9 chapter 9 PID controller

SYSTEM DYNAMICS & CONTROL CHAPTER PID CONTROLLER Dr Vo Tuong Quan HCMUT - 2011 PID CONTROLLER Ziegler–Nichols Unit step response of a plant  2011 – Vo Tuong Quan PID CONTROLLER Ziegler–Nichols S-shaped response curve:  2011 – Vo Tuong Quan PID CONTROLLER Ziegler–Nichols PID control : 𝐺𝐶 𝑠 = 𝐾𝑃 (1 + 𝑇𝐼 𝑠 + 𝑇𝐷 𝑠) - Frequency Domain Notice that the PID controller tuned by the first method of Ziegler–Nichols gives (𝑠 + )2 𝐿 𝐺𝑐 𝑠 = 0.6𝑇 𝑠 The PID controller has a pole at the origin and double zeros at 𝑠 =– 1/𝐿  2011 – Vo Tuong Quan PID CONTROLLER Ziegler–Nichols Ziegler–Nichols Tuning Rule Based on Step Response of Plant (First Method) 𝑲𝑷 𝑻𝑰 𝑻𝑫 𝑃 𝑇 𝐿 ∞ 𝑃𝐼 0.9 𝑇 𝐿 𝑃ID 1.2 𝑇 𝐿 𝐿 0.3 2𝐿 Type of Controller 0.5𝐿  2011 – Vo Tuong Quan PID CONTROLLER Ziegler–Nichols In the second method, we first set 𝑇𝐼 = ∞, 𝑇𝐷 = Using the proportional control action only (see Figure), increase 𝐾𝑃 from to a critical 𝐾𝑐𝑟 at which the output first exhibits sustained oscillations (If the output does not exhibit sustained oscillations for whatever value 𝐾𝑃 may take, then this method does not apply.) Thus, the critical gain 𝐾𝑐𝑟 and the corresponding period 𝑃𝑐𝑟 are experimentally determined Ziegler and Nichols suggested that we set the values of the parameters 𝐾𝑃 , 𝑇𝐼 , 𝑇𝐷 according to the formula shown in Table  2011 – Vo Tuong Quan PID CONTROLLER Ziegler–Nichols  2011 – Vo Tuong Quan PID CONTROLLER Ziegler–Nichols Notice that the PID controller tuned by the second method of Ziegler–Nichols rules gives 𝐺𝑐 𝑠 = 0.075𝐾𝑐𝑟 𝑃𝑐𝑟 𝑠+ 𝑃𝑐𝑟 𝑠 the PID controller has a pole at the origin and double zeros at 𝑠 = Type of Controller 𝑲𝑷 𝑻𝑰 𝑻𝑫 𝑃 0.5𝐾𝑐𝑟 ∞ 𝑃𝐼 0.45𝐾𝑐𝑟 𝐾𝑐𝑟 1.2 𝑃ID 0.6𝐾𝑐𝑟 0.5𝑃𝑐𝑟 0.125𝑃𝑐𝑟 − 𝑃𝑐𝑟  2011 – Vo Tuong Quan PID CONTROLLER Ziegler–Nichols – Example Design a PID controller to control a furnace providing the open loop characteristic of the furnace obtained from a experiment :  2011 – Vo Tuong Quan PID CONTROLLER Ziegler–Nichols – Example We can find: 𝐾 = 150, 𝑇1 = 𝑚𝑖𝑛 = 480 𝑠𝑒𝑐, 𝑇2 = 24 𝑚𝑖𝑛 = 1440 𝑠𝑒𝑐 𝑇2 1440 𝐾𝑃 = 1.2 = 1.2 = 0.024 𝑇1 𝐾 480 × 150 𝑇𝐼 = 𝑇1 = × 480 = 960 𝑠𝑒𝑐 𝑇𝐷 = 0.5 𝑇1 = 0.5 × 480 = 240 𝑠𝑒𝑐 𝐺𝑃𝐼𝐷 𝑠 = 0.024 + + 240𝑠 960𝑠 10  2011 – Vo Tuong Quan PID CONTROLLER Control system design in state space - Pole Placement Method Observability 𝑥 𝑡 = 𝐴𝑥 𝑡 + 𝐵𝑢(𝑡) Consider a system: 𝑦 𝑡 = 𝐶𝑥(𝑡) The system is complete state observable if the control law 𝑢(𝑡) and the output signal 𝑦(𝑡) in a finite time interval 𝑡0 ≤ 𝑡 ≤ 𝑡𝑓 It is possible to determine the initial states 𝑥 𝑡0 Qualitatively, the system is state observable if all state variable 