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7/28/2019 Cs - Assign Matlab Original 2013
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 512101120541
1. OBJECTIVE
2. METHODOLOGY
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 512101120542
3. DESIGN PROBLEM
A unity feedback control for a vehicle manoeuvring system is given in the following
diagram:
where K = 100 and G(s) =)8)(4(
12
s s s
The initial design of the system is not very good with regards to the step input. As an
engineer, you have been given the task to improve certain performance criteria of the
system. To complete your assignment, you are to analyse the system design and give some
recommendations for improvement. You are required to submit a complete report detailing
your work.
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 512101120543
3.1 TASK A (Problem Identification)
i. Calculation
T(s) =
()()
=
()
=
ii. Command
step calculates the step response of a dynamic system. For the state space case, zero
initial state is assumed. When it is invoked with no output arguments, this function plots
the step response on the screen. step(sys) plots the step response of an
arbitrary dynamic system model sys. This model can be continuous or discrete, and SISO
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 512101120544
or MIMO. The step response of multi-input systems is the collection of step responses for
each input channel. The duration of simulation is determined automatically, based on the
system poles and zeros.
iii. Result
a. time (step) response
The step function is one of most useful functions in MATLAB for control design. Given a
system representation, the response to a step input can be immediately plotted, without
need to actually solve for the time response analytically. A step input can be described as
a change in the input from zero to a finite value at time t = 0. By default, thestep command performs a unit step (i.e. the input goes from zero to one at time t = 0).
The basic syntax for calling the step function is the following, where sys is a defined LTI
object.
Peak Response
Rise Time
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 512101120545
b. frequency response (bode plot)
A Bode plot is a graph of the transfer function of a linear, time-invariant system
versus frequency, plotted with a log-frequency axis, to show the system's frequency
response. It is usually a combination of a Bode magnitude plot, expressing the
magnitude of the frequency response gain, and a Bode phase plot, expressing the
frequency response phase shift .
A Bode phase plot is a graph of phase versus frequency, also plotted on a log-
frequency axis, usually used in conjunction with the magnitude plot, to evaluate how
much a signal will be phase-shifted. For example a signal described by: Asin(ωt ) may
be attenuated but also phase-shifted. If the system attenuates it by a factor x and
phase shifts it by −Φ the signal out of the system will be ( A/ x ) sin(ωt − Φ). The phase
shift Φ is generally a function of frequency. Phase can also be added directly from the
graphical values, a fact that is mathematically clear when phase is seen as the
imaginary part of the complex logarithm of a complex gain.
Phase Margin
Gain Margin
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 512101120546
c. Root Locus
In addition to determining the stability of the system, the root locus can be used to
design the damping ratio and natural frequency of a feedback system. Lines of constant damping ratio can be drawn radially from the origin and lines of constant
natural frequency can be drawn as arcs whose center points coincide with the origin.
By selecting a point along the root locus that coincides with a desired damping ratio
and natural frequency a gain, K, can be calculated and implemented in the controller.
More elaborate techniques of controller design using the root locus are available in
most control textbooks: for instance, lag, lead, PI, PD and PID controllers can be
designed approximately with this technique.
The definition of the damping ratio and natural frequency presumes that the overall
feedback system is well approximated by a second order system; i.e. the system has a
dominant pair of poles. This is often not the case, so it is good practice to simulate the
final design to check if the project goals are satisfied.
RHSLHS
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 512101120547
3.2 TASK B (Design Solution)
Improve the system with new design to make the system are stable.
i. Command
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 512101120548
a. Step response
b. Bode Plot
Settling Time
Gain Margin
Phase Margin
Peak response
Rise Time
Steady State
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 512101120549
c. Root locus
LHS
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 5121011205410
3.3 TASK C (System Analysis)
From the task A, the system is not stable. The reason why system is not stable is because of
a given step input. However, the system can be improved to achieve stability.
Step response above not stable. After 30 seconds, step response overshoot. The system can
be improved by changing the value of K to decrease the percentage of overshoot. Bychanging the value of gain, the percentage of new system can less a bit from initial system.
