Lab Report Rotary Complete

7/30/2019 Lab Report Rotary Complete

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Table of Contents1.0 INTRODUCTION ........................................................................................................ 2

1.1 Background ............................................................................................................. 2

b. Position control ........................................................................................................ 2

1.2 Theory ..................................................................................................................... 3

2.0 OBJECTIVE ................................................................................................................ 5

3.0 APPARATUS .............................................................................................................. 5

4.0 PROCEDURE ............................................................................................................. 6

5.0 PRE LAB ASSIGNMENT ............................................................................................ 9

5.1 Derivations .............................................................................................................. 9

5.2 Controller Design ...................................................................................................12

6.0 RESULTS AND DISCUSSIONS ................................................................................14

7.0 CONCLUSIONS ........................................................................................................21

8.0 REFERENCES ..........................................................................................................22


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1.0 INTRODUCTION

1.1 Background

a. Servomechanism

A servomechanism, sometimes shortened to servo, is an automatic device that

uses error-sensing negative feedback to correct the performance of a mechanism.

The term correctly applies only to systems where the feedback or error-correction

signals help control mechanical position, speed or other parameters. For example,

an automotive power window control is not a servomechanism, as there is no

automatic feedback that controls positionthe operator does this by observation.

By contrast the car's cruise control uses closed loop feedback, which classifies it

as a servomechanism.

b. Position control

A common type of servo providesposition control. Servos are commonly electrical

or partially electronic in nature, using an electric motoras the primary means of

creating mechanical force. Other types of servos use hydraulics, pneumatics,

ormagnetic principles. Servos operate on the principle of negative feedback,

where the control input is compared to the actual position of the mechanical

system as measured by some sort oftransducerat the output. Any difference

between the actual and wanted values (an "error signal") is amplified (and

converted) and used to drive the system in the direction necessary to reduce or

eliminate the error. This procedure is one widely used application ofcontrol theory.
http://en.wikipedia.org/wiki/Negative_feedbackhttp://en.wikipedia.org/wiki/Feedbackhttp://en.wikipedia.org/wiki/Cruise_controlhttp://en.wikipedia.org/wiki/Control_theoryhttp://en.wikipedia.org/wiki/Electric_motorhttp://en.wikipedia.org/wiki/Forcehttp://en.wikipedia.org/wiki/Hydraulicshttp://en.wikipedia.org/wiki/Pneumaticshttp://en.wikipedia.org/wiki/Magnetichttp://en.wikipedia.org/wiki/Transducerhttp://en.wikipedia.org/wiki/Control_theoryhttp://en.wikipedia.org/wiki/Control_theoryhttp://en.wikipedia.org/wiki/Transducerhttp://en.wikipedia.org/wiki/Magnetichttp://en.wikipedia.org/wiki/Pneumaticshttp://en.wikipedia.org/wiki/Hydraulicshttp://en.wikipedia.org/wiki/Forcehttp://en.wikipedia.org/wiki/Electric_motorhttp://en.wikipedia.org/wiki/Control_theoryhttp://en.wikipedia.org/wiki/Cruise_controlhttp://en.wikipedia.org/wiki/Feedbackhttp://en.wikipedia.org/wiki/Negative_feedback


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1.2 Theory

We shall begin by examining the electrical component of the motor first. In Figure 1, you

see the electrical schematic of the armature circuit.

Figure 1.1: Armature circuit in the time-domain

Using Kirchhoffs voltage law, we obtain the following equation:

Since Lm


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Where is the load torque seen thru the gears. And is the efficiency of the

gearbox.

We now apply the 2nd law of motion at the load of the motor:

Where Beq is the viscous damping coefficient as seen at the output.

Substituting [3.4] into [ 3.5], we are left with:

that and (where is the motor efficiency),We know

we can re-write [3.6] as:

Finally, we can combine the electrical and mechanical equations by substituting [3.3] into

[3.7], yielding our desired transfer function:

Where:

This can be interpreted as the being the equivalent moment of inertia of the motor systemas seen at the output.


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2.0 OBJECTIVE

The objective in this experiment is to introduce the student to the fundamentals of control

using the PID family of compensators.

At the end of this session, you should know the following:

How to mathematically model the servo plant from first principles.

An understanding of the different tuning parameters in the controller.

