1. MESB374 System Modeling and Analysis Feedback Control Design Process
ME 375 – Fall 2002
MESB374 System Modeling and Analysis Feedback Control Design Process

2. Classical Feedback Control Structure
ME 375 – Fall 2002
Classical Feedback Control Structure
Disturbance D(s)
Reference
Input R(s)
Control Input
U(s)
E(s)
Output
Y(s) +
Gf(s)
G(s)
Gc(s) -
filter
Plant
Controller
H(s)
Ym(s)
Sensor Dyamics
Plant Equation (Transfer function model that we all know how to obtain ?!):
Control Law (Algorithm) (we will try to learn how to design):

3. Closed-Loop Transfer Function
ME 375 – Fall 2002
Closed-Loop Transfer Function
GC (s)
H(s) + -
Reference Input
R(s)
Error
E(s)
Output
Y(s)
Disturbance D(s)
Plant
G(s)
Control Input
U(s)
Gf (s)
Can you obtain it by block algebra?
What is Y(s) if disturbance dynamics and/or noise are taken into account?

4. Closed-Loop Transfer Function
ME 375 – Fall 2002
Closed-Loop Transfer Function
The closed-loop transfer functions relating the output y(t) (or Y(s)) to the reference input r(t) (or R(s)) and the disturbance d(t) (or D(s)) are:
The objective of control system design is to design a controller GC (s) and Gf (s), such that certain performance (design) specifications are met. For example:
we want the output y(t) to follow the reference input r(t), i.e., for certain frequency range. This is equivalent to specifying that
we want the disturbance d(t) to have very little effect on the output y(t) within the frequency range where disturbances are most likely to occur. This is equivalent to specifying that
Ideally
In reality
Ideally
In reality

5. Performance Specifications
ME 375 – Fall 2002
Performance Specifications
Given an input/output representation, GYR (s), for which the output of the system should follow the reference input, what specifications should you make to guarantee that the system will behave in a manner that will satisfy its functional requirements?
Input
R(s)
Output
Y(s)
GYR (s)
r(t)
Time
y(t)
Time
rss
yss=GYR(0)rss t0 t0

6. Unit Step Response ± X% yMAX OS tP tS tr Unit Step Response Time 1.6
ME 375 – Fall 2002
Unit Step Response
0.2
0.4
0.6
0.8 1
1.2
1.4
1.6
yMAX OS
± X%
Unit Step Response tP tS
Time tr

7. Performance Specifications
ME 375 – Fall 2002
Performance Specifications
Steady State Performance  Steady State Gain of the Transfer Function
Specifies the tracking performance of the system at steady state. Often it is specified as the steady state response, y(¥) (or ySS(t)), to be within an X% bound of the reference input r(t), i.e. the steady state error eSS(t) = ySS(t) - r(t) should be within a certain percent. For example, for step reference input r(t)=rss :
To find the steady state value of the output, ySS(t):
Sinusoidal references: use frequency response, i.e.
General references: use FVT, provided that sY(s) is stable, ...

8. Performance Specifications
ME 375 – Fall 2002
Performance Specifications
Transient Performance (Transient Response)
Transient performance of a system is usually specified using the unit step response of the system. Some typical transient response specifications are:
Settling Time (tS):
Specifies the time required for the response to reach and stay within a specific percent of the final (steady-state) value. Some typical settling time specifications are: 5%, 2% and 1%. For 2nd order systems, the specification is usually: 
% Overshoot (%OS):(2nd order systems)
Recall step response of second order system
Q: How can we link this performance specification to the closed-loop transfer function? (Hint) What system characteristics affect the system performance ?

9. Performance Specifications
ME 375 – Fall 2002
Performance Specifications
Transient Performance Specifications and CLTF Poles
Recall that the locations of TF poles directly affect the system output. For example, assume that the closed-loop transfer function of a feedback control system is:
The characteristic poles are:
Settling Time (2%):
 Puts constraint on the real part of the dominating closed-loop poles
%OS:
 Puts constraint on the damping ratio  or the angle  of the dominating closed-loop poles.

