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DC Motor Speed Modeling

Lecture delivered by:

V SitaramGupta

Session-3


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DC Motor Speed ModelingThe DC motor has been the workhorse in industry for many reasons including good

torque speed characteristics.

It is a common actuator in control systems. It directly provides rotary motion and,coupled with wheels or drums and cables, can provide transitional motion.

The electric circuit of the armature and the free body diagram of the rotor are

shown in the following Figure.

We develop here the transfer function of a separately excited armature controlled

DC motor.

MotorGenerator

Mechanical

energy

(T, )

VA

RA IA

VF

RF

LF

IF L

k

Jmotor & load

T

Field circuit Armature circuit


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Physical Parameters

Electrical ResistanceR= 1

Electrical InductanceL = 0.5 H

Input Voltage V

Electromotive Force Constant K = 0.01 nm/A

Moment of Inertia of the RotorJ= 0.01 kg.m2/s2

Damping Ratio of the Mechanical System b = 0.1 Nms

Position of the Shaft

The rotor and shaft are assumed to be rigid

The motor torque T is related to the armature current by a constant Kt

The back emf, e, is related to the rotational velocity

KVRidtdiL

KibJ

Ke

iKT

e

t


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Speed Control

Speed Control by Varying Circuit Resistance: The operating speed can only be

adjusted downwards by varying the external resistance,Rext

Speed Control by Varying Excitation Flux:

Speed Control by Varying Applied Voltage: Wide range of control 25:1; fast

acceleration of high inertia loads.

Electronic Control.

rad/s

22

a

exta

aa

aaa

mk

RR

k

Va

k

IRV

1

2

2

1

m

m


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Transfer Function

2

))((

)()()(

)()()(

KRLsbJs

K

V

sKsVsIRLs

sKIsbJss

Controller PlantR u


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A common actuator in control systems is the DC motor. It

directly provides rotary motion and, coupled with wheels or

drums and cables, can provide transitional motion.

The electric circuit of the armature and the free body diagram

of the rotor are shown in the following figure:

6

Physical setup and system equations


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Specifications

For this example, we will assume the following values for thephysical parameters.

moment of inertia of the rotor (J) = 0.01 kg.m^2/s^2

damping ratio of the mechanical system (b) = 0.1 Nms

electromotive force constant (K=Ke=Kt) = 0.01 Nm/Ampelectric resistance (R) = 1 ohm

electric inductance (L) = 0.5 H

input (V): Source Voltage

output (theta): position of shaft



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Modeling The motor torque, T, is related to the armature current, i, by a

constant factor Kt. The back emf, e, is related to the rotationalvelocity by the following equations:

In SI units (which we will use), Kt (armature constant) is equal

to Ke (motor constant).

From the figure above we can write the following equations based

on Newton's law combined with Kirchhoff's law:


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Transfer Function

Using Laplace Transforms, the above modeling equations canbe expressed in terms of s.

By eliminating I(s) we can get the following open-looptransfer function, where the rotational speed is the output and

the voltage is the input.


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State-Space

In the state-space form, the equations above can beexpressed by choosing the rotational speed and electric

current as the state variables and the voltage as an input.

The output is chosen to be the rotational speed


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Design Requirements

First, our uncompensated motor can only rotate at 0.1rad/sec with an input voltage of 1 Volt (this will be

demonstrated later when the open-loop response is

simulated).

Since the most basic requirement of a motor is that it

should rotate at the desired speed, the steady-state

error of the motor speed should be less than 1%.


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Design Requirements The other performance requirement is that the motor

must accelerate to its steady-state speed as soon as it

turns on.

In this case, we want it to have a settling time of 2

seconds. Since a speed faster than the reference maydamage the equipment, we want to have an overshoot

of less than 5%.

