Systems Analysis and Control


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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 SteadyState Error Response to ramps and parabolae Limits of PD control Integral Feedback Elimination of steadystate error PolePlacement 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 2ndOrder 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 steadystate 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 2ndorder system Lets look at the effect of PD control on a 2ndorder 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 2ndorder 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 2ndorder system. M. Peet Lecture : Control Systems / 32
12 PD Control Example: Suppose we have the 2ndorder 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 SteadyState 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 steadystate 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 SteadyState Error Step Response.4.2 Amplitude Time (sec) The steadystate 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 steadystate error! M. Peet Lecture : Control Systems 8 / 32
19 Recall: SteadyState Error Lets take another look at steadystate 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 steadystate 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 steadystate 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 steadystate response to a parabolic input is Ĝ(s) lim sŷ(s) = sĝ(s)û(s) = s 0 s 2 Note: The steadystate 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 steadystate 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 steadystate 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 steadystate error becomes finite. M. Peet Lecture : Control Systems 26 / 32
27 Integral Control The purpose of integral control is primarily to eliminate steadystate 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 reexamine 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: ProportionalIntegral t ) u(t) = K (e(t) + TI e(θ)dθ 0 ( ˆK(s) = K + ) T I s PID Feedback: ProportionalIntegralDifferential 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 secondorder 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 SteadyState 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 SteadyState 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 SteadyState Error Response to ramps and parabolae Limits of PD control Integral Feedback Elimination of steadystate error PolePlacement Next Lecture: Midterm Review M. Peet Lecture : Control Systems 32 / 32
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