MESB374
System Modeling and Analysis
Feedback Control Design Process
Classical Feedback Control Structure
Disturbance D(s)
Reference
Input R(s)
Gf(s)
+
E(s)
Gc(s)
filter
Control
Input
Output
G(s)
U(s)
Plant
Controller
H(s)
Ym(s)
Sensor
Dyamics
Plant Equation (Transfer function model that we all know how to obtain ?!):
Y  s   G  s  U  s   D  s  
Control Law (Algorithm) (we will try to learn how to design):


U  s   Gc  s  G f  s  R  s   H  s  Y  s  


Y
s


m


Y(s)
Closed-Loop Transfer Function
Disturbance D(s)
Reference
Input
R(s)
Gf (s)
+
Error
 E(s)
GC (s)
Control
Input +
U(s)
+
Output
G(s)
Y(s)
Plant
H(s)




Y  s   G  s  U  s   D  s    G  s  Gc  s  G f  s  R  s   H  s  Y  s    D  s  


U s


1

G
s
G
s
H
s
Y
s

G
s
G
s
G
s
R
s

G
s
D
s

  c       
  c  f    
   
Y ( s) 
G  s  Gc  s  G f  s 
1  G  s  Gc  s  H  s 
G  s
1  G  s  Gc  s  H  s 
GYR ( s )
GYD ( s )
 R( s ) 
 D( s)
Can you obtain it by block algebra?
What is Y(s) if disturbance dynamics and/or noise are taken into account?
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:
Y ( s) 
 R( s) 
GYR ( s)

Closed-Loop Transfer Function
From R( s ) to Y ( s )
 D( s)
GYD ( s)

Closed-Loop Transfer Function
From R( s ) to Y ( s )
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
Ideally
GYR ( s)  1
In reality
GYR ( j )  1,   low , high 
Frequency Range of Ref. Input
• 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
GYD ( s)  0
In reality
GYD ( j )  0,  
d _ low , d _ high 
Frequency Range of Disturbance Input
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
GYR (s)
r(t)
Y(s)
y(t)
rss
yss=GYR(0)r
t0
Time
ss
t0
Time
Unit Step Response
1.6
yMAX
1.4
Unit Step Response
OS
1.2
 X%
1
0.8
0.6
0.4
0.2
0
tP
tr
Time
tS
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 :
GYR  0  rss  rss
 X %,
ySS (t )  r(t )
rss
 X %  0.0 X
r (t )
1  X %  GYR  0   1  X %
To find the steady state value of the output, ySS(t):
– Sinusoidal references: use frequency response, i.e.
r  t   A  r  sin r t 
yss  t   GYR  jr  A  r  sin rt  GYR  jr  
GYR  jr  Ar  Ar
Ar
 X %,
1  X %  GYR  jr   1  X %
– General references: use FVT, provided that sY(s) is stable, ...
sY  s   lim sGYR  s  R  s 
Y ( s)  GYR ( s) R( s) yss  t   lim
s 0
s 0
lim sGYR  s  R  s   r  t 
s 0
r t 
 X%
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:
 4
Recall step
4
 4

  for 2% bound
 
n  T
 n

response


n
s
 Desired Settling Time (TS )


T



s
of second
 5 for 1% bound
 5
n  5
  n
 n

Ts
order
– % Overshoot (%OS):(2nd order systems)
%OS  100%  e

n
n 1 2
 100%  e


1 2
 X%
2



system
x




 ln 
 100  


2
x



2
  ln 

 100  


p1,2   n  j n ( 2  1)

Q: How can we link this performance specification to the closed-loop transfer function?
(Hint) What system characteristics affect the system performance ?
d
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:
Kn 2
GYR ( s )  GCL ( s )  2
s  2n s  n 2
The characteristic poles are:
s1,2  n  j n 1   2   n  jd    jd 

d
Settling Time (2%):
 Puts constraint on the real part of the dominating closed-loop poles
tS (2%) 
4
n

4

%OS:
 Puts constraint on the damping ratio  or the angle  of the dominating closed-loop poles.
%OS  100%  e


1
2
 100%  e
 
n
n 1
2
 100%  e
 

d
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:
p1,2    j d
Img.
Transient Performance Specifications:
– Settling Time (2 %) TS
tS (2%) 
4

