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7/31/2019 Time Respons
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System Response
This module is concern with the response ofLTI system.
L.T. is used to investigate the response of first
and second order systems. Higher ordersystems can be considered to be the sum ofthe response of first and second order system.
Unit step, ramp, and sinusoidal signal playimportant role in control system analysis. It istherefore we will investigate this signals.
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(1)
2. Convolution
Output of LTI system is the convolution
of its input and its impulse response:
h(t)r(t) c(t)
t
dthrthtrtc )()()(*)()(
t
dhtr )()(
Taking the L.T. of (1) yields
C(s) = R(s)H(s) (2)Where C(s)is the L.T. of c(t)
R(s) is the L.T. of r(t)
H(s) is the L.T. of h(t)
H(s) is called the transfer function (T.F)
3. Derivative
If the input is r(t) then the output is c(t)
wherer(t) denotes the derivative ofr(t)
Review of some LTI properties
We will express system as in figure below,system input is r(t), output is c(t), and
impulse response is h(t)
4. Integral
If the input is r(t)dtthen the output isc(t)dt
1. Impulse response
Impulse response, denoted by h(t), is the
output of the system when its input is
impulse (t). h(t) is called the impulse
response of the system or the weighting
function5. Poles and ZeroT.F.isusually rational and therefore can be
expressed as N(s)/D(s). Poles is the values
ofs resulting T.F to be infinite. Zeroes is
the values ofs resulting T.F to be zero
System Response
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Poles, Zeros, and System Response The output response of a system is the sum of two
responses: theforced response andthe natural response The poles of a transfer function are (1) the values of
the Laplace transform variable, s, that cause the
transfer function to become infinite or (2) any roots of
the denominator of the transfer function that are
common to roots of the numerator
The zeros of a transfer function are (1) the values of the
Laplace transform variable, s, that cause the transferfunction to become zero, or (2) any roots of the
numerator of the transfer function that are common to
roots of the denominator.
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Pole dan Zero Pada System Orde Satu
Fungsi alih sistem orde 1,
Untuk mencari A dan B,
Persamaan respon sistemnya,
Secara grafis,
INTRO
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Time Response of the First Order Systems.
We can found the differential equation
first we write (1) as
The diff. Eq. is the inverse L.T. of (2)
Now we take the L.T of (3) and include
the initial condition term to get
1)(
)()(
s
K
sR
sCsG (1)
)(
)(
1sR
KsCs
(2)
)()(1
)( tr
Ktc
tc (3)
)(
)(
1)0()( sR
KsCcssC (4)
Solving for C(s) yields
)/1(
)R(s)/(
)/1(
)0()(
s
K
s
csC (5)
Note that the initial condition as an
input has a Laplace transform ofc(0),which is constant.
The inverse L.T of a constant is impulse
(t). Hence the initial condition appearsas the impulse function
Here we can see that the impulse
function has a practical meaning, eventhough it is not a realizable signal
1
s
K
1
1
s
R(s)
c(0)
+
+
C(s)
Here we will investigate the time response
of the first order systems.
The transfer function of a general first
order system can be written as:
The eq. can be shown in the block diagram as
shown in the figure bellow.
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Time Response of the First Order Systems.
Unit step response
For unit step input R(s)=1/s, then
Since we usually ignore the initial
condition in block diagram, we use the
system block diagram as shown bellow.
)()/1(
)/()( sR
s
KsC
(1)
Suppose that the initial condition is
zero then
1
s
KR(s) C(s)
/1s
1
)/1(
)/()(
s
K
s
K
s
KsC (2)
Taking the inverse L.T of (2) yields
)1()( t
eKtc (3)
The first term originates in the pole of input
R(s) and is called the forced response or steady
state response The second term originates inthe pole of the transfer function G(s) and is
called the natural response Figure below plot
c(t)
t
c(t)
K1/
)1( t
eK
The final value or the steady state value ofc(t)
is K that is
lim c(t)= Kt
c(t) is considered to reach final value after
reaching 98% of its final value. The parameter is called the time constant. The smaller the time
constant the faster the system reaches
the final value.
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Time Response of Second Order System
22
2
2)(
)()(
nn
n
sssR
sCsG
(1)
The standard form second order system is
Where = damping ratio, n = naturalfrequency, or undamped frequency.
Consider the unit step response of this system
sssRsGsC
nn
n
)2()()()(
22
2
(2)
Case 1: 0
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General Second-Order System
Natural Frequency,n
The natural frequency of a second-order system is the frequency of
oscillation of the system without damping
Damping Ratio,
So,
= =
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PROBLEM:
Given the transfer function of Eq. , find and n
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Time Response Specification in Design
The typical of unit step response of a system
is as in figure below
nt
c(t)
Mpt
1.0
0.9
0.1
Tr Tp
1+ d
1d
css
Ts
Rise time Tr
is the time required for the
response to rise from 10% to 90% of the final
value css. Mpt is the peak value, and Tpis the
time required to reach Mpt.
Mptcsspercent overshoot =100%
css
Settling time, Ts,is the time required for the
output to settle within a certain percent of its
final value. Common values are 5% and 2%.
Settling time is proportional to the ,
kTs = k =
n
If 2% is used to specify the settling value thenk=4. T
r
, Ts
,and css
, are equally meaningful for,over, critically, or under-damped cases, whileTp,and Mp,are meaningful only for under-damped cases.
Under-damped cases.
To find Tp,we differentiate c(t) and equalizing
the result to zero and solve the equation for t.
We will find that2
1/1
1
eMT p
n
p
%100shootover%2
1/ e
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Time Response Specification in DesignRecall that
1
or
1
pn
n
p TT
and
%100shootover%2
1/ eBoth %over-shootand nTp are functionsonly of and can be plotted as in thefollowing figure
%over-shoot
n
Tp
100
80
60
40
20
0
5.0
4.6
4.2
3.8
3.4
3.00.2 0.4 0.6 0.8 1
%over-s
hoot
nTp
Example:
Servomotor is used to control the position of plotter
pen as in the following figure.
a
a
a
a
Kss
K
ssK
ssK
sT 5.02
5.0
)2(/5.01
)2(/5.0
)( 2
ann K5.0and222
Here we have
Suppose that we want to have =1, the fastest
response with no overshoot thenn= 1 and Ka=2.
Note that the settling time is
Tp= 4/ n= 4sIt is not fast enough. If we want the faster response
First we have to choose a different motor, then
Redesign the compensator.
0.5/s(s+2)
motorR(s)
Kp
Amplifier
+
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TRANSIENT RESPONS ANALYSIS WITH MATLAB(2)
Example 2 : find the unit impuls- respons for following equation,
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Sumber Pustaka
Norman S. Nise, Control Systems Engineering, Fourth Edition,
2004, Wiley & Sons..