To design and simulate a PV controller to meet the required specifications .

To implement your controller and evaluate its performance.

3.0 APPARATUS

To complete this lab, the following hardware is required:

[1] Quanser UPM 2405/1503 Power Module or equivalent.

[1] Quanser MultiQ PCI / MQ3 or equivalent.

[1] Quanser SRVO2-(E) servo plant.

[1] PC equipped with the required software as stated in the WinCon user manual.

Figure 1.2: Quanser Consulting Plant SRV-02 components


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Figure 1.3: Quanser Consulting Plant SRV-02

4.0 PROCEDURE

1. Wiring and Connections

The first task upon entering the lab is to ensure that the complete system is wired as

described in the SRV02 Experiment #0 - Introduction. If you are unsure of the wiring,

please refer to the SRV02 User Manualor ask for assistance from a TA assigned to

the lab.

Now that all the signals are connected properly, start-up MATLAB and start Simulink.

You are now ready to begin the lab.

2. Controller Specifications

This lab requires you to design a Proportional + Velocity (PV) controller to control the

position of the load shaft with the following specifications:

1. The Overshoot should be less than 5% ( 0.707).

2. The time to first peak should be 100ms (Tp = 0.100).

3. Simulation of the Plant

In Simulink, open a model called s_position_pv.mdl. This model includes the

modeled plant (SRV-02 Plant Model), as well as the PV controller. Kp and Kv are

both set by slider gains. Before you begin, you must run an M-File called

Setup_SRV02_Exp1.m. This file initializes all the motor parameters and gear ratios.

Click on Simulation->Start, and bring up the Simulated Position scope. As you

monitor the response, adjust Kp and Kv using the slider gains. Try a variety ofcombinations, and note the effects of varying each parameter.


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Make a table of system characteristics ( and ) with respect to ch anges in Kp

and

Kv. (Hold one variable constant while adjusting the other).

Does the system response react to how you had theorized in section 3.1?

Now that you are familiar with the actions of each parameter, enter in the designed

Kp and Kv that you had calculated to meet the system requirements.

*Note: the values should fallwithin the slider limits.

Does the response look like you had expected? What is your percent

overshoot?

Calculate yourTp. Does it match the requirements?

*Hint: To get a better resolution when calculatingTp, decrease the time range under

the parameters option of the scope.

If the simulated response is as expected, you can move on and implement your

controller, if you are close to meeting the requirements, try fine-tuning your

parameters to achieve the desired response. If the response is far from the

specifications, you should re-iterate your design process and re-calculate your

controller gains.

4. Implementation of the Controller

After successfully simulating your controller and achieving your desired response, you

are now ready to implement your controller and observe its effect on the physical

plant.

Open a Simulink model called q_position_pv_e.mdlor q_position_pv_pot- ask a

TAassigned to this lab if you are unsure which model is to be used in the lab . The

model has 2 identical closed loops; one is connected similar to the simulation block of

the previous section, and the other loop has the actual plant in it. To better familiarize

yourself with the model, it is suggested that you open both sub-systems to get abetter idea of the systems as well as take note of the I/O connections. In the SRV02

plant block (blue), you will see a gain of 1/K_Cable to normalize the system due to

our use of a gain cable (to enable a greater control signal being fed into the plant).

*Note: In place of a standard derivative block in the PVcontroller, we have place a

derivative with a filter in order to eliminate any high frequencies from reaching the

plant as high frequencies will in the long term damage the motor.


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Before running the model, you must set your final values of Kp and Kv in the

MATLAB workspace (type it in MATLAB). You can now build the system using the

WinCon->Buildmenu. You will see the model compile, and then you can use the

WinCon Server to run the system (click on the start button). Your plant should now be

responding and tracking a square wave to the commanded angle (Setpoint Amplitude

(deg)).

Plot the Measured Theta (deg) as well as the Setpoint Amplitude (deg) and the

SimulatedTheta (deg). This is done by clicking on the scope button in WinCon and

choosing Measured Theta (deg). Now you must choose the Setpoint Amplitude (deg)

and the Simulated Theta (deg) signals thru the Scope->File->Variables menu.

How does your actual plant response compare to the simulated response?

Is there a discrepancy in the results? If so, why?

