Systems Analysis and Control

Transcription

1 Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture : Different Types of Control

2 Overview In this Lecture, you will learn: Limits of Proportional Feedback Performance Specifications. Derivative Feedback Pros and Cons PD Control Pole Placement More on Steady-State Error Response to ramps and parabolae Limits of PD control Integral Feedback Elimination of steady-state error Pole-Placement M. Peet Lecture : Control Systems 2 / 32

3 Recall the Inverted Pendulum Problem Transfer Function Ĝ(s) = Js 2 Mgl 2 For a simple proportional gain: ˆK(s) = k Closed Loop Transfer Function: There 2.5 are two cases: 2 Impulse Response k Js 2 Mgl 2 + k 8 x 06 Impulse Response Amplitude Amplitude Time (sec) Time (sec) Figure: Case : k > Mgl 2 Figure: Case 2: k < Mgl 2 Both cases are unstable! M. Peet Lecture : Control Systems 3 / 32

4 Differential Control Now suppose we furthermore have a performance specification: Overshoot Rise Time Settling Time - u(s) + T D s G(s) y(s) Problem: There is no solution using proportional gain: ˆK(s) = k. Now we must consider a New Kind of Controller: Derivative Control: Choose ˆK(s) = T D s The controller is of the form u(t) = T D ė(t) The controller is called Differential/Derivative Control because it is proportional to the rate of change of the error. M. Peet Lecture : Control Systems 4 / 32

5 Differential Control Differential control improves performance by reacting quickly. Prediction: To measure ẏ(t), recall the definition of derivative: ẏ(t) = e(t + t) e(t) t The ẏ(t) term depends on both the current position and predicted position. A way to speed up the response. M. Peet Lecture : Control Systems 5 / 32

6 Differential Control: Using the Delay is Dangerous! Problem: Differential control is implemented using delay. y(t) is the measurement. ẏ(t) cannot be measured directly Approximate using the delayed response: ẏ(t) = e(t) e(t t) t Delay can cause instabilities. M. Peet Lecture : Control Systems 6 / 32

7 Differential Control: Noise is Dangerous! Noise Amplification: Measurement of ẏ(t) is heavily influenced by noise. ẏ(t) e(t) e(t t) = t Sensor measurements have error As t 0, the effect of noise, σ is amplified: ẏ(t) = e(t) e(t t) t + 2σ t M. Peet Lecture : Control Systems 7 / 32

8 Failure of Derivative Control Inverted Pendulum Controller: ˆK(s) = TD s Closed Loop Transfer Function: T D /Js s 2 + T D /Js Mgl 2J 2nd-Order System As we learned last lecture, stable iff both T D /J > 0 Mgl 2J > 0 Derivative Feedback Alone cannot stabilize a system. M. Peet Lecture : Control Systems 8 / 32

9 PD Control Differential Control is usually combined with proportional control. To improve stability To reduce steady-state error. To reduce the effect of noise. Controller: The form of control is u(t) = K [e(t) + T D ė(t)] or û(s) = K [ + T D s] ê(s) M. Peet Lecture : Control Systems 9 / 32

10 PD Control 2nd-order system Lets look at the effect of PD control on a 2nd-order system: Ĝ(s) = Controller: ˆK(s) = K [ + TD s] Closed Loop Transfer Function: s 2 + bs + c ˆK(s)Ĝ(s) + ˆK(s)Ĝ(s) = K [ + T D s] s 2 + bs + c + K [ + T D s] = K [ + T D s] s 2 + (b + KT D )s + (c + K) The poles of the system are freely assignable for a 2nd order system. T D and K allow us to construct any denominator we desire. M. Peet Lecture : Control Systems 0 / 32

11 PD Control 2nd-order system Suppose we want poles at s = p, p 2. Im(s) Re(s) We want the closed loop of the form: (s p )(s p 2 ) = (s 2 (p + p 2 )s + p p 2 ) Thus we want c + K = p p 2 which means K = p p 2 c. b + KT D = (p + p 2 ) which means T D = p+p2+b K = p+p2+b p p 2 c PD feedback gives Total Control over a 2nd-order system. M. Peet Lecture : Control Systems / 32

12 PD Control Example: Suppose we have the 2nd-order system Ĝ(s) = s 2 + s + Im(s) and performance specifications: Overshoot: M p,desired =.05 Rise Time: T r,desired = s Settling Time: T s,desired = 3.5s. Re(s) As we found in Lecture 9, these specifications mean that the poles satisfy: We chose the pole locations: σ <.9535ω, σ <.333, ω n >.8 s =.5 ±.4ı M. Peet Lecture : Control Systems 2 / 32

