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

2. 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 − +

3. 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

4. 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 +

5. 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 − +

6. 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.

7. 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.

8. 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.

9. 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 𝐺𝐷 compared to open-loop system.

10. 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

11. 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)

12. 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
= 𝐺𝐷(0) = 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.

13. 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

14. 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

15. 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

16. 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

17. 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)

18. 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)

19. 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

20. Example
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