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Dr. Mohd Azri Mohd Izhar
Rom No.: F306M
Email: [email protected]
SMJE 4243
Control Systems Design
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Outline
Introduction Transfer Function
Closed-Loop Control
Transient and Steady-State Response First-Order Systems
Second-Order Systems
Higher Order Systems
Stability Criterion
Steady-State Errors
Frequency Response
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Why Control System?
Good Improved control is a key enabling technology underpinning:
- generally
- Convenience and Comfort
- doctor
- Precision of tools
- industry
- Enhance product quality
- Waste minimization
- Higher safety margins
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What is Control System?
Open-Loop Control System (No feedback)
Input Response
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Closed-Loop Control System
Feedback Feedback is a key tool that can be used to modify
the behavior of a system. This behavior altering effect of
feedback is a key
mechanism that control engineers exploit deliberately to achieve
the objective of acting on a system to ensure that the desired
performance specifications are achieved.
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Closed-Loop Control System
Closed-Loop versus Open-Loop Control Systems?
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Example Control System
Autopilot Control System
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Design and Compensation of Control Systems
Performance Specifications
- Requirement on control system
- Can be in the form of transient response, steady-state
requirements or in frequency response
- Specifications must be given before the design process
begins
System Compensation
- Adjusting system for satisfactory performance
- A device inserted into the system for the purpose of
satisfying the specifications is called a compensator
- The compensator compensates for deficient performance of the
original system
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Design and Compensation of Control Systems
Design Procedures
- Set up a mathematical model of the control system and adjust
the parameters of a compensator
- The most time-consuming part of the work is the checking of
the system performance by analysis with each adjustment of the
parameters
- After that, designer must construct a prototype and test the
open-loop system
- If absolute stability of the closed loop is assured, the
designer closes the loop and tests the performance of the closed
loop
- Redesign until the system meets the specifications
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Transfer Function
Used to characterize the input-output relationships of
components or systems that can be described by linear,
time-invariant, differential equations
Definition: Ratio of the Laplace transform of the output
(response function) to the Laplace transform of the input (driving
function) under the assumption that all initial conditions are
zero
Inverse Laplace transform of G(s) gives the impulse response of
the system
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Transfer Function
Closed-loop transfer function
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Automatic Controllers
An automatic controller compares the actual value of the plant
output with the reference input (desired value), determines the
deviation, and produces a control signal that will reduce the
deviation to zero or to a small value.
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Disturbance Rejection
Assume reference input is zero, Assume disturbance is zero,
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The time response of a control system consists of two parts:
1. Transient response
- from initial state to the final
state purpose of control
systems is to provide a desired
response.
2. Steady-state response
- the manner in which the
system output behaves as t
approaches infinity the error
after the transient response has
decayed, leaving only the
continuous response.
Time Response
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First-Order Systems
Unit-Step Response Example
The smaller time constant T, the faster the system response
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First-Order Systems
Unit-Ramp Response Example
Unit-Impulse Response Example
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Second-Order Systems
Second-order systems exhibit a wide range of responses which
must be analyzed and described. Whereas for a first-order system,
varying a single parameter changes the speed of response, changes
in the parameters of a second order system can change the form of
the response. For example: a second-order system can display
characteristics much like a first-order system or, depending on
component values, display damped or pure oscillations for its
transient response.
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Second-Order Systems
Closed-loop transfer function:
Example: Servo System
Standard form of the second-order system
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Second-Order Systems
Standard form of the second-order system
n- referred to as the un-damped natural frequency of the second
order system, which is the frequency of oscillation of the system
without damping.
- referred to as the damping ratio of the second order system,
which is a measure of the degree of resistance to change in the
system output.
Poles; Poles are complex if < 1!
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Second-Order Systems
- According the value of , a second-order system can be set into
one of the four categories:
1. Overdamped - when the system has two real distinct poles (
>1). 2. Underdamped - when the system has two complex conjugate
poles (0
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Unit-step response curves with various values of
The fastest response?
