sos, fopz, sopz
DESCRIPTION
Process Dynamics and Control NotesTRANSCRIPT
SOS β General Form
β’ General second order system
Standard form
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π2
π2π¦(π‘)
ππ‘2+ π1
ππ¦(π‘)
ππ‘+ πππ¦(π‘) = ππ’(π‘)
π2π2π¦(π‘)
ππ‘2+ 2ππ
ππ¦(π‘)
ππ‘+ π¦(π‘) = πΎπ’(π‘)
π =π2
ππ, 2ππ =
π1
ππ, πΎ =
π
ππ
π2π 2π π + 2πππ π π + π π = πΎπ(π )
π π =πΎ
π2π 2 + 2πππ + 1π(π )
SOS Parameters
β’ Three key parameters
Gain K
Natural period Ο
Damping coefficient ΞΆ (Zeta)
β’ Existence and nature of oscillation are
characterized by ΞΆ and Ο
β’ Poles?
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π1,2 =β2ππ Β± 4π2π2 β 4π2
2π2=
βπ Β± π2 β 1
π
Lets look at 3 test cases and their
response to a unit stepβ¦
K 10 10 10
Ο2 40 42.25 13
2ΟΞΆ 25 13 3
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0 10 20 30 40 50 60 70 80 90 1000
5
10
15
time
y(t)
K=10,t2=40, 2tz=25
K=10,t2=42.25, 2tz=13
K=10,t2=40, 2tz=3
Which is :
Underdamped?
Overdamped?
Critically Damped?
SOS Dynamics
Damping Coefficient ΞΆ
β’ Damping coefficient characterizes qualitative
process response
β’ Case 1: 0 < ΞΆ < 1
Poles are real or imaginary?
Displays oscillation
Has OVERSHOOT
β’ progression past the final value,
β’ followed by a return to the
β’ steady state
Underdamped
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0 20 40 60 80 1000
5
10
15
time
y(t)
K=10,t2=40, 2tz=3
SOS Dynamics
Damping Coefficient ΞΆ
β’ Case 2: ΞΆ = 1
Poles are real or
imaginary?
Fastest approach to
final value w/o
overshoot
Critically damped
β’ Case 3: ΞΆ > 1
Poles are real or
imaginary?
Slower response than
case 2
Overdamped system
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0 20 40 60 80 1000
2
4
6
8
10
time
y(t)
Overdamped
Critically Damped
SOS Dynamics
Damping Coefficient ΞΆ
β’ Case 4: ΞΆ = 0
β’ Pole are real or imaginary?
β’ Oscillatory response
β’ with no damping
β’ Frequency of oscillation :
β’ 1/Ο => period = Ο
β’ Case 5: ΞΆ < 0
β’ UNSTABLE
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0 10 20 30 40 50 60 70 80 90 1000
5
10
15
20
25
30
time
y(t)
Oscillatory No Damping
Unstable
FOS vs. SOS
β’ Step response comparison
SOS (w/ no zeros) always more
sluggish than FOS
SOS has an βSβ shape response
TOS is even more sluggish
System FOS SOS
Final Value AK AK
Initial Value 0 0
Initial
SLOPE
Finite, non-
zero0
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Underdamped SOS
β’ Rise time
Time to the first crossing of the final steady state
value
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tr=12
π‘π =π
π½π β π
π½ = 1 β π2
π = tanβ1π½
πA=1, K=10, Ο2=40, 2ΟΞΆ=3
Underdamped SOS
β’ Period
Time between successive oscillation peaks
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π =2ππ
π½ππππππ ππ ππ ππππππ‘πππ
π =π½
ππππππ’ππππ¦ ππ ππ ππππππ‘πππ
A=1, K=10, Ο2=40, 2ΟΞΆ=3
Underdamped SOS
β’ Decay Ratio
A measure of the rate of oscillation decay
β’ Overshoot
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a1=4.64
a2=1.0π·π =π2
π1= π
β2πππ½
A=1, K=10, Ο2=40, 2ΟΞΆ=3ππ = π΄π
βππ
1βπ2
Underdamped SOS
β’ Settling Time β time at which the output
enters (and remains within) a percentage of
the final value
Often 90%, 95% or 99% settling time
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t90%=49
t95%=69
t99%=123
A=1, K=10, Ο2=40, 2ΟΞΆ=3
FOS - Lead-lag Systems
β’ System with a βproperβ transfer function
Gain K
Zero -1/ππ
Pole -1/Ο
Lead-to-lag ratio Ο = πππ
β’ Lead arises from the zero, lag from the pole.