10. Performance Specification ® CL Pole Locations
ME 375 – Fall 2002
Performance Specification ® CL Pole Locations
Transient Performance Specifications and CLTF Pole Locations
Transient performance specifications can be interpreted as constraints on the positions of the poles of the closed-loop transfer function. Let a pair of closed-loop poles be represented as:
Transient Performance Specifications:
Settling Time (2 %) £ TS
%OS £ X %
Img.
jwd
Real
-jwd

11. ME 375 – Fall 2002
Example
A DC motor driven positioning system can be modeled by a second order transfer function:
A proportional feedback control is proposed and the proportional feedback gain is chosen to be 16/3. Find the closed-loop transfer function, as well as the 2% settling time and the percent overshoot of the closed loop system when given a step input.
Draw block diagram:
Find closed-loop transfer function: K
H(s)=1 + -
E(s)
Plant
Output
controller

12. Example Find closed-loop poles: 2% settling time: %OS:
ME 375 – Fall 2002
Example
Find closed-loop poles:
2% settling time:
%OS:

13. Example Find Find closed-loop transfer function:
ME 375 – Fall 2002
Example
A DC motor driven positioning system can be modeled by a second order transfer function:
A proportional feedback control is proposed. It is desired that:
for a unit step response, the steady state position should be within 2% of the desired position,
the 2% settling time should be less than 2 sec, and
the percent overshoot should be less than 10%.
Find
(1) the condition on the proportional gain such that the steady state performance is satisfied;
(2) the allowable region in the complex plane for the closed-loop poles.
Find closed-loop transfer function:
Write down the performance specifications:
Assume that closed-loop is an under-damped second order system

14. Example Percent Overshoot (%OS) Steady state performance constraint:
ME 375 – Fall 2002
Percent Overshoot (%OS)
Example
Steady state performance constraint:
Transient performance constraint:
2% Settling Time
Assume that closed-loop is an under-damped second order system
Img.
Real -2
How to discuss the case of over-damped system?

15. Example Over-damped case Steady state performance constraint:
ME 375 – Fall 2002
Example
Over-damped case
Steady state performance constraint:
Stability Requirement
Transient performance constraint:
2% Settling Time
No Overshoot!!
In all, we have

16. Review of Structure of Closed-loop System
Disturbance D(s)
Disturbance Channel
GD (s)
Reference Input
R(s)
Control Input
U(s) +
Error
E(s)
Output
Y(s) +
Gf (s)
GC (s)
GA (s)
G(s) - +
filter
Control algorithm
Actuator
Plant +
H(s) +
Sensor
GN (s)
Noise Channel
Noise N(s)
How many transfer functions do we need to determine the output?
Do you know how to find them?

17. Feedback Control Design Process
ME 375 – Fall 2002
Feedback Control Design Process
A typical feedback controller design process involves the following steps:
Model the physical system (plant) that we want to control and obtain its I/O transfer function G(s). (Sometimes, certain model simplification should be performed)
Determine sensor dynamics (transfer function of the measurement system) H(s) and actuator dynamics (if necessary).
Draw the closed-loop block diagram, which includes the plant, sensor, actuator and controller transfer functions GC (s) and Gf (s).
Obtain the closed-loop transfer functions GYR (s) and GYT (s) .
Based on the performance specifications, find the conditions that the CLTFs, GYR (s) and GYT (s), have to satisfy.
Choose controller structure GC (s) and Gf (s) and substitute it into the CLTFs GYR (s) and GYT (s).
Select the controller parameters (e.g. the proportional feedback gain of a proportional control law) so that the design constraints established in (5) are satisfied.
(8) Verify your design via computer simulation (MATLAB) and actual implementation.

18. ME 375 – Fall 2002
In Class Exercise
You are the young engineer that is in charge of designing the control system for the next generation inkjet printer (refer the example discussed in previous lecture notes). During the latest design review, the following plant parameters are obtained:
LA = 10 mH
RA = 10 W
KT = 0.06 Nm/A
JE = 6.5 ´ Kg m2
BE = 1.4 ´ Nm/(rad/sec)
The drive roller angular position is sensed by a rotational potentiometer with a static sensitivity of KS = 0.03 V/deg.
The design specifications for the paper positioning system are:
The steady state position for a step input should be within 5% of the desired position.
The 2% settling time should be less than 200 msec, and
the percent overshoot should be less than 5%.
You are to design a controller that satisfies the above specifications.

