Colorado School of Mines CHEN403
Linear Open Loop Systems
Linear Open Loop Systems................................................................................................................................ 1
2nd Order Systems ............................................................................................................................................... 1
Dynamic Response of 2nd Order System ...................................................................................................... 1
Step Change Response ................................................................................................................................... 3
Descriptors for an Underdamped System .............................................................................................. 6
Natural 2nd Order Systems................................................................................................................................ 7
2nd Order Systems
Output modeled with a 2nd order ODE:
d2 y
dy
a2 2 + a1
+ a0 y = bf ( t )
dt
dt
If a0 ≠ 0 , then:
2
a2 d 2 y a1 dy
b
dy
2 d y
+
+ y = f (t ) ⇒ τ
+ 2ζτ + y = K p f ( t )
2
2
a0 dt
a0 dt
a0
dt
dt
where: τ is the natural period of oscillation.
ζ is the damping factor.
K p is the steady state gain.
For deviation variables, where y ( 0) = f ( 0 ) = 0 , the Laplace transform will be:
(τ s
2 2
)
+ 2ζτs + 1 y ( s ) = K p f ( s ) ⇒ G ( s ) =
y(s)
f (s)
=
Kp
τ s + 2ζτs + 1
2 2
2nd order systems arise from several situations:
•
Multicapacity processes — two or more 1st order systems in series.
•
Inherently 2nd order system — not typical in chemical-type processes.
•
Process with controller.
Dynamic Response of 2nd Order System
We can express the transfer function in terms of its roots, or poles:
2nd Order Systems
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December 21, 2008
Colorado School of Mines CHEN403
K p / τ2
K p / τ2
.
G(s) = 2 2
=
=
τ s + 2ζτs + 1 s 2 + 2ζ s + 1 ( s − p1 )( s − p2 )
τ
τ2
Kp
What are the poles associated with the 2nd order equation? Can determine the roots using
the quadratic equation:
2ζ
4ζ 2 4
±
− 2
2
ζ2 − 1
ζ
τ
τ
τ
=− ±
p1 , p2 =
τ
τ
2τ2
−
The nature of the solution depends upon the damping factor, ζ :
•
Over Damped Response. If ζ > 1 , then the poles are real & distinct, leading to the
terms in the transfer function:
G(s) =
K p C1
K C2
+ 2p
2
τ s − p1 τ s − p2
with the following terms in the solution:
y ( t ) = C1
•
Kp
τ
2
e tp1 + C2
Kp
etp2
τ2
Critically Damped Response. If ζ = 1 , then there is a multiple real pole,
p = −1/ τ , leading to the terms:
 1
C1  s +  + C 2
τ
G(s) = 2 
2
τ
 1
s + τ 


Kp
with the following terms in the solution:
y ( t ) = C1
•
Kp
τ
2
e − t / τ + C2
Kp
τ
2
te −t / τ
Under Damped Response. If ζ < 1 , then the poles are complex conjugates,
leading to the terms in the transfer function:
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December 21, 2008
Colorado School of Mines CHEN403
ζ

C1  s +  + C 2
τ

G(s) = 2
2
τ 
ζ  1 − ζ2
+
+ 2
s

τ 
τ

Kp
with the following terms in the solution:
y ( t ) = C1
Kp
τ2
e
−t ζ / τ
 t 1 − ζ2
cos 

τ


 t 1 − ζ2
Kp
−t ζ / τ
 + C2
sin 
e
2


τ
τ
−
ζ
1






Step Change Response
Let’s look at response to a unit step change. f ( t ) = α ⋅ H ( t ) ⇒ f ( s ) = α / s . So, for a 2nd
order system:
y(s) =
1
α
τ2 s + 2ζτ 
=
α
−
K
p

2 2
τ2 s 2 + 2ζτs + 1 s
 s τ s + 2ζτs + 1 
Kp
2ζ


s+
1

τ
y ( s ) = αK p  −
2ζ
1
 s s2 + s + 2 
τ
τ 


ζ ζ 

s+ +



1
τ τ 


∴ y ( s ) = αK p −
2
s 
ζ  1 − ζ2 
+
+ 2 
s


τ 
τ 


The form for inverting the 2nd term depends upon the value of the damping factor, ζ . For
an over damped 2nd order system, ζ > 1 , and:

ζ ζ 

s+ +



1
τ τ 


y ( s ) = αK p −
2
s 
ζ  ζ2 − 1 
+
− 2 
s


τ 
τ 


2nd Order Systems
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December 21, 2008
Colorado School of Mines CHEN403

 t ζ2 − 1 
 t ζ2 − 1 
ζ
−

y ( t ) = αK p 1 − e −t ζ / τ cosh 
e −t ζ / τ sinh 
2




τ
τ

1
ζ
−



 


 t ζ2 − 1 
 t ζ 2 − 1  
ζ
−t ζ / τ 

+
 
= αK p 1 − e
sinh 
cosh 
2



τ
τ

ζ −1



 


This type of response is termed over damped since it resembles a sluggish exponential
decay. It becomes more and more sluggish as ζ is increased.
Next, for a critically damped 2nd order system, ζ = 1 :



 1 1
s+ + 




1
1
τ τ
1 − 1 − 1

y ( s ) = αK p  − 
K
=
α
p
2
s
 s  1  τ  1 2 
 1 
s+ τ


s+ τ 
s+ τ 



 

