contr systems ppt07e (time domain analysis, 2nd order...
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Class 07“Time domain analysis”
parte II - 2nd order systems
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cbsas)s(G
2 ++α=
outputinput
Second order systems
of the type
S
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S cbsas2 ++
α
cbsas)s(G
2 ++α=
Second order systems
of the type
Time domain analysis – 2nd order systems ______________________________________________________________________________________________________________________________________________________________________________________
outputinput
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a
cs
a
bs
a
)s(R
)s(Y
2 ++
α
=
Second order systems:
Koωn2
ωn2
that is:
2ζωn
cbsas2 ++
α
cbsas)s(R
)s(Y2 ++
α=
Time domain analysis – 2nd order systems ______________________________________________________________________________________________________________________________________________________________________________________
outputinput
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the transfer function can be rewritten as:
Ko = gain of the system
ζ = damping coefficient
ωn = natural frequency
2
nn
2
2
no
s2s
K
)s(R
)s(Y
ω+ζω+ω=
outputinput
2o n
2 2n n
K
s 2 s
ω+ ζω + ω
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Besides these 3 above parameters, we also have
outputinput
2o n
2 2n n
K
s 2 s
ω+ ζω + ω
ωd = damping frequency
1012
nd ≤ζ<ζ−⋅ω=ω
Time domain analysis – 2nd order systems ______________________________________________________________________________________________________________________________________________________________________________________
Ko = gain of the system
ζ = damping coefficient
ωn = natural frequency
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Example 1:
1s12s4
3
)s(R
)s(Y2 ++
=
Ko = 3 ζ = 1 ωn = 1
Example 2:
1s2s
3
)s(R
)s(Y2 ++
=
Ko = 3 ζ = 3 ωn = 0,5
poles are
real and
distinct
poles are
real and
repeated
poles:
s = –2,914
s = –0,086
poles:
s = –1
(duple)
ωd = 0
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2s2s2
3
)s(R
)s(Y2 ++
=
Example 3:
Ko = 3 ζ = 0 ωn = ωd = 1
Example 4:
Ko = 1,5 ζ = 0,5 ωn = 1
1s
3
)s(R
)s(Y2 +
=
complex
conjugate
poles
complex
conjugate
poles
poles:
s = –0,5 ± 0,866j
poles:
s = ± j
(pure imaginary)
ωd = 0,866
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2
nn
2s2s)s(p ω+ζω+=
Characteristic equation:
∆ = 4 ζ2 ω2n – 4 ω2
n =
= 4 ω2n (ζ2 – 1)
ζ > 1 → poles are real and distinct
ζ = 1 → poles are real and repeated
0 < ζ < 1 → poles are complex conjugate
∆ > 0 → (ζ2 – 1) > 0 → ζ2 > 1 → ζ > 1
∆ = 0 → (ζ2 – 1) = 0 → ζ2 = 1 → ζ = 1
∆ < 0 → (ζ2 – 1) < 0 → ζ2 < 1 → ζ < 1
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What is the output?
