Feedback Control Systems
Lecture 4
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
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FCS 4/1
Feedback Control Systems
Chapter 4
A First Analysis of Feedback
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Chapter 4
A First Analysis of Feedback
Control Specifications
While maintaining the essential property of stability, the
control specifications include both static and dynamic
requirements such as the following:
 The permissible steady-state error
while performing regulation in the
presence of constant (= “bias”)
disturbance signal and sensor noise
 The permissible steady-state error
while tracking a polynomial
reference signal such as a step or a
ramp
 The sensitivity of the system
transfer function to changes in
model /plant parameters
 The permissible transient error in
response to a step in either the
reference or the disturbance input
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Steady-state
response, static
properties
Transient response,
dynamic properties
FCS 4/3
Chapter 4
A First Analysis of Feedback
The Basic Equations of Control
 Open-loop control system
can be given as:
 The output is given by:
Yol  H r DolGR  GW
W : Disturbance
 The error, the difference between reference input and system
output, is given by:
Eol  R  Yol
 R   H r DolGR  GW 
 1  Hr DolG R  GW
Eol  1  Tol  R  GW
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Chapter 4
A First Analysis of Feedback
The Basic Equations of Control
 Feedback control system
can be given as:
V : Sensor noise
≡
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W and V are taken to
be at the inputs of
the process and the
sensor
FCS 4/5
Chapter 4
A First Analysis of Feedback
The Basic Equations of Control
 Assuming the sensor to be fast and accurate, Hy can be
taken to be a constant. The input shaping Hr can be moved
inside the loop to become:
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Chapter 4
A First Analysis of Feedback
The Basic Equations of Control
 If Hy is constant, it is standard practice to select equal input
shaping factor so that Hr = Hy and a unity feedback gain is
obtained:
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Chapter 4
A First Analysis of Feedback
The Basic Equations of Control
 In case W= 0 and V= 0,
Ycl
DclG

R 1  DclG
 In case R= 0 and V= 0,
Ycl
G

W 1  DclG
 In case R= 0 and W= 0,
Ycl
 DclG

V 1  DclG
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Chapter 4
A First Analysis of Feedback
The Basic Equations of Control
 The equation of output is:
DclG
DclG
G
Ycl 
R
W
V
1  DclG
1  DclG
1  DclG
 The equation of error is:
 DclG

DclG
G
Ecl  R  
R
W
V
1  DclG
1  DclG 
1  DclG
 1

DclG
G

R
W
V
1  DclG
1  DclG 
1  DclG
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Chapter 4
A First Analysis of Feedback
The Basic Equations of Control
 1

DclG
G
Ecl  
R
W
V
1  DclG
1  DclG 
1  DclG
Or:
Ecl  S R  S GW  T V 
where:
1
S
1  DclG
DclG
T  1 S 
1  DclG
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S : Sensitivity function
T : Complementary sensitivity function
Erwin Sitompul
FCS 4/10
Chapter 4
A First Analysis of Feedback
Tracking: Following the Reference Input
 Tracking problem is to cause the output to follow the
reference input as closely as possible.
 For the special case, where only R is the input to the system,
the ability of the system to track a reference can be
compared.
 For open-loop control:
Yol  DolGR  Tol R
Eol  1  Tol  R
 For closed-loop control:
DclG
Ycl 
R  Tcl R
1  DclG
• R is said to be perfectly tracked
if E is zero
• In open-loop control, Dol (taking
Hr = 1) cannot be freely chosen
so that the error Eol is zero
• In closed-loop control, D = HrDcl
(taking Hr = 1) can be chosen to
be large so that Ecl approaches
zero
Ecl  1  Tcl  R
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Chapter 4
A First Analysis of Feedback
Regulation: Disturbance Rejection
 Regulation problem is to keep the error small when the
reference is at most a constant set point and disturbances
are present.
 Suppose that disturbance W interacts with applied input R.
 Now, compare open-loop control with feedback control on
how well each system maintains a constant steady-state
reference output in the face of external disturbance.
Yol  DolGR  GW
Disturbance directly
affect the output
DclG
G
Ycl 
R
W
1  DclG
1  DclG
Disturbance can be
substantially reduced by
assigning Dcl properly
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Chapter 4
A First Analysis of Feedback
Regulation: Sensor Noise Attenuation
 Only applies for closed-loop control, where sensor noise V is
defined as
DclG
DclG
Ycl 
R
V
1  DclG
1  DclG
 1