4
  
TS
 TS
– %OS X %
%OS  100%  e
 tan    
 

d
  jd
jd

 X%


ln  X % 



   arctan  

ln
X
%





Real
-jd
  jd
4
TS
Example
A DC motor driven positioning system can be
modeled by a second order transfer function:
3
G(s) 
s( s  6 )
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.
Find closed-loop transfer function:
Draw block diagram:
d  s  +
-
E(s)
K
controller
G(s) 
3
s( s  6 )
Plant
H(s)=1
KG ( s )
1  KG ( s )
3
K
s ( s  6)

3
1 K
s ( s  6)
16
 2
s  6 s  16
GCL ( s ) 
Output
  s

2n  6 
 n  4


4
n 2  16 

3

Example
Find closed-loop poles:
2% settling time:
s 2  6 s  16  0
s1,2   3  j

ts 
7
4


4
 1.333sec
3
d
Step Response
%OS:
1.4
%OS  100%  e
1.2
1
Amplitude
 100%  e
0.8
 100%  e
0.6
 100%  e
0.4
 2.84%
0.2
0
tp 
0
0.5
1
1.5
Time (sec)
2
2.5
3
 


1 2
n
n
1 2
 

d
 
3
7

 1.1874
d
Example
A DC motor driven positioning system can
be modeled by a second order transfer
function:
3
G(s) 
s( s  6 )
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:
KG ( s )
1  KG ( s )
3
K
s ( s  6)

3
1 K
s ( s  6)
3K
 2
s  6 s  3K
GCL ( s ) 
Assume that closedloop is an underdamped second order
system
s 2  6 s  3K  0
s1,2   3  j

3K  9
d
Write down the performance specifications:
 SS   d
 2%
d
Ts  2 sec
OS%  10%
Percent Overshoot (%OS)
Example
Steady state performance constraint:
1  2%  GCL  0   1  2%
GCL  0   1
%OS  100%  e
 100%  e
 
3
3 K 9

  arctan  

4


d
 10%
Assume that closed-loop
is an under-damped
second order system
3  K  8.5846

o
  53.76 ,
Img.
ln 10%  

  53.76o
Transient performance constraint:
2% Settling Time
ts 
 
-2
 2 sec
  53.76o
 2
How to discuss the case
of over-damped system?
Real
Example
Over-damped case
s 2  6s  3K  0
s1,2  3  9  3K , K  3
Steady state performance constraint:
1  2%  GCL  0   1  2%
GCL  0   1
K 0
Stability Requirement
Transient performance constraint:
2% Settling Time
No Overshoot!!
In all, we have
 4 4 
ts  max 
,


s
 1  s2 
4

 2sec
 9  3K  3
8
8
3
3
K 3
 K  8.5846
Review of Structure of Closed-loop System
Disturbance D(s)
GD (s)
Reference
Input
R(s)
Gf (s)
filter
+ Error

E(s)
GC (s)
GA (s)
Control
Input
+
U(s) +
Actuator
Control algorithm
+
+
GN (s)
H(s)
Sensor
Noise Channel
Noise N(s)
How many transfer functions do we need to determine the output?
Do you know how to find them?
Disturbance
Channel
Output
G(s)
Plant
Y(s)
Feedback Control Design Process
A typical feedback controller design process involves the following steps:
(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)
(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 transfer functions GC (s) and Gf (s).
(4) Obtain the closed-loop transfer functions GYR (s) and GYT (s) .
(5) Based on the performance specifications, find the conditions that the CLTFs,
GYR (s) and GYT (s), have to satisfy.
(6) Choose controller structure GC (s) and Gf (s) and substitute it into the CLTFs
GYR (s) and GYT (s).
(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.
(8) Verify your design via computer simulation (MATLAB) and actual
implementation.
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  10-6 Kg m2
BE = 1.4  10-5 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.
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.)
DC Motor
+ eRa 
+
ei(t)
_
R
+ eLa 
iA
LA
A
+
Eemf
_
tm
,
L,L
N2
JL
JA
BL
N1
B
From previous example, the DC motor driven paper positioning system can be modeled by
Ei(s)


1
LA s + RA
IA(s)
KT
Tm(s)
1
JE s + BE
 s
1
s
s
2
1
JE  JA     JL
N
2
Kb
Read textbook about the equivalent moment of inertia
and equivalent damping constant
1
BE  B     BL
N
N
N 2
N1
In Class Exercise
The plant transfer function G(s) can be derived to be:
G( s ) 
 ( s)
Ei (s)

