ContSys1 L5 SDOF RespAll
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Transcript of ContSys1 L5 SDOF RespAll
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Mechanical Engineering Science 8
Dr. Daniil Yurchenko
Response of a second
order system
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Remote position controller
A device which is used to position aninertia load according to a desired angular
position from a remote point.
Input, i
Angular
position,
Output, o
Position
transducer
Error unit
Electric
Motor Inertia, J
Viscous friction, F
Torque, mo
motor
Inertial load
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Remote position controller
Desired angular position input to error unit
Error unit subtracts actual angular position togive error signal
Error signal passed to electric motor whichgenerates a torque directly proportional to errorvoltage applied to it
Such a system is known as a proportional action
controllersince the controlling action (motortorque in this case) is directly proportional tosystem error
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Remote position controller Open loop descriptions of system elements
Motor
Where K is a constant of proportionality & m isthe motor torque
Thus motor torque is proportional to error.
Leads to
Km
Km
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Remote position controller
Applying Newtons second law to inertialload
Where F is viscous friction constant and Jis mass moment of inertia of load
Re-arranging
2
2
dt
dJ
dt
dFm OO
mdt
dF
dt
dJ OO
2
2
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Remote position controller
Applying L-transform and assuming zero ICs
This may be represented by the system blockdiagram in the next slide
FsJsM
2
0 1
][2
2
mLdt
dF
dt
dJL OO
)()()( 002 sMsFssJs
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Remote position controller
The system closed loop equation
GH1
GFunctionTransfer
is
K
+
-
o
Proportional
Controller
FsJs 21Mi
1;2
HFsJs
KG
Oi
KFsJs
K
FsJs
KFsJs
K
i
2
2
20
1
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Remote position controller
The relationship between system outputand the error is
and
Re-arrange to give system characteristicequation:
)()()( sEss iO
)()( 2 sFsJs KsO
)()( 22 sFsJssEKFsJs i
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Standard forms
Similar to the standard form concept westudied when we worked with vibrating
systems. In this case system naturalfrequencies and damping ratios will bemade up of a mixture of mechanical andelectronic components rather than springs,
masses & dampers
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Characteristic equation
Or
Leads to standard form
iFsJssEKFsJs 22 )(
)()(
22
ssJ
F
ssEJ
K
sJ
F
s i
)(2)(2 222
ssssEss innn
where
J
Kn
and
KJ
F
J
Fn
22
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Similarly the governing equation becomes
Or
Leading to the standard form
)()(2
sKsKFsJs iO
)()(2 222 ssss inOnn
)()(2
sJ
K
sJ
K
sJ
F
s iO
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Control system response
This can be defined using 2 parameters,the error and the output O.
The error response can be obtained fromsolving the characteristic equation for aspecified input and the output responsecomes from solving the governingequation for a specific input.
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Control system response
Both the error and the output can each beconsidered to be the sum of two components:
a transient response and a steady stateresponse.
The transient response is the complementaryfunction solution to the describing equationand the steady state response is the particularintegral solution (c.f. first order systems)
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Control system response
Thus for the error:
For a satisfactory control system weshould have:
And
statesteadytransienttotal _
0statesteady _
transientShould be of minimum duration and
may need to have limited overshoot
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Control system response 1)unit step
As before the input is defined as:
We will use the remote position controlleras an example.
First we will investigate the error responseof this system.
00 tti , 01 tti ,
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Control system response 1)unit step
Repeating the characteristic equation:
L-transform of the unit step response is:
Rearranging:
s
tULsi1
)()(
)()( 22 sFsJssEKFsJs i
sKFsJs
FsJssE
1)(
2
2
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Control system response 1)unit step
Thus:
The steady-state error can be easily obtainedby using Theorem 3 (Lecture 3):
Thus,
sKFsJsFsJssE 1)(
2
2
)(lim)()(lim0
ssQqtqst
0lim)(lim)(2
2
00
KFsJs
FsJsssEe
ssstst
0statesteady
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Control system response 1)unit step
The system response will depend on thevalue of
To show that lets get the characteristicequation, which is obtained by assumingzero input:
Thus:02 22 nnss
0)(2 22
sss Onn
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Control system response 1)unit step
This is a quadratic with solution:
Consider 3 cases:1
2 nns
10
1
Underdamped
Critical
1 Overdamped
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Control system response 1)unit step
Case
Both roots are real:
Then
1,1 222
1 nnnn ss
1
11
2)(
)(
22
2
22
2
nnnn
n
nn
n
i
O
ss
sss
s
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Control system response 1)unit step
For unit step response
Which can be written as
Using the table of inverse L-transform
1
ssss
nnnn
nO
1
11
)(22
2
bsasss nO
2
)(
btatnOO aebe
baab
sLt 1
1)()(2
1
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Control system response 1)unit step
Which after simplification gives:
Then the error is:
And the
1
212
21
121)(
se
set
tsts
nO
212
0
21
12)()()(
s
e
s
ettttsts
ni
0)(lim
tt
statesteady
No Vibrations
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Control system response 1)unit step
Then
Which leads to
And for unit step input
1
nnns 12
2,1
22
22
2
2)(
)(
n
n
nn
n
i
O
ssss
s
ss
s
n
nO
1)(
2
2
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Control system response 1)unit step
Then
Then the error is:
And
1
tettt nt
in 1)()()( 0
0)(lim
tt
statesteady
tesLt nt
OOn 11)()( 1
No Vibrations
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Control system response 1)unit step
Then we have two complex conjugate roots
Which leads to
And for unit step input
10
22
2
2)(
)(
nn
n
i
O
sss
s
ssss
nn
nO
1
2
)(22
2
222,1 111 nnnn is
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Control system response 1)unit step
Then
te
sLt n
t
OO
n
2
2
1 1sin1
1)()(
10
21 1tan
21
nd
The system response is oscillatory with
damped natural frequency and decays
exponentially
d
ttet dd
t
On
sin
1cos1)(
2
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Control system response 1)unit step
The error is
And
t
e
ttt n
t
i
n
2
20 1sin1)()()(
0)(lim
tt
statesteady
10
0
2sin1)(
tt nO
2
sin)( tt n
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Control system response 1)unit step
21 1tan
n
ncos
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Control system response 1)unit step
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samethestaysst
samethestayspM
samethestaysn
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Control system response 1)unit step
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Control system response 1)unit step
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Control system response 1)unit step
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Control system response 1)unit step
Mp - maximum percent overshoot
where tp - peak time
td - delay time. The time required for theresponse to reach half of the final value.
