Post on 05-Apr-2018
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AE 230 - Modeling and
Simulation Laboratory
8/2/2019 Q2 2 Second_order
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Second order systems
iqb
dt
idq
bdt
iqd
bo
qoa
dt
odq
adt
oqd
a012
2
212
2
2++=++
iqboqoadt o
dq
adt
o
qd
a 012
2
2 =++ iq
a
b
oq
dt
odq
a
a
dt
oqd
a
a
0
0
0
1
2
2
0
2 =++
iKq
oq
n
D
n
D=
++ 12
2
2
ω
ζ
ω
ty)(sensitivigainstatesteadysystem
essdimensionlratiodamping
timerad frequencynaturalundamped
where
Δ
0a
0b
ΔK
Δ
0a
2a
1a
Δζ
Δ
2a0
a
Δn
ω
Most general form of second order system
Special case and most practical in nature
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Cascaded Second order systems
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Cascaded Second order systems
( )
( )1
1
122222
1
1 iioq
D
K K qK q D
+
==+
τ
τ
Output of one system going to input of other;
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Cascaded Second order systems
( )
( )1
1
122222
1
1 iioq
D
K K qK q D
+
==+
τ
τ
1212]1)12(21[2)12)(11(
2
iqK K oq D Doq D D=+++=++
τ τ τ τ τ τ
Output of one system going to input of other; second ordersystem but not as a single system
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Second order systems
022s02012
0110111i
xK)x-x(B
xMxB-f
=
=
Bottom damper and spring canbe assumed to be a velocitysensor. Interested in finding X
02
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Second order systems
1
) /()(
12
21212
12
21
212
1
02
++
+=
D BK
B BK M D
BK
B M
K B B D
f
x
s
s
s
s
i
21
1
02 x)( TF TF D f
x
i=
1
/ 1
1
1
1
1+
= D
B
M
BTF
1
/
2
2
22
2+
= D
K
B
K BTF
s
s
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Second order systems
02202012
011020120111
)(
)(
xK x x B
x M x x B x B f
s
i
=−
=−−−
02
2
221
02
2
2220222
2
11 1)(
x D BK D B M
x D B
K D B B xK D B
B
B f
s
ssi
+=
−
+−+−
1)()(
))( /()(
212
21212
212
21
2122
1
02
++
+
++
+=
D B BK
B BK M
D B BK
B M
B BK B D
f
x
s
s
s
s
i
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Second order systems
1)()(
))( /()(
212
21212
212
21
2122
1
02
+++++
+=
D B BK B BK M D
B BK B M
B BK B D
f
x
s
s
s
s
i
1
) /()(
12
21212
12
21
212
1
02
++
+=
D BK
B BK M D
BK
B M K B B D
f x
s
s
s
s
i
Approximate (no interaction/no loading effect)
Exact (with interaction/loading effect)
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Second order systems
1)()(
))( /()(
212
21212
212
21
2122
1
02
++++
+
+=
D B BK B BK M D
B BK B M
B BK B D
f
x
s
s
s
s
i
If B2 is small compared to B1 then individual transferfunction will be a good approximation. System 2 is notsignificantly loading system1. Assuming all parameter areequal to 1, except B2 = 0.05
105.105.0
05.0)(
21
02
++=
D D D
f
x
i
10.10476.0
0476.0)(
2
1
02
++=
D D D
f
x
i
No Loading
Exact
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Second order systems
Two isolated first order system when joined togetherdraws power from one of the system, if it is significant,system equation cannot be developed using just two firstorder system
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Second order systems
o x M x B
o xs
K W W i f =−+−+
0)(
i f
o xs
K x Bo
x M =++0
iKf
oq
n
D
n
D=
++ 12
2
2
ω
ζ
ω
Force balance
Newton
meter
sK
KMs
K
B
M
sK
n
1,
2,
time
rad ∆∆∆ ζ
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Second order systems
iKf oqn
D
n
D
=
++ 12
2
2
ω
ζ
ω 12 and12 −−−−+− ζ ω ω ζ ζ ω ω ζ nnnn
)(
roots,0,0:Undamped
φ
ω
+=
±===
tn
CSinocx
niB
)21(
21roots ,0.10,20:dUnderdampe
φω
ζ
+−−
=
−±−=<<<<
tn
Sint
nCeocx
nin
Ms
KB
Two roots of the characteristic equation
