Dynamic%20Behavior%20of%20Closed-Loop%20Control%20Systems

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Dynamic Behavior of Closed-Loop Control Systems
Chapter 11
2
Chapter 11
3
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
Chapter 11
where
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Chapter 11
5
The symbol denotes the internal
set-point composition expressed as an equivalent
electrical current signal. is related
to the actual composition set point
by the composition sensor-transmitter gain Km
Chapter 11
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Current-to-Pressure (I/P) Transducer The
transducer transfer function merely consists of a
steady-state gain KIP
Chapter 11
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
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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 tm is small so it can be neglected.
Controller Suppose that an electronic
proportional plus integral controller is used.
Chapter 11
where and E(s) are the Laplace
transforms of the controller output and
the error signal e(t). Kc is dimensionless.
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Chapter 11
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1. Summer
2. Comparator
Chapter 11
3. Block

• Blocks in Series

are equivalent to...
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Chapter 11
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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)
Chapter 11
(11-26)

• Disturbance Changes (Regulator Problem)

Assume D ? 0 and Ysp 0 (constant set-point)
(11-29)
Note same denominator for Y/D, Y/Ysp.
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Chapter 11
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Chapter 11
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Chapter 11
Figure 11.16 Block diagram for level control
system.
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Chapter 11
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Chapter 11
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Chapter 11
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EXAMPLE 1 P.I. control of liquid level
Block Diagram
Chapter 11
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Assumptions 1. q1, varies with time q2 is
constant. 2. Constant density and x-sectional
area of tank, A. 3. (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).
Chapter 11
For these assumptions, the transfer functions are
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The closed-loop transfer function is
(11-68)
Substitute,
(2)
Chapter 11
Simplify,
(3)
Characteristic Equation
(4)
Recall the standard 2nd Order Transfer Function
(5)
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To place Eqn. (4) in the same form as the
denominator of the T.F. in Eqn. (5), divide by
Kc, KV, KM
Comparing coefficients (5) and (6) gives
Chapter 11
Substitute,
For 0 lt ? lt 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

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Stability of Closed-Loop Control Systems
Chapter 11
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Proportional Control of First-Order Process
Set-point change
Chapter 11
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Set-point change M
Chapter 11
Offset
See Section 11.3 for tank example
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Closed-Loop Transfer function approach
Chapter 11
First-order behavior closed-loop time constant
(faster, depends on Kc)
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Chapter 11
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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
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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).
Chapter 11
Kc is the controller function, i.e.,
.
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Let
If Kc 1,
Chapter 11
Since all of the factors are positive,
, the step response will be the sum
of negative exponentials, but will exhibit
oscillation.
If Kc 8,
Corresponds to sine wave (undamped), so this case
is marginally stable.
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If Kc 27
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.
Chapter 11
Offset with proportional control (disturbance
step-response D(s) 1/s )
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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
Chapter 11
For a unit step load-change and Kc1,
no offset
(note 4th order polynomial)
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PI Control
no offset
adjust Kc and ?I to obtain satisfactory response
(roots of equation which is 4th order).
Chapter 11
PID Control (pure PID)
No offset, adjust Kc, ?I , ?D to obtain
satisfactory result (requires solving for roots
of 4th order characteristic equation).
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Rule of Thumb Closed-loop response becomes less
oscillatory and more stable by decreasing Kc or
increasing tI .
General Stability Criterion Consider the
characteristic equation,
Note that the left-hand side is merely the
denominator of the closed-loop transfer function.
Chapter 11
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
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Chapter 11
Figure 11.25 Stability regions in the complex
plane for roots of the characteristic equation.
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Stability Considerations

• Feedback control can result in oscillatory or
  even unstable closed-loop responses.

Chapter 11

• Typical behavior (for different values of
  controller gain, Kc).

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Roots of 1 GcGvGpGm
Chapter 11
(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.
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Chapter 11
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Routh Stability Criterion
Characteristic equation
Chapter 11
(11-93)
Where an gt0 . 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
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Chapter 11
For stability, all elements in the first column
must be positive.
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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
(11-94)
Chapter 11
(11-95)
. .
(11-96)
(11-97)
(n1 rows must be constructed n order of the
characteristic eqn.)
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Application of the Routh Array
Characteristic Eqn is
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,
Chapter 11
Conditions for Stability
The important constraint is Kclt8. Any Kc ?8
will cause instability.
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Figure 11.29 Flowchart for performing a stability
analysis.
Chapter 11
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Additional Stability Criteria

• 1. Bode Stability Criterion
• Ch. 14 - can handle time delays
• 2. Nyquist Stability Criterion
• Ch. 14

Chapter 11
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Direct Substitution Method
Imaginary axis is the dividing line between
stable and unstable systems.

• Substitute s jw into characteristic equation
• 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 (11-101)
• set s jw 1 5jw 2Kce-jw 0
• 1 5jw 2Kc (cos(w) j sin(w)) 0

Chapter 11
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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)
Chapter 11
Solve for w wc 1.69 rad/min (96.87/min)
from (1) Kcm 4.25
(vs. 5.5 using Pade approximation in Example
11.11)
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Chapter 11
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