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MATLAB
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Time-Domain Specifications
(A) For transient response, we have 5 specifications:
(B) Steady State Response
(a) Steady state error
1. Delay time, td : Time required for the response to reach half
the final value for the first time
2. Rise time, tr : Time required for response to rise from 10%
to 90% of final value (overdamped) : Time required for response to
rise from 0% to 100% of final value (underdamped)
3. Peak time, tp : Time required for the response to reach the
first peak of the overshoot
4. Max. % overshoot , Mp : Maximum peak value of the response
curve measured from the steady state value. Indirectly related to
stability of system
5. Settling time, ts : Time required for the response curve to
reach and stay within a range about the final value of size
specified by absolute percentage of the final value (usually 2% or
5%)
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Time-Domain Specifications
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Time-Domain Specifications
(a) tr rise time =
12
(b) tp peak time =
(c) MP max overshoot
(d) ts settling time (2% error) =
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n
n
4
For unit-step response (for underdamped system):
If final value not unity,
=n is the attenuation
d =n 12 is the
damped natural frequency
For %, multiply with 100%
For rapid response, n must be large
All equations above valid only for the standard second-order
system (without zero) 26
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Exercise
Consider the system shown above, where =0.6 and n=5 rad/sec.
Find the rise time tr , peak time tp, maximum overshoot Mp, and
settling time ts when the system is subjected to a unit-step
input.
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MATLAB
num = [25]; den = [1 6 25]; step_time=0.005; t = 0:step_time:5;
y = step(num,den,t); r = 1; while y(r) < 1.0001; r = r + 1; end;
rise_time = (r - 1)*step_time [ymax,tp] = max(y); peak_time = (tp -
1)*step_time max_overshoot = ymax-1 s = length(t); while y(s) >
0.98 & y(s) < 1.02; s = s - 1; end; settling_time = (s -
1)*step_time
MATLAB can conveniently be used to obtain the rise time, peak
time, maximum overshoot, and settling time. Consider the system
defined by
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Mp versus curve
To limit the maximum overshoot Mp, the damping ratio should not
be too small
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Unit-impulse response curves
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Higher-Order System
After factorization,
The poles of C(s) consist of real poles and pairs of
complex-conjugate poles. R(s) is unit-step.
Composed of first and second-order terms! 31
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Higher-Order System
Inverse Laplace transform of C(s):
The response curve of a stable higher-order system is the sum of
a number of exponential curves and damped sinusoidal curves
Type of transient response is determined by the closed-loop
poles, while the shape of the transient response is primarily
determined by the closed loop zeros.
The zeros of C(s)/R(s) do not affect the exponents in the
exponential terms, but they affect the magnitudes and signs of the
residues.
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Stability in S Plane
The stability of a linear closed-loop system can be determined
from the location of the closed-loop poles in the s plane.
If poles lie in the right-half s plane, then the transient
response increases monotonically or oscillates with increasing
amplitude. Unstable system
If all closed-loop poles lie to the left-half s plane, any
transient response eventually reaches equilibrium Stable system
To guarantee fast, yet well-damped, transient response
characteristics, it is necessary that the closed-loop poles of the
system lie in a particular region in the complex plane.
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How about closed-loop poles on the j axis?
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Rouths Stability Criterion
It is a method for determining continuous system stability.
The Routh (or Routh-Hurwitz) criterion states that the number
of
roots of the characteristic equation with positive real parts is
equal to
the number of changes in sign of the first column of the Routh
array
Using this method, we can tell how many closed-loop system poles
are in the left-half plane, in the right-half plane, and on the
j-axis. The method requires two steps: 1. Generate a data table
called a Routh table. 2. Interpret the Routh table to tell how many
closed-loop system poles are in
the LHP, the RHP, and on the j -axis.
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Rouths Stability Criterion
The characteristic equation of the nth order continuous system
can be write as: The stability criterion is applied using a Routh
table which is defined as;
Where are coefficients of the characteristic equation.
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Generating a Basic Routh Table
1. First label the rows with powers of s from highest power of s
down to lowest power of s in
a vertical column.
2. Next form the first row of the Routh table, using the
coefficients of the denominator of
the closed-loop transfer function (characteristic equation).
3. Start with the coefficient of the highest power and skip
every other power of s.
4. Now form the second row with the coefficients of the
denominator skipped in the
previous step.
5. The table is continued horizontally and vertically until
zeros are obtained.
6. For convenience, any row can be multiplied or divide by a
positive constant before the
next row is computed without changing the values of the rows
below and disturbing the
properties of the Routh table.
The number of roots of the polynomial that are lies in the
right-half plane is equal to the number of sign changes in the
first column
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Example: Generating a Basic Routh Table.