If we simulate the reference input (r) by an unit step

input, then the motor speed output should have:

Settling time less than 2 seconds, Overshoot less than

5%, Steady-state error less than 1%

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MATLAB Representation

and open-loop response 1. Transfer Function We can represent the above transfer function into MATLAB by defining

the numerator and denominator matrices as follows:

Create a new m-file and enter the following commands:

J=0.01;

b=0.1;

K=0.01;

R=1;

L=0.5; num=K;

den=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)];

motor=tf(num,den);

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Output Response

Now let's see how the original open-loop system performs. Addthe following commands onto the end of the m-file and run it in

the MATLAB command window:

step(motor,0:0.1:3);

title('Step Response for the Open Loop System'); The output plot is as follows:


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Output Response

From the plot we see that when 1 volt is applied tothe system, the motor can only achieve a maximum

speed of 0.1 rad/sec, ten times smaller than our

desired speed.

Also, it takes the motor 3 seconds to reach its steady-

state speed; this does not satisfy our 2 seconds

settling time criterion

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State Space 2. State-Space

We can also represent the system using the state-space equations. Try thefollowing commands in a new m-file.

J=0.01;

b=0.1;

K=0.01;

R=1;

L=0.5;

A=[-b/J K/J -K/L -R/L];

B=[0 1/L];

C=[1 0];

D=0;

motor_ss=ss(A,B,C,D);

step(motor_ss)

Run this m-file in theMATLAB command window, and

you should get the same output as

the one shown above.


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PID Design Method for

DC Motor Speed Control


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Modeling

From the main problem, the dynamic equations andthe open-loop transfer function of the DC Motor are

and the system schematic looks like:


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Modeling

For the original problem setup and the derivation of the above equations, please

refer to the Modeling a DC Motor page.

With a 1 rad/sec step input, the design criteria are:

Settling time less than 2 seconds

Overshoot less than 5%

Steady-stage error less than 1%

Now let's design a PID controller and add it into the system. First create a new m-

file and type in the following commands.

J=0.01;

b=0.1;

K=0.01;

R=1;

L=0.5;

num=K;

den=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)];

motor=tf(num,den);19

Recall that the transfer function for a

PID controller is:


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Proportional control

Let's first try using a proportional controller with a gain of

100. To determine the closed-loop transfer function, we usethe feedback command. Add the following code to the end of

your m-file:

Kp=100;

contr=Kp;

sys_cl=feedback(contr*motor,1);

Now let's see how the step response looks. Add the following

to the end of your m-file, and run it in the command window:

t=0:0.01:5;

step(sys_cl,t)

title('Step response with Proportional Control')

You should get the following plot:20


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Response

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PID control

From the plot above we see that both the steady-stateerror and the overshoot are too large.

Recall from the PID tutorial page that adding an

integral term will eliminate the steady-state error and

a derivative term will reduce the overshoot.

Let's try a PID controller with small Ki and Kd.

Change your m-file so it looks like the following.

Running this new m-file gives you the following plot. J=0.01;

b=0.1;

K=0.01; 22


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Program

R=1; L=0.5;

num=K;

den=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)];

motor=tf(num,den);

Kp=100;

Ki=1;

Kd=1;

contr=tf([Kd Kp Ki],[1 0]); sys_cl=feedback(contr*motor,1);

step(sys_cl)

title('PID Control with small Ki and Kd')

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Result

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Tuning the gains

Now the settling time is too long. Let's increase Ki toreduce the settling time. Go back to your m-file and

change Ki to 200. Rerun the file and you should get a

plot like this:

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Gains

Now we see that the response is much faster thanbefore, but the large Ki has worsened the transient

response (big overshoot). Let's increase Kd to reduce

the overshoot. Go back to the m-file and change Kd

to 10. Rerun it and you should get this plot:

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So now we know that if we use a

PID controller with

Kp=100,

Ki=200,Kd=10,

all of our design requirements

will be satisfied.


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DC Motor Speed Modeling in Simulink

Physical setup

A common actuator in control systems is the DC

motor.

It directly provides rotary motion and, coupled with

wheels or drums and cables, can provide transitionalmotion.

The electric circuit of the armature and the free body

diagram of the rotor are shown in the following

figure:

27


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Specifications

For this example, we will assume the following values for thephysical parameters.

moment of inertia of the rotor (J) = 0.01 kg.m^2/s^2

damping ratio of the mechanical system (b) = 0.1 Nms

electromotive force constant (K=Ke=Kt) = 0.01 Nm/Ampelectric resistance (R) = 1 ohm

electric inductance (L) = 0.5 H

input (V): Source Voltage

output (theta): position of shaft


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Modeling

The motor torque, T, is related to the armaturecurrent, i, by a constant factor Kt.