DclG
Ecl  
R
V
1  DclG 
1  DclG
• To reduce steady-state error, Dcl is selected to
be large
• If Dcl is large, the transfer function of noise V
tends to unity and the sensor noise is not
reduced at all
• Resolution: Frequency separation between
reference R and disturbance W (very low
frequency content) and sensor noise V (mostly
high frequencies)
• Each of R and V is a function of frequency, but
with different frequency content
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Chapter 4
A First Analysis of Feedback
Sensitivity
 The parameter changes may be caused by external factors
from the surroundings such as temperature, pressure, etc.
 Internal factors may also contribute to parameter changes,
in the form of imperfections in the system components such
as static friction, amplifier drift, aging, deterioration, etc.
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Chapter 4
A First Analysis of Feedback
Sensitivity
 Suppose that parameter changes happen in an operation,
and make the gain of the plant G differs from its original
value to G + δG.
 As the result, the overall transfer function of the system T
will also change to to T + δT, and in time cause the change of
system gain.
 By definition, the sensitivity of system gain T with respect to
the plant gain G is given by:
T
G T
T
S

G T G
G
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Chapter 4
A First Analysis of Feedback
Sensitivity
 For the ideal case, only R is the input to the system, the
original system gain T can be calculated.
 For open-loop control
Yol  DolGR
Yol
 Tol 
 DolG
R
G  Tol
G
 Sol 

Dol  1
Tol  G
Tol
• The. output Yol is directly influenced by
the plant gain G
• Any change of the plant gain G will give
impact to system gain T in the same
fractional change (same magnitude)
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 Tol
Tol

G
T
G
G T
S T 
G T G
G
FCS 4/16
Chapter 4
A First Analysis of Feedback
Sensitivity
 For closed-loop control
DclG
Ycl
DclG
Ycl 
R  Tcl 

1  DclG
R 1  DclG
G  Tcl
 Scl 
Tcl  G
Dcl
G

DclG (1  DclG ) 2
1  DclG
1
 Tcl
1
G
Scl 

1  DclG
Tcl 1  DclG G
.
• The output Ycl is not directly influenced by
the plant gain G
• The fractional change of system gain can
be reduced by properly assigning Dcl to
make the sensitivity Scl low enough
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 Tcl Dcl (1  DclG )  Dcl  DclG

G
(1  DclG ) 2

Dcl
(1  DclG ) 2
T
G T
S T 
G T G
G
FCS 4/17
Chapter 4
A First Analysis of Feedback
Advantages of Feedback Control
+ The permissible steady-state error in the presence of a
constant disturbance signal (“bias”) can be reduced.
+ The permissible steady-state error while tracking a
polynomial reference signal such as a step or a ramp can be
reduced.
+ The sensitivity of the system transfer function to changes in
model /plant parameters can be reduced.
+ The transient response can be speeded up to maintain
permissible transient error in response to a step in either the
reference or the disturbance input.
+ Unstable plants can be stabilized.
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Chapter 4
A First Analysis of Feedback
Disadvantages of Feedback Control
– Feedback control requires a sensor that can be very
expensive and may introduce additional noise.
– Feedback control systems are often more difficult to design
and to operate than open-loop systems.
– Feedback changes the dynamic response and often makes a
system both faster and less stable.
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Chapter 4
A First Analysis of Feedback
System Type
 In the previous study we assumed both reference and
disturbance to be constants, and also D(0) and G(0) to be
finite constants.
 In this section, we will consider the possibility that either or
both of D(s) and G(s) have poles at s = 0.
 An example for D(s) is the well-known structure for the
control equation of the form:
de(t )
u (t )  k p e(t )  ki  e( )d  kd
dt
t
 With U ( s )  D( s ) E ( s ),
we can deduct the corresponding transfer function
ki
D( s)  k p   kd s
s
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Chapter 4
A First Analysis of Feedback
System Type
 To accommodate various inputs that may be fed to a control
system, the polynomial inputs of different degrees will now
be fed to the system and the resulting steady-state tracking
errors will be evaluated.
“
A stable system can be classified as a type k system,
with k defined to be the degree of the input
polynomial for which the steady-state system error is
a nonzero finite constant.
”
”
“
Stable systems are classified into “types”
according to the degree of the polynomial
that they can reasonably track.
 For example, a system that can track a polynomial of degree
1 with a constant error is called Type 1.
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Chapter 4
A First Analysis of Feedback
System Type: Unity Feedback Case
1
G
DG
E
R
W
V
1  DG
1  DG
1  DG
 If we consider only the reference input R alone, set W= V=0,
1
E
R  SR
1 L
where
L  DG
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Chapter 4
A First Analysis of Feedback
System Type: Unity Feedback Case
 To consider polynomial inputs, let
r (t )  t 1(t )
k
L

