Dynamic Behavior of Closed-Loop Control Systems
Next, we develop a transfer function for each of the five elements in the feedback control
loop. For the sake of simplicity, flow rate w1 is assumed to be constant, and the system is
initially operating at the nominal steady rate.
Process
In section 4.3 the approximate dynamic model of a stirred-tank blending system was
developed:
 K 
 K 
X   s    1  X1  s    2 W2  s 
 τs  1 
 τs  1 
where

Vρ
,
w
K1 
w1
, and
w
K2 
(11-1)
1 x
w
(11-2)

The symbol xsp
 t denotes the internal set-point composition expressed as an
 t is related to the actual composition set
equivalent electrical current signal. xsp
point xsp
  t  by the composition sensor-transmitter gain Km:
  t   K m xsp
 t 
xsp

(11-7)
Current-to-Pressure (I/P) Transducer
The transducer transfer function merely consists of a steady-state gain KIP:
Pt s 
 K IP
P  s 
(11-9)
Control Valve
As discussed in Section 9.2, control valves are usually designed so that the flow rate
through the valve is a nearly linear function of the signal to the valve actuator.
Therefore, a first-order transfer function is an adequate model
W2  s 
Kv

Pt s  τv s  1
(11-10)
Composition Sensor-Transmitter (Analyzer)
We assume that the dynamic behavior of the composition sensor-transmitter can be
approximated by a first-order transfer function, but τm is small so it can be neglected
X m  s 
 Km
X  s 
Controller
Suppose that an electronic proportional plus integral controller is used.
P  s 

1 
 K c 1 

E s
 τI s 
(11-4)
where P  s  and E(s) are the Laplace transforms of the controller output p  t  and the
error signal e(t). Kc is dimensionless.
1. Summer
2. Comparator
3. Block
 Y(s)  G(s)X(s)
Blocks in Series
are equivalent to…
“Closed-Loop” Transfer Functions
•Indicate dynamic behavior of the controlled process (i.e., process plus
controller, transmitter, valve etc.)
•Set-point Changes (“Servo Problem”)
Assume Ysp  0 and D = 0 (set-point change while disturbance
change is zero)

K M GC GV GP
Y (s)

Ysp( s) 1  GC GV GP GM
(11-26)
•Disturbance Changes (“Regulator Problem”)
Assume D  0 and Ysp = 0 (constant set-point)

Gd
Y (s)

D( s) 1  GC GV GP GM
(11-29)
*Note same denominator for Y/D, Y/Ysp.
EXAMPLE 1:
P.I. control of liquid level
Block Diagram:
Assumptions
1. q1, varies with time; q2 is constant.
2. Constant density and x-sectional area of tank, A.
3.
q3  f (h)
(for uncontrolled process)
4. The transmitter and control valve have negligible dynamics
(compared with dynamics of tank).
5. Ideal PI controller is used (direct-acting).
For these assumptions, the transfer functions are:

1 
GC ( s )   K C 1 

 Is 
GV ( s )  KV
GP ( s )  
GL ( s ) 
GM ( s )  K M
1
As
1
As
KC  0
The closed-loop transfer function is:
Y H
GL


D Q1 1  GC GV GP GM
Substitute,
(11-68)
1
Y
As

D

1   1 
 KV  
1  K C 1 
K M
  I s   As 
(2)
Simplify,
Y
Is

2
D A I s  K C KV K M  I s  K C K P K M
(3)
Characteristic Equation:
A I s 2  K C KV K M  I s  K C K P K M  0
(4)
Recall the standard 2nd Order Transfer Function:
G( s) 
K
 2 s 2  2 s  1
(5)
To place Eqn. (4) in the same form as the denominator of the T.F. in Eqn. (5),
divide by Kc, KV, KM :
A I
s2   I s  1  0
KC KV K M
Comparing coefficients (5) and (6) gives:
2 
A I
K CK VK M
2   I




I
2
A I
K CK VK M
0   1
Substitute,

1 K C K V K M I
2
A
For 0 <  < 1 , closed-loop response is oscillatory. Thus decreased degree of
oscillation by increasing Kc or I (for constant Kv, KM, and A).
•unusual property of PI control of integrating system
•better to use P only
Stability of Closed-Loop Control Systems
Proportional Control of First-Order Process
Set-point change:
Y
Ysp
Y
Ysp
K C KV K P K M
s  1

K K K K
1 C V P M
s  1
K1

 1s  1
K1 
K OL
1  K OL
1 

K OL  K C KV K P K M
1  K OL
Set-point change = M

y (t )  K1M 1  e t 1
Offset =
ysp     y    

M
1  K OL
See Section 11.3 for tank example
Closed-Loop Transfer function approach:
KKC
Y


