Esa 251 Control System Theory

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ESA 251 CONTROL SYSTEM THEORY

SEMESTER II 2014

PROJECT LAB

OPEN LOOP CONTROL FOR MOTOR SPEED

GROUP MEMBER :

a.MOHD NAZREEN BIN ABAS 115902b.MUHAMMAD NIZAM BIN ABU BAKAR 117931c. MUHAMMAD AIZAT BIN MOHD AKHIR 117924d.SHARMENDRAN A/L KUMARASAMY 115911

PENSYARAH : DR ELMI ABU BAKAR


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RESULT AND DISCUSSION

Graph tabulation

Sensor data

Output Graph - Graph of step against time

Motor Data

Input graph - Graph of voltage against time


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Output Graph - Graph of step against time (Gain=1)

Motor Gain data

Input graph - Graph of voltage against time


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Output Graph - Graph of step against time (Gain=5)

Open Loop Data

Output Graph - Graph of step against t ime


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Output Graph - Graph of mo tor speed against t ime

Eight different graph are plotted from SensorData, MotorData, MotorGainData and

OpenLoopData. Fundamentally, the result can be correspond or obtained by the graphs plotted as

shown above, which are SensorData, MotorData, MotorGainData and OpenLoopData graphs

respectively.

SensorData graph is a graph of step against time (Graph 1). The step that is mention in

this part refers to the angular displacement of the shaft because the encoder is used to convert the

angular displacement to become a code. In this experiment, we were given 60 second to save the

data. However, there is a delay of time in turning the motor that can be seen in the graph that

indicates no rotations. Based on the graph that the motor remain static from 407 to 412 second.

After this short period. The graph starts to increase gradually from 412 to 430 seconds. The

graph took 1900 steps raised. This indicated the motor turns for a duration of 18 seconds in one

revolution with 1900 steps. After 1900 steps, the graph show a constant line that lasted for 36


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second. The graph is shows a constant line because there is no change in the step which show no

motion.

The next graph is based on MotorData with the gain value is set to 1. Two graph is

plotted from this data. The first graph is voltage against time which is the input graph (Graph 2)

and the second is step against time which is the output graph (Graph 3). For our first graph the

scale is too big. The input graph shows that the voltage is first stepped up from -0.5V to 0.5V

and then stepped down from 0.5V to -0.5V after 1 second. This process was repeated for every 1

second interval exactly as shown in the graph. However, for the output graph, there is an increase

of step from -5412.5 to -3625 for 1 second and then decrease to -6120 after another second. After

that, the graph have an increase in step from -6120 to -3625 for a second. The graph repeat thispattern for 2 second and at the last 0.5 second, the graph have a decrease of step from -3625 to -

4250. Although, the shape for the both of the graph are different, but the pattern are similar. The

pattern is there is an increase in step and then a decrease in the step. In conclusion, the step up of

the voltage will result in the increase of the step which cause the motor to rotate in clockwise

direction. Similarly, when the voltage is step down, the step will decrease and the motor will

rotate counter clockwise.

Similar to MotorData, MotorGainData also have 2 graph plotted with gain value 5, which

are the graph of voltage against time (Graph 4) and the graph of step against time (Graph 5).

Graph voltage against time is the input graph while graph step against time is the output graph.

The pattern of the graphs for the MotorGainData file are exactly the same as the MotorData file.

For the input graph, the time interval for each step up and step down of the voltage is 1 second.

The scale for the input graph is also too big. For the output graph, there is a slight different in the

shape of the graph but the pattern is the same. The graph step down from -5700 to -12700 for 0.5second and then step up to 1000 for 1 second. This pattern is repeated until the sixth second.

Form 6th - 6.5th second, the graph step up from -12700 to -7000. Since the gain of the motor is

5, therefore the output graph of the MotorGainData is actually the amplification of the step value

of the output graph of the MotorData. Besides that, the change of step in this file is 5 times larger

compare to the step of the MotorData. As a result, the angular displacement of the motor is 5


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times larger as compared to the previous data. Since the angular displacement is directly

proportional to angular velocity, so the angular velocity of motor is also 5 times larger.