(step response)
unit step input
outputinput
2o n
2 2n n
K
s 2 s
ω+ ζω + ω
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2o n
2 2n n
KY(s) R(s)
s 2 s
ω= ⋅+ ζω + ω
and since r(t) = unit step:
2o n
2 2n n
K 1Y(s)
ss 2 s
ω= ⋅+ ζω + ω
2o n
2 2n n
K
s 2 s
ω+ ζω + ω
outputinput
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[ ])s(Y)t(y 1−= Lthe unit step response depends on the value of ζ:
a) 0 < ζ < 1 (under damping)
b) ζ = 1 (critical damping)
c) ζ > 1 (over damping)
2o n
2 2n n
K
s 2 s
ω+ ζω + ω
outputinput
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0t,tsen1
tcos1K)t(y d2
d
tn
o >
ω⋅
ζ−ζ+ω−= ζω−
e
So, in the case 0 < ζ < 1 (under damping) the unit step response
is:
[ ])s(Y)t(y 1−= L
2
nd 1 ζ−⋅ω=ωwhere
( damping frequency )
2o n
2 2n n
K
s 2 s
ω+ ζω + ω
outputinput
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ω⋅
ζ−ζ+ω−= ζω−
tsen1
tcos1K)t(y d2
d
tn
o e
unit step response :
2o n
2 2n n
K
s 2 s
ω+ ζω + ω
outputinput
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unit step response:
ω⋅
ζ−ζ+ω−= ζω−
tsen1
tcos1K)t(y d2
d
tn
o e
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( )[ ] 0t,t11K)t(y n
tn
o >ω+⋅−= ζω−e
In the case ζ = 1 (critical damping) the unit step response is:
2o n
2 2n n
K
s 2 s
ω+ ζω + ω
[ ])s(Y)t(y 1−= L
outputinput
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2o n
2 2n n
K
s 2 s
ω+ ζω + ω
unit step response :
outputinput
( )[ ]t11K)t(y n
tn
o ω+−= ζω−e
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unit step response :
( )[ ]t11K)t(y n
tn
o ω+−= ζω−e
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0t,pp12
1K)t(y2
tp
1
tp
2
n21
o >
−⋅
−ζω+= ee
In the case ζ > 1 (over damping) the unit step response is:
[ ])s(Y)t(y 1−= L
( )11p 2
n
2
nn2,1 −ζ±ζω−=−ζωζω−= m
whereSystem has
real poles
2o n
2 2n n
K
s 2 s
ω+ ζω + ω
outputinput
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unit step response:
−⋅
−ζω+=
2
tp
1
tp
2
n
pp121K)t(y
21
o
ee
2o n
2 2n n
K
s 2 s
ω+ ζω + ω
outputinput
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unit step response:
−⋅
−ζω+=
2
tp
1
tp
2
n
pp121K)t(y
21
o
ee
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ζ = 1
ζ = 2
unit step
response
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ζ = 4
ζ = 12
unit step
response
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The under damping case
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In the case 0 < ζ < 1,
the unit step response
can have many different forms, depending on the
values of ζ (damping coefficient),
ωn
(natural frequency) e Ko (gain)
0t,tsen1
tcos1K)t(y d2
d
tn
o >
ω⋅
ζ−ζ+ω−= ζω−
e
Observe that ωd
depends on ζ and ωn
2
nd 1 ζ−⋅ω=ω (damping frequency)
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ζ = 0,132
ωn = 0,57
ζ = 0,1
ωn = 2
unit step
response
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ζ = 0,25
ωn = 1
ζ = 0,5
ωn = 0,2
unit step
response
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ζ = 0,65
ωn = 2
ζ = 0,72
ωn = 0,8
unit step
response
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ζ = 0,8
ωn = 1,4
ζ = 0,85
ωn = 0,7
unit step
response
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ζ = 0
ωn = 0,2
unit step
response
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Now let us concentrate in the
case 0 < ζ < 1 and calculate
some parameters.
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calculate some parame-
ters/variables for y(t)
the unit step response
of the 2nd order system.
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steady state output
yss
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yss = steady state response
( steady state output )
)s(Rs2s
K)s(Y
2
nn
2
2
no ⋅ω+ζω+
ω=
o
2
nn
2
2
no
0s
ostss
K
s
1
s2s
sKlim
)s(Yslim)t(ylimy
=
=⋅ω+ζω+
ω=
=⋅==
→
→∞→
yss = Ko
s
1)s(R =
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oss Ky =
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rising time
tr
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tr rising timetime it takes the step response y(t)
to reach the final value yss = Ko for
the first time.
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oo Ktsen1
tcos1K)t(y rd2
rd
t
rrn =
ω⋅
ζ−ζ+ω−= ζω−
e
tr = rising time
0tsen1
tcos rd2
rdrtn =
ω⋅
ζ−ζ+ωζω−
e
1
ζζ−−=ω
2
rd
1)t(tg
n
d
ζωω−=
Is the instant of time that y(t) reaches the final value Ko
for the first time.
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tr = rising time
Depends on the values of ζ ( damping coefficient ),
and of ωn ( natural frequency )
d
2
d
n
d
r
1arctgarctg
tω
ζζ−−
=ω
ζωω−
=
tr = arctg(– ωd /ζω n) / ωd
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d
n
d
r
arctg
tω
ζωω−
=
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peak time
tp
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tp peak time is the instant of time in
which the step response y(t) reaches
the first peak.