Calculate your system Tp and %OS. Are the values what you had expected?

*You can calculate these parameters by saving these traces as an m-file and making

our calculations in MATLAB. You could also make your calculations directly from the

WinCon scope by zooming in on the signals. It is suggested to make these

calculations thru MATLAB as this method will provide greater accuracy.

If you are sufficiently happy with your results and your response looks similar to

Figure 4.1 below, you can move on and begin the report for this lab. Remember,

there is no such thing as a perfect model, and your calculated parameters were

based on the plant model. A control design will usually involve some form of fine-

tuning, and will more than likely be aniterative process.

At this point, you should be fine-tuning yourKp and Kv based on your findings from

above (use the table created in section 4.3of this lab as a guide) to ensure your

responsematches the system requirements as seen in Figure 3.


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Figure 4.1: Step Response to a Command of 30 degrees

We can see by looking at Figure 3, that the position response has a 100ms time to

peak and an overshoot of less than 5%. The system requirements have been met andimplemented using a PV controller.

5.0 PRE LAB ASSIGNMENT

5.1 Derivations

The schematics for the electric motor is given as below,

Figure 5.1: Electric Motor Schematics


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The mechanical torque on the shaft is proportional to the current Im passing through the

system,

Tm =mKtIm

Where m is the motor efficiency and Kt is the motortorque constant that represents thelumped electrical terms. The backemf is produced as the armature rotates in the magneticfield, resulting in the following relationship,

Eemf=

Where is the backemf constant and is the motor shaft velocity. Using Kirchoffsvoltage law, which states that the voltage drop across a close circuit must be zero, the

following equation is obtained.

Vm RmIm Lm( ) Eemf= 0

Since = 100 H(The small value can also be deducted from the relative small size of thestator.),and


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Note that = and = , the above equation can be rewritten as

= - -

Differentiate

to obtain

and plug in the equation for

, the equation above

becomes

=

- -

Take the Laplace transforms and rearranges the terms, the desired transfer function

becomes

=

Where ( ) = 0.0052, 0.1894, , andb = 0.333. The parameters used are listed in the table below,

Table 5.1: Motor Parameters


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5.2 Controller Design

The design specifications for this system are

Zero steady state error

5% 0.1 sec

The form of transfer function for the system as derived in the previous section is,

G (s) =

=

With 0, this becomes a type 1 system, and will have zero steady state error for a stepinput, thus no additional integrator is needed.

A PD controller of the form will be used to obtain desired transientresponse for the system.

The closed loop transfer function becomes

=

=

To approximate the 2nd order system response, disregard the additional zero in the

numerator and assume the following form for the denominator,

( ) --1

Using the design criterions, the damping ratio and natural frequency can be calculatedas

0.1 =


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Take the natural log of each side for and solve for, obtaining 0.698. Plug this valueinto the equation for to obtain = 43.8.

Compare the 2nd order response with the closed loop transfer function, values for and are obtained,

= 29.96

= 0.389

Note that these values are only approximations of the transient behavior; the additional zero

is likely to decrease the rise time but also increase the overshoot in the process.

In order to satisfy requirements for the actual system, the controller gains must be tuned. The

following diagram can be sketched to give an idea of required location of the poles.

Figure 5.2: Design Requirement


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6.0 RESULTS AND DISCUSSIONS

Table 6.1: Results From The Experiment (Kp fixed)

Kp Kv Peak Time, Tp (s)Maximum Overshoot,

MO (%)

25 -0.1 0.0875 60

25 0 0.07 46.67

25 0.2 0.1 20

25 0.3 0.1 10

25 0.4 0.1 3.33

The table 6.1 above shows the results from the experiment when value of Kp is fixed and

value of Kv adjusted. From the table, it can be observed that when the value of the required

design specifications obtained when value of Kp is 25 and value of Kv is 0.4. It is also

indicated that when the value of Kv increased, the value of Tp also increased while the value

of MO decreased. This is consistent with the assumptions made in pre-lab assignment.

Figure 6.1: Block Diagram

The figure 6.1 shows the block diagram of the PV controller Simulation at Kp 25 and Kv 0.4.

after being simulated the graphs below obtained.