The back emf, e, is related to the rotational velocity by

the following equations:

In SI units (which we will use), Kt (armature constant)

is equal to Ke (motor constant).

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Building the Model

This system will be modeled by summing the torquesacting on the rotor inertia and integrating the

acceleration to give the velocity.

Also, Kirchoff's laws will be applied to the armature

circuit.

Open Simulink and open a new model window.

First, we will model the integrals of the rotational

acceleration and of the rate of change of armaturecurrent.

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Modeling

Insert an Integrator block (from the Linear blocklibrary) and draw lines to and from its input and

output terminals.

Label the input line "d2/dt2(theta)" and the output

line "d/dt(theta)" as shown below. To add such a

label, double click in the empty space just above the

line.

Insert another Integrator block above the previousone and draw lines to and from its input and output

terminals.

Label the input line "d/dt(i)" and the output line "i"31


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Simulink

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Model

Next, we will start to model both Newton's law and Kirchoff'slaw. These laws applied to the motor system give the

following equations:

The angular acceleration is equal to 1/J multiplied by the sum

of two terms (one pos., one neg.). Similarly, the derivative of

current is equal to 1/L multiplied by the sum of three terms(one pos., two neg.).

Insert two Gain blocks, (from the Linear block library) one

attached to each of the integrators.

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Model

Edit the gain block corresponding to angular acceleration by

double-clicking it and changing its value to "1/J".

Change the label of this Gain block to "inertia" by clicking on

the word "Gain" underneath the block.

Similarly, edit the other Gain's value to "1/L" and it's label to

Inductance.

Insert two Sum blocks (from the Linear block library), one

attached by a line to each of the Gain blocks.

Edit the signs of the Sum block corresponding to rotation to

"+-" since one term is positive and one is negative.

Edit the signs of the other Sum block to "-+-" to represent the

signs of the terms in Kirchoff's equation.

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Simulink

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Procedure

Now, we will add in the torques which are represented inNewton's equation. First, we will add in the damping torque.

Insert a gain block below the inertia block, select it by single-

clicking on it, and select Flip from the Format menu (or type

Ctrl-F) to flip it left-to-right. Set the gain value to "b" and rename this block to "damping".

Tap a line (hold Ctrl while drawing) off the rotational

integrator's output and connect it to the input of the damping

gain block. Draw a line from the damping gain output to the negative input

of the rotational Sum block.

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Procedure

Next, we will add in the torque from the armature. Insert a gain block attached to the positive input of

the rotational Sum block with a line.

Edit it's value to "K" to represent the motor constantand Label it "Kt".

Continue drawing the line leading from the current

integrator and connect it to the Kt gain block.

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Simulink

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Procedure

Now, we will add in the voltage terms which are representedin Kirchoff's equation. First, we will add in the voltage drop

across the coil resistance.

Insert a gain block above the inductance block, and flip it left-

to-right. Set the gain value to "R" and rename this block to

"Resistance".

Tap a line (hold Ctrl while drawing) off the current integrator's

output and connect it to the input of the resistance gain block. Draw a line from the resistance gain output to the upper

negative input of the current equation Sum block.

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Procedure

Next, we will add in the back emf from the motor. Insert a gain block attached to the other negative

input of the current Sum block with a line.

Edit it's value to "K" to represent the motor constantand Label it "Ke".

Tap a line off the rotational integrator output and

connect it to the Ke gain block.

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Simulink

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Procedure

The third voltage term in the Kirchoff equation is the

control input, V. We will apply a step input.

Insert a Step block (from the Sources block library)

and connect it with a line to the positive input of the

current Sum block. To view the output speed, insert a Scope (from the

Sinks block library) connected to the output of the

rotational integrator.

