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EC - Control Systems
Laboratory 01 Transfer Functions
Objective
The main purpose of this lab session is to be familiar with
characteristics of transfer functions using
MATLAB software
TASK 1
To construct a transfer function as bellow (System variable
conversions)
G(S) = 3(+7)
2+2+2
In MATLAB command window, type
(Assigning the numerator and denominator coefficient
vectors)
>> num = [3 21];
>> den = [1 2 2];
>> G = tf(num,den)
OR
>>s = tf(s);
>>G = ((3*s+21)/(s^2+2*s+2))
Try to obtain the following transfer functions:
G(S) = 5(+2)
(+1)(2+6)
G(S) = 2(s+3)(s+5)2
(s+2)(s2+4)2
Time response of a system
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The time response represents how the state of a dynamic system
changes in time when subjected to a
particular input. The time response of a linear dynamic system
consists of the sum of the transient
response which depends on the initial conditions and the
steady-state response which depends on the
system input.
Frequency response of a system
In linear time invariant (LTI) systems have the extremely
important property that if the input to the
system is sinusoidal, then the steady-state output will also be
sinusoidal at the same frequency but in
general with different magnitude and phase. These magnitude and
phase differences as a function of
frequency comprise the frequency response of the system.
The frequency response of a system can be found from the
transfer function in the following way: create
a vector of frequencies (varying between zero or "DC" to
infinity) and compute the value of the plant
transfer function at those frequencies. If () is the open-loop
transfer function of a system and is the
frequency vector, we then plot () versus . Since ()is a complex
number, we can plot both its
magnitude and phase (the Bode Plot) or its position in the
complex plane (the Nyquist Diagram). Both
methods display the same information in different ways.
Stability of a system
The transfer function representation is especially useful when
analyzing system stability. If all poles of
the transfer function (values of s at which the denominator
equals zero) have negative real parts, then
the system is stable. If any pole has a positive real part, then
the system is unstable. If we view the poles
on the complex s-plane, then all poles must be in the left half
plane (LHP) to ensure stability. If any
pair of poles is on the imaginary axis, then the system is
marginally stable and the system will oscillate.
The poles of a LTI system model can easily be found in MATLAB
using the pole command.
>>s = tf(s);
>>G = 1/(s^2 + 2*s + 5)
>>pole(G)
ans =
-1.0000 + 2.0000i
-1.0000 - 2.0000i
If we use pole-zero map using the pzmap command on MATLAB:
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>>pzmap(G)
Thus this system is stable since the real parts of the poles are
both negative.
Exercise 1
Check the following transfer functions stability
1. = 1
22+9
2. = 1
2 + 1
3. = 1
( 6)( 4)
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System Order
The order of a dynamic system is the order of the highest
derivative of its governing differential
equation. Equivalently, it is the highest power of s in the
denominator of its transfer function.
First Order System
The first order system can take the general form
() =
( + )+
(. + 1)
DC gain: , is the ratio of the magnitude of the steady-state
step response to the magnitude of the
step input. From the Final Value Theorem, for stable transfer
functions the DC gain is the value of the
transfer function when s=0. For first order systems equal to
=
.
Time constant t: is the time to reach 63% of the steady state
value for a step input or to decrease to
37% of the initial value and = 1
is found. It is special for the first order system only.
Rise Time (Tr): = .
Settling Time (Ts): =
The first order system has no overshooting but can be stable or
not depending on the location of its pole.
The first order system has a single pole at -a. If the pole is
on the negative real axis (LHP), then the
system is stable. If the pole is on the positive real axis
(RHP), then the system is not stable. The zeros
of a first order system are the values of s which makes the
numerator of the transfer function equal to
zero.
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Second Order System
Second order systems are commonly encountered in practice, and
are the simplest type of dynamic
system to exhibit oscillations. In fact many real higher order
systems are modeled as second order to
facilitate analysis.
The general form of second order system is:
() =
(2 + + )=
. 2
(2 + 2 + 2)
Natural frequency is the frequency of oscillation of the system
without damping.
Damping Ratio
The damping ratio is a dimensionless quantity characterizing the
energy losses in the system due to
such effects as viscous friction or electrical resistance.
=
Poles and zeros
Note that the system has a pair of complex conjugate poles
at:
= =
: damped frequency of oscillation.
DC gain
The DC gain, =
, again is the ratio of the magnitude of the steady-state step
response to the
magnitude of the step input, and for stable systems it is the
value of the transfer function when s = 0.
For second order systems
=
Percent Overshoot
The percent overshoot is the percent by which a system exceeds
its final steady-state value. For a second
order under damped system, the percent overshoot is diretly
related to the damping ratio by the
following equation:
% =
12 100
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Settling Time
The settling time, , is the time required for the system ouput
to fall within a certain percentage of the
steady state value for a step input or equivalently to decrease
to a certain percentage of the initial value
for an impulse input. For a second order, underdamped system,
the settling time can be approximated
by the following equation:
=
=
Rise Time: = . +.
