Chapter 11
Dynamic Behavior of
Closed-Loop Control
Systems
Chapter 11
Chapter 11
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 4.1 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 
(11-1)
1 x
K2 
w
(11-2)
where
Vρ
 ,
w
w1
K1  , and
w
Chapter 11
Chapter 11
  t  denotes the internal set-point composition
The symbol xsp
  t  is
expressed as an equivalent electrical current signal. xsp
  t  by the
related to the actual composition set point xsp
composition sensor-transmitter gain Km:
  t   K m xsp
 t 
xsp
(11-7)
Current-to-Pressure (I/P) Transducer
Chapter 11
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)
Chapter 11
We assume that the dynamic behavior of the composition sensortransmitter can be approximated by a first-order transfer function,
but τm is small so it can be neglected.
Controller
X m  s 
 Km
X  s 
Suppose that an electronic proportional plus integral controller is
used.
P  s 

1 
 Kc 1 

E s
τ
s
I 

(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.
Chapter 11
1. Summer
Chapter 11
2. Comparator
3. Block
 Y(s)  G(s)X(s)
•Blocks in Series
are equivalent to...
Chapter 11
“Closed-Loop” Transfer Functions
•Indicate dynamic behavior of the controlled process
(i.e., process plus controller, transmitter, valve etc.)
Chapter 11
•Set-point Changes (“Servo Problem”)
Assume Ysp  0 and D = 0 (set-point change while disturbance
change is zero)
K mGcGvG p
Y (s)


Ysp ( s ) 1  GcGvG p Gm
(11-26)
•Disturbance Changes (“Regulator Problem”)
Assume D  0 and Ysp = 0 (constant set-point)
Gd
Y ( s)


D( s) 1  GcGvG pGm
*Note same denominator for Y/D, Y/Ysp.
(11-29)
Chapter 11
Chapter 11
Chapter 11
Figure 11.16 Block diagram for level control system.
Chapter 11
Chapter 11
Chapter 11
Chapter 11
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.
Chapter 11
3. q 3  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 

G C (s)   K C 1 
G M (s)  K M
  Is 
G V (s)  K V
1
As
1
G L (s) 
As
G P (s)  
KC  0
The closed-loop transfer function is:
Gd
Y H


D Q1 1  GC GV GPGM
(11-68)
Chapter 11
Substitute,
Simplify,
1
Y
As

D

1   1 

 KV  
1  K C 1 
K M
  I s   As 
Y
Is

D A I s 2  K C KV K M  I s  K C K P K M
(2)
(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
 2s 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
s 2   Is  1  0
KCK VK M
Chapter 11
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
Chapter 11
Stability of Closed-Loop
Control Systems
Chapter 11
Proportional Control of First-Order Process
Set-point change:
K C KV K P K M
Y
s  1

Ysp 1  K C KV K P K M
s  1
Y
K1

Ysp  1s  1
K1 
K OL
1  K OL
1 
GP 
KP
s  1
GV , GC , GM constant gains
(K V , K C , K m )

1  K OL
K OL  K C KV K P K M
Set-point change = M

Chapter 11
y (t )  K1M 1  e t 1

M
Offset = ysp     y    
1  K OL
See Section 11.3 for tank example
Chapter 11
Closed-Loop Transfer function approach:
KKC
Y


Ysp s  1  KKC
First-order behavior
closed-loop time constant
KKC
1  KKC

1  KKC

(faster, depends on Kc)
s 1

1  KK C
Chapter 11
Chapter 11
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.
Chapter 11
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 third-order system (disturbance change).
8
Y (s)  3
D( s)
2
s  6s  12s  8  8K C
GV  1
GM  1
GP  Gd 
D (s ) is unspecified
8
 s  2
3
Kc is the controller function, i.e., G C (s)  K C .
Let ( s)  s 3  6s 2  12s  8  8K C
Chapter 11
If Kc = 1,




( s)  s  4 s 2  2s  4  s  4 s  1  3 j s  1  3 j

 1

 e at 

sa

Since all of the factors are positive,
the step response will be 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



