Presentation is loading. Please wait.

Presentation is loading. Please wait.

Control Systems EE 4314 Lecture 29 May 5, 2015

Similar presentations


Presentation on theme: "Control Systems EE 4314 Lecture 29 May 5, 2015"β€” Presentation transcript:

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

2 Numerical Integration
What is the equivalent of the different operator ( 𝑑 𝑑𝑑 or 𝑠) in terms of 𝑧? Consider π‘ˆ(𝑠) 𝐸(𝑠) =𝐷 𝑠 = 1 𝑠 𝑒 (π‘˜+1)𝑇 = 0 𝑇 𝑒 𝑑 𝑑𝑑+ 𝑇 (π‘˜+1)𝑇 𝑒 𝑑 𝑑𝑑 =𝑒 𝑇 +area under e t over the last T

3 Numerical Integration

4 Numerical Integration

5 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

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

7 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.

8 State Space Formulation
Find the state space model described by difference equation 𝑦 π‘˜+2 =𝑒 π‘˜ +1.7𝑦 π‘˜+1 βˆ’0.72𝑦(π‘˜)

9 Discrete State Space Equation
Find the state space model described by difference equation 𝑦 π‘˜+2 =𝑒 π‘˜ +1.7𝑦 π‘˜+1 βˆ’0.72𝑦(π‘˜)

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

11 Solutions of Discrete State Space Equation
Continue

12 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

13 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) οƒ  D(z) Use discrete analysis, simulation or experimentation to verify the design

14 Digitization Technique: Tustin’s Method
Consider π‘ˆ(𝑠) 𝐸(𝑠) =𝐷 𝑠 = 1 𝑠 𝑒 π‘˜π‘‡ = 0 π‘˜π‘‡βˆ’π‘‡ 𝑒 𝑑 𝑑𝑑+ π‘˜π‘‡βˆ’π‘‡ π‘˜π‘‡ 𝑒 𝑑 𝑑𝑑 =𝑒 π‘˜π‘‡βˆ’π‘‡ +area under e t over the last T 𝑒 π‘˜ =𝑒 π‘˜βˆ’1 + 𝑇 2 [𝑒 π‘˜βˆ’1 +𝑒 π‘˜ ] οƒŸ trapezoidal integration Taking z-transform π‘ˆ 𝑧 𝐸 𝑧 =𝐷(𝑧)= 𝑇 𝑧 βˆ’1 1βˆ’ 𝑧 βˆ’1 𝑠= 2 𝑇 1βˆ’ 𝑧 βˆ’1 1+ 𝑧 βˆ’1 π‘ˆ(𝑠) 1 𝑠 𝐸(𝑠) Trapezoidal integration

15 Digitization Technique: Tustin’s Method
MATLAB command 𝐷(𝑧)= 𝑇 𝑧 βˆ’1 1βˆ’ 𝑧 βˆ’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.

16 Relationship between s and z
Consider 𝑓 𝑑 = 𝑒 βˆ’π‘Žπ‘‘ , 𝑑>0 Laplace transform 𝐹 𝑠 = 1 𝑠+π‘Ž , and it has a pole at 𝑠=βˆ’π‘Ž Z-transform 𝐹 𝑧 = 𝑧 π‘§βˆ’ 𝑒 βˆ’π‘Žπ‘‡ , and it has a pole at 𝑧= 𝑒 βˆ’π‘Žπ‘‡ A pole at 𝑠=βˆ’π‘Ž in the s-plane corresponds to a pole at 𝑧= 𝑒 βˆ’π‘Žπ‘‡ S-plane Re Im 𝑠=βˆ’π‘Ž Im Z-plane 1 Re 𝑧= 𝑒 βˆ’π‘Žπ‘‡

17 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+π‘Ž 𝑠+𝑏 is 𝐷 𝑧 = 𝐾 𝑑 π‘§βˆ’ 𝑒 βˆ’π‘Žπ‘‡ π‘§βˆ’ 𝑒 βˆ’π‘π‘‡

18 Digitization Technique: Pole-Zero (MPZ) Method
Adjusting DC gain of D(z) 𝐷 𝑠 = 𝐾 𝑐 𝑠+π‘Ž 𝑠+𝑏 𝐷 𝑧 = 𝐾 𝑑 π‘§βˆ’ 𝑒 βˆ’π‘Žπ‘‡ π‘§βˆ’ 𝑒 βˆ’π‘π‘‡ Using the Final Value Theorem 𝐾 𝑐 π‘Ž 𝑏 = 𝐾 𝑑 1βˆ’ 𝑒 βˆ’π‘Žπ‘‡ 1βˆ’ 𝑒 βˆ’π‘π‘‡ 𝐾 𝑑 = 𝐾 𝑐 π‘Ž 𝑏 1βˆ’ 𝑒 βˆ’π‘π‘‡ 1βˆ’ 𝑒 βˆ’π‘Žπ‘‡ The difference equation is 𝑒 π‘˜ =𝑒 π‘˜βˆ’1 + 𝐾 𝑑 [𝑒 π‘˜ βˆ’ο‘π‘’ π‘˜βˆ’1 ]

19 Final Value Theorem Final value theorem for continuous system
lim π‘‘β†’βˆž π‘₯ 𝑑 = π‘₯ 𝑠𝑠 = lim 𝑠→0 𝑠𝑋(𝑠) Final value theorem for discrete system lim π‘˜β†’βˆž π‘₯ π‘˜ = π‘₯ 𝑠𝑠 = lim 𝑧→1 (1βˆ’ 𝑧 βˆ’1) 𝑋(𝑧)

20 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+π‘Ž 𝑠(𝑠+𝑏) 𝐷 𝑧 = 𝐾 𝑑 (𝑧+1)(π‘§βˆ’ 𝑒 βˆ’π‘Žπ‘‡ ) (π‘§βˆ’1)(π‘§βˆ’ 𝑒 βˆ’π‘π‘‡ )

21 Digitization Technique: Matched Pole-Zero (MPZ) Method
Example: Design a digital controller to have a closed-loop natural frequency  𝑛 β‰…0.3π‘Ÿπ‘Žπ‘‘/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 π‘Œ βˆ’ + 𝐷(𝑠) 𝐸 π‘ˆ

22 Digitization Technique: Matched Pole-Zero (MPZ) Method
Example: Design a digital controller to have a closed-loop natural frequency  𝑛 β‰…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œ‹  𝑠 =1sec. MPZ digitization of 𝐷 𝑠 =0.81 𝑠+0.2 𝑠+2 is 𝐷 𝑧 =0.389 π‘§βˆ’0.82 π‘§βˆ’0.135 = 0.389βˆ’0.319 𝑧 βˆ’1 1βˆ’0.135 𝑧 βˆ’1 The difference equation is 𝑒 π‘˜ =0.135𝑒 π‘˜βˆ’ 𝑒 π‘˜ βˆ’0.319𝑒(π‘˜βˆ’1) >> T=1; numD=[1 0.2]; denD=[1 2]; Ds=0.81*tf(numD,denD); Dz=c2d(Ds,T,'matched') Dz = z z

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

24 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)οƒ  G(z) Design the compensator directly in its discrete form D(z) π‘Œ(𝑧) A practical approach is to start the design using discrete equivalents, then tune up the result using discrete design.

25 Discrete Design For a plant described by G(s) and precede by a ZOH, the discrete transfer function is 𝐺 𝑧 = 1βˆ’ 𝑧 βˆ’1 𝑍 𝐺(𝑠) 𝑠 The closed-loop transfer function π‘Œ(𝑧) 𝑅(𝑧) = 𝐷 𝑧 𝐺(𝑧) 1+𝐷 𝑧 𝐺(𝑧) 𝑍𝑂𝐻 𝑠 = 1βˆ’ 𝑒 βˆ’π‘ π‘‡ 𝑠 Mixed control system Pure discrete system

26 Discrete Root Locus Consider 𝐺 𝑠 = π‘Ž 𝑠+π‘Ž 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.

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

28 Relationship b/w z-plane and s-plane

29 Discrete Controllers Proportional Derivative Integral
Lead Compensation

30 Discrete Design Example: Design a digital controller to have a closed-loop natural frequency  𝑛 β‰…0.3rad/s and a damping ratio =0.7. Use a discrete design method. From 𝐺 𝑧 = 1βˆ’ 𝑧 βˆ’1 𝑍 𝐺(𝑠) 𝑠 𝐺 𝑧 = 𝑇 𝑧+1 (π‘§βˆ’1) 2 When T=1, 𝐺 𝑧 = 𝑧+1 (π‘§βˆ’1) 2 1 𝑠 2 𝑅 π‘Œ βˆ’ + 𝐷(𝑠) 𝐸 π‘ˆ οƒŸ Z-transform table 1 𝑠 3 β†’ 𝑇 𝑧(𝑧+1) (π‘§βˆ’1) 3

31 Discrete Design Example: Design a digital controller to have a closed-loop natural frequency  𝑛 β‰…0.3π‘Ÿπ‘Žπ‘‘/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 (π‘§βˆ’0.85) 𝑧

32 Digital Control Continuous control sysGs=tf(1,[1 0 0]);
1 𝑠 2 𝑅 π‘Œ βˆ’ + 0.81 𝑠+0.2 𝑠+2 𝐸 π‘ˆ 𝐷(𝑠) 𝐺(𝑠) 𝑇 𝑧+1 (π‘§βˆ’1) 2 𝑅(𝑧) π‘Œ(𝑧) βˆ’ + 𝐸 π‘ˆ(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)

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


Download ppt "Control Systems EE 4314 Lecture 29 May 5, 2015"

Similar presentations


Ads by Google