13 PD Control Example: The desired system is (s 2 (p + p 2 )s + p p 2 ) The closed loop is K [ + T D s] s 2 + (b + KT D )s + (c + K) To get the pole locations: p,2 =.5 ±.4ı we choose The gain K = p p 2 c = (.5 +.4ı)(.5.4ı) + = = 3.2 The derivative gain T D = p + p 2 + b = 3 + = 2 K =.623 This gives the controller: ˆK(s) = K( + T D s) = s M. Peet Lecture : Control Systems 3 / 32

14 PD Control Problem with Steady-State Error 3 x 06 Step Response Step Response Amplitude 2.5 Amplitude Time (sec) Time (sec) Figure: Open Loop Figure: Closed Loop Although the PD controller gives us control of the pole locations, the steady-state value is y ss = K c + K = =.7625 M. Peet Lecture : Control Systems 4 / 32

15 PD Control Inverted Pendulum Lets look at the effect of PD control on the inverted Pendulum: Ĝ(s) = /J s 2 Mgl 2J Controller: K [ + T D s] Closed Loop Transfer Function: ˆK(s)Ĝ(s) + ˆK(s)Ĝ(s) = K/J [ + T D s] s 2 Mgl + K/J [ + T Ds] = 2J K/J [ + T D s] s 2 + K/JT D s + (K/J Mgl 2J ) M. Peet Lecture : Control Systems 5 / 32

16 PD Control Inverted Pendulum To achieve the performance specifications: Overshoot: M p,desired =.05 Rise Time: T r,desired = s Settling Time: T s,desired = 3.5s. We want poles at s =.5 ±.4ı Im(s) Re(s) Thus we want c + K = p p 2 which means K = p p 2 c. b + KT D = (p + p 2 ) which means T D = p + p 2 + b K = p + p 2 + b p p 2 c M. Peet Lecture : Control Systems 6 / 32

17 PD Control Inverted Pendulum The closed loop is K/J [ + T D s] s 2 + K/JT D s + (K/J Mgl 2J ) To get the pole locations p,2 =.5 ±.4ı we choose The gain K/J = p p 2 c = Mgl 2J The derivative gain T D = p + p 2 + b 3 = p p 2 c Mgl 2J This gives the controller: ) ˆK(s) = K( + T D s) = 4.2J + ( 2 Mgl Mgl s 2J M. Peet Lecture : Control Systems 7 / 32

18 PD Control Inverted Pendulum: Problem with Steady-State Error Step Response.4.2 Amplitude Time (sec) The steady-state error with this controller is (K = J = M = g = l = ) y ss = K/J (K/J Mgl 2J ) = =.35 Derivative Control has No Effect on the steady-state error! M. Peet Lecture : Control Systems 8 / 32

19 Recall: Steady-State Error Lets take another look at steady-state error Recall: Figure: Suspension Response for k = Problems: If target is moving, we may never catch up. Even if we can catch a moving target, we may not catch an accelerating target. For these problems, the step response is not appropriate. We measured steady-state error using the step response. ess = lim t y(t) Sometimes this doesn t work. Assumes objective doesn t move. M. Peet Lecture : Control Systems 9 / 32

20 Ramp and Parabolic Inputs There are other types of response we can consider. Ramp response tracks error for a target with constant velocity. Parabolic response tracks error for a target with a constant acceleration. M. Peet Lecture : Control Systems 20 / 32

21 Ramp and Parabolic Inputs We can use the final value theorem to find the response to ramp and parabolic inputs: Ramp Response: Recall the ramp input: u(t) = t û(s) = s 2 The steady-state error to a ramp input is e ss = lim sê(s) = lim s( Ĝ(s))û(s) = lim Ĝ(s) s 0 s 0 s 0 s M. Peet Lecture : Control Systems 2 / 32

22 Ramp and Parabolic Inputs We can use the final value theorem to find the response to parabolic inputs: Parabolic Response: Recall the parabolic input: u(t) = t 2 û(s) = s 3 The steady-state response to a parabolic input is Ĝ(s) lim sŷ(s) = sĝ(s)û(s) = s 0 s 2 Note: The steady-state error to a parabolic input is usually infinite. M. Peet Lecture : Control Systems 22 / 32

23 Ramp and Parabolic Inputs The effect of the numerator For steady-state error, the numerator of the transfer function becomes important: for Ĝ(s) = n(s) d(s) Steady state error is ( d(s) lim( Ĝ(s))sû(s) = lim s 0 s 0 d(s) n(s) ) sû(s) d(s) d(s) n(s) = lim sû(s) s 0 d(s) û(s) is the test signal Step Input: sû(s) = Ramp Input: sû(s) = s Parabolic Input: sû(s) = s 2 M. Peet Lecture : Control Systems 23 / 32