General Form of a Denominator: S2 + 2nS + n2 (Second order
system)
Then, S = -n (n)2 n2
= (n)2 n2
Step time response:
If the input is step function then the output or the response is
called step time response. The system can
be represented by a transfer function which has poles (values
make the denominator equal to zero),
depending on these poles the step response divided into four
cases:
1. Underdamped response:
In this case the response has an overshooting with a small
oscillation which results from complex poles
in the transfer function of the system. i.e. < 0 or <
1
2. Critically response:
In this case the response has no overshooting and reaches the
steady state value (final value) in the
fastest time. In other words it is the fastest response without
overshooting and is resulted from the
existence of real & repeated poles in the transfer function
of the system. i.e. = 0 or =1
3. Overdamped response:
In this case no overshooting will appear and reach the final
value in a time larger than critically case.
This response is resulted from the existence of real &
distinct poles in the transfer function of the system.
i.e. > 0 or > 1
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4. Undamped response:
In this case a large oscillation will appear at the output and
will not reach a final value and this because
of the existence of imaginary poles in the transfer function of
the system and the system in this case is
called "Marginally stable".
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MATLAB Work
Step and Impulse Responses of a Transfer Function
To get the step response of a transfer function: use,
1 . = 2
+ 3
>> G = tf ([2], [1 3])
>> step (G)
To obtain impulse response: use,
>> impulse (G)
stepinfo(sys): this command is used to Compute
step response characteristics.
For the following transfer functions we will find
the settling time, rise time, overshoot and steady
state error:
>> stepinfo(G)
Result
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Underdamped Response
= 12
(2 + 2 + 9)
>> G = tf ([12], [1 2 9])
>> step (G)
>> figure, impulse (G)
>> stepinfo(G)
Critically damped Response
= 12
(2 + 6 + 9)
>> G = tf ([12], [1 6 9])
>> step (G)
>> figure, impulse (G)
>>stepinfo(G)
Over damped Response
= 15
(2 + 7 + 12)
>> G = tf ([15], [1 7 12])
>> step (G)
>> figure, impulse (G)
>>stepinfo(G)
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Undamped Response
= 9
2 + 3
>> G = tf ([9], [1 0 3])
>> step (G)
>> figure, impulse (G)
>>stepinfo(G)
Exercise 2
Obtain the step response of the following transfer functions
given below and find the step info using
MATLAB and calculations
1. = 21
22 + 16 + 21
2. = 2
32 + 4 + 6
3. = 6
2+8+ 6
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Step Response using Matlab Simulink
Starting Simulink
Simulink is started from the MATLAB command prompt by entering
the following command:
>>Simulink
Alternatively, you can hit the Simulink button at the top of the
MATLAB window as shown here
When it starts, Simulink brings up a single window, entitled
Simulink Library Browser which can be
seen here.
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Then goto NewSimulink Model
Construct the following Simulink model using Simulink library
components as follows
The simple model consists of three blocks: Step, Transfer
Function, and Scope. The Step is
a Source block from which a step input signal originates. This
signal is transferred through the line in
the direction indicated by the arrow to the Transfer Function
Continuous block. The Transfer
Function block modifies its input signal and outputs a new
signal on a line to the Scope. The Scope is
a Sink block used to display a signal much like an
oscilloscope.
Modifying blocks
A block can be modified by double-clicking on it. For example,
if you double-click on the Transfer
Function block in the Simple model, you will see the following
dialog box.
Modify the transfer function block as follows
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Running simulation
To run the simulation hit the play button
The simulation should run very quickly and the scope window will
appear as shown below.
Note that the simulation output (shown in yellow) is at a very
low level relative to the axes of the scope.
To fix this, hit the auto scale button (binoculars), which will
rescale the axes as shown below.
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Exercise 3
1. Obtain the step response of the following transfer functions
using Simulink.
1. = 12
(2+6+9)
2. = 15
(2+7+12)
3. = 9
2+3
The effect of varying damping ratio on a second-order system
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Exercise 4
Review Questions
1. What is the different between steady state response and
transient response of a control
system?
2. The pole-zero plot for the two transfer functions are given
below, obtain their unit step
response and comment on it.
3. How damping ratio (zeta) affects the time response of a
second order system?
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NOTE
What is time response?
It is an equation or a plot that describes the behavior of a
system and contains much information about
it with respect to time response specification as overshooting
setting time, peak time, rise time and
steady state error. Time response is formed by the transient
response and the steady state response.
= +
Transient time response describes the behavior of the system in
its first short time until arrives the
steady state value and this response will be our study
focus.
If the input is step function then the output or the response is
called step time response and if the input
is ramp, the response is called ramp time response etc.
Delay Time (Td): is the time required for the response to reach
50% of the final value.
Rise Time (Tr): is the time required for the response to rise
from 0 to 90% of the final value.
Settling Time (Ts): is the time required for the response to
reach and stay within a specified tolerance
band (2% or 5%) of its final value.
Peak Time (Tp): is the time required for the underdamped step
response to reach the peak of time
response (Yp) or the peak overshoot.
Percent Overshoot (OS%): is the normalized difference between
the response peak value and the
steady value This characteristic is not found in a first order
system and found in higher one for the
underdamped step response.
Steady State Error (ess): indicates the error between the actual
output and desired output ast tends to
infinity, and is defined as: