1 development of empirical dynamic models from step response data some processes too complicated...
DESCRIPTION
3 Step Input Step response is the easiest to use but may upset the plant manager Other methods –impulse - dye injection, tracer –random - PRBS (pseudo random binary sequences) –sinusoidal - theoretical approach –frequency response - modest usage (incl. pulse testing) –on-line (under FB control)TRANSCRIPT
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Development of Empirical Dynamic Models from Step Response Data
Some processes too complicated to model using physical principles
• material, energy balances• flow dynamics• physical properties (often unknown)• thermodynamics
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Black Box Models
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Step Input
• Step response is the easiest to use but may upset the plant manager
• Other methods– impulse - dye injection, tracer– random - PRBS (pseudo random binary sequences)– sinusoidal - theoretical approach– frequency response - modest usage (incl. pulse testing)– on-line (under FB control)
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Fitting of 1st-Order Model
/
0
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0.632
1 1
t
t
K MG s U ss s
y t KM e
y KM
dyKM dt
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(θ = 0)
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FOPDT and SOPDT Models
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First-Order-Plus-Dead-Time (FOPDT) Model
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Second-Order-Plus-Dead-Time (SOPDT) Model
2 1
s
s
KeG ss
KeG ss s
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For a 1st order model, we note the following characteristics in step response:
1. The response attains 63.2% of its final response at one time constant (t = ).
2. The line drawn tangent to the response at maximum slope (t = ) intersects the 100% line at (t = ).
There are 3 generally accepted graphical techniques for determining the first-order system parameters and .
( )1
sKeG ss
Fitting of FOPDT ModelC
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Method 1: Sundaresan & Krishnaswany (1978)
1. Find K from stead-state response.2. Normalize step response by dividing all data with KM (t =
0, y = 0; t →∞, y = 1)3. Use 35.3% and 85.3 % response times (t1 and t2), i.e.
4. Calculate = 1.3 t1 – 0.29 t2
= 0.67 (t2 – t1)
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1
2
0.353
0.853
y t KM
y t KM
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Method 2: Numerical Fitting
(1) Find and in ( ) 1
to fit data of vs.
(2) Find and in ln
to fit data of ln vs.
t
y t KM e
y t
KM y t tKM
KM y tt
KM
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7Method 3: Fitting an Integrator Model
to Step Response Data
In Chapter 5 we considered the response of a first-order process to a step change in input of magnitude M:
/1 1 ty t KM e
For short times, t < , the exponential term can be approximated by
/ τ 1τ
t te
so that the approximate response is:
1 1 1τt KMy t KM t
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22 (7-23)KG s
s
In the time domain, the step response of an integrator is
2 2 (7-24)y t K Mt
Hence an approximate way of modeling a first-order process is to find the single parameter
2 (7-25)τKK
that matches the early ramp-like response to a step change in input.
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Figure 7.10. Comparison of step responses for a FOPTD model (solid line) and the approximate integrator plus time delay model (dashed line).
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Fitting 2nd-Order Models
1 21 1
MU ss
KG ss s
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7Harriot’s Method
1.3
0.73
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0.39
0.26
1 2
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Harriot’s Method
0.73
0.73
0.731 2
0.5 1 2
0.5 0.5
1) Determine experimentally to satisfy 0.73
2) Calculate 1.3
3) Calculate 0.5
4) Determine from experimental data
4) From Figure 7.6, det
ty t KM
t
t
y y t
1 2.ermine and then calculate
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Smith’s Method
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20
20
60
1) Determine t and t experimentally so that 0.6
0.2
2) Fig 7.7 ,
y t KM
y t KM
tt
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1.3=
1.79= 8.260
t
84.081.3
2
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1 2 Sum of squares S 3.81 0.84 0.0757
NLR (θ=0) 2.99 1.92 0.000028 FOPTD (θ = 0.7) 4.60 - 0.0760
Smith’s Method20% response: t20 = 1.8560% response: t60 = 5.0t20 / t60 = 0.37from graph
Solving,
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