tr - raise time. The time required for theresponse to rise from 10% to 90% or from0 to 100% of its final value
%100)(
)()(
0
00
pp
t
M
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Control system response 1)unit step
tssettling time. The time required for theresponse curve to reach and stay within arange about the final value
(usually 2% or 5%)
In control system design we are largely
interested in obtaining a rapid responsewhilst minimising overshoot.
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Control system response 1)unit step
The peak time may be obtained bydifferentiation
tetetete
ttette
ttedt
d
dt
td
dnt
dd
t
d
tnd
t
n
d
d
dd
t
dd
t
n
dd
tO
nnnn
nn
n
cos1
1sinsin
1cos
cos1sinsin1cos
sin1
cos1)(
2
2
2
2
22
2
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Control system response 1)unit step
Equating to zero yields
The first peak corresponds to
p
nn
ttdd
t
d
tnO tete
dt
td
sinsin
1
0)(
2
2
0sin1 2
2
pdd
n t
0sin pdt ,...2,1,0, nntpd
d
pt
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Control system response 1)unit step
The first peakd
pt
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Control system response 1)unit step
The maximum overshoot occurs at tp=/d
sin1
sin1
1sin1
1)()(
22
2
d
n
d
n
dn
ee
et
d
dOO
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Control system response 1)unit step
2
22
221 1tan
1tan
1tan
22
2
2
2
2
2
2
2 1sin11
1
1
tan1tansin
dn
dn
eet OO
sin1
)()(2
%100
21
eMp
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Control system response 1)unit step
By increasing K we increase the overshoot
%10021
eMp KJF
2 K
K
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Control system response 1)unit step
The rise time tr ,
rd
t
rO t
e
t
rn
sin111)( 2
1)( rO t
0sin1 2
rd
t
te rn
0sin rdt
,...2,1,0, nntrd 21
nd
rt
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Control system response 1)unit step
By increasing K we decrease the rise time
K
21
nd
rt
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Control system response 1)unit step
In MATLAB
dcsbsa
s
s
i
O
2
)(
)(
][
]00[
dcbden
anum
),( dennumstep
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Unit ramp response of RPC
In this case we present the system with aunit ramp input:
The L-transformtti )(
2
1)()(
s
tLs ii
22
2 1)(
sKFsJs
FsJssE
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Unit ramp response of RPC
222
2 1
2
2)(
sss
sssE
nn
n
J
Kn
KJ
F
2
02
2
2lim
)(lim)(
222
2
0
0
nnn
n
s
sstst
s
s
ss
ss
ssEe
This is aconstant
steady
state error
)(lim)()(lim0
ssQqtqst
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Unit ramp response of RPC
Transient error response.
To find this we need to examine thecomplementary solution to the characteristic
equation. i.e. the solution to:
This is identical to the equation we used to find
the complementary function solution for the stepinput. Therefore the complementary solution isidentical to that case
02 22 nnss
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Unit ramp response of RPC
For two real roots
No oscillatory response.
1
2222 1
11)(
ssss
nnnn
nO
22
2
2)(
)(
nn
n
i
O
sss
s
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Unit ramp response of RPC
For two real roots
It results in
No oscillatory response
1
nnn
s 122,1
22
2 1
)( sss n
n
O
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Unit ramp response of RPC
Then we have two complex conjugate roots
Which leads to
Partial fraction expansion
10
222
2 1
2
)(
sss
s
nn
nO
222,1 111 nnnn is
222222
2
2
1
2 nnnn
n
ss
DCs
s
BAs
sss
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Unit ramp response of RPC
10
222
2322
222
2
)2(
)2)((
2 sss
DsCsssBAs
sss nn
nn
nn
n
220
21
2
3
:
02:
02:
0:
nn
nn
n
Bs
BAs
DBAs
CAs
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Unit ramp response of RPC
10
222
2 2
242)(
nnn
nn
n
nO
ss
s
s
ss
212
2
2
2
2
2222
2
2
1
1tan,1sin
1
2
1sin1
142
2
2
2
1421)(
te
tet
ss
s
ssssLt
n
t
n
n
t
nn
nnnnnn
O
n
n
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Unit ramp response of RPC
When t goes to infinity the second term
vanishes
10
12cos
,1sin1
2)(
21
2
2
te
tt n
n
t
n
O
n
te
ttt nn
t
n
i
n
2
20 1sin
1
2)()()(
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Unit ramp response of RPC
The error is
And
n
tstatesteady t
2)(lim
10
te
ttt nn
t
n
i
n
2
2
0 1sin
1
2)()()(
J
Kn
KJ
F
2
K
F
K
J
KJ
F
n
stst
2
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Unit ramp response of RPC
At t=0
If then
0sin1
120 2
nn
0 0t
te
ttt nn
t
n
i
n
2
20 1sin
1
2)()()(
)1(4sin,12cos 222222
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Unit ramp response of RPC
Either the output response or the errorresponse may be plotted against time