Assignment: MatLab
DESIGN PROBLEM A unity feedback control for a vehicle
manoeuvring system is given in the following diagram:
where K = 100 and G(s) =
1 s ( s 4)( s 8)2
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. TASK A (Problem
Identification) Analyse the system as follows. Use Matlab to find
the time (step) response, frequency response (bode plot) and root
locus of the system. Obtain the three plots and clearly identify
the problems with this initial design. List down as many problems
as possible. For each one of the problem, explain why it is not
acceptable in an engineering product. (Note: Problems refer to
performance criteria and the number is not more than 10).
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Assignment: MatLab
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Assignment: MatLab
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Assignment: MatLab
TASK B (Design Solution) Improve the system as follows. Find out
what cause the problems. Suggest the actions you may try to solve
the problems. Try out your solutions using Matlab. Once you have
obtained an acceptable design, reproduce the step response, bode
plot and root locus. Make sure that all three plots agree with each
other in terms of performance criteria. Your new performance
criteria must be clearly printed on the plots. Compare your new
design with the initial one. (Note: You are improving the system,
not changing it e.g. moving the poles).
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Assignment: MatLab
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Assignment: MatLab
TASK C (System Analysis) For each problem identified in TASK A,
explain in detail the reason of using a certain parameter value,
how it affects the system, and how much improvement you get as a
result of this new design parameter. For example, increasing the
damping factor from 0.25 to 0.6 has reduced the overshoot by half,
from 20% to 10%. Use suitable method to justify the new design
parameter, e.g Routh-Hurwitz method to check the stability, etc.
(Note: these are just an examples, not necessarily the solutions to
this assignment). TASK C (SYSYTEM ANALYSIS) 1) PEAK RESPONSE
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Assignment: MatLab
The value in Task A was infinity, so there was no value.
Compared to Task B, the value of peak amplitude was at 1.18 and
overshoot at 17.8. Peak amplitude is the highest value of the
amplitude, called the peak value. Peak value in Task A
Peak amplitude in Task B
2) 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
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Assignment: MatLab
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 I got in Task A
was unstable and the value was infinity. Its also had low
performance in the system. But in Task B, the settling time was
stable and the value was 28.4s.
3) 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. Thats why in this assignment 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.
4) 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.
5) 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
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Assignment: MatLab
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. 6) 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. 7) 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 stabilize 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 more severe 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.
CLOSED-LOOP TRANSFER FUNCTION FOR A PARTICULAR SYSTEM:
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Assignment: MatLab
Where N is the numerator polynomial and D is the denominator
polynomial of the transfer functions, respectively. Now, we know
that to find the poles of the equation, we must set the denominator
to 0, and solve the characteristic equation. In other words, the
locations of the poles of a specific equation must satisfy the
following relationship:
FROM THIS SAME EQUATION, WE CAN MANIPULATE THE EQUATION AS
SUCH:
And finally by converting to polar coordinates:
Now the 2 equations that govern the locations of the poles of a
system for all gain values: [The Magnitude Equation]
[The Angle Equation]
FROM THE RESULT IN TASK B, WE CALCULATE THE % OVERSHOOT AND ALSO
DAMPING FACTOR, : ,
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Assignment: MatLab
TRIAL AND ERROR SOLUTION We want to examine how 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 Response
K=1
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Assignment: MatLab
K=10
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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