To provide a appropriate unit step input at t=0,

double-click the Step block and set the Step Time to

"0".42


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Simulink

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Open-loop response

To simulate this system, first, an appropriate simulation timemust be set. Select Parameters from the Simulation menu and

enter "3" in the Stop Time field. 3 seconds is long enough to

view the open-loop response. The physical parameters must

now be set. Run the following commands at the

MATLAB prompt:

J=0.01;

b=0.1; K=0.01;

R=1;

L=0.5;44


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Response

Run the simulation (Ctrl-t or Start on the Simulation

menu).

When the simulation is finished, double-click on the

scope and hit its autoscale button.

You should see the following output.

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Extracting a Linear Model into MATLAB

A linear model of the system (in state space ortransfer function form) can be extracted from a

Simulink model into MATLAB.

This is done through the use of In and Out

Connection blocks and the ATLAB function linmod.

First, replace the Step Block and Scope Block with an

In Connection Block and an Out Connection Block,

respectively (these blocks can be found in theConnections block library).

This defines the input and output of the system for the

extraction process.46


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Simulink

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Procedure

Save your file as "motormod.mdl" (select Save Asfrom the File menu). MATLAB will extract the linear

model from the saved model file, not from the open

model window. At the MATLAB prompt, enter the

following commands:

[A,B,C,D]=linmod('motormodel')

[num,den]=ss2tf(A,B,C,D)

You should see the following output, providing bothstate-space and transfer function models of the

system.

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Matrix values

A = -10.0000 1.0000-0.0200 -2.0000

B = 0 2

C = 1 0

D = 0

num = 0 0.0000 2.0000

den = 1.0000 12.0000 20.0200

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Procedure

To verify the model extraction, we will generate anopen-loop step response of the extracted transfer

function in MATLAB. Enter the following command

in MATLAB.

step(num,den);

You should see the following plot which is equivalent

to the Scope's output.

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Response

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Implementing Lag Compensator Control

In the motor speed control root locus example a LagCompensator was designed with the following

transfer function.

To implement this in Simulink, we will contain the

open-loop system from earlier in this page in a

Subsystem block.

Create a new model window in Simulink.

Drag a Subsystem block from the Connections blocklibrary into your new model window.

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Simulink

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Procedure

Double click on this block. You will see a blank window

representing the contents of the subsystem (which is currentlyempty).

Open your previously saved model of the Motor Speed

system, motormod.mdl.

Select Select All from the Edit menu (or Ctrl-A), and selectCopy from the Edit menu (or Ctrl-C).

Select the blank subsystem window from your new model and

select Paste from the Edit menu (or Ctrl-V). You should see

your original system in this new subsystem window. Close this window. You should now see input and output

terminals on the Subsystem block.

Name this block "plant model".54


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Simulink

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Procedure

Now, we will insert a Lag Compensator into a closed-loop around the plant model. First, we will feed back

the plant output.

Draw a line extending from the plant output.

Insert a Sum block and assign "+-" to it's inputs.

Tap a line of the output line and draw it to the

negative input of the Sum block.

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Simulink

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Procedure

The output of the Sum block will provide the errorsignal. We will feed this into a Lag Compensator.

Insert a Transfer Function Block after the summer

and connect them with a line.

Edit this block and change the Numerator field to

"[50 50]" and the Denominator field to "[1 0.01]".

Label this block "Lag Compensator".

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Simulink

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d


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Procedure

Finally, we will apply a step input and view theoutput on a scope.

Attach a step block to the free input of the feedback

Sum block and attach a Scope block to the plant

output.

Double-click the Step block and set the Step Time to

"0".

60

Si li k


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Simulink

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You can download our version of the closed-loop

system here

Cl d l


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Closed-loop response

To simulate this system, first, an appropriate simulation timemust be set. Select Parameters from the Simulation menu and

enter "3" in the Stop Time field.

The design requirements included a settling time of less than 2

sec, so we simulate for 3 sec to view the output. The physical

parameters must now be set. Run the following commands at

the MATLAB prompt:

J=0.01;

b=0.1;

K=0.01;

R=1;

L=0.5;62

C l i


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Conclusion

Run the simulation (Ctrl-t or Start on the Simulationmenu).

When the simulation is finished, double-click on the

scope and hit its autoscale button. You should see the

following output

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