Transcript Mathematical Models of Systems

‫بسم ا‪ ...‬الرحمن الرحيم‬
‫سیستمهای کنترل خطی‬
‫پاییز ‪1389‬‬
‫دکتر حسین بلندي‪ -‬دکتر سید مجید اسما عیل زاده‬
Recap.
Signal Flow Graph:
– Algebra,
– Mason Rule,
Introduction to Time Domain
Analysis
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• Time Domain Performance
Specification
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Outline
•
•
•
•
•
Introduction
Test Input Signals
Performance of a first-order system
Performance of a second-order system
Effects of a Third Pole and a Zero on the SecondOrder System Response
• Estimation of the Damping Ratio
• The s-plane Root Location and the Transient
Response
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Introduction/1
• Objectives
The ability to adjust the transient and
steady-state response of a feedback
control system is a beneficial outcome
of the design of control systems.
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Introduction/2
• One of the first steps in the design
process is to specify the measures of
performance.
We introduce the common time-domain
specifications such as percent
overshoot, settling time, time to
peak, time to rise, and steady-state
tracking error.
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Introduction/3
• We will use selected input signals
such as the step and ramp to test the
response of the control system.
• The correlation between the system
performance and the location of the
system transfer function poles and
zeros in the s-plane is discussed.
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Introduction/4
• We will develop valuable relationships
between the performance specifications
and the natural frequency and damping
ratio for second-order systems.
• Relying on the notion of dominant poles,
we can extrapolate the ideas associated
with second-order systems to those of
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higher order.
Test Input Signal
• Since the actual input signal of the system
is usually unknown, a standard test input
signal is normally chosen. Commonly
used test signals include step input, ramp
input, and the parabolic input.
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• For example in a radar tracking system for
antiaircraft missiles, the position and speed of
the target to be tracked may vary in an
unpredictable manner.
• Mobile robot with collision avoidance, …
• From the step function to the parabolic
function, they become progressively faster
with respect to time.
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• For the purposes of analysis and design, it is
necessary to assume some basic types of
test inputs so that the performance of a
system can be evaluated (CLASSFICATION
OF INPUTS).
• The responses due to these inputs allow the
prediction of the system’s performance to
other more complex inputs.
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General form of the standard test
signals
r(t) = t
n
n+1
R(s) = n!/s
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Test signals r(t) = A tn
n=0
r(t) = A
R(s) = A/s
n=1
r(t) = At 2
R(s) = A/s
n=2 2
r(t) = At 3
R(s) = 2A/s
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• The ramp signal is the integral of the
step input, and the parabola is the
integral of the ramp input. The unit
impulse function is also useful for test
signal purposes.
• The responses due to these inputs
allow the prediction of the system’s
performance to other more complex
inputs.
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• Step-for representation a stationary
target
• Ramp- for track a constant angular
position (first derivatives are constant)
• Parabolas- can be used to represent
accelerating targets (second derivatives
are constant)
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Test Signal Inputs
Test Signal
Step
position
Ramp
velocity
Parabolic
acceleration
r(t)
R(s)
r(t) = A, t > 0
R(s) = A/s
= 0, t < 0
r(t) = At, t > 0 R(s) = A/s2
= 0, t < 0
r(t) = At2, t > 0 R(s) = 2A/s3
= 0, t < 0
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Test inputs vary with target type
step
ramp
parabola
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Steady-state error
• Is a difference between input and the
output for a prescribed test input as
t 
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Application to stable systems
• Unstable systems represent loss of
control in the steady state and are
not acceptable for use at all.
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Steady-state error:
a) step input, b) ramp input
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Time response of systems
c(t) = ct(t) + css(t)
The time response of a control system is divided
into two parts:
• ct(t) - transient response
• css(t) - steady state response
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Transient response
• All real control systems exhibit transient
phenomena to some extend before steady
state is reached.
lim ct(t) = 0
for t 
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Steady-state response
• The response that exists for a long time
following any input signal initiation.
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Poles and zeros of a first order system
Css(t)
Ct(t)
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Poles and zeros
•
•
A pole of the input function generates the form of the
forced response ( that is the pole at the origin generated
a step function at the output).
The zeros and poles generate the amplitudes for both
the transit and steady state responses
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Effect of a real-axis pole upon transient
response
A pole on the real axis generate an exponential response
of the form Exp[-t] where - is the pole location on real axis.
The farther to the left a pole is on the negative real axis,
the faster the exponential transit response will decay to zero.
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Evaluating response using poles
K1 K2
K3
K4
C(s)  