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ω⋅
ζ−ωζ+ωω−−
+
ω⋅
ζ−ζ+ω⋅ζω==′
ζω−
ζω−
rd2
drdd
t
rd2
rd
t
n
tcos1
tsen
tsen1
tcosKdt
dyy
n
n
o
e
e
peak time yss = Ko ( gain )
ζ ( damping coefficient ),
ωn ( natural frequency )
ωn
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)tcostsen
tsen1
tcosKdt
dy
dndd
d2
n
2
dn
tn
o
ω⋅ζω−ωω+
+ω⋅
ζ−ωζ+ωζω⋅= ζω−
e
ζ−ζ−ω+
ζ−ωζ⋅ω⋅⋅= ζω−
2
2
n
2
n
2
d
t
1
)1(
1tsenK n
o e
01
tsenK2
nd
tn
o =ζ−
ω⋅ω⋅⋅= ζω−e
Time domain analysis – 2nd order systems ______________________________________________________________________________________________________________________________________________________________________________________
peak time
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0dt
dy =
d
ptωπ=
0tsen d =ω
L,3,2,,0td πππ=ω
tp = π / ωd
Time domain analysis – 2nd order systems ______________________________________________________________________________________________________________________________________________________________________________________
peak time
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d
ptωπ=
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overshoot
Mp
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Mp overshoot
it is the percentage above of
the final value yss that is
reached by the first peak.
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The overshoot Mp can be expressed as a
value between 0 and 1.
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Or, instead, it can be expressed as a
value between 0% and 100%.
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0 ≤ overshoot Mp ≤ 1
or
0% ≤ overshoot Mp ≤ 100%
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o
o
ζ1
t
o
pK
K)(senζ
)(cos1K
M2
pnζ −
π−π−
=−
ω−e
overshoot
ss
ssmaxp
y
yyM
−=o
op
pK
K)t(yM
−=ou
( )
o
o
/
oop
K
KKKM
dnζ −+=
ωπω−e
– 1 0
yss = Ko ( gain )
ζ ( damping coefficient ),
ωn ( natural frequency )
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o
2
op
K
KM
ζ1
πζ
−
−
= e
2
p
ζ1
πζ
M−
−
= e
Mp depends only on ζ ( damping coefficient )
Time domain analysis – 2nd order systems ______________________________________________________________________________________________________________________________________________________________________________________
overshoot
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%100M2
p
ζ1
πζ
×= −
−
e
overshoot
Mp depends only on ζ ( damping coefficient )
2
p
ζ1
πζ
M−
−
= eor
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2
p
ζ1
πζ
M
−
−
= e
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settling time
ts
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ts settling time
is the time required for the step
response y(t) to reach and remain
within a given error band around
the final value yss.
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This error band can
be of 5% above and
5% below of the
final value yss.
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or, instead, from
2% above and 2%
below the final
value yss.
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The settling time is obtained from the equations of ye(t), the
curves that encompass y(t).
[ ] 0t,1K)t(yt
en
o >±= ζω−e
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n
s
3%)5(t
ζω=
n
s
4%)2(t
ζω=
That is, the settling time ts is obtained by calculating
ye(ts) ≈ 1,05 Ko
for the case of ts with 5% tolerance, and
ye(ts) ≈ 1,02 Ko
for the case of ts with 2% tolerance,
obtaining the following values:
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n
s
3%)5(t
ζω=
n
s
4%)2(t
ζω=
ts(5%) = 3 / ζωn
ts(2%) = 4 / ζωn
hence:
and
Análise no domínio do tempo - Sistemas de 2ª ordem______________________________________________________________________________________________________________________________________________________________________________________
Note that the settling time
tr is inversely proportional
to ζωn, which is the
distance of the real part of
the poles to the origin.
settling time
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n
s
3%)5(t
ζω=
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n
s
4%)2(t
ζω=
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However, if ζ < 0 then:
Note that we have seen cases in which
the system is unstable
0 < ζ < 1
ζ = 1
ζ > 1
that is:
ζ > 0
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ζ < 0 → unstable system (an example)
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ζ < 0 → unstable system (another example)
Time domain analysis – 2nd order systems ______________________________________________________________________________________________________________________________________________________________________________________