19. ME 375 – Fall 2002
In Class Exercise
(1) Model the physical system (plant) that we want to control and obtain its I/O transfer function G(s). (Sometimes, certain model simplification should be performed.)
From previous example, the DC motor driven paper positioning system can be modeled by iA
+ eLa - LA
+ eRa - RA +
ei(t) _
Eemf JA B tm
q, w JL N1 N2
qL, wL BL
DC Motor
Ei(s) +
IA(s) 1
LA s + RA
Tm(s) 1
JE s + BE 1 s KT - Kb
Read textbook about the equivalent moment of inertia
and equivalent damping constant

20. ME 375 – Fall 2002
In Class Exercise
The plant transfer function G(s) can be derived to be:
As discussed in the previous example, we can further simplify the plant model by neglecting the electrical subsystem dynamics (i.e., by letting LA = 0 ):
Substituting in the numerical values, we have our plant transfer function:

21. In Class Exercise G (s) Gf (s) GC (s) H(s) + -
ME 375 – Fall 2002
In Class Exercise
(2) Determine sensor dynamics (transfer function of the measurement system) H(s) and actuator dynamics (if necessary).
(3) Draw the closed-loop block diagram, which includes the plant, sensor, actuator and controller GC (s) transfer functions.
Reference Input +
Error
E(s)
G (s)
Output
q (s)
Input
Ei (s)
Gf (s)
GC (s) -
H(s)

22. ME 375 – Fall 2002
In Class Exercise
(4) Obtain the closed-loop transfer function GYR (s).

23. In Class Exercise -20 Transient performance Region
ME 375 – Fall 2002
In Class Exercise
(5) Based on the performance specifications, find the conditions that GYR (s) has to satisfy.
Steady State specification:
For step reference input,
Transient Specifications:
Settling Time Constraint:
Overshoot Constraint:
Real
Imag. jw
- jw
-20
Transient performance
Region

24. ME 375 – Fall 2002
In Class Exercise
(6) Choose controller structure GC (s) and Gf (s) substitute it into the CLTF GCL (s).
Let’s try a simple proportional control, where the control input to the plant is proportional to the current position error:
In s-domain (Laplace domain), this control law can be written as:
Substitute the controller transfer function into GCL (s):

25. ME 375 – Fall 2002
In Class Exercise
(7) Select the controller parameters (e.g., the proportional feedback gain of a proportional control law) so that the design constraints established in (5) are satisfied.
Steady State Constraint:
Want:
Transient Constraints:
To satisfy transient performance specifications, we need to choose KP such that the closed-loop poles are within the allowable region on the complex plane. To do this, we first need to find an expression for the closed-loop poles:

26. ME 375 – Fall 2002
In Class Exercise
For every KP , there will be two closed-loop poles (closed-loop characteristic roots). It’s obvious that the two closed-loop poles change with the selection of different KP . For example:
KP =0 ® p1,2 =0,
KP =0.26 ® p1,2 =-8.4, -50
KP =0.475 ® p1,2 =-20.1, -37.4
KP =0.52 ® p1,2 =-28.7,-28.7
KP =0.7 ® p1,2 = j
KP =1.08 ® p1,2 = j
KP =1.75 ® p1,2 = j
By inspecting the root-locus, we can find
that if
then the closed-loop poles will be in the
allowable region and the performance
specifications will be satisfied.
-60
-50
-40
-30
-20
-10
Real Axis 10 20 30
Img. Axis

27. ME 375 – Fall 2002
In Class Exercise
(8) Verify your design via computer simulation (MATLAB) and actual implementation.
>> num = 16*Ks*Kp;
>> den = [tauM 1 16*Ks*Kp];
>> T = (0:0.0002:0.25)’;
>> y = step(num,den,T);
>> plot(T,y);
Kp=Kp*180/pi
0.05
0.1
0.15
0.2
0.25
Time (sec)
0.4
0.6
0.8 1
Unit Step Response
KP = 100
KP = 40
KP = 29.93
KP = 15
(Kp=1.75)
(Kp=0.7)
(Kp=0.52)
(Kp=0.26)

28. ME 375 – Fall 2002
In Class Exercise
(9) Check the Bode Plots of the open loop and closed loop systems:
Frequency (rad/sec)
Phase (deg); Magnitude (dB)
-80
-60
-40
-20 10 -1 1 2 3
-180
-135
-90
-45
KP = 100
KP = 40
KP = 29.93
KP = 15
Open Loop