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Copyright © 2006 by Douglas J. Cooper All rights reserved
Deriving "τp = 63.2% of Process Step Response" Rule By Doug Cooper
Chemical, Materials and Biomolecular Engineering University of Connecticut, Unit 3222
Storrs, CT 06269-3222
Email: [email protected] Academic: http://www.engr.uconn.edu/control
Editor: http://www.controlguru.com To derive the “time constant, τP = 63.2% of Process Step Response” rule used in model fitting, we must solve a linear first order ordinary differential equation (ODE):
Time, (t)0 τP
0.632 AKP
AKP
A
Pro
cess
Var
iabl
e, y
(C
ontro
ller O
utpu
t, u
(t)
AKP
0
0
A
CO
= u
(t)
P
V =
y(t)
Time, (t)τP Time, (t)0 τP
0.632 AKP
AKP
A
Pro
cess
Var
iabl
e, y
(C
ontro
ller O
utpu
t, u
(t)
AKP
0
0
A
CO
= u
(t)
P
V =
y(t)
Time, (t)τP
Figure 1 – Response of true first order process to a step change in controller output
The figure above shows the open loop step response of a true first order process model:
( ) ( ) ( )τ + =P P
dy t y t K u tdt
where y(0) = 0
As shown in Fig 1, the measured process variable, y(t), and controller output signal, u(t), are initially at steady state with y(t) = u(t) = 0 for t < 0. Other parameters include the process time constant, τP, and process gain, KP. At time t = 0, the controller output is stepped to u(t) = A, where it remains for the duration of the experiment. Hence, the first order model becomes
( ) ( )τ + =P P
dy t y t AKdt
Copyright © 2006 by Douglas J. Cooper All rights reserved
To solve this ODE, rearrange as:
( ) 1 ( )
τ τ+ = P
P P
AKdy t y tdt
First compute the integrating factor,(1/ )τ
µ = ∫ p dte /τ= pte . Solving the ODE using this integrating
factor yields
/1/
1( ) ττ τ⎡ ⎤⎛ ⎞
= +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
∫ P
P
t Pt
P
AKy t e dt ce
/ /1
τ τ
τ− ⎡ ⎤
= +⎢ ⎥⎣ ⎦
∫P Pt tP
P
AKe e dt c
Recall that ∫ = axax ea
dxe 1
and thus ∫ = PP t
Pt edte ττ τ //
so / /
1( ) τ τ− ⎡ ⎤= +⎣ ⎦P Pt t
Py t e AK e c
/
1 τ−= + PtPAK c e
Next, apply the initial condition: @ t = 0, y = 0, 0 = AKP + c1, and thus c1 = −AKP. Substituting and rearranging, we obtain the solution to the ODE:
( ) 1 τ−⎡ ⎤= −⎣ ⎦Pt/
Py t AK e
After the passage of one time constant, time t = τP, the solution becomes
/ 1( ) 1 1τ ττ − −⎡ ⎤ ⎡ ⎤= − = −⎣ ⎦⎣ ⎦P P
P P Py AK e AK e
Therefore, the measured process variable step response at time t = τP is
( ) 0.632τ =P Py AK As we set out to show, at time t equals one time constant,τP, the measured process variable, y(t), will have traveled to 63.2% of the total change that it will ultimately experience.