k!
R ( s )  k 1
s
k 1
k 0
1(t )
1
step input t
(“position” input)
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r (t )
k 2
p (t )
1
1
1
t
ramp input
(“velocity” input)
Erwin Sitompul
parabola input t
(“acceleration” input)
FCS 4/23
Chapter 4
A First Analysis of Feedback
System Type: Unity Feedback Case
 Applying the Final Value Theorem,
lim e(t )  ess  lim s  E ( s )
t 
s 0
1
 lim s 
R( s)
s 0
1 L
1 k!
 lim s 
s 0
1  L s k 1
 We consider first a system for which L has no pole at the
origin. For this system, only step input, R(s) = 1/s (or k=0),
will guarantee that the system error is a nonzero finite
constant.
1
1 1
ess  lim s 

s 0
1  L s 1  L(0)
Type 0
K p  L(0)
“Position error
constant”
Nonzero & finite
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Chapter 4
A First Analysis of Feedback
System Type: Unity Feedback Case
 For general expression of the steady-state errors, we collect
all terms except the pole(s) at the origin into a function L0(s)
L0 ( s )
L( s )  n
s
Finite at s = 0
and define
00 = 1
01 = 0
02 = 0, …
Kn  L0 (0)
 Substituting this expression to calculate the steady-state
error,
1
1
ess  lim s 
s 0
L0 ( s ) s k 1
1 n
s
1
sn
 lim n
s 0 s  L ( s ) s k
0
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• If n > k,
• If n < k,
• If n = k,
•n=k
•n=k
Erwin Sitompul
then ess = 0
then ess → ∞
then ess is nonzero finite
= 0, ess = 1/(1+Kp)
≠ 0, ess = 1/Kn
FCS 4/25
Chapter 4
A First Analysis of Feedback
System Type: Unity Feedback Case
 A unity feedback system is defined to be of type k if
1
ess  0 for R ( s )  k
s
1
0  ess   for R ( s )  k 1
s
 For type 0 system, the error
constant Kp, position
constant, is given by
K p  lim L( s ), k  0
 For type 1 system, the error
constant Kv, velocity
constant, is given by
K v  lim s  L( s ), k  1
 For type 2 system, the error
constant Ka, acceleration
constant, is given by
Ka  lim s 2  L(s), k  2
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s 0
s 0
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s 0
FCS 4/26
Chapter 4
A First Analysis of Feedback
System Type: Unity Feedback Case
The higher the constant, the smaller the
steady-state error
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Chapter 4
A First Analysis of Feedback
System Type: Unity Feedback Case
“
Classifying a system as type k indicates the
ability of the system to achieve zero
steady-state error to any polynomial input
r(t) of degree less than k.
”
“
The system is of type k if the steady-state
error is zero to all polynomials r(t) of degree
less than k, nonzero finite for all polynomial
of degree k, and infinite for all polynomial of
degree more than k.
”
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Chapter 4
“
A First Analysis of Feedback
System Type: Unity Feedback Case
An unity feedback system is of type k if the
open-loop transfer function of the system
has k poles at s=0.
With pi ≠ 0,
K ( s  z1 )( s  z2 )
L( s ) 
 L0 ( s )
( s  p1 )( s  p2 )
L0 ( s)
K ( s  z1 )( s  z2 )
L( s ) 

