2. Chapter 4
A First Analysis of Feedback
Feedback Control
A Feedback Control seeks to bring the measured quantity to its desired value or set-point (also called reference trajectory) automatically.
There are two possible causes of difference between measured value and desired value:
Disturbance and noise
Change of set point, where the control system must act to bring the measured quantity to the new set point

3. The Three-Term Controller: PID Control
Chapter 4
A First Analysis of Feedback
The Three-Term Controller: PID Control
The PID Controller is the most popular feedback control algorithm used in process control and industries.
It is robust yet simple, easy to understand, and can provide excellent control performance despite the variation of dynamic characteristics of the process.
As the name suggests, the PID Controller consists of three basic terms: the Proportional term, the Integral term, and the Derivative term.
Each term can be activated separately (P, I, or D) or combined to obtain the desired controller.
Depending on the characteristics of the process, the following controller are generally used: P, PI, PD, or PID.

4. The Three-Term Controller: PID Control
Chapter 4
A First Analysis of Feedback
The Three-Term Controller: PID Control
Provides excitation to the process
The control variable or manipulated variable
To be controlled
Measured variable
The input to the controller, the tracking error, will vary due to the change in measured value or desired value.
In response, the controller produces control signal in order to minimize the error.

5. first step to close a feedback loop
Chapter 4
A First Analysis of Feedback
Proportional Term
P – Term
Proportional Gain,
first step to close a feedback loop

6. Chapter 4
A First Analysis of Feedback
Integral Term
I – Term
Integral Gain, to assure zero error to constant reference and disturbance inputs
How if P and I are combined together?

7. Derivative Term D – Term
Chapter 4
A First Analysis of Feedback
Derivative Term
D – Term
Derivative Gain, to improve (or even realize) stability and good dynamic response
How if P and D are combined together?
How if P, I, and D?

8. PID Control Altogether, the PID Controller may be represented as:
Chapter 4
A First Analysis of Feedback
PID Control
Altogether, the PID Controller may be represented as:
Another way to formulate PID Controller in time domain is:
Integral time constant
Derivative time constant

9. Chapter 4
A First Analysis of Feedback
Effects of Each Term
The proportional term makes the response of the system faster and reduces error due to disturbance by increasing the overall gain of the system, but it alone may still allow a steady-state error.
The integral term may eliminate the steady-state error to a constant input, although at the cost of deteriorating the dynamic response of the system by causing slower response of the system to varying set point.
The derivative term typically makes the system faster and better damped, although at the cost of less stability and of high sensitivity to high frequency disturbance and sensor noise.
Note that the effects of each term may not be accurately identifiable because P, I, and D gains are dependent of each other

10. Problem Example Consider the mass-spring-damper system below.
Chapter 4
A First Analysis of Feedback
Problem Example
Consider the mass-spring-damper system below.
The dynamic model of the process can be derived as:
Taking the Laplace Transform yields:
The transfer function is then given by:

11. Chapter 4
A First Analysis of Feedback
Problem Example
Let m = 1 kg, b = 10 N.s/m, and k = 20 N/m, while the applied force F = 1 N.
The transfer function can now be written as:
The process will now be controlled by a PID-Controller.
The contribution of each controller parameter kp, ki, and kd will be observed on an effort to obtain an overall system with small rise time, minimum overshoot, small settling time, and zero steady-state error.

12. Problem Example: Open-Loop
Chapter 4
A First Analysis of Feedback
Problem Example: Open-Loop
The open-loop transfer function of the process is:
The steady-state output value of the system is:
The gain of the process is 0.05

13. Problem Example: Open-Loop
Chapter 4
A First Analysis of Feedback
Problem Example: Open-Loop
The open-loop unit-step response of the process can be shown as:
No overshoot
Settling time is around 1.5 s

14. Problem Example: Closed-Loop
Chapter 4
A First Analysis of Feedback
Problem Example: Closed-Loop
The process is now placed into a closed-loop system as follows: ≡
Now, the gain of the process is 1/21 = (decrease)

15. Problem Example: Closed-Loop
Chapter 4
A First Analysis of Feedback
Problem Example: Closed-Loop
The closed-loop unit-step response of the system can be shown as:
Set point
0.0476
ess =

16. Problem Example: Using P-Controller
Chapter 4
A First Analysis of Feedback
Problem Example: Using P-Controller
With the P-Controller in the loop, the transfer function of the system is given as:
The parameter kp can now be tuned to obtain the desired closed-loop response of the overall system.

17. Problem Example: Using P-Controller
Chapter 4
A First Analysis of Feedback
Problem Example: Using P-Controller
Let kp = 300, the closed-loop transfer function of the system is:
Final value 300/320 =
 Steady-state error reduced
The unit-step response is now:
Overshoot increased
Effects of P-Controller
Rise time reduced
Settling time slightly reduced

18. Problem Example: Using PD-Controller
Chapter 4
A First Analysis of Feedback
Problem Example: Using PD-Controller
With the PD-Controller in the loop, the transfer function of the system is given as:
The parameter kp and kd can now be tuned to obtain the desired closed-loop response of the overall system.

19. Problem Example: Using PD-Controller
Chapter 4
A First Analysis of Feedback
Problem Example: Using PD-Controller
Let kp = 300, kd = 10, the closed-loop transfer function of the system is:
Final value 300/320 =
 Unchanged
 Steady-state error still
The unit-step response is now:
Overshoot reduced
Effects of PD-Controller
Settling time reduced even more
Rise time almost unchanged

20. Problem Example: Using PI-Controller
Chapter 4
A First Analysis of Feedback
Problem Example: Using PI-Controller
With the PI-Controller in the loop, the transfer function of the system is given as:
The parameter kp and ki can now be tuned to obtain the desired closed-loop response of the overall system.