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Figure 6.2: Graph Obtained after Simulation (Kp fixed)

The figure 6.2 above shows the result graph obtained after the simulation. The graph

obtained is similar to the graph sample in the procedure. The peak time obtained is 0.1

second while the percentage of maximum overshoot is 3.33 %.


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Figure 6.3: Graph of System Response (Simulation)

Figure 6.4: Graph of System Response (Actual)

Figure 6.3 shows the system response of simulated plant. While, figure 6.4 shows the

system response of actual plant after the controller is implemented. From the results, it canbe observed that there is discrepancy between the actual and simulation. It can be seen that


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in actual system response, the maximum overshoot and peak time are decreased. this can

be explain by the action of PD controller. The proportional controller will decrease the rise

time and also the peak time of the system. While, the derivative controller will decrease the

maximum overshoot of the system. Even though the proportional controller will also increase

the maximum overshoot, the increment will be eliminated by the action of derivative

controller.

Table 6.2: Results From The Experiment (Kp fixed)

No. Kv Kp Peak time, Tp (s) Maximum overshoot, Mo (%)

1 0 8 0.20 5.00

2 0 7 0.10 3.33

Table 6.2 above shows the data for simulation of PV controller with K v is fixed to 0 and Kp is

varies. If Kp is keep decreasing, the values of Tp and Mo will decrease as per pre lab

assignment had theorized.

Figure 6.5: Block diagram

Figure 6.5 above shows the block diagram for the simulation of PV controller with at Kp=7

and Kv is fixed to 0. After being simulated the graphs below obtained.


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Figure 6.6: Graph obtained from the simulation of PV controller (Kv fixed)

Figure 6.6 above shows the response of simulation PV controller with the calculated Kp = 7

and Kv = 0. The response that obtained matched what we had expected. The percentage

overshoot Mo = 3.33% and Tp = 0.1 both are less than Mo = 5% and Tp=0.1 respectively.


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Figure 6.7: Graph of System Response (Simulation)

Figure 6.8: Graph of System Response (Actual)


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Figure 6.7 shows the system response of simulated plant. While, figure 6.8 shows the

system response of actual plant after the controller is implemented. From the results, it can

be observed that there is discrepancy between the actual and simulation. It can be seen that

in actual system response, the maximum overshoot and peak time are decreased. this can

be explain by the action of PD controller. The proportional controller will decrease the rise

time and also the peak time of the system. While, the derivative controller will decrease the

maximum overshoot of the system. Even though the proportional controller will also increase

the maximum overshoot, the increment will be eliminated by the action of derivative

controller.

There are several limitations during the course of this lab. First, the lack of information and

knowledge about PD controller. To overcome the problem, we did some research through

internet and books. Second, there are some procedures that we do not understand. In order

to understand the procedure, we ask the technician's guidance during the experiment is

conducted. Lastly, the main problem is to obtain the mathematical model of the plant. We

overcome this problem, we refer to notes, textbook and advice from the lecturer.

After completing this lab, we know that the controller could not be arbitrarily applied to any

system. The specification and design of the need to be considered first, before implementing

any types of controller to avoid any unwanted result. Wrongly applied the controller can result

in instability of the system.

When the integral control action applied to the system, it could bring benefits to the system.

Despite decreasing the rise time, it also can eliminate the steady state error, which will bring

more stability of the system and reduced system response error.


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7.0 CONCLUSIONS

From the experiment, it can be concluded that after fine-tuning the controller, there are two

combinations of Kp and Kv could be use which are Kp=25,Kv=0.4 and Kp=7, Kv=0. The

proportional and derivates control action effect can concluded as in table below.

Parameter Rise Time Overshoot Settling TimeSteady state

error

Kp Decrease Increase Small change Decrease

Kd Small Changes Decrease Decrease None

It also can be concluded that, effect of adjusting Kp and Kv is consisten with the pre-lab

assignment.


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8.0 REFERENCES

1. Chang, A., Gurocak, H., Karam, P., Levis, M., & Apkarian, J. (2011). Quanser Course

Material Sample. Quanser.

2. Jalili, D. N. (2002). Modelling and Control of Electromechanical Systems.

3. Muthuswamy, D. (2012). Position Control Design:Position and Rate Feedback.

4. Ogata, K. (2012). Modern Control Engineering. New Jersey: Pearson Education Inc.

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