Control Systems EE 4314 Lecture 12 March 17, 2015 Spring 2015 Indika Wijayasinghe

Chapter 4: A First Analysis of Feedback Control: process of making a system variable converge to a reference value Tracking control (servo): reference value = changing Regulation control: reference value = constant (stabilization) Open-loop system Closed-loop system R: reference input, U: control input, W: disturbance, Y: output, V: sensor noise Controller 𝐷(𝑠) Plant G(𝑠) 𝑊 𝑅 𝑌 𝑈 + Controller 𝐷(𝑠) Plant G(𝑠) 𝑊 𝑅 𝑌 𝑈 𝑉 − +

Feedback Control Role of feedback: Reduce sensitivity to system parameters (robustness) Disturbance rejection Track desired inputs with reduced steady state errors, overshoot, rise time, settling time (performance) Make system stable

Open-Loop System Output 𝑌=𝐺𝐷𝑅+𝐺𝑊 Control input U=𝐷𝑅 (Feedforward control input) Error E=𝑅−𝑌=𝑅− 𝐺𝐷𝑅+𝐺𝑊 = 1−𝐺𝐷 𝑅−𝐺𝑊 Transfer function 𝑇 𝑜𝑝𝑒𝑛_𝑙𝑜𝑜𝑝 =𝐷𝐺 Can not reduce the effect of disturbance Controller 𝐷(𝑠) Plant G(𝑠) 𝑊 𝑅 𝑌 𝑈 +

Closed-Loop System Output 𝑌= 𝐺𝐷 1+𝐺𝐷 𝑅+ 𝐺 1+𝐺𝐷 𝑊− 𝐺𝐷 1+𝐺𝐷 𝑉 Error E=𝑅−𝑌=𝑅− 𝐺𝐷 1+𝐺𝐷 𝑅+ 𝐺 1+𝐺𝐷 𝑊− 𝐺𝐷 1+𝐺𝐷 𝑉 = 1 1+𝐺𝐷 𝑅− 𝐺 1+𝐺𝐷 𝑊+ 𝐺𝐷 1+𝐺𝐷 𝑉 Transfer function 𝑇 𝑐𝑙𝑜𝑠𝑒𝑑_𝑙𝑜𝑜𝑝 = 𝐺𝐷 1+𝐺𝐷 Can not reduce the effect of noise Reduce the effect of disturbance if 𝐺𝐷 is large Controller 𝐷(𝑠) Plant G(𝑠) 𝑊 𝑅 𝑌 𝑈 𝑉 − +

Stability: Open-Loop vs. Closed-Loop The requirement for stability: all poles of the transfer function must be in the left half plane (LHP). Define 𝐺 𝑆 = 𝑏(𝑠) 𝑎(𝑠) and D 𝑆 = 𝑐(𝑠) 𝑑(𝑠) , and 𝑎 𝑠 , 𝑏 𝑠 , 𝑐 𝑠 , 𝑑 𝑠 are polynomials. For open-loop system Transfer function 𝑇 𝑜𝑝𝑒𝑛_𝑙𝑜𝑜𝑝 =𝐷𝐺= 𝑏(𝑠)𝑐(𝑠) 𝑎(𝑠)𝑑(𝑠) 𝑎 𝑠 and 𝑑 𝑠 should have roots in the LHP. For closed-loop system Transfer function 𝑇 𝑐𝑙𝑜𝑠𝑒𝑑_𝑙𝑜𝑜𝑝 = 𝐺𝐷 1+𝐺𝐷 Characteristic equation 1+𝐺𝐷=0, 𝑎 𝑠 𝑑 𝑠 +b s c s =0 Even for unstable 𝑎 𝑠 , all poles can be in the LHP.

Stability: Open-Loop vs. Closed-Loop Example: plant 𝐺 𝑆 = 1 (𝑠+1)(𝑠−1) , controller 𝐷(𝑆)= 𝐾(𝑠+𝛾) (𝑠+𝛿) Open-loop system Transfer function 𝐺 𝑆 𝐷(𝑆)= 𝐾(𝑠+𝛾) (𝑠+1)(𝑠−1)(𝑠+𝛿) Unstable because one pole (𝑠=1) is in the RHP. Closed-loop system Transfer function 𝐺𝐷 1+𝐺𝐷 = 𝐾(𝑠+𝛾) 𝑠+1 𝑠−1 𝑠+𝛿 +𝐾(𝑠+𝛾) Characteristic equation 𝑠+1 𝑠−1 𝑠+𝛿 +𝐾 𝑠+𝛾 =0 We can select δ, 𝛾, and 𝐾 so that all the poles are in the LHP. Closed-loop system can be stable by proper controller design even though open-loop system is unstable.

Regulation: Open-Loop vs. Closed-Loop Regulation is to keep the error small when the reference is constant and disturbances are present. Open-loop system Error E= 1−𝐺𝐷 𝑅−𝐺𝑊 Controller 𝐷 is useless for regulation. It does not reduce the effect of disturbance 𝑊. Closed-loop system Error E= 1 1+𝐺𝐷 𝑅− 𝐺 1+𝐺𝐷 𝑊+ 𝐺𝐷 1+𝐺𝐷 𝑉 Controller 𝐷 can be effective to reduce the effect of disturbance 𝑊, but is not effective to reduce the effect of noise 𝑉. Large 𝐷 makes 𝐺 1+𝐺𝐷 𝑊 smaller.