𝑥 𝑡 influences the output 𝑦(𝑡) 35  2011 – Vo Tuong Quan PID CONTROLLER Control system design in state space - Pole Placement Method Observability 𝑥 𝑡 = 𝐴𝑥 𝑡 + 𝐵𝑢(𝑡) Consider a system: 𝑦 𝑡 = 𝐶𝑥(𝑡) It is necessary to estimate the state 𝑥 𝑡 , from mathematical model of the system and the input-output data 𝐶 Observability matrix: 𝜕 = 𝐶𝐴 𝐶𝐴2 ⋮ 𝐶𝐴𝑛−1 The necessary and sufficient condition for the observability is: 𝑟𝑎𝑛𝑘 𝜕 = 𝑛 36  2011 – Vo Tuong Quan PID CONTROLLER Control system design in state space - Pole Placement Method Observability 𝑥 𝑡 = 𝐴𝑥 𝑡 + 𝐵𝑢(𝑡) Consider a system: 𝑦 𝑡 = 𝐶𝑥(𝑡) where: 𝐴 = −2 1 ,𝐵 = ,𝐶 = −3 Evaluate the observability of the system 𝐶𝐴 −6 −8 det 𝜕 = 10 → 𝑟𝑎𝑛𝑘 𝜕 = → 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑖𝑠 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑏𝑙𝑒 Solution: Observability matrix: 𝜕 = 𝐶 → 𝜕= 37  2011 – Vo Tuong Quan PID CONTROLLER Control system design in state space - Pole Placement Method State Feedback Control Consider a system described by the state-space equation: 𝑥 𝑡 = 𝐴𝑥 𝑡 + 𝐵𝑢(𝑡) 𝑦 𝑡 = 𝐶𝑥(𝑡) The state feedback controller: 𝑢 𝑡 = 𝑟 𝑡 − 𝐾𝑥(𝑡) The characteristic equation of the closed-loop system: 𝑥 𝑡 = [𝐴 − 𝐵𝐾]𝑥 𝑡 + 𝐵𝑟(𝑡) 𝑦 𝑡 = 𝐶𝑥(𝑡) If the system is controllable then it is possible to determine the feedback gain 𝐾 so that the closed-loop system has the poles at any location Step 1: the characteristic equation of the closed-loop system: det 𝑠𝐼 − 𝐴 + 𝐵𝐾 = (1) 38  2011 – Vo Tuong Quan PID CONTROLLER Control system design in state space - Pole Placement Method State Feedback Control Step 2: the desired characteristic equation: 𝑛 𝑖=1 𝑠 − 𝑝𝑖 = (2) where 𝑝𝑖 , (𝑖 = 1, 𝑛) are the desired poles Step 3: balance the coefficient of the equation (1), (2), we can find the state feedback gain 𝐾 39  2011 – Vo Tuong Quan PID CONTROLLER Control system design in state space - Pole Placement Method State Feedback Control Given a system described by the state-state equation: 𝑥 𝑡 = 𝐴𝑥 𝑡 + 𝐵𝑢(𝑡) 𝑦 𝑡 = 𝐶𝑥(𝑡) 0 𝐴 = 0 , 𝐵 = , 𝐶 = [0 1] −4 −7 −3 Determine the state feedback controller 𝑢 𝑡 = 𝑟 𝑡 − 𝐾𝑥(𝑡) so that the closed-loop system has complex poles with ξ = 0.6, 𝜔𝑛 = 10 and the third pole at -20 40  2011 – Vo Tuong Quan PID CONTROLLER Control system design in state space - Pole Placement Method State Feedback Control The characteristics equation is: det 𝑠𝐼 − 𝐴 + 𝐵𝐾 = 0 0 → 𝑑𝑒𝑡 𝑠 − 0 + 𝑘1 𝑘2 𝑘3 = 0 −4 −7 −3 → 𝑠3 + + 3𝑘2 + 𝑘3 𝑠2 + + 3𝑘1 + 10𝑘2 − 21𝑘3 𝑠 + + 10𝑘1 − 12𝑘3 = (1) The desired characteristic equation: 𝑠 + 20 𝑠2 + 2ξ𝜔𝑛 𝑠 + 𝜔𝑛 = → 𝑠3 + 32𝑠2 + 340𝑠 + 2000 = (2) + 3𝑘2 + 𝑘3 = 32 From (1), (2) we have: + 3𝑘1 + 10𝑘2 − 21𝑘3 = 340 + 10𝑘1 − 12𝑘3 = 2000 41  2011 – Vo Tuong Quan PID CONTROLLER Control system design in state space - Pole Placement Method State Feedback Control 𝑘1 = 220.578 We have: 𝑘2 = 3.839 𝑘3 = 17.482 So the state feedback gain 𝐾 𝐾 = 220.578 3.839 17.482 42  2011 – Vo Tuong Quan PID CONTROLLER Design PID controller with computational optimization approach Consider the PID-controlled system shown in Figure below The PID controller is given by (𝑠 + 𝑎)2 𝐺𝐶 𝑠 = 𝐾 𝑠 It is desired to find a combination of K and a such that the closed-loop system will have 10%, maximum overshoot in the unit-step response 43  2011 – Vo Tuong Quan PID CONTROLLER Design PID controller with computational optimization approach Assume that the region to search for 𝐾 and 𝑎 is: ≤ 𝐾 ≤ 3, 0.5 ≤ 𝑎 ≤ 1.5 K = [2.0 2.2 2.4 2.6 2.8 3.0]; a = [0.5 0.7 0.9 1.1 1.3 1.5]; t = 0:0.01:5; g = tf([1.2],[0.36 1.86 2.5 1]); k = 0; for i = 1:6; for j = 1:6; gc = tf(K(i)*[1 2*a(j) a(j)^2], [1 0]); % controller G = gc*g/(1 + gc*g); % closed-loop transfer function y = step(G,t); m = max(y); if m < 1.10 k = k+1; solution(k,:) = [K(i) a(j) m]; end end end solution % Print solution table sortsolution = sortrows(solution,3) 44  2011 – Vo Tuong Quan PID CONTROLLER Design PID controller with computational optimization approach K = sortsolution(k,1) a = sortsolution(k,2) gc = tf(K*[1 2*a a^2], [1 0]); G = gc*g/(1 + gc*g); 1.4 step(G,t) 1.2 grid Step Response Amplitude 0.8 0.6 0.4 0.2 0 0.5 1.5 2.5 Time (sec)  2011 – Vo Tuong Quan 3.5 4.5 45 PID CONTROLLER Design PID controller with computational optimization approach K = sortsolution(11,1) a = sortsolution(11,2) gc = tf(K*[1 2*a a^2], [1 0]); 1.4 G = gc*g/(1 + gc*g); step(G,t) 1.2 grid Step Response Amplitude 0.8 0.6 0.4 0.2 0 0.5 1.5 2.5 Time (sec)  2011 – Vo Tuong Quan 3.5 4.5 46 PID CONTROLLER Design PID controller with computational optimization approach Consider the PID-controlled system shown in Figure below The PID controller is given by (𝑠 + 𝑎)2 𝐺𝐶 𝑠 = 𝐾 𝑠 We want to find all combinations of K and a values such that the closed-loop system has a maximum overshoot of less than 15%, but more than 10%, in the unit-step response In addition, the settling time should be less than sec 47  2011 – Vo Tuong Quan PID CONTROLLER Design PID controller with computational optimization approach Assume that the region to search for 𝐾 and 𝑎 is: ≤ 𝐾 ≤ 5, 0.1 ≤ 𝑎 ≤ 3.0 t = 0:0.01:8;k = 0; for K = 3:0.2:5; for a = 0.1:0.1:3; num = [4*K 8*K*a 4*K*a^2]; den = [1 8+4*K 4+8*K*a 4*K*a^2]; y = step(num,den,t); s = 801;while y(s)>0.98 & y(s)