24 Ramp and Parabolic Inputs Systems in Feedback When in feedback, the closed loop has the form Ĝ ˆK + Ĝ ˆK Hence steady-state error has the form ( ê(s) = Ĝ ˆK ) ( ) + Ĝ ˆK sû(s) = + Ĝ ˆK sû(s) Step Response: Ramp Response: Parabolic Response: e ss,step = lim + Ĝ(s) ˆK(s) s 0 e ss,ramp = lim + Ĝ(s) ˆK(s) s 0 s = lim s 0 sĝ(s) ˆK(s) e ss,parabola = lim s 0 + Ĝ(s) ˆK(s) s 2 = s2ĝ(s) ˆK(s) M. Peet Lecture : Control Systems 24 / 32

25 Example of Ramp Response Consider the Suspension Example: Open Loop: Ĝ(s) = s 2 + s + s 4 + 2s 3 + 3s 2 + s + Ĝ(s) = s4 + 2s 3 + 3s 2 + s + s 2 s s 4 + 2s 3 + 2s 2 s 4 + 2s 3 + 3s 2 = + s + s 4 + 2s 3 + 3s 2 + s + Ramp Response: lim Ĝ(s) s 0 s = lim s 0 What happens when we close the loop? Closed Loop Transfer Function: Ramp Response: s 3 + 2s 2 + 2s s 4 + 2s 3 + 3s 2 + s + = 0 k(s 2 + s + ) s 4 + 2s 3 + (3 + k)s 2 + ( + k)s + ( + k) e ss,ramp = lim = lim sĝ(s) ˆK(s) s 0 s 0 s 4 + 2s 3 + 3s 2 + s + k(s 2 + s + ) s = Proportional response isn t capable of controlling a ramp input M. Peet Lecture : Control Systems 25 / 32

26 Example of Ramp Response The only way to control a ramp input using feedback is to put a pole at the origin: Controller: ˆK(s) = T I s Ramp Response: e ss,ramp = lim = lim sĝ(s) ˆK(s) s 0 s 0 s 4 + 2s 3 + 3s 2 + s + T I s s 2 + s + s = T I By including /s in the controller, the steady-state error becomes finite. M. Peet Lecture : Control Systems 26 / 32

27 Integral Control The purpose of integral control is primarily to eliminate steady-state error. Controller: The form of control is or, in the Laplace transform u(t) = t e(θ)dθ T I 0 û(s) = T I sê(s) One must be careful when using integral feedback By itself, an integrator is unstable. A pole at the origin. M. Peet Lecture : Control Systems 27 / 32

28 Integral Control Suspension Problem Again Now lets re-examine the suspension problem Controller: ˆK(s) = T I s Closed Loop Transfer Function: Ĝ(s) ˆK(s) + Ĝ(s) ˆK(s) = s 2 + s + T I s 5 + 2T I s 4 + 3T I s 3 + (T I + )s 2 + (T I + )s + If we set T I =., then the transfer function has poles at p,2 =.55 ±.89ı, p 3 = 2.26, p 4,5 =.6384 ±.877ı Integral feedback can Destabilize the system where proportional feedback couldn t! M. Peet Lecture : Control Systems 28 / 32

29 Integral Control Integral Feedback is destabilizing! Integral feedback is always combined with proportional or differential feedback: PI Feedback: Proportional-Integral t ) u(t) = K (e(t) + TI e(θ)dθ 0 ( ˆK(s) = K + ) T I s PID Feedback: Proportional-Integral-Differential t ) u(t) = K (e(t) + TI e(θ)dθ + T D ė(t) 0 ( ˆK(s) = K + ) T I s + T Ds M. Peet Lecture : Control Systems 29 / 32

30 PID Control Example Finally, lets see the effect of PID control on a second-order system: ( Ĝ(s) = ˆK(s) s 2 = K + ) + bs + c T I s + T Ds Closed Loop: ( ) Ĝ ˆK K + + Ĝ ˆK = T I s + T Ds ( ) s 2 + bs + c + K + T I s + T Ds ( ) K s + T I + T D s 2 = ( ) s 3 + bs 2 + cs + K s + T I + T D s 2 Steady-State Response: y ss,step = KT D s 2 + Ks + K T = I s 3 + (b + KT D )s 2 + (c + K)s + K T I K T I K = No Steady-State Error! T I M. Peet Lecture : Control Systems 30 / 32

31 PID Control Pole Placement: The three pole locations can be determined exactly. Given three poles: p, p 2, p 3. Construct Desired denominator: Three equations: b + KT D = a d c + K = b d K T I = c d Which can be solved as K = b d c T I = K c d T D = a d b K (s p )(s p 2 )(s p 3 ) = s 3 + a d s 2 + b d s + c d M. Peet Lecture : Control Systems 3 / 32

32 Summary What have we learned today? In this Lecture, you learned: Limits of Proportional Feedback Performance Specifications. Derivative Feedback Pros and Cons PD Control Pole Placement More on Steady-State Error Response to ramps and parabolae Limits of PD control Integral Feedback Elimination of steady-state error Pole-Placement Next Lecture: Midterm Review M. Peet Lecture : Control Systems 32 / 32