21. Problem Example: Using PI-Controller
Chapter 4
A First Analysis of Feedback
Problem Example: Using PI-Controller
Let kp = 300, ki = 160, the closed-loop transfer function of the system is:
Final value 160/160 = 1
 Zero steady-state error
The unit-step response is now:
Overshoot increased
Effects of PI-Controller
Settling time increased
Actually, both P and I terms tend to increase overshoot and reduce rise time at the same time (double effect)
Rise time almost unchanged

22. Problem Example: Using PID-Controller
Chapter 4
A First Analysis of Feedback
Problem Example: Using PID-Controller
With the PID-Controller in the loop, the transfer function of the system is given as:
The parameter kp, ki, and kd can now be tuned to obtain the desired closed-loop response of the overall system.

23. Problem Example: Using PID-Controller
Chapter 4
A First Analysis of Feedback
Problem Example: Using PID-Controller
Let kp = 300, ki = 160, kd = 10, the closed-loop transfer function of the system is:
Final value 160/160 = 1
 Zero steady-state error
The unit-step response is now:
Overshoot significantly reduced
Effects of PID-Controller
Settling time unchanged
Small rise time, almost unchanged

24. Chapter 4
A First Analysis of Feedback
Problem Example
To ease the comparison, all the unit-step responses are plotted on the same figure.
P-Controller
PD-Controller
PI-Controller
PID-Controller
Original Process

25. Chapter 4
A First Analysis of Feedback
Conclusions
Proportional control can reduce steady-state errors, but the high gains almost always destabilize the system.
Integral control provides total reduction in steady-state errors, but often makes the system less stable.
Derivative control usually increases damping and improves stability, but has almost no effect on the steady-state error.
These 3 kinds of control combine to form the classical PID-Controllers which provide reasonable control for most industrial processes, provided that the performance demand is not too high.

26. Control of 1st Order System Using P-Controller
Chapter 4
A First Analysis of Feedback
Control of 1st Order System Using P-Controller
For a unit-step reference trajectory, the steady-state error:
 In order to reduce ess, kp must be increased

27. Control of 1st Order System Using P-Controller
Chapter 4
A First Analysis of Feedback
Control of 1st Order System Using P-Controller
Close-loop sensitivity
At high frequency, S≈1
At low frequency, if kp increased, S decreases, T increases
If kp is increased to reduce ess, then the system is less vulnerable to disturbance with low frequency, but more vulnerable to sensor noise with low frequency

28. Control of 1st Order System Using PI-Controller
Chapter 4
A First Analysis of Feedback
Control of 1st Order System Using PI-Controller

29. Control of 1st Order System Using PI-Controller
Chapter 4
A First Analysis of Feedback
Control of 1st Order System Using PI-Controller
For a unit-step reference trajectory, the steady-state error:
At high frequency, S≈1, T≈0
At low frequency, S≈0, T≈1
The system is very sensitive to disturbance with high frequency and sensor noise with low frequency
 No steady-state error

30. Control of 2nd Order System Using P-Controller
Chapter 4
A First Analysis of Feedback
Control of 2nd Order System Using P-Controller

31. Control of 2nd Order System Using P-Controller
Chapter 4
A First Analysis of Feedback
Control of 2nd Order System Using P-Controller
For R(s) = 1/s,
For R(s) = 1/s2,
 No steady-state error
 The error ess can be reduced, but cannot equal zero

32. Control of 2nd Order System Using P-Controller
Chapter 4
A First Analysis of Feedback
Control of 2nd Order System Using P-Controller
In order to reduce the steady-state error ,
Decrease B
 ζ decreased
Increase kp
 ωn increased while ζ decreased
In any way, ζ will decrease
 maximum overshoot increases

33. Control of 2nd Order System Using PD-Controller
Chapter 4
A First Analysis of Feedback
Control of 2nd Order System Using PD-Controller

34. Control of 2nd Order System Using PD-Controller
Chapter 4
A First Analysis of Feedback
Control of 2nd Order System Using PD-Controller
For R(s) = 1/s2,

35. Control of 2nd Order System Using PD-Controller
Chapter 4
A First Analysis of Feedback
Control of 2nd Order System Using PD-Controller
In order to reduce the steady-state error ,
Decrease B
 ζ decreased, but can be compensated by increasing kd
Increase kp
 ωn increased and ζ decreased, but can be compensated by increasing kd
Steady-state response (ess) can be tuned without affecting transient response (Mp)

36. Chapter 4
A First Analysis of Feedback
Homework 5
No.1.
The performance of P- and PD-Controller to control the 2nd order system are already analyzed on the previous pages.
Now, conduct further analysis for PI-Controller. Explain about its ability to reduce steady-state error in the presence of step input and ramp input.
No.2.
A motor used for laser surgery requires high accuracy for position and velocity response. Consider the system shown below.

37. Homework 5 No.2. (continued)
Chapter 4
A First Analysis of Feedback
Homework 5
No.2. (continued)
The amplifier gain K must be adjusted so that the steady- state error for a ramp input r(t) = At (where A = 1 mm/s) is less than or equals to 0.1 mm, while a stable response is maintained.
(a) Determine the value of K to meet the requirement.
(b) In MATLAB Simulink, build the motor system as given previously. Now, prove the answer of (a), that ess = 0.1 mm can be achieved for that certain value of K.
(c) A technician claims that he can obtain ess = 0.02 mm. What is your opinion? Is his claim possible to do or not? How? Why? Explain.
Deadline: , at 07:30.