. 5 / 26

6 First-Order Specs: Step : Pole First-order systems cannot track a ramp command. The tracking error is always equal to the time constant. 6 / 26

7 First-Order Specs: Step : Pole The standard form of a second-order system is: C(s) R(s) = w 2 n s 2 +2ζw n s+w 2 n where w n and ζ are called the natural frequency (in rad/s) and damping ratio (dimensionless), respectively. The response will depend on the nature of the poles: Underdamped: 0 < ζ < 1 Marginally Stable: ζ = 0 Critically Damped: ζ = 1 Overdamped: ζ > 1 7 / 26

8 Case 1: Underdamped (0 < ζ < 1) First-Order Specs: Step : Pole The poles are complex conjugates. The response to a unit step input is given by ( ) c(t) = 1 e ζw nt 1 ζ sin w d t+tan ζ 2 where w d is the damped natural frequency: w d = w n 1 ζ 2 The damped natural frequency is the one obtained by counting the cycles per unit time in an experimental (and underdamped) response and converting the result to radians per second. ζ 8 / 26

9 Underdamped Case First-Order Specs: Step : Pole Poles: ζw n ± w d j cos β = ζ ζw n β w n Im w d = w n 1 ζ 2 0 Re w d 9 / 26

10 Marginally Stable and Overdamped Cases First-Order Specs: Step : Pole Marginally Stable Case (ζ = 0) When ζ = 0 the poles lie on the imaginary axis and the response to a step input is c(t) = 1 cosw n t that is, the oscillation does not die out. The system oscillates with its natural frequency w n. Critically Damped Case (ζ = 1) When ζ = 1 we get a real pole of multiplicity two at s = w n. The response is given by c(t) = 1 e w nt (1+w n t). No oscillations occur for a step input. 10 / 26

11 Overdamped Case (ζ > 1) First-Order Specs: Step : Pole When ζ > 1 the poles lie are real and unequal. The response features two decaying exponential terms (no oscillations): c(t) = 1+ where the poles are w n 2 ζ 2 1 ( e s 1 t s 1 e s2t s 2 s 1,2 = (ζ ± ζ 2 1)w n ) 11 / 26

12 Normalized First-Order Specs: Step : Pole 12 / 26

13 First-Order Specs: Step : Pole Many practical design cases call for time-domain specifications Step response specs are often demanded (worst-case, drastic situation) Common specifications 1. Rise time (T r ) 2. Peak time (T p ) 3. Percent overshoot (P.O.) 4. Settling time (T s ) 13 / 26

14 Specs: Step First-Order Specs: Step : Pole FV Setpoint 1.02*F.V 0.98*F.V Second Order Step Characteristics Overshoot 0 0 Tr Tp Tset 2% Time (sec) Setpoint Steady Error Final Value 14 / 26

15 First-Order Specs: Step 2nd-Order : Pole 2 % settling time: t s = 4 ζw n Rise time: t r = π β w d Peak time: t p = π w d Peak value: M pt = (e ( ζ 1 ζ 2)π +1)y( ) Percent overshoot: P.O.= 100e ( ζ 1 ζ 2)π 15 / 26

16 First-Order Specs: Step 2nd-Order : Pole 16 / 26

17 First-Order Specs: Step Pole Locations and : Pole 17 / 26

18 First-Order Specs: Step : Pole The above design formulas are valid when no zeros are present. In general, zeros spoil the response by increasing the overshoot. If zeros are found on the r.h.p., the system is said to be nonminimum phase, and the response is worsened. Zeros may be neglected in some cases (more about this later in this lecture) 18 / 26

19 : Pole First-Order Specs: Step : Pole When more than two poles are present, or when two distinct real poles are present, it can be difficult to predict the response without computer simulation Frequently, one or two poles are dominant. Remember that each single real pole gives rise to a term of the form a s(s+a) when a step input is applied. If we invert the term, the contribution of the pole is 1 e at. So if the time constant of the pole (1/a) is very small, the transient is very short and does not contribute much to the overall response. On the other hand, poles with large time constants are the ones dictating the response. We say that such poles are dominant. The same applies to complex conjugate pairs, where the time 1 constant taken into consideration is ζw n. 19 / 26

20 Pole Criterion First-Order Specs: Step : Pole Begin by finding all poles of the TF and finding their time constants. Remember: For a real pole at s = a, the time constant is 1/a, and for a complex conjugate pair with 0 < ζ < 1, it is 1 ζw n. Sort out the time constants: (high) τ 1,τ 2...,τ n (low). Attempt to find a gap of a factor of eight or more between two consecutive time constants, starting with the highest. If such a gap is found, neglect all poles on the fast side of the gap (small time constants). Careful! when reassembling the TF. You may be inadvertently changing the TF gain (a.k.a. DC gain): Make sure that the responses to a step input have the same final value. If not, correct the reduced TF with an appropriate constant. 20 / 26

21 Pole : Example First-Order Specs: Step : Pole G(s) = 3 s s s s+32 Poles: s 1 = 2, s 2 = 4, s 3,4 = 0.2 ± 1.989i(ζ = 0.1, w n = 2). Time constants for real poles: τ 1 = 0.5 and τ 2 = For the complex poles : τ = 5. The factored TF is: G(s) = 3 (s+2)(s+4)(s s+4). Time constants: 5,0.5,0.25. A gap of a factor of 10 is found, therefore we neglect the fast poles at s 2 = 4 and s 1 = 2. If we simply eliminate the factors (s + 2) and (s + 4), we have a gain mismatch. In order to have the same final value, we correct the gain to obtain: G r (s) = 3/8 s s + 4 Check with the final value theorem: lim s 0 sg r (s) = lim s 0 sg(s). 21 / 26

22 First-Order Specs: Step : Pole For second-order systems, we can predict the effect of a single real zero on the overshoot. Consider G(s) = (w2 n /a)(s + a) s 2 + 2ζw n s + wn 2 If a/ζw n > 8, neglect the zero due to dominance: G r (s) = wn 2 s 2 +2ζwns+wn 2. If not, use the chart: 22 / 26

23 Example First-Order Specs: Step : Pole Sketch the response of the following TF when the input is a step of size 2: 3(s+5) G(s) = (s+23)(s 2 +5s+16) 23 / 26

24 Solution First-Order Specs: Step : Pole 24 / 26

25 Solution First-Order Specs: Step : Pole 25 / 26

26 Matlab Simulation First-Order Specs: Step : Pole c(t) c(t p )= t s =1.3 s c( )= Overshoot=12.76% t 26 / 26

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