Control Systems EE 4314 Lecture 29 May 5, 2015

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Control Systems EE 4314 Lecture 29 May 5, 2015 Spring 2015 Indika Wijayasinghe

Numerical Integration What is the equivalent of the different operator ( 𝑑 𝑑𝑡 or 𝑠) in terms of 𝑧? Consider 𝑈(𝑠) 𝐸(𝑠) =𝐷 𝑠 = 1 𝑠 𝑢 (𝑘+1)𝑇 = 0 𝑇 𝑒 𝑡 𝑑𝑡+ 𝑇 (𝑘+1)𝑇 𝑒 𝑡 𝑑𝑡 =𝑢 𝑇 +area under e t over the last T

Numerical Integration

Numerical Integration

Numerical Integration Example: Using three (forward difference, backward difference, Tustin method) approximation methods to find the discrete equivalent for 𝐷 𝑠 = 10𝑠+1 𝑠+1

Numerical Integration Example: Using three (forward difference, backward difference, Tustin method) approximation methods to find the discrete equivalent for 𝐶 𝑠 = 10𝑠+1 𝑠+1

Numerical Integration Frequency responses for sampling periods T=0.5 and 1 Approximation is better for higher sampling rate (T=0.5). The sampling rate should be at least 10 times higher than the highest frequency of interest. Tustin’s method is the best approximation.

State Space Formulation Find the state space model described by difference equation 𝑦 𝑘+2 =𝑢 𝑘 +1.7𝑦 𝑘+1 −0.72𝑦(𝑘)

Discrete State Space Equation Find the state space model described by difference equation 𝑦 𝑘+2 =𝑢 𝑘 +1.7𝑦 𝑘+1 −0.72𝑦(𝑘)

Solutions of Discrete State Space Equation x k+1 =Ax k +Bu k y k =Cx k +Du k Recursive solution

Solutions of Discrete State Space Equation Continue

Digital Controller Design There are two techniques for finding the difference equations for the digital controller Discrete equivalent: Design D(s) first, and then obtain equivalent D(z) using Tustin’s method, Matched Pole-Zero (MPZ) method. Discrete design: directly obtain the difference equation without designing D(s) first. Obtain G(z) and design D(z). Difference equations D/A and hold sensor 1 r(t) u(kT) u(t) e(kT) + - r(kT) plant G(s) y(t) clock A/D T y(kT) Digital controller

Design Using Discrete Equivalent Design by discrete equivalent Design a continuous compensation D(s) using continuous controller design methods such as PID, lead/lag compensator. Digitize the continuous compensation: D(s)  D(z) Use discrete analysis, simulation or experimentation to verify the design

Digitization Technique: Tustin’s Method Consider 𝑈(𝑠) 𝐸(𝑠) =𝐷 𝑠 = 1 𝑠 𝑢 𝑘𝑇 = 0 𝑘𝑇−𝑇 𝑒 𝑡 𝑑𝑡+ 𝑘𝑇−𝑇 𝑘𝑇 𝑒 𝑡 𝑑𝑡 =𝑢 𝑘𝑇−𝑇 +area under e t over the last T 𝑢 𝑘 =𝑢 𝑘−1 + 𝑇 2 [𝑒 𝑘−1 +𝑒 𝑘 ]  trapezoidal integration Taking z-transform 𝑈 𝑧 𝐸 𝑧 =𝐷(𝑧)= 𝑇 2 1+ 𝑧 −1 1− 𝑧 −1 𝑠= 2 𝑇 1− 𝑧 −1 1+ 𝑧 −1 𝑈(𝑠) 1 𝑠 𝐸(𝑠) Trapezoidal integration

Digitization Technique: Tustin’s Method MATLAB command 𝐷(𝑧)= 𝑇 2 1+ 𝑧 −1 1− 𝑧 −1 Dz=c2d(Ds,1,'tustin') Dz = 0.5 z + 0.5 ----------- z - 1 Sample time: 1 seconds Discrete-time transfer function. 𝐷 𝑠 = 1 𝑠 >> numD=[1]; denD=[1 0]; Ds=tf(numD,denD) Ds = 1 - s Continuous-time transfer function.