KT
s( LA JE s 2  ( BE LA  RA J E )s  ( RA BE  Kb KT ))
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 ):
G( s) 
 ( s)
Ei ( s )
KT
RA BE  K b KT

KT

s ( RA J E s  ( RA BE  K b K T ))
s(
KM
RA J E
s  1)
RA BE  K b K T
tM

KM
s (t M s  1)
Substituting in the numerical values, we have our plant transfer function:
G( s) 
KM
16

s(t M s  1) s(0.017 s  1)
In Class Exercise
(2) Determine sensor dynamics (transfer function of the measurement system) H(s) and actuator
dynamics (if necessary).
H ( s)  K s  0.03v / deg  1.719v / rad
(3) Draw the closed-loop block diagram, which includes the plant, sensor, actuator and controller
GC (s) transfer functions.
Reference
Input
Gf (s)
+ Error

E(s)
Input
GC (s)
Ei (s)
G (s)
H(s)
Output
(s)
In Class Exercise
(4) Obtain the closed-loop transfer function GYR (s).
( s )  G  s  Ei  s 
 G s G
s E s
 G  s  G c  s   dv  s    v ( s ) 
 G  s  G c  s  G f  s  d ( s )  H  s  ( s ) 
 G  s  G c  s  H  s  ( s )  G  s  G c  s  G f  s  d ( s )
c
1  G  s  G c  s  H  s   ( s )  G  s  G c  s  G f  s  d ( s )
G s G c s G f s
( s ) 
d ( s )
1  G s G c s H s
GCL  s 
In Class Exercise
(5) Based on the performance specifications, find the conditions that GYR (s) has to satisfy.
Steady State specification:
d  t   d 0  constant,
For step reference input,
 SS   d 0
 5%
d 0
ss  GCL  0 d 0
Imag.
0.95  GCL  0  1.05
j
Transient Specifications:
Settling Time Constraint:
ts 
4

 0.2 sec
  20
-20
Real
Overshoot Constraint:

  arctan  

  46.4o
  46.4o

o
  46.4 ,
ln  5%  

 j
Transient performance
Region
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:
ei (t )  KP  e V (t )  KP  ( dV (t )   V (t ))
In s-domain (Laplace domain), this control law can be written as:
Gc  s   K p
Bi (t )  K P E ( s ),
G f  s   H  s   1.719
Substitute the controller transfer function into GCL (s):
16
K p 1.719
s  0.017 s  1
G f s
GCL  s  
G c s
G s 
16
1
K p 1.719
s  0.017 s  1
H s
G s 

1580.5K p
s 2  57.47 s  1580.5K p
2n
n2
G c s

27.5K p
0.017 s 2  s  27.5K p
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:
0.95  GCL  0  1.05
GCL  0  1
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:
s 2  57.47s  1580.5K p  0
p1,2  28.7  826  1580.5K p
In Class Exercise
Img. Axis
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, -57.47
30
KP =0.26
 p1,2 =-8.4, -50
KP =0.475  p1,2 =-20.1, -37.4
20
KP =0.52
 p1,2 =-28.7,-28.7
KP =0.7
 p1,2 =-28.7+17j
10
KP =1.08
 p1,2 = -28.7+29.7j
0
KP =1.75
 p1,2 = -28.7+44j
By inspecting the root-locus, we can find -10
that if
0.475  K p  1.08
-20
then the closed-loop poles will be in the
allowable region and the performance
specifications will be satisfied.
-30
-60
-50
-40
-30
-20
Real Axis
-10
0
In Class Exercise
(8) Verify your design via computer simulation (MATLAB) and actual implementation.
>>
>>
>>
>>
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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
Unit Step Response
1
KP = 100
KP = 40 (Kp=0.7)
(Kp=1.75)
0.8
KP = 29.93 (Kp=0.52)
0.6
KP = 15
(Kp=0.26)
0.4
0.2
0
0
0.05
0.1
0.15
Time (sec)
0.2
0.25
In Class Exercise
(9) Check the Bode Plots of the open loop and closed loop systems:
10
0
KP = 100
Phase (deg); Magnitude (dB)
-20
KP = 40
-40
KP = 29.93
-60
KP = 15
-80
Open Loop
0
KP = 100
-45
KP = 40
-90
KP = 29.93
KP = 15
-135
Open Loop
-180
10
-1
10
0
1
10
Frequency (rad/sec)
10
2
10
3