Sensitivity When plant 𝐺 changes to 𝐺+𝛿𝐺 Open-loop system 𝑇+𝛿𝑇= 𝐺+𝛿𝐺 𝐷 𝛿𝑇=𝛿𝐺𝐷 where 𝑇 : transfer function Closed-loop system 𝑇+𝛿𝑇= 𝐺+𝛿𝐺 𝐷 1+ 𝐺+𝛿𝐺 𝐷 ≅ 𝐺𝐷 1+𝐺𝐷 + 𝛿𝐺𝐷 1+𝐺𝐷 𝛿𝑇= 𝛿𝐺𝐷 1+𝐺𝐷 Sensitivity S= 1 1+𝐺𝐷 *Closed-loop system is less sensitive by a factor of 1 1+𝐺𝐷 compared to open-loop system.

Steady-State Error In the unity feedback system, error equation 𝐸=𝑅−𝑌=𝑅− 𝐺𝐷 1+𝐺𝐷 𝑅= 1 1+𝐺𝐷 𝑅=𝑆𝑅 where 𝑆: sensitivity Let input 𝑟 𝑡 = 𝑡 𝑘 𝑘! , which is 𝑅 𝑆 = 1 𝑠 𝑘+1 For 𝑘=0, step input or position input For 𝑘=1, ramp input or velocity input For 𝑘=2, acceleration input Using Final Value Theorem lim 𝑡→∞ 𝑒(𝑡) = 𝑒 𝑠𝑠 = lim 𝑠→0 𝑠 𝐸(𝑠) = lim 𝑠→0 𝑠 1 1+𝐺𝐷 𝑅(𝑠) = lim 𝑠→0 𝑠 1 1+𝐺𝐷 1 𝑠 𝑘+1

Steady-State Errors Steady-state errors as a function of system type Type Input Step (position) Ramp (velocity) Parabola (acceleration) Type 0 1 1+ 𝐾 𝑝  Type 1 1 𝐾 𝑣 Type 2 1 𝐾 𝑎 Position error constant 𝐾 𝑝 = lim 𝑠→0 𝐺 𝐷(𝑠) Velocity error constant 𝐾 𝑣 = lim 𝑠→0 𝑠𝐺 𝐷(𝑠) Acceleration error constant 𝐾 𝑎 = lim 𝑠→0 𝑠 2 𝐺 𝐷(𝑠)

Steady-State Error Consider a system for which 𝐺𝐷 has no pole at the origin and a step input for which 𝑅 𝑆 =1/𝑠. Then, 𝑒 𝑠𝑠 = lim 𝑠→0 𝑠 1 1+𝐺𝐷 1 𝑠 = 1 1+𝐺𝐷(0) = 1 1+ 𝐾 𝑝 We define this system is Type 0 and 𝐾 𝑝 is “position error constant.” Notice that if the polynomial input is higher than 1 degree (𝑘≥1, 1 𝑠 𝑘+1 ), then steady-state error becomes infinity. A polynomial of degree 0 is the highest degree a system of Type 0 can track.

State-State Error Example: Determine the system type and relevant error for speed control with proportional control given by 𝐷 𝑆 = 𝑘 𝑝 and plant transfer function 𝐺= 𝐴 𝜏𝑠+1

State-State Error Example: Determine the system type and relevant error for speed control with proportional control given by 𝐷 𝑆 = 𝑘 𝑝 and plant transfer function 𝐺= 𝐴 𝜏𝑠+1

State-State Error Example: Determine the system type and relevant error for speed control with proportional and integral feedback given by 𝐷 𝑆 = 𝑘 𝑝 + 𝑘 𝐼 𝑠 and plant transfer function 𝐺= 𝐴 𝜏𝑠+1

State-State Error Example: Determine the system type and relevant error for speed control with proportional and integral feedback given by 𝐷 𝑆 = 𝑘 𝑝 + 𝑘 𝐼 𝑠 and plant transfer function 𝐺= 𝐴 𝜏𝑠+1

State-State Error For a feedback system with sensor gain 𝐻, Transfer function becomes 𝑇 𝑠 = 𝑌(𝑠) 𝑅(𝑠) = 𝐺𝐷 1+𝐺𝐷𝐻 Error 𝐸=𝑅−𝑌=𝑅−𝑇 𝑠 𝑅= 1−𝑇 𝑠 𝑅 Applying the Final Value Theorem, 𝑒 𝑠𝑠 = lim 𝑡→∞ 𝑒(𝑡) = lim 𝑠→0 𝑠 𝐸(𝑠) = lim 𝑠→0 𝑠 1−𝑇 𝑠 𝑅(s) For 𝑅 𝑆 = 1 𝑠 𝑘+1 , 𝑒 𝑠𝑠 = lim 𝑠→0 1−𝑇(𝑠) 𝑠 𝑘 Controller 𝐷(𝑠) Plant G(𝑠) 𝑊 𝑅 𝑌 𝑈 𝑉 − + Sensor H(𝑠)

State-State Error Example: Consider an electric motor control problem including a non-unity feedback system. System parameters are 𝐺 𝑠 = 1 𝑠(𝜏𝑠+1) , 𝐷 𝑠 = 𝑘 𝑝 , 𝐻 𝑠 =1+ 𝑘 𝑡 𝑠 Determine the system type and relevant error constant. Controller 𝐷(𝑠) Plant G(𝑠) 𝑊 𝑅 𝑌 𝑈 𝑉 − + Sensor H(𝑠)

State-State Error Example: Consider an electric motor control problem including a non-unity feedback system. System parameters are 𝐺 𝑠 = 1 𝑠(𝜏𝑠+1) , 𝐷 𝑠 = 𝑘 𝑝 , 𝐻 𝑠 =1+ 𝑘 𝑡 𝑠 Determine the system type and relevant error constant. Answer: For 𝑅 𝑆 = 1 𝑠 𝑘+1 , 𝑒 𝑠𝑠 = lim 𝑠→0 1−𝑇(𝑠) 𝑠 𝑘

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