Example: DC Motor Speed Modeling in Simulink

Physical setup
Building the model
Open-loop response
Extracting the Model
Implementing PI control
Closed-loop response

## 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 transitional motion. The electric circuit of the armature and the free body diagram of the rotor are shown in the following figure:

For this example, we will assume the following values for the physical 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/Amp
* electric resistance (R) = 1 ohm
* electric inductance (L) = 0.5 H
* input (V): Source Voltage
* output (theta): position of shaft
* The rotor and shaft are assumed to be rigid

The motor torque, T, is related to the armature current, 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).

## Building the Model

This system will be modeled by summing the torques acting 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 armature current.
• Insert an Integrator block (from the Linear block library) 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 previous one and draw lines to and from its input and output terminals.
• Label the input line "d/dt(i)" and the output line "i".

Next, we will start to model both Newton's law and Kirchoff's law. 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.
• 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.

Now, we will add in the torques which are represented in Newton'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.
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 constant and Label it "Kt".
• Continue drawing the line leading from the current integrator and connect it to the Kt gain block.

Now, we will add in the voltage terms which are represented in 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.
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 constant and Label it "Ke".
• Tap a line off the rotational integrator output and connect it to the Ke gain block.

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".

## Open-loop response

To simulate this system, first, an appropriate simulation time must 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;
```
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.

## Extracting a Linear Model into MATLAB

A linear model of the system (in state space or transfer 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 MATLAB 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 the Connections block library). This defines the input and output of the system for the extraction process.

Save your file as "motormod.mdl" (select Save As from 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 both state-space and transfer function models of the system.
```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
```
To verify the model extraction, we will generate an open-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.

## Implementing Lag Compensator Control

In the motor speed control root locus example a Lag Compensator 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 block library into your new model window.

• Double click on this block. You will see a blank window representing the contents of the subsystem (which is currently empty).
• Open your previously saved model of the Motor Speed system, motormod.mdl.
• Select Select All from the Edit menu (or Ctrl-A), and select Copy 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".

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.

The output of the Sum block will provide the error signal. 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".

Finally, we will apply a step input and view the output 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".

## Closed-loop response

To simulate this system, first, an appropriate simulation time must 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;
```
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.