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7. Reduction of Multiple Subsystem
8. System Response – Poles/ Zeros, Second Order
System, Steady State Error, Stability Analysis
Muhammad Faizal bin Ismail
Dept. of Electrical Engineering
PPD, UTHM
013-7143106
7. Reduction of Multiple Subsystem
• Subsystem is represented as a block with an input, output and transfer function.
• Many systems are composed of multiple subsystems.
• When multiple subsystem are interconnected, new element must be added like summing junction and pickoff points.
• The main purpose of reduction multiple subsystem is to reduce more complicated systems to a single block.
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Component of a block diagram for a
linear, time-invariant system
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Example 1Cascade form
7. Reduction of Multiple Subsystem
Example 2Parallel form
7. Reduction of Multiple Subsystem
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Example 3
Cascade
Eliminate feedback
Cascade
Example 4
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8. System response
• Time is an independent variable mostly used in a control system evaluation, also referred as time response.
• In order to find a system response, 2 steps are commonly used:
1. differential equation
2. inverse Laplace transformation
8. System response (cont.)
• In general, system response contains 2 parts:
1. Transient response (natural response)
• Part of the time response that goes to zero as time becomes very large.
2. Steady state response (forced response)
• Part of the time response that remains after the transient has died out
0000(t)(t)(t)(t)yyyytttt limlimlimlim tttt ====∞∞∞∞→→→→
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Example 1
Find the response of the system for a step input .
Solution :
Poles, Zeros and System
Response
• Poles is the values of the Laplace transform variable(s) that cause the transfer function to become infinite or any roots of the denominator of the transfer function .
• Zeros is the values of the Laplace transform variable(s) that cause the transfer function to become zero, or any roots of the numerator of the transfer function .
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Example 2
Given the transfer function G(s) in Figure below, a pole exists
at s = -5, and a zero exists at -2. These values are plotted on
the complex s-plane, using x for the pole and О for the zero.
Example 2 (cont.)
Solution:
To show the properties of the poles and zeros, lets applied
unit step response of the system. Multiplying the transfer
function by a step function yields :
( )
( )5ss
2sC(s)
+
+=
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Second Order System
• A system where the closed loop transfer function possesses
two poles is called a second-order system
Second Order System (cont.)
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• Step
input
Second Order System (cont.)
Second Order System (cont.)
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1. Rise Time, Tr, Time required for response to to rise
from 10% to 90% of its final value.
2. Peak Time, Tp, Time required for response the reach
the first peak of overshoot.
3. Max overshoot, Mp, Maximum peak value of the
response curve measured from unity.
4. Settling time, Ts, Time required for the response
curve to reach and stay within the range about the
final value of sized specified by percentage of the
final value( 2% or 5%)
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Second Order System (cont.)
Example 3
The transfer function of a position control system is given by:
Determine:
Rise time, tr
Peak time, tp
Percent overshoot, %Mp
Settling time ts for 2% criterion
7505.27
750
)(
)(2
++=
sss
s
i
o
θ
θ
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Solution:
27.39 rad/s
2 x 27.39
Example 3
Example 3
Solution:
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Steady State Error
• Steady-state error is the difference between the
input and the output for prescribed test input as t �
∞.
• Test inputs used for steady-state error analysis are:
– Step Input
– Ramp Input
– Parabolic Input
• Figure has a step input and two possible output. Output 1 has
zero steady-state error, and Output 2 has a finite steady-state
error e2(∞).
Steady State Error
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• The formula to obtain steady-state error is as
follow:
Steady State Error (cont.)
Step input
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Ramp Input
Parabolic Input
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Stability Analysis
• A system is stable if every bounded input yields a bounded output.
• A system is unstable if any bounded input yields an unbounded output.
• A system is marginally stable/neutral if the system is stable for some bounded input and unstable for the others (as undamped) but remains constant or oscillates.
Stability Analysis in the Complex
s-Plane
There have 3 condition of poles location that indicates transient response, which is:
1. Stable systems have closed-loop transfer function with poles only in the left half-plane(LHP).
2. Unstable systems have closed loop transfer function with at least one pole in the right half-plane (RHP).
3. Marginally stable systems have closed loop transfer function with only imaginary axis poles.
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Stability Analysis in the Complex
s-Plane
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Example 4
Determine whether the unity feedback system below is stable
or unstable.
Solution:
Let equation = 0 to find the poles,
)2)(1(
31
)2)(1(
3
)()(1
)()(
+++
++=
+=
sss
sss
sHsG
sGsT
3)2)(1(
3
+++=
sss323
323
+++=
sss
032323
=+++ sss
Thus, the roots of characteristic equation are
-2.672
-0.164 + j1.047
-0.164 - j1.047
Plot pole-zero on the s-plane and sketch the response of the system >
all of the roots are locate at left half-plane, therefore the system is
stable.
Example 4 (cont.)
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Determine the stability of the system shown below.
Solution:
)2)(1(
71
)2)(1(
7
)()(1
)()(
+++
++=
+=
sss
sss
sHsG
sGsT
7)2)(1(
7
+++=
sss 723
723
+++=
sss
Example 5
Let equation = 0 to find the poles,
Thus, the roots of characteristic equation are,
-3.087
0.0434 + j1.505
0.164 - j1.505
Plot pole-zero on the s-plane and sketch the response of the system >
only one of the roots are locate at left half-plane and the others 2
roots locate at right half-plane, therefore the system is unstable
072323
=+++ sss
Example 5 (cont.)