Systems Analysis and Control

Transcription

1 Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 8: Response Characteristics

2 Overview In this Lecture, you will learn: Characteristics of the Response Stability Real Poles Steady-State Error Rise Time Settling Time Complex Poles Complex Pole Locations Damped/Natural Frequency Damping and Damping Ratio M. Peet Lecture 8: Control Systems 2 / 22

3 Feedback Control Recall the Feedback Interconnection - u(s) + K(s) G(s) y(s) Feedback: Controller: u i = K(u y) Plant: y = Gu i The output signal is ŷ(s), ŷ(s) = Ĝ(s) ˆK(s) 1 + Ĝ(s) ˆK(s)û(s) M. Peet Lecture 8: Control Systems 3 / 22

4 Controlling the Inverted Pendulum Model Open Loop Transfer Function 1 Ĝ(s) = Js 2 Mgl 2 Controller: Static Gain: ˆK(s) = K Input: Impulse: û(s) = 1. Closed Loop: Lower Feedback ŷ(s) = Ĝ(s) ˆK(s) = 1 + Ĝ(s) ˆK(s)û(s) K Js 2 Mgl K Js 2 Mgl 2 = K Js 2 Mgl 2 + K M. Peet Lecture 8: Control Systems 4 / 22

5 Controlling the Inverted Pendulum Model Closed Loop Impulse Response: Lower Feedback K ŷ(s) = Js 2 Mgl 2 + K Traits: Infinite Oscillations Oscillates about 0. Amplitude Impulse Response Open Loop Impulse Response: Time (sec) 18 x 105 Impulse Response ŷ(s) = 1 2J 1 J Mgl s Mgl 2J 1 s + Mgl 2J Amplitude Unstable! Time (sec) M. Peet Lecture 8: Control Systems 5 / 22

6 Controlling the Suspension System Open Loop Transfer Function: Set m c = m w = g = c = K 1 = K 2 = 1. x 1 m c Ĝ(s) = s 2 + s + 1 s 4 + 2s 3 + 3s 2 + s + 1 x 2 m w Controller: Static Gain: ˆK(s) = k u Closed Loop: Lower Feedback Ĝ(s) ˆK(s) ŷ(s) = 1 + Ĝ(s) ˆK(s)û(s) = k(s 2 + s + 1) s 4 + 2s 3 + (3 + k)s 2 + (1 + k)s + (1 + k) M. Peet Lecture 8: Control Systems 6 / 22

7 Controlling the Suspension Problem Effect of changing the Feedback, k Closed Loop Step Response: ŷ(s) = k(s 2 + s + 1) 1 s 4 + 2s 3 + (3 + k)s 2 + (1 + k)s + (1 + k) s High k: Overshot the target Quick Response Closer to desired value of f Low k: Slow Response No overshoot Final value is farther from 1. Questions: Which Traits are important? How to predict the behaviour? Figure: Step Response for different k M. Peet Lecture 8: Control Systems 7 / 22

8 Stability The most basic property is Stability: Definition 1. A system, G is Stable if there exists a K > 0 such that kgukl2 KkukL2 Note: Although this is the true definition for systems defined by transfer functions, it is rarely used. Bounded input means bounded output. Stable is y(t) 0 when u(t) 0. M. Peet Lecture 8: Control Systems 8 / 22

9 Stability Definition 2. The Closed Right Half-Place, CRHP is the set of complex numbers with non-negative real part. {s C : Real(s) 0} Im(s) CRHP Re(s) Im(s) Figure: Unstable Re(s) Theorem 3. A system G is stable if and only if it s transfer function Ĝ has no poles in the Closed Right Half Plane. Check stability by checking poles. x is a pole o is a zero M. Peet Lecture 8: Control Systems 9 / 22

10 Predicting Steady-State Error Definition 4. Steady-State Error for a stable system is the final difference between input and output. e ss = lim t u(t) y(t) Usually measured using the step response. Since u(t) = 1, e ss = 1 lim t y(t) Figure: Suspension Response for k = 1 M. Peet Lecture 8: Control Systems 10 / 22

11 Predicting Steady-State Error Recall: For any system G, by partial fraction expansion: So which means and hence ŷ(s) = Ĝ(s)1 s = r 0 s + r 1 s p y(t) = r 0 1(t) + r 1 e p1t r n e pnt lim t y(t) = r 0 e ss = 1 r 0 r n s p n M. Peet Lecture 8: Control Systems 11 / 22

12 Predicting Steady-State Error The steady-state error is given by r 0. e ss = 1 r 0 Recall: The residue at s = 0 is r 0 and is found as r 0 = Ĝ(s) s=0 = lim s 0 Ĝ(s) Thus the steady-state error is e ss = 1 lim s 0 Ĝ(s) This can be generalized to find the limit of any signal: Theorem 5 (Final Value Theorem). lim y(t) = lim sŷ(s) t s 0 Assumes the limit exists (Stability) Can be used to find response to other inputs Ramp, impulse, etc. M. Peet Lecture 8: Control Systems 12 / 22

