Transcript
Chapter 2 – Analog Control System (cont.)
Reduction of Multiple SubsystemSystem Response – Poles/ Zeros, Second Order System, Steady State Error, Stability Analysis
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.
Component of a block diagram for a linear, time-invariant system
Example 1
Cascade form
Example 2
Parallel form
Example 3
Example 4
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 equation2. inverse Laplace transformation
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
0(t)yt
limt
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 .
Example 1
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.
cont.
Solution:Solution:
To show the properties of the poles and zeros, lets applied unit step 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 response of the system. Multiplying the transfer function by a step function yields :yields :
Thus :Thus :
5s
3/5s
2/55ss2s
C(s)
5te53
52
c(t)
cont.
From the development summarize, the following From the development summarize, the following conclusions can be drawn:conclusions can be drawn:
Second Order System
A system where the closed loop transfer function possesses two poles is called a second-order system
cont.
cont.
Step input
cont.
cont.
cont.
Example 1
The transfer function of a position control system is given by:
Determine:
Rise time, tr
Peak time, tpPercent overshoot, %Mp
Settling time ts for 2% criterion
7505.27
750
)(
)(2
sss
s
i
o
cont.
Solution:
27.39 rad/s
2 x 27.39
cont.
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
cont.
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().
cont.
The formula to obtain steady-state error is as follow:
Step input
Ramp Input
Parabolic Input
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.
cont.
Example 1
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
ssssHsG
sGsT
3)2)(1(
3
sss 323
323
sss
0323 23 sss
cont.
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 2
Determine the stability of the system shown below.
Solution:
)2)(1(
71
)2)(1(
7
)()(1
)()(
sss
ssssHsG
sGsT
7)2)(1(
7
sss
723
723
sss
cont.
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
0723 23 sss
top related