TU Angular Speed of a DC Motor Project

DescriptionIn this project, we will design feedback control for angular speed of a DC motor. DC motors are
often the plant/process models used in vehicle position control and robotics. See figure below for
a photograph of a DC motor, its equivalent model, and parameter definitions for the model
( , , , ). The armature circuit is placed in a fixed magnetic field. When current passes
through the armature circuit, a force will be exerted on the circuit causes it to rotate about the rotor.
The rotation of the armature circuit also causes and induced voltage or back emf indicated as .
DC motor and its equivalent model
Derive the transfer function from voltage to rotational velocity
expression in the following standard forms.
( ) =
Θ( )
( )
=
2
0
.
2 + 1 + 0
Θ( )
( )
. Write your final
Enter your final expressions for 0 , 1 , and 0 in terms of DC motor parameters
( , , , ) in the table below:
0
1
3
0
Suppose we have the following parameter values
J=0.01;
b=0.1;
K=Kt=Ke=0.01;
R=1;
L=0.5.
Find the step response analytically in for the DC motor transfer function ( ) =
Θ( )
( )
. Show all your analytical work using Equation editor (no hand drawn equations
or screenshots).
4
Assess the DC motor open-loop model analytically. Is the step response
overdamped, critically damped, underdamped, and undamped? Explain why using
analytical results (not simulation). What is the steady-state error to a step input? Be
sure to show and clearly discuss all your work for this analysis.
5
Verify your assessment of the step response using Matlab simulation. Plot the step
response and measure steady-state error from the simulated step response data.
Show your equation used for calculating steady-state error from simulated step
response data (not the analytical expression). Substitute numerical data from your
simulated step response into this equation to demonstrate the measured step
response. Is this consistent with your previous assessment? What about damping?
Submit any Matlab scripts and Simulink diagrams used for simulation and results.
Provide the step response figure so it is clearly readable.
6
Let us now decide on desired specifications for our control design. The DC motor
at this link is rated for maximum speed of 30 (Can you figure out why from the
lists specifications?). It is recommended that the motor does not continuously
operate as this speed because it may stall. So in our control design we can set the
maximum speed at 25 with an error margin of 5 . Thus we can say our target
percent overshoot is
% =
5
= 16.67%.
30
It would also be useful to change the speed relatively often. Let’s say we want to
change the motor speed every 1 . Then in our control design we can set out
settling time to 0.75 to allow enough time for the motor to reach its steadystate value before a change in motor speed happens.
The last specification is steady-state error to a step input or motor speed change. If
we select steady-state error to a step input to be less than 0.15
to approximately 8.6

, this amounts
. The minimum encoder measurements are estimated
as 7.5 . So at a 1 update rate this steady-state error is close to the
minimum encoder measurement. Therefore we will select (∞) = ≤
0.15

In summary, the desired specifications for our control design are
1. %OS = 16.67%
2. Settling Time = 0.75
3. Steady-state error to a step input (∞) = ≤ 0.15

Here is some guidance on the control design process:
• Determine desired closed-loop poles from specifications.
• Plot root locus of DC motor model in series with three candidate control
designs : PID, Phase-lead, Phase-lag
o ( ) = ( ) ( ), where
▪ ( ) = DC motor transfer function
▪ ( ) = candidate control design
7






o Candidate control designs
▪ PID controller : 2 zeros, 1 pole
▪ Phase-lead : 1 zero, 1 pole
▪ Phase-lag : 1 zero, 1 pole
Iterate on pole/zero placements of these controllers to observe how it affects
the root locus.
Select control design structure (PID, Phase-lead, Phase-lag) based on which
controller can be designed to have the root locus intersect the desired pole
locations.
Calculate specific pole/zero locations of controller so root locus intersects
desired pole locations.
Plot root locus and find gain value at desired pole.
Plot step response of closed-loop system and measure specifications from
simulated step response data (not just stepinfo) to assess how well your control
design meets specifications
Iterate on control design selection and / or pole/zero placement if necessary to
improve performance.
For your report, please




Show calculations of desired pole location from specifications
State which control design was selected and why
Explain why you may have selected pole/zero placements of controllers
Show calculations for pole/zero of controllers to intersect desired closedloop pole
• Show clearly readable root locus plot with gain value at desired pole.
• Show clearly readable step response plot of closed-loop system. Measure
specifications from simulated step response data (not just stepinfo) to assess
how well your control design meets specifications. To receive full credit, you
must show the %OS, , and measurements from the simulated step data
(Not stepinfo). You can report stepinfo as an extra validation but you are
required to show this from the step response plot.
• Compare open and closed-loop performance of the system in a table. Show
%OS, , and measurements in a table for the open and closed-loop
cases. How do they compare? What were the difference between open loop
8
and closed loop performance ? How will that impact the motor in it’s
potential applications?
9
Capacitor Relationships
( ) =
( )

1
( ) = ( )
= ( ) =
( ) = ( )
= ( ) =
( ) = ( )
= ( ) =

Inductor Relationships
( ) =
( )

Resistor Relationships
( ) = ( )
KVL, KCL
Spring Relationships
( ) = ( )
( ) = ( )
Damper Relationships
( ) =
( )

= ̇ ( )
( ) = ( )
Newton’s Law
2
∑ = ∗ = ̈ ( ) = 2

10
General Second Order System
2
( ) = 2
+ 2 + 2
∶ 1,2 = − ± √ 2 − 1
Under-damped Step Response Specifications for Second Order System
1,2 = − ± √1 − 2
=
=
% =
−(
4
4
=
| { 1,2 }|

√1 − 2
=

| { 1,2 }|
| { 1,2 }|

)
−(
)
2
√1−
× 100 = | { 1,2}| × 100
=
−ln (% /100)
√ 2 + [ln (% /100)]2
11
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