Ysp s  1  KKC
KKC
1  KKC

1  KKC
s 1
First-order behavior closed-loop time constant


1  KK C
(faster, depends on Kc)
General Stability Criterion
Most industrial processes are stable without feedback control. Thus, they are said to be
open-loop stable or self-regulating. An open-loop stable process will return to the
original steady state after a transient disturbance (one that is not sustained) occurs. By
contrast there are a few processes, such as exothermic chemical reactors, that can be
open-loop unstable.
Definition of Stability. An unconstrained linear system is said to be stable if the output
response is bounded for all bounded inputs. Otherwise it is said to be unstable.
Effect of PID Control on a Disturbance Change
For a regulator (disturbance change), we want the disturbance effects to
attenuate when control is applied.
Consider the closed-loop transfer function for proportional control of a thirdorder system (disturbance change).
C (s) 
8
D( s )
s  6s  12s  8  8 K C
3
GV  1
D (s ) is unspecified
2
GM  1
GP  Gd 
8
 s  2
3
Kc is the controller function, i.e., GC ( s)  KC
Let
( s)  s 3  6s 2  12s  8  8K C
If Kc = 1,




( s)  s  4 s 2  2s  4  s  4 s  1  3 j s  1  3 j

 1

 e  at , the step response will be
Since all of the factors are positive, 
sa

the sum of negative exponentials, but will exhibit oscillation.
If Kc = 8,
(s)  s3  6s 2  12s  72  ( s  6)( s 2  12)
Corresponds to sine wave (undamped), so this case is marginally stable
If Kc = 27


(s)  s 3  6s 2  12s  224  s  8 s 2  2s  28


  s  8 s  1  3 3 j s  1  3 3 j

Since the sign of the real part of the root is negative, we obtain a positive
exponential for the response. Inverse transformation shows how the controller
gain affects the roots of the system.
Offset with proportional control (disturbance step-response; D(s) =1/s )
8
1
Y ( s)  3

2
s  6s  12s  8  8K C s
y (t  )  lim sY ( s) 
s 0
8
1

8  8K C 1  K C
Therefore, if Kc is made very large, y(t) approaches 0, but does not equal zero.
There is some offset with proportional control, and it can be rather large when
large values of Kc create instability.
Integral Control:
P
KC
I

t
0
e  t   dt 
P(s) 
KC
E (s)
Is
GC ( s ) 
KC
Is
For a unit step load-change and Kc=1,
Y ( s) 
8s
s  s  2  8
3
I

1
s
lim sY ( s)  0  y()
no offset
s 0
(note 4th order polynomial)
PI Control:

1 
GC ( s)  KC 1 

 Is 
8s
Y ( s) 
8K
s ( s  2)3  C
1
 8K C s s

I
lim sY ( s)  0
s 0
no offset
adjust Kc and I to obtain satisfactory response (roots of equation which is
4th order).
PID Control: (pure PID)


1
GC ( s)  KC 1 
 Ds 
 Is

No offset, adjust Kc, I , D to obtain satisfactory result (requires solving for roots
of 4th order characteristic equation).
 Analysis of roots of characteristic equation is one way to
analyze controller behavior
1  GC GV GPGM  0
Rule of Thumb:
Closed-loop response becomes less oscillatory and more stable by
The
roots (poles)
of the characteristic
equation (s - pi) determine
decreasing
Kc or increasing
 .
the type of response that occurs:
General Stability Criterion
Consider
the “characteristic
equation,”
Complex
roots  oscillatory
response
All real roots  no oscillations
1  GC GV GPGM  0
***All roots in left half of complex plane = stable system
Note that the left-hand side is merely the denominator of the closed-loop
transfer function.
Stability Considerations
• Feedback control can result in oscillatory or even unstable closed-loop responses.
• Typical behavior (for different values of controller gain, Kc).
Roots of 1 + GcGvGpGm
(Each test is for different value of Kc)
(Note complex roots always occur in pairs)
GOL ( s) 
2 KC
( s  1)( s  2)( s  3)
Routh Stability Criterion
Characteristic equation
an s n  an 1s n 1    a1s  a0  0
(11-21)
Where an . According to the Routh criterion, if any of the coefficients a 0, a1,
…, an-1 are negative or zero, then at least one root of the characteristic
equation lies in the RHP, and thus the system is unstable. On the other hand,
if all of the coefficients are positive, then one must construct the Routh Array
shown below:
For stability, all elements in the first column must be positive.
The first two rows of the Routh Array are comprised of the coefficients in the
characteristic equation. The elements in the remaining rows are calculated
from coefficients by using the formulas:
b1 
b2 
a n-1a n-2  a n a n-3
a n-1
a n 1a n  4  a n a n 5
a n 1
.
.
(11-94)
(11-95)
c1 
b1a n-3  a n-1b 2
b1
(11-96)
c2 
b1a n 5  a n 1b 3
b1
(11-97)
(n+1 rows must be constructed; n = order of the characteristic eqn.)
Application of the Routh Array:
GP  GL 
8
( s  2)3
Characteristic Eqn is
1
GV  GM  1
GC  KC
1  GC GV GPGM  0
8
0
(s  2)3
(s  2)3  8K C  0
s 3  6s 2  12s  8  8K C  0
We want to know what value of Kc causes instability, I.e., at least one root of
the above equation is positive. Using the Routh array,
1
12
6
8  8KC
6(12)  1 8  8 K C 
6
8  8KC
n3
0
0
Conditions for Stability
72  8  8KC   0
KC  8
8  8KC  0
KC  1
The important constraint is Kc<8. Any Kc 8 will cause instability.
Figure 11.29: Flowchart for
performing stability analysis
Additional Stability Criteria
1.
Bode Stability Criterion
•
Ch. 14 - can handle time delays
2.
Nyquist Stability Criterion
•
Ch. 14