s s2 s 4 s 5
Css(t
)
Ct(t)
2t
c(t)  K1  K 2e
4 t
 K 3e
5t
 K4e
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First order system
a
C(s)  R(s) G(s) 
s(s  a)
c(t)  1 e
at
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First-order system response to a unit step
Transient response specification:
1. Time-constant, 1/a
2. Rise time, Tr
3. Settling time, Ts
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Transient response specification
for a first-order system
1.
Time-constant, 1/a
Can be described as the time for (1 - Exp[- a t])
to rise to 63 % of final value.
1.
Rise time, Tr = 2.2/a
The time for the waveform to go from 0.1 to 0.9
of its final value.
3.
Settling time, Ts = 4/a
The time for response to reach, and stay within,
2% of its final value
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Transfer function via laboratory testing
K
G(s) 
(s  a)
K
K
K
a
C(s) 
 a
s(s  a)
s
(s  a)

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Identify K and a from testing
The time for amplitude to reach 63% of its final value:
63 x 0.72 = 0.45, or about 0.13 sec , a = 1/0.13 = 7.7
From equation, we see that the forced response reaches
a steady-state value of K/a =0.72 .
K= 0.72 x 7.7= 5.54
G(s) = 5.54/(s+7.7) .
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Exercise
A system has a transfer function
G(s)= 50/(s+50).
Find the transit response specifications
such as Tc, Tr, Ts.
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Steady-state response
• If the steady-state response of the output does not
agree with the steady-state of the input exactly, the
system is said to have a steady-state error.
• It is a measure of system accuracy when a specific type
of input is applied to a control system.
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Y(s) = R(s) G(s)
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Steady-state error
T(s) = 9/(s + 10)
Y(s) = 9/s(s+10)
-10t
y(t) = 0.9(1- e
)
y(∞) = 0.9
E(s) = R(s) - Y(s)
ess = lim s E(s) = 0.1
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E(s) 
R(s)
E(s) 
K
1
s
R(s)
1 K
1
e(t) 
1 K

e(t)  eK t

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Performance of a second-order system
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Numerical example of the secondorder system
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• A second order system exhibits a wide
range of response.
• The general case: which has two finite
poles and no zeros. The term in the
numerator is simple scale. Changing a
and b we can show all possible
transient responses.
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• Changes in the parameters of a secondorder system can change the form of
the response. Can display characteristic
much like a first-order system or
display damped or pure oscillation for
its transient response.
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• The unit step response then can be
found using C(s)= R(s) G(s),
where R(s) = 1/s, followed by a partialfraction expansion and inverse Laplace
transform
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Overdamped
9
9
C(s)  2

s(s  9s  9) s(s  7.854)(s  1.146)

c(t)  1 0.171e7.854 t  1.17e1.146t
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Underdamped
9
9
C(s)  2

s(s  2s  9) s(s  1 j 8)(s  1 j 8)
8
c(t)  1 e (cos 8t 
sin 8t)
8
t

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Undamped
9
C(s) 
s(s 2  9)
c(t) 1cos3t

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Critically damped
9
9
C(s)  2

s(s  6s  9) s(s  3) 2
3t
c(t)  1 3te

e
3t
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fig_04_11
Step response for second order system
cases
damping
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Summary
• Overdamped
Poles: Two real at - 1, - 2
• Underdamped
Poles: Two complex at - d + jd, - d - jd
• Undamped
Poles: Two imaginary at + j1, - j1
• Critically damped
Poles: Two real at - 1,
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