s( s  p1 )( s  p2 )
s
L0 ( s )
K ( s  z1 )( s  z2 )
L( s )  2
 2
s ( s  p1 )( s  p2 )
s
K ( s  z1 )( s  z2 )
L( s )  n
s ( s  p1 )( s  p2 )
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L0 ( s )
 n
s
Erwin Sitompul
”
Type 0
Type 1
Type 2
Type n
FCS 4/29
Chapter 4
A First Analysis of Feedback
System Type: Unity Feedback Case
Example:
A temperature control system is found to have zero steadystate error to a constant tracking input and a steady-state
error of 0.5 °C to a ramp tracking input, rising at the rate of
40 °C/s.
r (t ) C
 
 What is the system type?
Finite error of ess = 0.5 °C to
a ramp tracking input  Type 1
 What is the error constant?
For type 1, ess = 1/Kv
ess = 40 /Kv
40
1
for unit ramp input
for the input in this example
Kv = 40 /ess = 80 s–1
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t  s
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FCS 4/30
Chapter 4
A First Analysis of Feedback
System Type: Unity Feedback Case
Example:
Determine the system type and the relevant error constant for
the speed-control with unity feedback and PI-control, if the
plant transfer function is G= A/(τs+1) and the controller
transfer function is Dc = kp +ki/s.
 System type?
L(s) = D(s)·G(s)
= A(kps + ki)/{s(τs+1)}
 One pole at origin  Type 1
 Error constant?
For type 1, Kv = lim s·L(s)
s→0
= Aki
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• Try to check using Final
Value Theorem, ess = 1/Kv
Erwin Sitompul
FCS 4/31
Chapter 4
A First Analysis of Feedback
System Type: Unity Feedback Case
Example:
Find the steady-state error in terms of K and Kt when the
system above are subjected to a unit step and a unit ramp
function.
≡
Type 1
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Chapter 4
A First Analysis of Feedback
System Type: Unity Feedback Case
≡
1st Way
E ( s )  R( s )  Y ( s )
100K
 R( s )  2
R( s )
4s  100 Kt  1 20s  100 K


4s 2  (100 Kt  1)20s
 2
 R( s )
 4s  (100Kt  1)20s  100 K 
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Chapter 4
A First Analysis of Feedback
System Type: Unity Feedback Case
The steady-state error for unit step input R(s) = 1/s:

 1
4s 2  (100 Kt  1)20s
ess  lim s  E (s)  lim s   2


s 0
s 0
 4s  (100 Kt  1)20s  100 K  s
0
The steady-state error for unit ramp input R(s) =1/s2:

 1
4s 2  (100 Kt  1)20s
ess  lim s  E ( s)  lim s   2
 2

s 0
s 0
 4s  (100 Kt  1)20s  100 K  s
(100 K t  1)20

100 K
100 K t  1

5K
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Chapter 4
A First Analysis of Feedback
System Type: Unity Feedback Case
2nd Way
1 pole at the origin owned by
L(s) = D(s)·G(s)
System type: Type 1
Steady-state error for a unit
step input:
ess  0
Steady-state error for a unit
ramp input:
100 K
20s(0.2s  100 Kt  1)
K v  lim s  L( s)
1
100 K t  1
ess 

Kv
5K
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L( s ) 
s 0
100 K
 lim s 
s 0
20s(0.2 s  100 K t  1)
5K

100 K t  1
Erwin Sitompul
FCS 4/35
Chapter 4
A First Analysis of Feedback
Homework 4
 No.1.a
For a plant having the transfer function 1/(s2+3s+9), it is
proposed to use a controller in a unity feedback system and
having the transfer function (c2s2+c1s+c0)/(s2+d1s).
Solve for the parameters of this controller {c0, c1, c2, d1} so
that the closed-loop will have the characteristic equation
(s+6)(s+3)(s2+3s+9).
 No.1.b
Show that if the reference input to the system of the above
problem is a step of amplitude A, then steady-state error will
be zero.
Hint: Error is the difference between the desired output and
the actual one. Steady-state error is an error that still
persists in steady-state condition, at t →∞.
 No.2, FPE (5th Ed.), 4.20.
 Deadline: 02.10.2012, at 07:30.
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