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πΊ π =πΎ(πππ + 1)
ππ + 1
FOS β Lead-Lag Systems
β’ Partial Fraction Expansion
β’ General Form
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(πππ + 1)
ππ + 1= π΄ +
π΅
ππ + 1
πππ + 1 = π΄ ππ + 1 + π΅
π 1: πππ = π΄ππ
π΄ =ππ
π= π
π 0: 1 = π΄ + π΅π΅ = 1 β π΄ = 1 β π
πΊ π =πΎ(πππ + 1)
ππ + 1= πΎπ +
πΎ 1 β π
ππ + 1
Lead-Lag Step Response
β’ Observations
For very small t (t->0, i.e., use initial value
theorem) y(0)=KΟA
β’ Discontinuous jump in the output signal
β’ Effect of the zero
For very large t (t->β, i.e., use final value
theorem) y(β)=KA
β’ Effect of lag term
Behavior of y(t) is a big function of Ο
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π π =πΎ πππ + 1
ππ + 1π π =
πΎππ΄
π +
πΎ 1 β π π΄
π (ππ + 1)
Lead-Lag Systems β Effect of Ο
β’ Case 1: 0 < ππ < Ο (0 < Ο < 1)
Discontinuous Jump @ moment of
step
ππ-> 0, becomes more of a βpureβ
FOS and lag dominates
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A=1, K=5, ππ=0.5, Ο=1y(t
)
0
Lead-Lag Systems β Effect of Ο
β’ Case 2: ππ = Ο (Ο = 1)
Pole-zero cancellation
Pure gain system G(s)=K
Discontinuous jump @
moment of step
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A=1, K=5, ππ=0.5, Ο=0.5
y(t
)
0
Lead-Lag Systems β Effect of Ο
β’ Case 3: ππ > Ο (Ο > 1)
Overshoot
Lead dominates
Discontinuous jump @
moment of step
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A=1, K=5, ππ=1, Ο=0.5
y(t
)
0
Lead-Lag Systems β Effect of Ο
β’ Case 4: ππ < 0 < Ο (Ο < 0)
Inverse response
Initial move away from the SS Value
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A=1, K=5, ππ=-1, Ο=0.5
y(t
)
t0
0
FOS In Parallel
πΊ1 π =πΎ1
π1π + 1
πΊ2 π =πΎ2
π2π + 1
π π = π1 π + π2 π
= πΊ1 π π π + πΊ2 π π π
= π π πΊ1 π + πΊ2 π
= π π πΊ(π )
πΊ π = πΊ1 π + πΊ2 π =πΎ1
π1π + 1+
πΎ2
π2π + 1
SOS with
Zeroes!
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=πΎ1(π2π + 1) + πΎ2(π1π + 1)
(π1π + 1)(π2π + 1)
=(πΎ1 + πΎ2)
(πΎ1π2 + πΎ2π1)πΎ1 + πΎ2
π + 1
(π1π + 1)(π2π + 1)
=(πΎ1π2 + πΎ2π1)π +(πΎ1 +πΎ2)
(π1π + 1)(π2π + 1)=
πΎ1π2π + πΎ1 + πΎ2π1π + πΎ2
(π1π + 1)(π2π + 1)
G2
G1
U(s) Y(s)+
+
SOS with Zeroes
β’ 2-pole, 1 zero system:
β’ Output step response:
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πΊ π =πΎ πππ + 1
π1π + 1 π2π + 1
π π =πΎ πππ + 1
π1π + 1 π2π + 1
π΄
π= πΎπ΄
π΅
π +
πΆ
π1π + 1+
π·
π2π + 1
π΅ = 1
πΆ = βπ1 π1 β ππ
π1 β π2
π· =π2 π2 β ππ
π1 β π2
π¦ π‘ = π΄πΎ 1 βπ1 β ππ
π1 β π2π
βπ‘π1 +
π1 β ππ
π1 β π2π
βπ‘π2
SOS with Zeroes β Step Response
A=1, K=10, π1=5, π2=10
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>
SOS β Case Evaluations
β’ Let Ο1
< Ο2, Ο
a> 0
Unless otherwise noted
β’ Case 1: Οa
> Ο2
Overshoot
β’ Case 2: Οa
= Ο1
or Οa
= Ο2
Pole-zero cancellation
Yields a FOS
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πΊ π =πΎ πππ + 1
π1π + 1 π2π + 1=
πΎ
πππ + 1
>
SOS β Case Evaluations
β’ Case 3: 0 < Οa
< Ο2
Resembles a FOS until Οa
<< Ο1
β’ Case 4: Οa
< 0
Always displays inverse response
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Case Summary β FOS Οa
Values Key Observations
0 < Οa < ΟJump at t=0 toward
y(β)
Οa = Ο
Pure gain system
(pole-zero
cancellation)
Οa > Ο Overshoot
Οa < 0 < Ο Inverse response
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β’ Discontinuous jump @ t=0 for all cases
β’ Let Ο1
< Ο2, Ο
a> 0
β’ Unless otherwise noted
β’ No discontinuous jump @ t=0
Case Summary β SOS Οa
Values Key Observations
0 < Οa < Ο2 Similar to FOS
Οa = Ο1 or Ο2
FOS
(pole-zero cancellation)
Οa > Ο2 Overshoot
Οa < 0 Inverse response
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Case Summary β SOS ΞΆ
Values Key Observations
ΞΆ < 0 Unstable
ΞΆ = 0Underdamped
oscillates forever
0 < ΞΆ < 1Overshoot and
underdamped
ΞΆ = 1 Critically damped
ΞΆ > 1Overdamped β
sluggish
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Inverse Respone
β’ When a process output initially moves in a
direction opposite to its steady state value
followed by a return to steady state
Effect of (at least) two opposing processes are
different timescales
Occurs when Οa
< 0 in single-zero systems
y(t) crosses the zero axis (in deviation variables)
in response to a step input
β’ Where does this show up?
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A Two Timescale Exercise
β’ Given the following block diagram and
transfer functions, calculate G(s)
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πΊ1 π =5
10π + 1πΊ2 π =
β1
1π + 1
πΊ π = πΊ1 π + πΊ2 π
=5
10π + 1β
1
π + 1
=5 π + 1 β (10π + 1)
(10π + 1)(π + 1)
=4 β
54
π + 1
10π 2 + 11π + 1
G2
G1
U(s) Y(s)+
+
FOS in Parallel
β’ Two FOS in Parallel
β’ Let:
|K1|> |K
2|
K1
and K2
be opposite signs
Ο1
> Ο2
(G2
is faster than G1)
β’ Consequences
Fast process => Initial response
Slow Process => final response (due to higher
gain)
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πΊ(π ) =(πΎ1 + πΎ2)
(πΎ1π2 + πΎ2π1)π πΎ1 + πΎ2
π + 1
(π1π + 1)(π2π + 1)
Real Inverse Response
β’ Drum Boiler
Used for steam generation
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Heat
Source
Steam
Cold Water