ABSTRACT

A video projector is studied with intent of designing a control system so that we can improve it

to meets the desired performance specification. Size of image projected by the projector is

controlled by controlling the distance from the projector to the slide. This can be achieved by

varying the projector position via DC motor using some gears and timing belt. Feedback is

achieved by measuring the actual position of the projector using sensor. The information related

to this project is firstly obtained. Experiment to determine the relationship between the sizes of

image projected and the distance between the slide and projector was conducted. Result of the

experiment is then recorded in the form of mathematical equation. Later, the block and

systematic diagram of this system is drawn and by applying all the physical law, the

mathematical modeling is done to obtain the transfer function for our system. The modeling

process is then continued by writing the m. file for the transfer function. This transfer function

was used as the initial step to do the analysis.

Analysis process is then take part on the system using LTI viewer and step response was found

and analyzed. Later, the desired performance specification that is settling time, peak time and

percent overshoot were being identified. After that, the root locus techniques are applied to

design the controlled in order to meet the desired performance specification by using 2 methods:

a) Manually,

b) SISO Design Tools.

Lastly, the performance of the proposed closed-loop system is being verified through simulation

using MATLAB Simulink.

1.0 INTRODUCTION

A projector is used to project an image directly from the computer. There are lots of types of

projector in the market, such as image projector, video projector and slide projector. Here

however, we will focus on video projector as this is the most common type of projector used.

Video projector is a device that projects a video signal from computer, home theatre system and

etc. This type of projector is used widely in in all kind of settings such as, conference room

presentations, meeting, classroom, and even as a home theatre.

Video projector is one of the greatest inventions ever. It helps lots of educator deliver and

sharing their knowledge more effectively. As educators nowadays store lots of education related

materials in their notebook, the used of projector is a must. Projector here becomes a medium

that connect the educators with the students.

Problem arises as projector is usually fixed to a certain spot. This limit the users ability to

control the screen size. University students are different from school students. They does not sit

in bulk, they scattered. As the projector is fixed to a certain spot, this means that the size of the

virtual screen produce is also fixed. If the size is too small, the students at the back of the lecture

hall will have very difficult times to see and understand the lecture.

Through thorough observation on how the projector system works, we came up with a

solution. This adjustable projector will be of great help to both lecturer and students. As the

distance of the projector to the white screen is adjustable, hence so thus the size of the screen

produces. Students that sit at the back of the lecture hall can now see the lecture clearly as much

as the students that sit in front and thus, improving their understanding of the lecture.

1.1 Objective

The objective of this project is to design a control system for an adjustable projector.

User can control the desired size of image at the slide by buttons or similar control system. This

control system is designed to meet several specifications that is 10% of percentage overshoot and

10s of settling time, Ts. If the settling time is too long, it will take a very long time to make the

size of image as desired and can make the user become impatient. But, if the settling time is too

fast, it might damage the projector.

2.0 MODELING 2.1 Schematic Diagram

Figure 1- Systematic Diagram

Figure 2- Block Diagram

2.2 Mathematical Model

Data

Motor (this data we got from internet for our type of motor).

Tstall= 0.318 Nm

Motor resistance (armature) = Rm=0.175 ohms

Back EMF constant = Kb = 0.0239

Motor torque constant = KT = 0.024

Armature damping = Da = 5.7296 x 10-5 Nms/rad

Armature inertia = Ja = 1.92 x 10-5 kgm2

Spring constant of system (motor and load) = ks = 0.01 N/m

Rotational load

Damping rotational load = DL = 80 Nms/rad

Inertial rotational load = JL = 800 kgm2

Radius of gear = r = 0.05 m

Translational load

Damping of load = C = fv = 0.5 Ns/m

Mass (projector plus rail) = 5 kg

Calculate the gear ratio.

Total torque needed to move projector = Mgr = Td = 5(10)(0.05)= 2.5 Nm

!!"#$$! !

=!!! !

! !318 !"2.! !!" = ! .!"# !

!!! !

DC motor

We know such that; the DC motor run by using magnetic field phenomenon. A magnetic field is

created by stationary electromagnet called fixed field. The armature (rotating circuit); which

current ia(t) passed through the magnetic field at right angles and feels a force. Than this will

turn the rotor. The stator is a magnet which will not move.