Only the first 2 rows of the array are obtained from the
characteristic eq. the remaining are calculated as follows;
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Three Cases or Configurations in the First Column Array of the
Rouths Table:
1. Case-I: No element in the first column is zero.
2. Case-II: A zero in the first column but some other elements
of the row
containing the zero in the first column are nonzero.
3. Case-III: A zero in the first column and the other elements
of the row
containing the zero are also zero.
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The characteristic polynomial of a third order system is given
below The Routh array is Where The requirement for a stable third
order system is that the coefficients be positive
and
Case-I: No element in the first column is zero.
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Example: Find the stability of the continues system having the
characteristic polynomial of a third order system is given below
The Routh array is
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Because TWO changes in sign appear in the first column, we find
that two roots of the characteristic equation lie in the right hand
side of the s-plane. Hence the system is unstable.
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Exercise
Find the stability of the system shown below using Rouths
criterion.
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If only one element in the array is zero, it may be replaced w
ith a small positive
number that is allow ed to approach zero after completing the
array.
q s( ) s5
2s4
2s3
4s2
11s 10
The Routh array is then:
w here:
b12 2 1 4
20 c1
4 2 6
12
d1
6 c1 10
c16
There are two sign changes in the first column due to the large
negative number
calculated for c1. Thus, the system is unstable because two
roots lie in the
right half of the plane.
0010
00
010
06
1042
1121
0
1
1
1
2
1
3
4
5
s
ds
cs
bs
s
s
Case-II: A zero in the first column but some other elements of
the row containing the zero in the first column are nonzero.
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Case-III: A zero in the first column and the other elements of
the row containing the zero are also zero.
If all the coefficients in any derived row are zero, it
indicates that there are roots of equal magnitude lying radially
opposite in the s plane In such a case, the evaluation of the rest
of the array can be continued by forming an auxiliary polynomial
with the coefficients of the last row and by using the coefficients
of the derivative of this polynomial in the next row.
One root with a +ve real part
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Steady-State Errors In Unity Feedback Control Systems
Errors can be attributed to many factors, for example: 1)
Changes in reference input 2) Imperfection in system component
(static friction, deterioration and etc.)
In this part, the discussion is more on the type of steady-state
error that is caused by the incapability of a system to follow
particular types of inputs
It involves the term sN in the denominator, representing a pole
of multiplicity N at the origin. A system is called type 0, type 1,
type 2,p , if N=0, N=1, N=2, p , respectively. This classification
is different from that of the order of a system. As the type number
is increased, accuracy is improved; however, increasing the type
number aggravates the stability problem.
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Steady-State Errors In Unity Feedback Control Systems
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Static Position Error Constant
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Static Velocity Error Constant
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Static Velocity Error Constant
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Exercise
Find the steady-state error for the unit parabolic input
with =1
22.
Hint:
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Steady-State Error in Terms of Gain K
Error=0 for Type 3 or higher systems
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Frequency Response
Steady-state response of a system to a sinusoidal input
Vary the frequency of the input signal over a certain range and
study the resulting response
One advantage: can use the data obtained from measurements on
the physical system without deriving its mathematical model
Replacing s=j in transfer function, G(s):
The function G(j) is called the sinusoidal transfer function
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Frequency Response
There are three commonly used representations of sinusoidal
transfer functions: 1. Bode diagram or logarithmic plot 2. Nyquist
plot or polar plot 3. Log-magnitude-versus-phase plot (Nichols
plots)
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Example Bode Plot for G(j)=1/(1+jT).
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Example Polar Plot Example Nichols Chart
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Additional Exercises
(2) Determine the range of parameter K for which the system with
below closed-loop transfer function is unstable.
()
()=
4 + 3 + 2 + +
(1) Consider the system shown below. Determine the value of k
such that the damping ratio is 0.5. Then obtain the rise time tr ,
peak time tp, maximum overshoot Mp, and settling time ts in the
unit-step response.
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(3)
Additional Exercises
(4)
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Reflection
What we have learnt today:
1. Introduction a) Transfer Function
b) Closed-Loop Control
2. Transient and Steady-State Response a) First-Order
Systems
b) Second-Order Systems
c) Higher Order Systems
d) Stability Criterion
e) Steady-State Errors
3. Frequency Response
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