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Second order systems
Mass M = 1 kg; K = 1 N/m and B = 0.2 N/(m/sec)
sec / 1time
rad rad
M
sK
n=∆ω 1.0
2=∆
M s
K
Bζ
The system is subjected to
i) Initial displacement (1m)
ii) Initial velocity (1 m/sec)iii) Step force (1 N)
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Second order systems - underdamped
Response for initial displacement, initial velocity, step force input
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Second order systems – Step input
Step response of undamped second order system ζ =0
)cos1( t nisKf o x ω −=
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Second order systems – Step input
Step response of critically second order system ζ =1
))1(1(t n
et nisKf o xω
ω −
+−=
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Second order systems - Overdamped
21
21
2
/
2
/
1
1,11roots ,0.1,2:dampedOver
τ
τζ
teC
teC
ocx
nMs
KB
−+
−=
=−±−=>>
Mass M = 1 kg; K = 1 N/m and B = 20 N/(m/sec)
0.102
=∆ M s
K
Bζ sec / 1
time
rad rad
M
sK
n=∆ω
−
−
−++
−
−
−+−= 2
2
2
1
2
2 /
12
1 /
12
11
τ
ζ
ζ ζ τ
ζ
ζ ζ t e
t eKf
o x is
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Second order systems
Mass M = 1 kg; K = 1 N/m and B = 20 N/(m/sec)
0.102
=∆ M s
K
Bζ sec / 1
time
rad rad
M
sK
n=∆ω
−+−−=
t e
t eKf
o x nn
is
ω ω 95.19002.0
05.0002.11
−
−
−−+
−
−
−+−= 2
2
2
1
2
2 /
12
1 /
12
11
τ
ζ
ζ ζ τ
ζ
ζ ζ t e
t eKf
o x is
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Second order systems
Mass M = 1 kg; K = 1 N/m and B = 20 N/(m/sec)
0.102
=∆ M s
K
Bζ sec / 1
time
rad rad
M
sK
n=∆ω
−−≈
−+−−=
t eKf
t e
t eKf
o x
nis
nnis
ω
ω ω
05.0002.11
95.19002.0
05.0002.11
−
−
−−+
−
−
−+−= 2
2
2
1
2
2 /
12
1 /
12
11
τ
ζ
ζ ζ τ
ζ
ζ ζ t e
t eKf
o x is
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Second order systems – Step input
Non-dimensional step response of second order system
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Second order systems – Step input
Effect of damping on overshoot
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Significance of K, ζ, ωn
Steady state gain is only dependent of K
ωn largely governs the speed of response due to product (ωnt).Doubling the natural frequency will half the response time. Tospeed up by a factor n, natural frequency has to be increased by
a factor of n (when ζ is constant).
For step response, when ζ < 1.0 overshooting. To control theovershoot ζ should be adjusted. when ζ = 1.0, least time toreach steady state without overshoot. when ζ > 1.0, no
overshoot, time to reach steady state is more than when ζ = 1.0
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Lab testing of second order systems – Step inputs
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Lab testing of second order systems – Step inputs
etcd nd nd n ,,,
6,
4,
2,0
ω π
ω π
ω π
d n,
2
ω
π
Peaks occur at
Time period
Amplitude ratio of two successive peak is constant for anunder damped system, this can be used for finding thedamping ratio.
21
2
,
1, ζ
πζ
−
−
+= e
x
x
n p
n p
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Assignment
Write system equation and transfer function for any twomechanical and two electrical systems given insubsequent slides
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Second order systems
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Second order systems
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Second order – Electrical systems
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Second order – Electrical systems