Relationship between s and z Consider 𝑓 𝑡 = 𝑒 −𝑎𝑡 , 𝑡>0 Laplace transform 𝐹 𝑠 = 1 𝑠+𝑎 , and it has a pole at 𝑠=−𝑎 Z-transform 𝐹 𝑧 = 𝑧 𝑧− 𝑒 −𝑎𝑇 , and it has a pole at 𝑧= 𝑒 −𝑎𝑇 A pole at 𝑠=−𝑎 in the s-plane corresponds to a pole at 𝑧= 𝑒 −𝑎𝑇 S-plane Re Im 𝑠=−𝑎 Im Z-plane 1 Re 𝑧= 𝑒 −𝑎𝑇

Digitization Technique: Matched Pole-Zero (MPZ) Method MPZ technique applies the relation 𝑧= 𝑒 𝑠𝑇 . This digitization method is an approximation Map poles and zeros according to the relation 𝑧= 𝑒 𝑠𝑇 . If the numerator is of lower order than the denominator, add powers of (z+1) to the numerator until numerator and denominator are of equal. Set the DC or low-frequency gain of D(z) equal to that of D(s). The MPZ approximation of 𝐷 𝑠 = 𝐾 𝑐 𝑠+𝑎 𝑠+𝑏 is 𝐷 𝑧 = 𝐾 𝑑 𝑧− 𝑒 −𝑎𝑇 𝑧− 𝑒 −𝑏𝑇

Digitization Technique: Pole-Zero (MPZ) Method Adjusting DC gain of D(z) 𝐷 𝑠 = 𝐾 𝑐 𝑠+𝑎 𝑠+𝑏 𝐷 𝑧 = 𝐾 𝑑 𝑧− 𝑒 −𝑎𝑇 𝑧− 𝑒 −𝑏𝑇 Using the Final Value Theorem 𝐾 𝑐 𝑎 𝑏 = 𝐾 𝑑 1− 𝑒 −𝑎𝑇 1− 𝑒 −𝑏𝑇 𝐾 𝑑 = 𝐾 𝑐 𝑎 𝑏 1− 𝑒 −𝑏𝑇 1− 𝑒 −𝑎𝑇 The difference equation is 𝑢 𝑘 =𝑢 𝑘−1 + 𝐾 𝑑 [𝑒 𝑘 −𝑒 𝑘−1 ]

Final Value Theorem Final value theorem for continuous system lim 𝑡→∞ 𝑥 𝑡 = 𝑥 𝑠𝑠 = lim 𝑠→0 𝑠𝑋(𝑠) Final value theorem for discrete system lim 𝑘→∞ 𝑥 𝑘 = 𝑥 𝑠𝑠 = lim 𝑧→1 (1− 𝑧 −1) 𝑋(𝑧)

Digitization Technique: Matched Pole-Zero (MPZ) Method For D(s) with a higher-order denominator, adds (z+1) to the numerator 𝐷 𝑠 = 𝐾 𝑐 𝑠+𝑎 𝑠(𝑠+𝑏) 𝐷 𝑧 = 𝐾 𝑑 (𝑧+1)(𝑧− 𝑒 −𝑎𝑇 ) (𝑧−1)(𝑧− 𝑒 −𝑏𝑇 )

Digitization Technique: Matched Pole-Zero (MPZ) Method Example: Design a digital controller to have a closed-loop natural frequency  𝑛 ≅0.3𝑟𝑎𝑑/𝑠 and a damping ratio =0.7. First step is to find the proper D(s) 𝐷 𝑠 =0.81 𝑠+0.2 𝑠+2 1 𝑠 2 𝑅 𝑌 − + 𝐷(𝑠) 𝐸 𝑈

Digitization Technique: Matched Pole-Zero (MPZ) Method Example: Design a digital controller to have a closed-loop natural frequency  𝑛 ≅0.3rad/s and a damping ratio =0.7. Second step is to obtain D(z) Select sampling time T so that sample rate should be about 20 times  𝑛 . Thus  𝑠 =20  𝑛 =6rad/sec. Since sampling time 𝑇= 2𝜋  𝑠 =1sec. MPZ digitization of 𝐷 𝑠 =0.81 𝑠+0.2 𝑠+2 is 𝐷 𝑧 =0.389 𝑧−0.82 𝑧−0.135 = 0.389−0.319 𝑧 −1 1−0.135 𝑧 −1 The difference equation is 𝑢 𝑘 =0.135𝑢 𝑘−1 +0.389𝑒 𝑘 −0.319𝑒(𝑘−1) >> T=1; numD=[1 0.2]; denD=[1 2]; Ds=0.81*tf(numD,denD); Dz=c2d(Ds,T,'matched') Dz = 0.3864 z - 0.3163 ----------------- z - 0.1353

Digitization Technique: Matched Pole-Zero (MPZ) Method Example: Design a digital controller to have a closed-loop natural frequency  𝑛 ≅0.3rad/s and a damping ratio =0.7.