Figure 3-DC motor

Now we will discuss the other concept of motor.

Figure 4-DC Servo Motor Circuit Diagram

Based on the figure above; the source voltage (EMF) will be supplied through the electric circuit

that consists of resistor, induction and armature current into the motor that will produce current.

Thus, the equation of the voltage will be supplied is,

Ea = Ea1 +Ea2 +Ea3

Note that the Ea1 ,Ea2 ,Ea3 is the voltage used (separated) from the EMF (Ea).

Note that the EMF is total voltage supplied to the motor; and the voltage lost by resistance and

inductance in the motor. Than the total voltage left for the motor to rotate the rotor is back emf

(em) or in this case is Ea3.

E! = ! !!" !d!+ ! !i! + ! !

From here we will Laplace transform;

! ! s = ! !SI! ! R!!! ! + ! !! s!

Thus the block diagram for the motor and including the mechanical load will be.

Ea1 Tm

+

Ea2 Tm +

+

+

Ea3

G11

G21

G3

G1/2

Torque Developed

In the motor system, when the current pass through the electric circuit, torque will be produced

resulting of the magnetic field and the current armature. The current and the magnetic field will

be constant. Thus the torque developed by the motor is proportional to the armature current.

T ! = ! !!!! t!

T m = K! I! (t!

! ! (! )! !

= !!!_!_!_!_!_!_!_!(! )

KT is torque constant which depend on the motor and magnetic field characteristic.

We also know such that the Tm is a torque developed by the motor to rotate the load (J) as shown

in figure 2.2.

Thus we can mature that Tm

Tm(s) = (JmS2 + DmS + K )

By ignoring stiffness of the load

Tm(s) = (JmS2 + DmS+ks)--------------- (2)

!!!!

= !!! ! ! !! ! + ! !!_!__! ! ! !_! !_! !_! ! )

! !! !

=!

!! !! +!!! + ! !_!_!_!_!_!_!_!_!__!(! ! /! )

Note that equation (3) is equal to G1/2

Back electromotive force

Back emf (em) is directly proportional to the angular velocity of the motor

e! ! = ! !!! !!"

e! ! ! ! !! !"

em(s) = KbSm-------------(4)

K b is back emf constant

For G11

! !SI! = ! !1---------------------(5)

Substituting the equation (1) into (5)

!! !!!!!= ! !!

!!!!"

=! !!! !

!_! !__!_!_!_!_!!""!

For G21

R! I! s ! !! !!_! ! !_!__!_!__!_!(! )

Substituting the equation (1) into (6)

!!! !!!

! !! !

!!! !"

!!!!!

_____!__!__!_!(! !" )

Summation of G21 and G11

! !!!"

!!!! !"

!!!!!

+! !! !!

Now G21 and G11 will multiplied with the G1/2

! !! !!

!!!!!

+!!! !!

!!!!!

!! !! !

!!!!! !

! (!

!!! ! ! !! ! ! !!)

! !! !"

!!!!!"

=!!! !

+! !

! !!!!

!!!! ! ! !! ! ! ! !

!

For G3

E!" = ! !

From equation 4

em(s) = KbSm

Thus

! !! !"

!!!! !

_!_!_!_!_!_!(!! ! )

Summation for all

!!!!"

+! !! !"

+!!! !"

!!!!!

+! !

! !!!

!!! !! ! !! ! ! !!

+!!!!

!!!! !

!!! !! !

!!!!!!

=! !!!+

! !!!!

!!! ! ! + ! ! ! ! !!

1 !!! !!

Since

Ea = Ea1 +Ea2 +Ea3

Thus

!!! !

!!!!!+

! !!!!

!!!! ! ! !!! + ! !

1 +! !!1

And also know such that the

Lm

Ea1 Tm

+

Ea Ea2 Tm +

+

+

Ea3

For motor and load

Ea m

Since we know Jm and Dm is based on the rotational and translational part of the system.

For the rotating load, we assume all inertia and damping has been reflected to the load, Thus we

will get the translational and rotational torque equation.

T1(s) = L (JeqS2 + DeqS)

T2(s) = F(s)r

T1 + T2 = Teq

!!! !!

