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The output response of a system is the sum of two responses: the forced response (steady-state response) and the natural response (zero input 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 transfer function to become zero, or (2) any roots of the numerator of the transfer function that are common to roots of the denominator.
Out response, Poles, and Zeros
Figure 4.1a. System showing
input and output;b. pole-zero plot
of the system;c. evolution of
a system response.
1. A pole of the input function generates the form of the forced response.
2. A pole of the transfer function generates the form of the natural response.
3. A pole on the real axis generates an exponential response.
4. The zeros and poles generate the amplitudes for both the forced and natural responses.
)5)(4)(2(
3
sss
sssR
1)( )(sC
542)( 4321
s
k
s
k
s
k
s
ksC
Natural response
Forced response
ttt ekekekktc 54
43
221)(
First-Order Systems
Figure 4.4a. First-order system
b. pole plot
Figure 4.5First-order system
response to a unit step
The time constant can be described as the time for to decay to 37% of its initial value. Alternately, the time is the time it takes for the step response to rise to of its final value.
The reciprocal of the time constant has the units (1/seconds), or frequency. Thus, we call the parameter the exponential frequency.
)()()()(
ass
asGsRsC
ate
atnf etctctc 1)()()(
37.01
1
eeat
at
63.037.011)(1
1
at
at
atetx
a
Rise time: Rise time is defined as the time for the waveform to go from 0.1 to 0.9 of its final value.
Settling time: Settling time is defined as the time for the response to reach, and stay within, 2% (or 5%) of its final value.
Second-Order Systems
Figure 4.7Second-order
systems, pole plots,and step responses
1. Overdamped response:
Poles: Two real at
Natural response: Two exponentials with time constants equal to the reciprocal of the pole location
2. Underdamped responses:
Poles: Two complex at
Natural response: Damped sinusoid with and exponential envelope whose time constant is equal to the reciprocal of the pole’s real part. The radian frequency of the sinusoid, the damped frequency of oscillation, is equal to the imaginary part of the poles
21,
tt ekektc 2121)(
dd j
)cos()( tAetc dtd
3. Undamped response:
Poles: Two imaginary at
Natural response: Undamped sinusoid with radian frequency equal to the imaginary part of the poles
4. Critically damped responses:
Poles: Two real at
Natural response: One term is an exponential whose time constant is equal to the reciprocal of the pole location. Another term is the product of time and an exponential with time constant equal to the reciprocal of the pole location
1j
)cos()( 1 tAtc
1
tt tekektc 1121)(
Figure 4.10Step responses for second-order
system damping cases
Figure 4.8Second-order step response components
generated by complex poles
Natural Frequency: The natural frequency of a second-order system is the frequency of oscillation of the system without damping.
Damping Ratio: The damping ratio is defined as the ratio of exponential decay frequency to natural frequency.
Consider the general system:
Without damping, bass
bsG
2)(
bbs
bsG n
2)(
b
a
radfrequencyNatural
frequencydecaylExponentia
n
2/||
sec)/(
2
4
2,
2
4
2
2
2
2
1
abj
as
abj
as
na 2 22
2
2)(
nn
n
sssG
Figure 4.11Second-order response as a
function of damping ratio
Underdamped Second-Order Systems
)1()(
1)(
)2()(
222
2321
22
2
nn
nn
nn
n
s
KsK
s
K
ssssC
)1()(
11
)(11
222
2
2
nn
nn
s
s
s
ttetc nn
tn 2
2
2 1sin1
1cos1)(
Step response
Taking the inverse Laplace transform
te ntn 2
21cos
1
11
where
2
1
1tan
Figure 4.13Second-order underdamped
responses for damping ratio values
Peak time: The time required to reach the first, or maximum, peak. Percent overshoot: The amount that the waveform overshoots the
steady-state, or final, value at the peak time, expressed as a percentage of the steady-state value.
Settling time: The time required for the transient’s damped oscillations to reach and stay within (or ) of the steady-state value.
Rise time: The time required for the waveform to go from 0.1 of the final value to 0.9 of the final value.
Evaluation of peak time:
%2 %5
22
2
2)()]([
nn
n
ssssCtcL
)1()(
11
222
2
2
nn
nn
s
tetc ntn n 2
21sin
1)(
Setting the derivative equal to zero yields
Peak time: 2
2
11
n
n
ntornt
21
n
pt
Figure 4.14Second-order underdamped
response specifications
Evaluation of percent overshoot ( ):
Evaluation of settling time:
The settling time is the time it takes for the amplitude of the decaying
sinusoid to reach o.o2, or
,
where is the imaginary part of the pole and is called the damped
frequency of oscillation, and is the magnitude of the real part of the
pole and is the exponential damping frequency.
100% max
final
final
c
ccOS
)1/(max
2
1)( etcc p 100% )1/( 2
eOS
02.01
12
tne
nnst
4102.0ln( 2
dn
pt
21
dnst
44
d
d
Figure 4.17Pole plot for an underdamped
second-order system
)1)(1(2)(
22
2
22
2
nnnn
n
nn
n
jsjssssG
Figure 4.18Lines of constant peak time,Tp , settlingtime,Ts , and percent overshoot, %OS
Note: Ts2 < Ts1 ;Tp2 < Tp1; %OS1 <%OS2
Figure 4.19Step responses of
second-orderunderdamped systems
as poles move:a. with constant
real part;b. with constant imaginary part;
c. with constant damping ratio
dn
pt
21 100% )1/( 2
eOS
2173 nndd jjj
dnst
44
Find peak time, percent overshoot, and settling time from pole location.
, ,
Figure 4.21Rotational mechanical system
Design: Given the rotational mechanical system, find J and D to yield 20% overshoot and a settling time of 2 seconds for a step input of torque T(t).
Under certain conditions, a system with more than two poles or with zeros can be approximated as a second-order system tat has just two complex dominant poles. Once we justify this approximation, the formulae for percent overshoot, settling time, and peak time can be applied to these higher-order systems using the location of the dominant poles.
rdn
dn
s
D
s
CsB
s
AsC
22)(
)()(
Figure 4.23Component responses of a three-pole system:
a. pole plot;b. component responses: nondominant pole is near dominant second-order pair (Case I),
far from the pair (Case II), and at infinity (Case III)
Remark:If the real pole is five timesfarther to the left than the dominant poles, we assumethat the system is represented by its dominant second-order pair of poles.
cs
bcac
bs
cbab
cs
B
bs
A
csbs
assT
)/()()/()(
))(()(
System response with zeros
If the zero is far from the poles, then is large comared to and . ca b
))((
)/(1)/(1)(
csbs
a
cs
bc
bs
cbasT
Figure 4.25Effect of adding
a zero to a two-pole system
)828.21( j
Figure 4.26Step response of a nonminimum-phase system
)()()()( saCssCsCas
Time Domain Solution of State Equations
)()()( tButAxtx
t tAAt dBuexetx0
)( )()0()(
t
dButxt0
)()()0()(
where is called the state-transition matrix.Atet )(
)0()()]0()([)]([ 1xAsIxttx LL
)(])[( 1 tAsI -1L