 


 

t
y ( t ) = αK p 1 −  1 +  e −t / τ 
τ
 

This is called critically damped because this is the smallest value of ζ to retain purely
exponential decay behavior.
Now, for an under damped 2nd order system, ζ < 1 :

ζ ζ 

s+ +



1
τ τ 


y ( s ) = αK p −
2
s 
ζ  1 − ζ2 

 s + τ  + τ2 





 t 1 − ζ2
y ( t ) = αK p 1 − e −t ζ / τ cos 

τ



 t 1 − ζ2
ζ
−
e −t ζ / τ sin 


τ
1 − ζ2



  t 1 − ζ 2
= αK p 1 − e −t ζ / τ cos 
τ

 


 t 1 − ζ2
ζ
+
sin 


τ
1 − ζ2





 
 
 

 
Notice that this response shows an exponential decay but there is also oscillatory behavior.
We can better see what the oscillatory behavior looks like if we combine the sine & cosine
terms into a single sine term with a phase angle offset. Remember:
2nd Order Systems
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December 21, 2008
Colorado School of Mines CHEN403
p cos θ + q sin θ = r sin ( θ + φ )
r = p2 + q 2
where:
tan φ =
p
q
So:
r = 1+
ζ2
1
=
2
1−ζ
1 − ζ2
tan φ =
1 − ζ2
ζ
Defining:
1 − ζ2
ω≡
τ
then:


e −t ζ / τ
y ( t ) = αK p 1 −
sin ( ωt + φ ) 


1 − ζ2
The following figure shows how the response to a unit step change of a 2nd order system
depending upon the damping factor. Note that all curves have an initial slope of zero —
this is different from a 1st order system. Note that as ζ gets smaller and smaller, the
oscillations get larger and take longer to damp out.
2nd Order Systems
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December 21, 2008
Colorado School of Mines CHEN403
2.0
1.8
1.6
ζ = 0.1
1.4
1.2
ζ = 0.5
y/α
α Kp 1.0
ζ = 1.1
ζ = 0.9
0.8
0.6
ζ= 1
ζ=5
0.4
ζ = 10
0.2
0.0
0
2
4
6
8
10
12
t/ττ
Descriptors for an Underdamped System
There are several terms defined to describe the characteristics of an underdamped
system’s response. These are also shown in the following figure:
•
Overshoot. A measure of how far the response exceeds the ultimate value. From
the figure, this is A / α . From the response curve, the value is:
 −πζ 

Overshoot = exp 
 1 − ζ2 


•
Decay Ratio. The ratio of the overshoot of two successive peaks, C / A . It can be
shown that this is:
 −2πζ
2
Decay Ratio = ( Overshoot ) = exp 
 1 − ζ2

•




Period of oscillation. The time between peaks. Since the frequency of oscillation
is:
ω=
2nd Order Systems
1 − ζ2
τ
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December 21, 2008
Colorado School of Mines CHEN403
then the period of oscillation is:
T=
1 2π
2πτ
=
=
f ω
1 − ζ2
•
Rise time. The time it takes the response to 1st get to the ultimate value.
•
Response time. The time it takes the response to stay within 5% of the ultimate
value.
A
C
y/Kp
Response Time
Rise Time
t
Natural 2nd Order Systems
There are four natural 2nd order systems described in Appendix 11A of Stephanopolous’s
text (pp. 205 - 211):
•
Simple manometers
•
Externally mounted level indicators
•
Variable capacitance differential pressure transducers
•
Pneumatic valves
We will go through the derivation of the transfer function for the externally mounted level
indicators since it is not given in the text.
2nd Order Systems
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December 21, 2008
Colorado School of Mines CHEN403
F0
hm
h
F1
We want to look at the flow of fluid within the external tube of length L having constant
cross-sectional area Am . Using a similar derivation as that done to get Equation 11A.5, let
us look at the forces along the plane at the bottom of the displacement chamber:
Ftank − Ftube − Ffriction =
where: Ftank
Ftube
mtube
atube
gc
force at the entrance of the displacement tube due to the static head in the
tank.
force at the bottom of the displacement tube due to the static head in the
displacement tube.
Ffriction frictional losses in the displacement tube.
mtube
mass of fluid in the tube.
atube
acceleration of the fluid in the tube.
As in the text, if we assume laminar flow in the displacement tube, then:
Ffriction = Am
8µL
8µL dhm
vave = Am 2
2
R gc
R gc dt
The acceleration of the fluid in the tube is:
atube =
dvave d 2hm
= 2
dt
dt
so:
2nd Order Systems
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December 21, 2008
Colorado School of Mines CHEN403
g
g
8µL dhm ρAm L d 2hm
ρ Amh − ρ Amhm − Am 2
=
gc
gc
R gc dt
gc dt 2
ρL d 2hm 8µL dhm ρg
ρg
+ 2
+ hm = h
2
gc dt
R gc dt
gc
gc
L d 2hm 8µL dhm
+
+ hm = h
g dt 2 R 2ρg dt
Putting this ODE into the standard 2nd order form gives:
τ2 =
L
⇒ τ=
g
2τζ =
and:
L
g
8µL
4µL 1 8µL g 8µ
⇒ ζ= 2 ⋅ = 2
=
2
R ρg
R ρg τ R ρg L ρR 2
L
g
Kp = 1
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December 21, 2008