!!!!

!

!!!

!!! !! ! !! ! ! !!

! ! !

! !!!! !

!!!!!

)

! ! + !!!!!!! !!!!! !

! + ! !!!

!!

!! + !" + !!

(Jeq s2+Deq s)L(s) + F(s)r = Teq(s)________(7)

We know that the

!!" ! !! (! !"2! !

! + !!!_!_!_!_!_!_!_!_(! )

!!" ! !! !!!"!!

)!! ! ! !!_! !_!_!_!_!_!_! 9)

Jeq = Inertia at the load of the rotational load and the armature

JL = Inertia at the load of the rotational load

Ja = Armature inertia

Deq = Damping at the load of the rotational load and the armature

DL = Damping at the load of the rotational load

Da = Armature damping

r= Radius of rotational load

We also know such that the

! !!! = ! ! !__!__!_!_!_! !_!_!10!

! ! ! !! !! + !" ! ! ! ! !_!_!_!_!_! !_!_!_!_! ! ! !! !

Put (7) into (8) and assuming stiffness (K) = 0

! ! ! !! !! + !" ! ! !_________ (12)

Put (8), (9) and (12) into (7)

!!" ! [ ! ! + ! !! ! ! !! ! ! ! !! + !!! ! ! !! ! ! ! !!!

Thus

! ! = !!!!!

)! (! ! ! !!! ! ! !! !!

!! ! ! .127!(! .!"#$ !!" !! + !80 ! .!" ! ! !! .! 0!05! )

Dm = 3.273 x 10-3 Nms/rad

!! = (!!! !

! !! !! ! !! !! + ! !! )

!! = ! .!"# ! (! .92!10!! ! 800 ! .!" ! ! ! 0!05! )

Jm=0.03272 kgm2

Now we will find the value of

! =!!!!

+! !! !!!! !

! !3!!"# !!!10 ! 3!

0!03272 !(! !024! (! .!"#$ )

!0!03272! (! .!"# )!

! = ! .!

! =! !!!

! =! .!"

0.!"#$#

! = ! .!"#$

! = !! !

!! !!!!!!

! !! .024

(0!!"#$# ! !0.!"# )!0!0239! )

! ! !""! .!!

! ! !7337.77

! ! ! 0!2! ! 0!3056

i(s)

+ Ve(s) Ea(s) m(s) L(s) X(s)

Vi(s)

Vt +

Potentiometer (input)

Voltage produced by potentiometer (Vi) ! desired angle (i)

Vi (s) =K1i (s)

! ! !!!!!(!!

= ! ! ! 0!318

K1=based on the book 0.318

When we put it after summation the Vi=>Vi+Vt

K1 K2 Gm N1/N2

r

K3

i=>Ve

Preamplifier

Total voltage (EMF); Ea ! Ve(s)

Ea(s) = K2 Ve(s)

!! (!!!! (! )

! !! = ! .!"#

Ve

Motor and Load (Plant)

Angle of motor load (ms) ! Total voltage (EMF) Ea

ms = Gm Ea

! !!!)! !!!!

= ! ! !7337!77

! ! ! 0!2! + ! .3056

Changing gear from motor to load

Angle of load (Ls) ! angle of motor load (ms)

Ls=(Kx) ms

Since the changing of angle rotation is based on the gear ratio. Thus

Kx=N1/N2

! ! (!)! !(!!

=!!!!

And we also know such that

Tmotor Td

!!"#$$!!

=! !!!= ! .!"#

The rotational to translational

The relationship is

X(s) = r L (s)

!! ! !!! (! )

! ! ! 0!05

Displacement sensor (Feedback)

The voltage produced by the sensor is directly proportional to distance

(Vi) ! X(s)

Vi = K3 X(s)

! !! (! ) ! !! = 0!146

Overall open loop transfer function

N2/N1

! (! !!!

!!! ! !!!!!!)

!!! (! )

! ! ! !!!!!!

! !!!!!! (! )

!!! !!!!! (! )

!!!!!!! !!!

! ! !!!!

! !! !!!!!!! !

!