Gain value must less than given value. Additional, if gain value is changed it better to
change the value of natural frequency also. By changing it, overshoot can be decrease.
Damping factor also give an effect of the step response. If the damping factor more than 1,
the system is overdamped, damping factor less than 1 the system are underdamped while
for equal to 1 is under the critically damped condition.
For frequency response,
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 5121011205411
For the system, frequency response is not stable because value of gain margin was
0rad/sec and phase margin was 1.68rad/sec. What make it happen? It happened because
of pole and zero placement and value. For frequency response, it just consider at the
system without looking at the feedback. Frequency response are depends on step input
value (pole and zero). However, the order to change the value of step order gain will
affected.
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 5121011205412
For root locus,
For the very simple system of this problem, there were many ways to find how the roots
varied as we varied the gain of the system. For a more complicated system this is not easy.
The root locus plot gives us a graphical way to observe how the roots move as the gain, K, is
varied.
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 5121011205413
Comparison between Task A and Task B
STEP RESPONSE COMPARISON
TASK A
The system is not stable
because it not achieved
constants. That why the
system is not stable. It has
overshoot at after it consta
and the settling time are bi
From this system, can
recognize the condition of
system whether it
underdamped or so on.
TASK B
The system is near to achie
stability because after it
overshoots, the graph isready to constant. The
settling time for the system
are quickly comparing the
initial one. The graph is
under the underdamped
condition with the value of
damping factor is 0<ζ<1. B
change the value of gain an
zeros, the stability are
achieved.
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 5121011205414
FREQUENCY RESPONSE COMPARISON
TASK A
The system is not stablebased on the value of gain
margin and phase margin.
TASK B
Gain margin and phase
margin are in the positive
value. The stability of syste
can determine by look at
value positive at gain and
phase margin. Without
change the poles value, the
stability can achieved
because it just depends on
zeroes and gain value.
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 5121011205415
ROOT LOCUS COMPARISON
TASK A
Number of branches = 4
Stability = Unstable
TASK B
Number of branches = 3
Stability = stable
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 5121011205416
Description Task A Task B
Transfer Function
Root Locus Dragging the locus to the
right hand side (RHS) of the
graph
The locus to the left hand
side (LHS) of the graph
Bode diagram Gain margin (Gm) for
bode diagram is negative
infinity in magnitude (db)
Phase margin (Pm), -34.6
deg at (1.68 rad/sec)
Gain margin (Gm) for
bode diagram is 46.6 dB
(at 3.08 rad/sec).
Phase margin (Pm), 65.7
deg at (0.323 rad/sec)
Step Response Peak response, the graph
is undamped and also
inversely
Settling Time, unstable
and the value was infinity
Peak amplitude was at
1.18 and %OS at 17.8
The rise time is 3.93s
The settling time was
stable which is 28.4s
The final value which isthe steady state are 1.
1. SETTLING TIME
The settling time of an amplifier or other output device is the time elapsed from the
application of an ideal instantaneous step input to the time at which the amplifier output
has entered and remained within a specified error band, usually symmetrical about the
final value. Settling time includes a very brief propagation delay, plus the time required
for the output to slew to the vicinity of the final value, recover from the overload
condition associated with slew, and finally settle to within the specified error. The
settling time that we’ve got in Task A was unstable and the value was infinity. It’s also
had low performance in the system. But in Task B, the settling time was stable and the
value was 28.4s.
2. RISE TIME
The effect of additional pole in the left-half s-pane (LHP) tends to slow the system down,
this make the rise time of the system, for example, will become larger. When pole is far
to the left of the imaginary axis, it is effect tend to be small. The effect becomes more
pronounced as the pole moves toward the imaginary axis. That’s why in this assignment
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 5121011205417
the additional on the poles are not included, but by doing additional on the zeros it give
a huge different for stabilizing. The effect on extra zeros in the LHP has the opposite
effect, as it tend to speed the system up. According the result in Task B, the value for rise
time was 3.93s compared to Task A that was infinity.