For the last two graphs, they were plotted based on the OpenLoopData file which are the

filtered output graph of step against time (Graph 6) and also the derivative output graph of motor

speed against time (Graph 7). From the filtered output graphs, we know that the motor is rotating

in both direction which are clockwise and counter clockwise. Besides that, from the filtered

output graph of step against time, the step increase from -6.5 to -6.0 and then decrease to -9.0

this pattern continues until the last 0.5 seconds. At the last 0.5th

second, there is a step down from

-6.0 to -8.5. Furthermore, the angular displacement for both clockwise and counter clockwise

rotation are the same. The value of derivative output graph is expected to be constant since thegradient of the filtered output graph is constant. Yet, this estimated situation does not happen as

the motor speed increase fast to the targeted speed which is almost a straight vertical line as

show in the graph. If the motor rotates in another direction, the motor speed will change its

direction.

Question answer

a. How many steps the encoder has rotated baesd on SensorData

Based on the graph obtained in the project, we knew that the value have been

increased from 100 to 2000 when the motor undergo 1 complete turn. As the result, we

can conclude that it take 1900 steps for the encoder to rotate one complete turn.

b. Why the graph plotted using SensorData is not linear?

This might be because we turn the motor by just using our bare hand. Thus, this might

cause fluctuation in angular velocity of the motor that cause the graph to be not linear.

c. What happen to the graph plotted using SensorData if the rotation of the encoder is

reversed?

If the rotation of the encoder is reversed, the graph plotted will be inverted. Then, the

step value will start to decrease.


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d. Why the pattern of input and output graphs plotted using MotorData are different?

The pattern of the input graph is square waveform while for the output data is

sinusoidal waveform. A negative gradient result is shown once the negative input

voltage is applied. In this case, the gradient of the graph refers to the speed of the

motor. Since the step is dependent on the rotation of the motor, thus the step increases

linearly when the motor rotates as shown in the graph. The encoder can only measure

the steps taken and cannot convert it into velocity automatically. Therefore, the output

graph shows linear increment and decrement according to the input voltage which thenresulting in different graph pattern.

e. Calculate the rotation speed (in degree/s) by finding the difference between the

maximum and minimum step from MotorData graphs (the encoder has 2000 steps in

one revolution).

Based on the graph, the gradient, m = 2350 step/s

Since 2000 steps = 360 degrees ; 1 steps = 0.18 degree

Therefore,

m = 2350 x 0.18 = 423 degree/srotation speed =

= 7.38 rad/ s

f. What is the difference between MotorData graphs and MotorGainData graphs?

One of the different between this two graphs is the gain of MotorData file is 1 while the

gain of MotorGainData file is 5. Besides that, there is another difference which isobvious. It is the gradient of the graph. The gradient of the MotorGainData is steeper

than the graph of the MotorData graph, which means, the MotorGainData motor is

faster.


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g. Based on the rotational speed calculated using MotorData graphs and MotorGainData

graphs, determine the conversion ratio of the voltage to speed assuming the motor

voltage is proportional to the speed.

Based on the MotorGainData graph, we can conclude that that the

Gradient, m = 14 000 step/s ; 1 step = 0.18 degree

Therefore, m = 2520 degree/s

Asume that the voltage is proportional to speed,

Speed = k X G X VoltageFor Gain, G= 1

468 = k X 1 X 5

k = 93.6

k 100

For Gain, G = 5

2520=k X 5 X 5

k = 100.8

k 100

h. What is the difference between derivative output graph and filter output graph from

OpenLoopData?

For the filtered output graph, the line (step output) increases and decrease linearly and

uniformly. However, for the derivative output graph, the motor speed shows

approximate constant speed at both minimum and maximum motor speeds.

i. What is the shape pattern of the graph that you expected in plotting OpenLoopData

after taking derivative of motor angle?

Based on our knowledge, we expected that the shape of our OpenLoopData graph have

sinusoidal pattern which is not is the result of it. The graph shows square waveform

pattern. Since the gradient of the output graph is constant, the derivative output graph

should also be constant, but such a result cannot be acquired due to some limitations.

j. Looking at the graph plotted using OpenLoopData after filtering, the pattern shows

that the speed response is overdamped or underdamped?


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From the graph, we can conclude that the speed response is consider as underdamped.

The underdamped behavior can be observed or noticed from occur of overshoot and

fluctuation of the line.

Esa 251 Control System Theory

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