!! ! ! .!"# ! .!"#7337!77

! ! ! 0!! ! ! 0!3056 ! .!"# ! ! .05)

! ! !0!2094

! ! ! 0!2! + ! .!"#$

The feedback will be

!! =! !! !

!! !0!146! .318 = 0!4594

+

+

! ! =! !!"#$

!! + ! .! ! ! 0!3056

!! =0!146! .!"#

= ! .!"#!

Open loop transfer function

! ! =! .!"#$

!! ! !!!" ! !. !"#$

Closed Loop transfer function

!! =! !

! ! !! !

! ! !

0!2094!! ! 0!2! + ! !3056

! ! .!"#$!! + ! .! ! ! 0!3056 !0!4594!

!! =! . !"#$

! ! ! !! !" ! !. !"#$

Input Output

2.3 Matlab m.file

! ! !!!!"#$

!! + ! . !" + ! . !"#$

3.0 ANALYSIS BY LTI VIEWER

Figure 5- Step Response

Figure 6- Bode Diagram

Figures above show Step Response and Bode Diagram of our system. By using the open-loop

transfer function from modeling part,

! ! = !0!2094

!! !! .! ! + ! .!"#$

we analyses and then proceed using LTIviewer. From the step response graph above, we can

observe that the step response is under damped. The performance specification of our system is,

Percent overshoot, OS% = 56%

Settling time, Ts = 36.3s

There are some differ for the value of settling time. This is due to the formula used for manual

calculation which is conservative. So, the value might be manipulated by neglecting the

oscillating component. This will be compared with the desired specification.

In order to get a better response, the desired specification should be set to obtain new poles

which known as desired poles. Thus, we had set the desired specification for our system that is

30% for percent overshoot, OS% and 10s for settling time, Ts.

Figure 7- Pole/zero Diagram Pole-zero plots is a graphical representation of a transfer function in the complex plane or

also known as Z plane. Z plane consist an imaginary and real axis referring to the complex

valued of variable z. In the Z plane, poles are denoted by x sign and zeros are denoted by

o sign. In general, a transfer function has the form of X(s) =P(s)/Q(s). The two

polynomials P(s) and Q(s) can be used to determine poles and zeros of the system. Poles of a

transfer function happen when the values of the Laplace transform variable, s that cause the

Q(s) to become zero or cause the X(s) to become infinite. Zeros of a transfer function

happen when the values of the Laplace transform variable, s that cause the P(s) or X(s) to

become zero.

Pole-zero plots is used to determine the stability of the system. The system can be said stable

when all the zeros and poles are located at the left hand side of the Z plane.

For our system, we use open-loop transfer function T(s) to perform the analysis of the

system using LTIviewer. Figure above is a pole-zero diagrams that we get from the

LTIviewer.

4.0 DESIGN 4.1 Manually Calculation

Designing the controller manually

Determination of poles

! ! = !0!2094

! ! ! 0!2! + 0!3036

Using,

! ! ! ! 4!"2!

Poles, s = 0.! ! .!"#$

0.5418

0.5418

0.5418

- 0.5418

!

- 0.1

j!

From the formula:

Peak time, !T! = !

!! = !

! .!"#$ = 5.7984s

Settling time, ! ! = !

!" = !

!! = !

! .! = 40s

Time constant, Ta = !! = !

!!!"#$ = 4.7755s

Desired performance:

Desired percentage overshoot %OS = 30%

Desired settling time Ts = 10s

Damping ratio,

! !! ln (! !" !""# )

!! ! !"! (! .!" )

Damping ratio, != 0.3579

Settling time,

! ! !! ln!(0!02 ! ! !!

!! !

Natural frequency, !! = 1.1122 rad/s

Desired Poles, !! ! !! !! + !! ! ! ! !

Desired Poles, !! = ! 0!3981 ! 1!0385!!

From the formula:

Peak time, T! = !

!! = !

!!!"#$ = 3.0251 s

Settling time, ! ! = !!"

= !! !

= !! .!"#$

= 11.1783 s

Sketching the root locus

! ! =!

! ! ! 0!25! + !

1.0385

0.5418

0.5418

- 1.0385

!

- 0.3579

j!

