Cs - Assign Matlab Original 2013

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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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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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

http://www.mathworks.com/help/ident/ug/dynamic-system-models.html

http://www.mathworks.com/help/ident/ug/dynamic-system-models.html



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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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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

http://en.wikipedia.org/wiki/Plot_(graphics)

http://en.wikipedia.org/wiki/Transfer_function

http://en.wikipedia.org/wiki/LTI_system_theory

http://en.wikipedia.org/wiki/Frequency

http://en.wikipedia.org/wiki/Frequency_response


http://en.wikipedia.org/wiki/Gain

http://en.wikipedia.org/wiki/Phase_(waves)





http://en.wikipedia.org/wiki/Gain



http://en.wikipedia.org/wiki/Frequency

http://en.wikipedia.org/wiki/LTI_system_theory

http://en.wikipedia.org/wiki/Transfer_function

http://en.wikipedia.org/wiki/Plot_(graphics)



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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

http://en.wikipedia.org/wiki/Damping_ratio

http://en.wikipedia.org/wiki/Natural_frequency

http://en.wikipedia.org/wiki/Lead-lag_compensator

http://en.wikipedia.org/wiki/PID_controller



http://en.wikipedia.org/wiki/Pole_(complex_analysis)

http://en.wikipedia.org/wiki/Pole_(complex_analysis)



http://en.wikipedia.org/wiki/PID_controller

http://en.wikipedia.org/wiki/Lead-lag_compensator





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3.2 TASK B (Design Solution)

Improve the system with new design to make the system are stable.

i. Command



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a. Step response

b. Bode Plot

Settling Time

Gain Margin

Phase Margin

Peak response

Rise Time

Steady State



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c. Root locus

LHS



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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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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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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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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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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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ROOT LOCUS COMPARISON

TASK A

Number of branches = 4

Stability = Unstable

TASK B

Number of branches = 3

Stability = stable



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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

http://en.wikipedia.org/wiki/Amplifier

http://en.wikipedia.org/wiki/Error

http://en.wikipedia.org/wiki/Propagation_delay

http://en.wikipedia.org/wiki/Slew_rate

http://en.wikipedia.org/wiki/Slew_rate

http://en.wikipedia.org/wiki/Propagation_delay

http://en.wikipedia.org/wiki/Error

http://en.wikipedia.org/wiki/Amplifier



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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.

http://en.wikipedia.org/wiki/System

http://en.wikipedia.org/wiki/System



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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.

http://en.wikibooks.org/w/index.php?title=S-plane&action=edit&redlink=1






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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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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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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.

Cs - Assign Matlab Original 2013

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