control system Lab 01-introduction to transfer functions

Nalan Karunanayake 16/03/2015 1
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(𝑠+7)
𝑠2+2𝑠+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(𝑠+2)
(𝑠+1)(𝑠2+𝑠−6)
G(S) =
2(s+3)(s+5)2
(s+2)(s2+4)2
Time response of a system

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 𝐺(𝑠) is the open-loop transfer function of a system and is the
frequency vector, we then plot 𝐺(𝑗𝜔) versus 𝜔. Since 𝐺(𝑗𝜔)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:

>>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. 𝐺𝑠 =
1
𝑠2−2𝑠+9
2. 𝐺𝑠 =
1
𝑠2 + 1
3. 𝐺𝑠 =
1
(𝑠 − 6)(𝑠 − 4)

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
𝐺(𝑆) =
𝑏
(𝑠 + 𝑎)
+
𝐾𝑑𝑐
(𝑡. 𝑠 + 1)
DC gain:𝑲𝒅𝒄 , 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 𝐾𝑑𝑐 =
𝑏
𝑎
.
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 𝑡 =
1
𝑎
is found. It is special for the first order system only.
Rise Time (Tr): 𝑻 𝒓 =
𝟐.𝟐
𝒂
Settling Time (Ts): 𝑻 𝒔 =
𝟒
𝒂
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.

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:
𝐺(𝑠) =
𝑎
(𝑠2 + 𝑏𝑠 + 𝑐)
=
𝐾 𝑑𝑐. 𝜔 𝑛
2
(𝑠2 + 2𝜀𝜉𝜔 𝑛 𝑠 + 𝜔 𝑛
2)
Natural frequency 𝝎 𝒏 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.
𝝃 =
𝒃
𝟐𝝎 𝒏
Poles and zeros
Note that the system has a pair of complex conjugate poles at:
𝑆 = −𝝃𝝎 𝒏 ± 𝒋𝝎 𝒏√𝟏 − 𝝃 𝟐 = −𝝈 ± 𝒋𝝎
𝝎 : damped frequency of oscillation.
DC gain
The DC gain,𝐾𝑑𝑐 =
𝑎
𝑐
, 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
𝐾𝑑𝑐 =
𝑎
𝑐
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:
𝑂𝑆% = 𝑒
−
𝜋𝜉
√1−𝜉2
∗ 100

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:
𝑻 𝒔 =
𝟒
𝝃𝝎 𝒏
=
𝟒
𝝈
Rise Time: 𝑻 𝒓 =
𝟏 − 𝟎.𝟒𝟏𝟔𝟕𝝃 +𝟐.𝟗𝟏𝟕𝝃 𝟐
𝝎 𝒏
 General Form of a Denominator: S2
+ 2ςωnS + ωn2
(Second order system)
Then, S = -ςωn ± √(ςωn)2 − ωn2
∆ = (ςωn)2
− ωn2
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 ς < 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 ς =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 ς > 1

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".

MATLAB Work
Step and Impulse Responses of a Transfer Function
To get the step response of a transfer function: use,
1 . 𝐺𝑠 =
2
𝑠 + 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

Underdamped Response
𝐺𝑠 =
12
(𝑠2 + 2𝑠 + 9)
>> G = tf ([12], [1 2 9])
>> step (G)
>> figure, impulse (G)
>> stepinfo(G)
Critically – damped Response
𝐺𝑠 =
12
(𝑠2 + 6𝑠 + 9)
>> G = tf ([12], [1 6 9])
>> step (G)
>>stepinfo(G)
Over – damped Response
𝐺𝑠 =
15
(𝑠2 + 7𝑠 + 12)
>> G = tf ([15], [1 7 12])
>> step (G)
>>stepinfo(G)

Undamped Response
𝐺𝑠 =
9
𝑠2 + 3
>> G = tf ([9], [1 0 3])
>> step (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. 𝐺𝑠 =
21
2𝑠2 + 16𝑠 + 21
2. 𝐺𝑠 =
2
3𝑠2 + 4𝑠 + 6
3. 𝐺𝑠 =
√6
𝑠2+8𝑠+ √6

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.

Then goto NewSimulink 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

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.

Exercise 3
1. Obtain the step response of the following transfer functions using Simulink.
1. 𝐺𝑠 =
12
(𝑠2+6𝑠+9)
2. 𝐺𝑠 =
15
(𝑠2+7𝑠+12)
3. 𝐺𝑠 =
9
𝑠2+3
The effect of varying damping ratio on a second-order system

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?

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.
𝑻𝒊𝒎𝒆 𝒓𝒆𝒔𝒑𝒐𝒏𝒔𝒆 = 𝑻𝒓𝒂𝒏𝒔𝒊𝒆𝒏𝒕 𝒓𝒆𝒔𝒑𝒐𝒏𝒔𝒆 + 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 as‘t’ tends to
infinity, and is defined as:

control system Lab 01-introduction to transfer functions

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