Lecture Notes of Control Systems I - ME 431/Analysis and Synthesis of Linear Control System
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Lecture Notes of Control Systems I - ME 431/Analysis and Synthesis of Linear Control System - ME862
Note 12
Design via Root Locus Part II
Department of Mechanical Engineering, University Of Saskatchewan, 57 Campus Drive, Saskatoon, SK S7N 5A9, Canada 1 Lecture Notes of Control Systems I - ME 431/Analysis and Synthesis of Linear Control System - ME862
4. Improving Steady-State Error – Using PI Controller
4.1 Review the steady-state error
Steady-state error is the difference between the input and the output for a prescribed test input (such as step, ramp, or parabola) as t . Consider the following unity feedback system
Controller Plant R(s) + E(s) C(s) Gc (s) G(S) _
The closed-loop transfer function of the system, T(s), is
C(s) G (s)G(s) T(s) c R(s) 1 Gc (s)G(s)
The error, E(s), is the difference between the system input, R(s), and the system output, C(s), and given by
E(s) R(s) C(s) R(s) R(s)T(s)
R(s)G (s)G(s) R(s) R(s) c 1 Gc (s)G(s) 1 Gc (s)G(s)
Using the final value theorem, the steady-state error is given by
sR(s) e() lim sE(s) lim s0 s0 1 Gc (s)G(s)
Department of Mechanical Engineering, University Of Saskatchewan, 57 Campus Drive, Saskatoon, SK S7N 5A9, Canada 2 Lecture Notes of Control Systems I - ME 431/Analysis and Synthesis of Linear Control System - ME862
Example
Develop the expression of the steady-state error for the following P control system if the input is a step (or R(s)=1/s).
Plant P Compensator R(s) + 1 C(s) K _ (s 1)(s 2)(s 10)
4.2 Improving Steady-State Error – Using PI Controller
Objective: design a control system to eliminate the steady-state error without affecting the transient response.
The steady-state error of P control system can be eliminated by adding an integrator (i.e., 1/s) to the controller. The resultant controller is called an integral (I) controller.
Problem: Using the I controller, Point A that represents the desired transient response is no longer on the root locus, as shown in the following figures.
Department of Mechanical Engineering, University Of Saskatchewan, 57 Campus Drive, Saskatoon, SK S7N 5A9, Canada 3 Lecture Notes of Control Systems I - ME 431/Analysis and Synthesis of Linear Control System - ME862
Solution: Add a zero that is close to the pole at the origin, as shown in following figure, such that the new system has a root locus going through Point A.
-a
The resultant controller is called an proportional-plus-integral (PI) controller or ideal integral compensator. It has the transfer function
K(s a) Ka G (s) K c s s
In the design of PI controller, the zero must be chosen to be close to the pole at the origin so that the angular contributions from the zero and the pole cancel out, i.e., zc pc 0 , as shown in the above figure. As a result, Point A is still on the root locus; meanwhile the steady-state error is eliminated. Typically, the value of a is set as 0.1.
Department of Mechanical Engineering, University Of Saskatchewan, 57 Campus Drive, Saskatoon, SK S7N 5A9, Canada 4 Lecture Notes of Control Systems I - ME 431/Analysis and Synthesis of Linear Control System - ME862
Design Problem 3
Given the system shown in the following figure, which operates with a damping ratio of 0.174. Design a PI controller to reduce the steady-state error of the step response to zero, and then compare the system specifications before and after using the PI controller.
R(s) + 1 C(s) K _ (s 1)(s 2)(s 10)
Department of Mechanical Engineering, University Of Saskatchewan, 57 Campus Drive, Saskatoon, SK S7N 5A9, Canada 5 Lecture Notes of Control Systems I - ME 431/Analysis and Synthesis of Linear Control System - ME862
5. Improving Steady-State Error and Transient Response – Using PID Controller
We now combine the design techniques discussed previously to obtain improvement in both steady-state error and transient response. Basically, we first improve the transient response by using a PD controller and then eliminate the steady-state error to 0 by using a PI controller. The resultant controller is called a proportional-plus-integral-plus derivative (PID) controller.
In other words, a PID controller is combination of a PD controller and a PI controller. Therefore, it has the following transfer function of
K(s a )(s a ) G (s) 1 2 c s
The PID controller has two zeros plus a pole at the origin. One zero (-a1) is designed as the PD controller; the other zero (-a2, typically taking a value of -0.1) and the pole at the origin are designed as the PI controller. Gc(s) can also be expressed as
K(s a )(s a ) Ks 2 (a a )Ks a a K K G (s) 1 2 1 2 1 2 K 2 K s c s s 1 s 3
where K1 (a1 a2 )K , K 2 a1a2 K , and K 3 K . The block diagram of the PID control system is shown in following figure.
Department of Mechanical Engineering, University Of Saskatchewan, 57 Campus Drive, Saskatoon, SK S7N 5A9, Canada 6 Lecture Notes of Control Systems I - ME 431/Analysis and Synthesis of Linear Control System - ME862
Design Problem 4
Consider the system shown in the following figure. Design a PID controller that will yield a peak time of 0.785 sec. and a settling time of 0.8 sec., with the zero steady-state error for a step input.
R(s) + 1 C(s) K _ (s 1)(s 5)
Department of Mechanical Engineering, University Of Saskatchewan, 57 Campus Drive, Saskatoon, SK S7N 5A9, Canada 7 Lecture Notes of Control Systems I - ME 431/Analysis and Synthesis of Linear Control System - ME862 Summary of P, PI, PD, and PID control systems Requirements Transient response Transient response specified Steady-state error is 0 Transient response specified by specified by the desired by two of the following without affecting the two of the following parameters damping ratio or parameters transient response 1. desired damping ratio or percent percent overshoot 1. desired damping ratio or overshoot percent overshoot 2. settling time 2. settling time 3. peak time 3. peak time Steady-state error is 0 Controllers P PD PI PID
Controller Trans. Func. K K(s+a1) K(s a2 ) K(s a1 )(s a2 ) s s Open-Loop Trans. Func. KG(s) K(s a )G(s) K(s a ) K(s a )(s a ) 1 2 G(s) 1 2 G(s) KG(s)H (s) s s Im Root Locus Im Im Im for the system with the plant’s transfer function of 1 G(s) (s 1)(s 3)(s 10)
Re × × × × O × × Re × × × O× × O × × O× Re -1 -10 -a -3 -1 -3 -1 -a 0 -a -1 -a 0 -10 -3 1 -10 2 -10 1 -3 2
Design 1. find the dominate pole 1. find the dominate pole, 1. typically set a2 as 0.1 1. design a PD controller to i.e., the intersection of defined by the desired (close to 0), achieve the desired transient root locus and the transient response, 2. find the value of K response desired damping ratio 2. find the zero -a1 based on based on the magnitude 2. design a PI controller to reduce line, the angle criterion, i.e., criterion, i.e., the steady-error to 0. 2. find the value of K, (s a )G(s) 180o 1 3. find the value of K, i.e., 1 K based on the (s a ) 1 3. find the value of K based on 2 G(s) K magnitude criterion, the magnitude criterion, i.e., s at the pole (s a1 )(s a2 ) 1 G(s) K 1 s G(s) K at the pole at the pole (s a )G(s) i.e., 1 at the pole
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