Discrete Design Discrete design is an exact design method and avoids the approximations inherent with discrete equivalent. The design procedures are Finding the discrete model of the plant G(s) G(z) Design the compensator directly in its discrete form D(z) 𝑌(𝑧) A practical approach is to start the design using discrete equivalents, then tune up the result using discrete design.

Discrete Design For a plant described by G(s) and precede by a ZOH, the discrete transfer function is 𝐺 𝑧 = 1− 𝑧 −1 𝑍 𝐺(𝑠) 𝑠 The closed-loop transfer function 𝑌(𝑧) 𝑅(𝑧) = 𝐷 𝑧 𝐺(𝑧) 1+𝐷 𝑧 𝐺(𝑧) 𝑍𝑂𝐻 𝑠 = 1− 𝑒 −𝑠𝑇 𝑠 Mixed control system Pure discrete system

Discrete Root Locus Consider 𝐺 𝑠 = 𝑎 𝑠+𝑎 and 𝐷 𝑧 =𝐾, discuss the implications of the loci. Z-transform table Continuous system remains stable for all values of K, but the discrete system becomes oscillatory with decreasing damping ratio as z goes from 0 to -1 and eventually becomes unstable.

Relationship b/w z-plane and s-plane 𝑧= 𝑒 𝑠𝑇 wn increase z increase

Relationship b/w z-plane and s-plane

Discrete Controllers Proportional Derivative Integral Lead Compensation

Discrete Design Example: Design a digital controller to have a closed-loop natural frequency  𝑛 ≅0.3rad/s and a damping ratio =0.7. Use a discrete design method. From 𝐺 𝑧 = 1− 𝑧 −1 𝑍 𝐺(𝑠) 𝑠 𝐺 𝑧 = 𝑇 2 2 𝑧+1 (𝑧−1) 2 When T=1, 𝐺 𝑧 = 1 2 𝑧+1 (𝑧−1) 2 1 𝑠 2 𝑅 𝑌 − + 𝐷(𝑠) 𝐸 𝑈  Z-transform table 1 𝑠 3 → 𝑇 2 2 𝑧(𝑧+1) (𝑧−1) 3

Discrete Design Example: Design a digital controller to have a closed-loop natural frequency  𝑛 ≅0.3𝑟𝑎𝑑/𝑠 and a damping ratio =0.7. Use a discrete design method. Becomes unstable as K increases Z-plane locus with proportional controller D z =K Z-plane locus with PD controller D z =𝐾 (𝑧−0.85) 𝑧

Digital Control Continuous control sysGs=tf(1,[1 0 0]); 1 𝑠 2 𝑅 𝑌 − + 0.81 𝑠+0.2 𝑠+2 𝐸 𝑈 𝐷(𝑠) 𝐺(𝑠) 𝑇 2 2 𝑧+1 (𝑧−1) 2 𝑅(𝑧) 𝑌(𝑧) − + 𝐸 𝑈(𝑧) 𝐷(𝑧) 𝐺(𝑧) Continuous control sysGs=tf(1,[1 0 0]); sysDs=tf(0.81*[1 0.2],[1 2]); sysGDs=sysGs*sysDs; sysCLs=feedback(sysGDs,1); step(sysCLs); Discrete equivalent sysGs=tf(1,[1 0 0]); sysDs=tf(0.81*[1 0.2],[1 2]); T=1; sysDz=c2d(sysDs,T,'matched') sysGz=c2d(sysGs,T,'zoh'); sysDGz=sysGz*sysDz; sysCLz=feedback(sysDGz,1) step(sysCLz) Discrete design sysGs=tf(1,[1 0 0]); T=1; sysGz=c2d(sysGs,T,'zoh'); sysDz=tf(0.374*[1 -.85],[1 0],T) sysDGz=sysGz*sysDz; sysCLz=feedback(sysDGz,1) step(sysCLz)

Step Responses of the Continuous and Digital Systems Discrete equivalent Discrete design