3. STEADY STATE
Steady state error is defined as the difference between the input and the output for a
prescribed test input as time (t) goes to infinity. A system in a steady state has numerous
properties that are unchanging in time.
4. GAIN MARGIN
The gain margin is the amount of gain increase or decrease required to make the loop
gain unity at the frequency where the phase angle is –180° (modulo 360°). In other
words, the gain margin is 1/g if g is the gain at the –180° phase frequency. Similarly, the
phase margin is the difference between the phase of the response and –180° when the
loop gain is 1.0. The frequency at which the magnitude is 1.0 is called the unity-gain
frequency or gain crossover frequency . It is generally found that gain margins of three or
more combined with phase margins between 30 and 60 degrees result in reasonable
trade-offs between bandwidth and stability. Gain margin is the amount you can increase
the gain of a system before achieve the 0 dB gain. In conclusion the gain is a measured of
how far from instability a system is.
5. PHASE MARGIN
The phase margin Pm is in degrees. The gain margin Gm is an absolute magnitude. Phase
margin is the amount of phase the system can lag before achieve the 180° phase lag. In
conclusion the gain is a measured of how far from instability a system is.
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 5121011205418
6. ROOT LOCUS
Based on the experiment that I had done, the root locus was not stable on the s-plane
was because the value of gain was high for system to generate. So, the stability on the
damping factor on step respond can be stabilizing by reducing the gain from 100 to 1.
Next, by adjusting the value and add the zeros from [1] to [ 1 0 10 1 ], the root locus
diagram show by dragging the locus to the negative side of the graph can be stabilize but
if the locus is more to the positive side, it show that it was not stable.
On the other hand, the effect of a zero far away to the left of the imaginary axis tends to
be small. It becomes more pronounced as the zero moves closer to the imaginary axis.
For the additional zero in the right-half s-plane (RHP) has a delaying effect much moresevere than the addition of a LHP pole. The RHP zero causes the response to start
toward the wrong direction. It will move down first and become negative. System with
RHP zeros are called non minimum Phase system (for reasons that will become clearer
after the discussion of the frequency design methods) and are typically difficult to
control. System with only LHP poles (Stable) and LHP zeros are called minimum phase
system.
The changing gain that the system poles and zeros actually move around in the S-
plane. This fact can make life particularly difficult, when to solve higher-order
equations repeatedly, for each new gain value. The solution to this problem is a
technique known as Root-Locus graphs. Root-Locus allows graph the locations of the
poles and zeros for every value of gain, by following several simple rules.
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 5121011205419
Trial and Error Solution
Here is some example that we had to examine the behavior of the system varies as K
changes, so let's try several values of K. Let's arbitrarily try K=1, 10 and 100 so that we
have a wide range of K values.
K Xfer Function Step ResponseK=1
K=10
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
TUN AYUNI DIYANA BT ARIFFIN 5121011205420
K=100
The response with K=1 was too slow, the response with K=100 was too oscillatory, and the
response with K=10 is almost just right, though we may want to adjust K to get a little bit
less overshoot. Clearly this method is rather "hit-or-miss" and it may take us a long time to
find a suitable value for K.
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MATLAB ASSIGNMENT
NURUL HAFIZAH SAAD 51211211425
4. CONCLUSION
From this assignment, we can conclude that by determining the Step Response, Bode Plot
and Root Locus will helps us analyzing a control system performance. Through this
graphical method, direct portrayal of the system can be obtained as a whole especially
referring to transient response.
Thus, from this assignment we’re need to investigate the poles and zeros’ position and in
using software (Matlab) is much better from manually analyzing the system. Through the
information of the graphical method we’re can
Apart from that, the stability of a system can also be known by referring to the route of the
root locus. Techniques of drawing for Step Response, Bode Plot and Root Locus have been
discussed in systematically which involve number of rules that need to be understood.Relevant examples are included to ensure that readers can understand and use them to
solve problems for other system as well.