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Unformatted text preview: Fitting First and SecondOrder Models
Using Step Tests Development of Empirical Models
From Process Data • Simple transfer function models can be obtained graphically
from step response data. Chapter 7 • In some situations it is not feasible to develop a theoretical
(physicallybased model) due to:
1. Lack of information
2. Model complexity
3. Engineering effort required.
• An attractive alternative: Develop an empirical dynamic model
from inputoutput data.
• Advantage: less effort is required • If the process of interest can be approximated by a first or
secondorder linear model, the model parameters can be
obtained by hand calculations using the response curve.
• The response of a firstorder model, Y(s)/U(s)=K/(s+1),
to
a step change of magnitude M is: / y t KM (1 e t )
(518) • Disadvantage: the model is only valid (at best) for the
range of data used in its development.
i.e., empirical models usually don’t extrapolate very
well. • A plot of the output response of a process to a step change in
input is sometimes referred to as a process reaction curve or
bump test. 1 2 y (t )
1/ t 1 e
KM
• The initial slope is given by: d y 1 dt KM t 0 τ (715) • The gain can be calculated from the steadystate changes
in u and y:
K y y u M Figure 7.3 Step response of a firstorder system and
graphical constructions used to estimate the time constant, τ. where Δy is the steadystate change in y. 3 4 FirstOrder Plus Time Delay Model
G (s) K e θs
τs 1 For this FOPTD model, we note the following characteristics of its step response:
1. The response attains 63.2% of its final response
at time, t = .
2. The line drawn tangent to the response at
maximum slope (t = ) intersects the y/KM=1
line at (t = ).
3. The step response is essentially complete (99.3%) at
t = 5 , the “settling” time. Figure 7.5 Graphical analysis of the process reaction curve
to obtain parameters of a firstorder plus time delay model.
5 6 Sundaresan and Krishnaswamy’s Method There are two generally accepted graphical techniques for
determining model parameters , , and K. • They proposed that two times, t1 and t2 , be found from a
step response curve, corresponding to the times for 35.3%
and 85.3% of the final response, respectively. Method 1: Slopeintercept method First, a slope is drawn through the inflection point of the
process reaction curve in Fig. 7.5. Then and are
determined by inspection. • The time delay and time constant are then estimated from the
values of t1 = t35.3 and t2 = t85.3 using the following equations:
θ 1.3t1 0.29t2
τ 0.67 t2 t1 Alternatively, can be found from the time that the
normalized response is 63.2% complete. (719) • These values of and approximately minimize the difference
between the measured response and the model, based on a
correlation for many data sets to yield the 5 “parameter
numbers” of the method: 35.3, 85.3, 1.3, 0.29, 0.67. Method 2. Sundaresan and Krishnaswamy’s Method This method avoids use of the point of inflection
construction entirely to estimate the time delay.
7 8 Estimating Secondorder Model Parameters
Using Graphical Analysis
• “In general”, a better approximation to an experimental step
response can be obtained by fitting a secondorder model to
the data (firstorder models are a “subset” of 2ndorder ones).
• Figure 7.6 shows the range of shapes that can occur for the
2ndorder overdamped step response model,
K
G s (539) τ1s 1 τ 2 s 1
• Figure 7.6 includes two limiting cases: τ 2 / τ1 0 , where the
system becomes first order, and τ 2 / τ1 1 , the critically
damped case. Note dimensionless time used. Figure 7.6 Step response for several overdamped secondorder systems. • The larger of the two time constants, τ1 , is called the
dominant time constant.
9 10 11 12 Smith’s Method – not just overdamped
• Assumed model:
G s Ke θs
τ 2 s 2 2ζτs 1 • Procedure: 1. Determine t20 and t60 from the step response and
their ratio t20/t60 – after eliminating time delay.
2. Find ζ and t60/ from Fig. 7.7 (2 yvalues).
3. Using t60/ from Fig. 7.7 and then calculate (since
t60 is known) ...
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