Chapter 11
 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 stepresponse; D(s) =1/s )
Y ( s) 
8
1

s 3  6s 2  12 s  8  8 K C s
y (t  )  lim sY ( s ) 
s 0
8
1

8  8K C 1  K C
Chapter 11
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:
K
P C
I

t
0
et dt 
P(s) 
KC
E(s)
 Is
G C (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()
s 0
no offset
(note 4th order polynomial)
Chapter 11
PI Control:

1 

G C (s)  K C 1 
  Is 
8s
Y(s) 
8K
s(s  2) 3  C
1
 8K Cs 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
G C (s)  K C 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  G CG V G P G M  0
Rule of Thumb:
Closed-loop response becomes less oscillatory and more stable by
decreasing Kc or increasing I .
Chapter 11
General Stability Criterion
Consider the “characteristic equation,”
1  G CG V G P G M  0
Note that the left-hand side is merely the denominator of the
closed-loop transfer function.
The roots (poles) of the characteristic equation (s - pi) determine
the type of response that occurs:
Complex roots  oscillatory response
All real roots  no oscillations
***All roots in left half of complex plane = stable system
Chapter 11
Figure 11.25 Stability regions in the complex plane for roots of the
characteristic equation.
Chapter 11
Stability Considerations
• Feedback control can result in oscillatory or even
unstable closed-loop responses.
• Typical behavior (for different values of controller
gain, Kc).
Chapter 11
Roots of 1 + GcGvGpGm
(Each test is for different
value of Kc)
(Note complex roots
always occur in pairs)
Figure 11.26 Contributions of characteristic equation roots to closed-loop response.
Chapter 11
GOL ( s ) 
2KC
( s  1)( s  2)( s  3)
Routh Stability Criterion
Chapter 11
Characteristic equation
an s n  an 1s n 1    a1s  a0  0
(11-93)
Where an 0 . According to the Routh criterion, if any of
the coefficients a0, 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:
Chapter 11
For stability, all elements in the first column must be
positive.
Chapter 11
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:
a n-1a n-2  a n a n-3
b1 
a n-1
a n 1a n  4  a n a n 5
b2 
a n 1
(11-94)
(11-95)
.
.
b1a n 3  a n 1b 2
b1
b1a n 5  a n 1b 3
c2 
b1
c1 
(11-96)
(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
Chapter 11
1
8K C
0
3
(s  2)
GV  GM  1
GC  KC
1  GC GV GPGM  0
(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  8K C
6(12)  18  8K C 
0
6
8  8K C
0
n 3
Conditions for Stability
72  8  8K C   0 K C  8
8  8K C  0
K C  1
The important constraint is Kc<8. Any Kc 8 will cause instability.
Chapter 11
Figure 11.29
Flowchart for
performing a
stability analysis.
Chapter 11
Additional Stability Criteria
1.
Bode Stability Criterion
• Ch. 14 - can handle time delays
2.
Nyquist Stability Criterion
• Ch. 14
Direct Substitution Method
Imaginary axis is the dividing line between stable and unstable systems.
Chapter 11
1. Substitute s = jw into characteristic equation
2. Solve for Kcm and wc
(a) one equation for real part
(b) one equation for imaginary part
Example (cf. Example 11.11)
characteristic equation: 1 + 5s + 2Kce-s = 0
set s = jw
(11-101)
1 + 5jw + 2Kce-jw = 0
1 + 5jw + 2Kc (cos(w – j sin(w) = 0
Chapter 11
Direct Substitution Method (continued)
Re:
1 + 2Kc cos w = 0
(1)
Im:
5w – 2Kc sin w = 0
(2)
solve for Kc in (1) and substitute into (2):
5w 
sin w
 5w  tan w  0
cos w
Solve for w:
from (1)
wc = 1.69 rad/min (96.87°/min)
Kcm = 4.25
(vs. 5.5 using Pade approximation in Example 11.11)
Chapter 11
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