Example: DC Motor Speed Modeling
Physical setup and system equations
Design requirements
Matlab representation and open-loop response
Physical setup and system equations
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. These values were derived by experiment from an actual motor
in Carnegie Mellon's undergraduate controls lab.
* 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).
From the figure above we can write the following equations based on
Newton's law combined with Kirchhoff's law:
1. Transfer Function
Using Laplace Transforms, the above modeling equations can be expressed in terms of s.
By eliminating I(s) we can get the following open-loop transfer function, where
the rotational speed is the output and the voltage is the input.
2. State-Space
In the state-space form, the equations above can be expressed 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.
Design requirements
First, our uncompensated motor can only rotate at 0.1 rad/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%. 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 may damage 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%
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)];
Now let's see how the original open-loop system performs. Add the
following commands onto the end of the m-file and run it in the Matlab
command window:
step(num,den,0:0.1:3)
title('Step Response for the Open Loop System')
You should get the following plot:
From the plot we see that when 1 volt is applied to the 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.
2. State-Space
We can also represent the system using the
state-space equations. Try the following 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;
step(A, B, C, D)
Run this m-file in the Matlab command window, and you should get the same output as the one shown above.
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