13 Predicting Steady-State Error Numerical Example Ĝ(s) = The steady-state response is k(s 2 + s + 1) s 4 + 2s 3 + (3 + k)s 2 + (1 + k)s + (1 + k) y ss = lim s 0 sŷ(s) = lim s 0 Ĝ(s) = lim s 0 k(s 2 + s + 1) s 4 + 2s 3 + (3 + k)s 2 + (1 + k)s + (1 + k) = k 1 + k The steady-state error is When k = 0, e ss = 1 As k, e ss = 0 e ss = 1 y ss = 1 = k k 1 + k M. Peet Lecture 8: Control Systems 13 / 22

14 Dynamic Response Characteristics Two Types of Response By now, you know that motion is dominated by the poles! Simplify the response by considering response of each pole. Allows quantitative analysis Figure: Real Pole Figure: Complex Pair of Poles We start with Real Poles M. Peet Lecture 8: Control Systems 14 / 22

15 Step Response Characteristics Real Poles Consider a real pole step response: ŷ(s) = r 1 s p s = r r p s p p s y(t) = r p ( e pt 1 ) Assume stable, so p < 0 Cases: p > 0 implies y(t) p < 0 implies y(t) r p Steady-State Error: e ss = 1 r p M. Peet Lecture 8: Control Systems 15 / 22

16 Step Response Characteristics Rise Time Besides the final value: How quickly will the system respond? Definition 6. The rise time, T r, is the time it takes to go from.1 to.9 of the final value. t 1 when y(t 1 ) =.1 r p is found as.1 = e pt1 1 ln(1.1) = pt 1 ln.9 t 1 = p Likewise for y(t 2 ) =.9 r p t 2 = ln.1 p = 2.31 p =.11 p we get Thus rise time for a Single Pole is: T r = t 2 t 1 = 2.31 p.11 p = 2.2 p M. Peet Lecture 8: Control Systems 16 / 22

17 Step Response Characteristics Settling Time Will it stay there: How fast does it converge? More important for complex poles. Definition 7. The Settling Time, T s, is the time it takes to reach and stay within.99 of the final value. The time at y(t s ) =.99 r p is found as.99 = e pts 1 ln(.01) = pt s ln.01 T s = p = 4.6 p The settling time for a Single Pole is: T s = 4.6 p M. Peet Lecture 8: Control Systems 17 / 22

18 Solution for Complex Poles ŷ(s) = ω 2 d + σ2 s 2 + 2σs + ω 2 d + σ2 1 s = ω 2 n s 2 + 2ζω n s + ω 2 n The poles are at s = σ ± ω d ı and s = 0. The solution is: y(t) = 1 e (cos(ω σt d t) σ ) sin(ω d t) ω d = 1 e σt ω n ω d sin (ω d t + φ) Where σ = ζω n, ω d = ω n 1 ζ2 and φ = tan 1 ( ωd ζω n ). The result is oscillation with an Exponential Envelope. Envelope decays at rate σ Speed of oscillation is ω d, the Damped Frequency 1 s = k 1 s + k 2 s 2 + 2ζω n s + ωn 2 + r 2 s M. Peet Lecture 8: Control Systems 18 / 22

19 Damping We use several adjectives to describe exponential decay: Undamped Oscillation continues forever, σ = 0 Underdamped Oscillation continues for many cycles. Damped Critically Damped No oscillation or overshoot. ω = 0 M. Peet Lecture 8: Control Systems 19 / 22

20 Step Response Characteristics Damping Ratio Besides ω, there is another way to measure oscillation Im(s) Definition 8. The Natural Frequency of a pole at p = σ + ıω d is ω n = σ 2 + ω 2 d. ω n Re(s) 1 1 for ŷ(s) = s 2 +as+b s, ω n = b. Radius of the pole in complex plane. Frequency of least damping. Also known as resonant frequency M. Peet Lecture 8: Control Systems 20 / 22

21 Step Response Characteristics Damping Ratio Besides σ, there are other ways to measure damping Definition 9. The Damping Ratio of a pole at p = σ + ıω is ζ = σ ω n. 1 1 for ŷ(s) = s 2 +as+b s, ζ = The amount the amplitude decreases per oscillation. a 2ω n. The angle that the pole makes in the complex plane. M. Peet Lecture 8: Control Systems 21 / 22

22 Summary What have we learned today? Characteristics of the Response Real Poles Steady-State Error Rise Time Settling Time Complex Poles Complex Pole Locations Damped/Natural Frequency Damping and Damping Ratio Continued in Next Lecture M. Peet Lecture 8: Control Systems 22 / 22