Pole location as function of gain for the system

K Pole 1 Pole 2 0 -0.2 0

0.01 -0.1 -0.1 1 -0.1+0.99 -0.1-0.99 2 -0.1+1.41 -0.1-1.41 3 -0.1+1.73 -0.1-1.73 4 -0.1+1.99 -0.1-1.99 5 -0.1+2.23 -0.1-2.23

First of all, the poles are real for gains less than 0.01. When K less than 0.01, the system becomes overdamped. At gain of 0.01 the poles are real and multiple and hence critically damped. For gains above 0.01, the system is underdamped.

Since the settling time is inversely proportional to the real part of the complex pole for this second-order system. The conclusion is that regardless of the value of gain, the settling time for the system remains the same under all conditions of underdamped response.

As we increase the gain, the damping ratio diminishes, and the percent overshoot increases. The damped frequency of oscillation, which is equal to the imaginary part of the pole, also increases with an increase in gain, resulting in a reduction of peak time. Finally, since the root never crosses over into the right half-plane, the system is always stable, regardless of the value gain and can never break into a sinusoidal oscillation.

Designing desired poles.

Desired percentage overshoot ! !" ! !" !

Desired settling time !! = !"#

Damping ratio, ! =! ln !"# /100%

! + !" ! !"# /!""#

!"#$%& !!"#$%, ! =! ln0!3

! ! ! ln! (! .! )

!"#$%&' !"#$%! ! ! ! . !"#

!"##$%&!time,T! =! ln ! .!" 1 ! !

! ! !

!" !ln ! .!" 1 ! 0!358 !

0!358 ! !

!"#$%"&!!"#$%#&'(!!! ! !! !! !"#! !

!"#$%"& Poles!S! = ! ! ! + !! 1 ! !

Desired!Poles, S! = ! 0.358 1!11 ! ! !11 1 ! .!"# !

!"#$%"& !"#$%! ! ! = ! !!!"# ! !!!"#$%

Designing the Compensator

Desired Poles, ! ! ! ! . !"# + ! . !"#$%

Determination of angle deficiency:

To calculate designed compensator, we assume zero compensator, Zc=0.397

tan x !! .0385 0.!"#$

! !397 ! 0!1

! = !"# !! 1!672

! ! !"!!"

! !=180 59.12 =120.880

tan y =! !!"#$ + 0!5418

! .397 ! .!

! = tan! ! ! .!"

! = !" .!"

! !=180 79.36 =100.640

! !"# + ! ! + !!

! = ! !"# ! !"# .!! + !"" .!"

! !"# + 221!52

= !"!!"

!"# ! =x

1!0385

x ! tan! ! !0385

! ! !"#$%.!" 1!0385

! = ! .!"!#

!"#$ !" !"#$%&()"* , ! ! ! ! .!"# ! .!"!# = ! ! . !"#

Designed Compensator

!"#$%&()"* ! ! s !S ! Z!S ! P!

!"#$%&" !"#$%&'()"*! ! ! !! ! ! . !"#! ! !!!"#

4.2 SISO Design Tools

SISO Design Tool application can be used to design the appropriate compensator. In the

design, a lead compensator was added. The design requirement of 10s settling time was applied

along with 30% overshoot. The final compensator acquired is :

! ! = ! .!"#!(! ! ! .!"# !!(! + 1!69! )

In this case, a percent overshoot of 60.4% and settling time of 9.19s is obtained.

Figure 8- SISO Design

5.0 SIMULINK

After design of compensator, Simulink is used to simulate the system response to verify that the

control system is valid and the compensator is able to achieve the desired specifications. The

objectives of this step are:

(i) to verify that the equivalent unity-feedback system acquired is correct, and

(ii) to set a limit to the maximum and minimum opening of duct to represent the real

physical limits.

Simulink setup is shown below.

Figure 9-Simulink Setup

Figure 10- Step response by Simulink

6.0 CONCLUSION

In this project, we have modeled and analyzed a projector systems response to a step user input.

The open loop system is a second-order control system. Two different lead compensators have

been designed to allow the system to meet the desired performance specification. The first

compensator is designed manually, and is able to offer 30% overshoot and 10s settling time. The

second compensator is designed using SISOTools optimization method and achieved 67.3%

overshoot and 33s settling time.