Selective Laser Sintering 3D Printer - Muchen HeSelective Laser Sintering 3D Printer Part 1 ELEC 341...
Transcript of Selective Laser Sintering 3D Printer - Muchen HeSelective Laser Sintering 3D Printer Part 1 ELEC 341...
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Selective Laser Sintering 3D Printer
Part 1
ELEC 341 Design Project
Muchen He 44638154
Ou Liu 18800152
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Coordinates (X, Y, Z)
Inverse Kinematic
Controller
Motors
Direct Kinematic
Laser Control
System OverviewThe 3D printer uses a laser to build parts. The laser is
oriented using a two-axis motor assembly.
The inverse kinematic module converts the
Cartesian coordinates to the desired motor angle.
The joint controller module handles the system
response and controller.
The direct kinematic converts actual motor angles
back to Cartesian coordinates
The laser control module adjusts the intensity of the
laser based on how far the target is
The motors control the orientation of the laser
Sensor & Feedback
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I.K. implementation in SimuLink D.K. implementation in SimuLink
๐0: angle of motor 0
๐1: angle of motor 1
๐ฅ
โ๐ฆ
๐ง
Desired ๐ฅ
Desired ๐ง
Desired ๐ฆ
๐งโฒ: Distance between laser and top of part
Inverse & Direct KinematicThe inverse kinematic module converts the
Cartesian coordinates to the desired motor angle.
๐1 = tanโ1
๐ฆ
๐งโฒ๐0 = tan
โ1๐ฅ
๐งโฒ๐งโฒ = height โ ๐ง
The direct kinematic converts actual motor angles
back to Cartesian coordinates๐ฅ = ๐งโฒ tan ๐0๐ฆ = ๐งโฒ tan ๐1
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๐๐
๐๐Power AmplifierThe power amplifier takes the role
of a filter and a voltage follower.
The transfer function is the output
divided by input.
Since there is negative feedback
coupled by the capacitor, ๐๐ = ๐๐.
MNA is used to solve the circuit
Transfer Function: ๐(๐ )
๐(๐ )=
Vout
Vin=
CR2R1โL
LCR1s+CR1R2
MNA:Vin โ Vn
๐ฟ๐ =Vn๐ 2
Vin โ Vn๐ 1
= Vn โ Vout ร s๐ถ
Amplifier circuit SimuLink implementation
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Electric Motor Dynamics
๐ = ๐พ๐๐The DC motor has relationships between the
torque and the current
๐๐๐๐ = ๐พ๐๐
๐ =๐๐๐ โ ๐๐๐๐๐ฟ๐ + ๐
The two motors for ๐0 ๐๐๐ ๐1 are identical, so the relationships are the
same for both motors
The current is given by Ohmโs law
The back-EMF is proportional to the
motor speed
Equivalent circuit
๐๐๐ ๐๐๐๐
๐ ๐ฟ
Implementation
๐๐ ๐๐
Response
Torque Gain
๐ผ ๐
Back EMF Gain
๐ ๐
๐(๐ )
๐(๐ )=
1
๐ฟ๐ + ๐
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๐(๐ )
๐(๐ )=๐
๐=
๐
๐ฝ๐ 2 + ๐ต๐ + ๐พ
Mechanical Motor Dynamics
Implementation
ฯ ฯ
Mechanical motor dynamics is a transfer function that converts
torque to angular speed. It is given as:
Where ๐ฝ is moment of inertia, ๐ต is the kinetic friction constant, and ๐พ is the spring constant.
Kinetic friction is the due to the mass imposed on the
motion given as
๐ต =๐ผnoLoad ๐พ๐๐noLoad
Spring constant is 0 for motor 1 and positive for motor 0
Moment of inertia for motor 1 is ๐ฝrotor since the laser has negligible mass
๐ฝ๐1 = ๐ฝrotor
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Mechanical Motor DynamicsThe moment of inertia component for motor 1 is the
superposition of four parts
๐1
๐2โmotor๐
โlink
Motor 0 rotor: Inertia due to the internal rotor of the outer motor
๐ฝlink =masslink
123 ๐2
2 + ๐12 + โlink
2
Aluminium link: Hollow cylinder. Mass is its volume multiplied by
6061 aluminium density
Motor 1 & counter weight: mass๐.๐ค. = mass๐1
mass๐1
mass๐.๐ค.
โ ๐
Extended
Imaginary
๐ฝ๐0 = ๐ฝlink + ๐ฝrotor + ๐ฝ๐0weight + ๐ฝcounter weight
๐ฝ๐1weight + ๐ฝcounter weight = ๐ฝextended โ ๐ฝimaginary
Motor 1
Counter weight
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Static Friction
๐น๐ ๐ก๐๐ก๐๐ = ๐๐ ๐น๐น๐๐น๐ ๐ก๐๐ก๐๐ = ๐๐ ๐น๐(masslink + massq1 +massc.๐ค.)
Static friction works against the applied force and turns into
dynamic friction after motor starts moving. It is given by:
Motor 1 does not experience any noticeable static
friction since massrotor โ 0
anglevoltage
Sensor Feedback
If ๐applied < ๐static then ๐net = 0
Implementation
The sensor maps the actual angle of the motors to voltage linearly
Gain: โ๐, ๐ โ [โ5,5]
5V
2.5V
๐
2
๐
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Electrical Response Amplifier Response
The step response
suggests that the
integrating amplifier act
as a voltage follower with
a gain of 1 ๐
๐
Performance
Rise time: 3 ร 10โ4s
The electrical system
produces current given a
voltage
๐(๐ )
๐(๐ )=
2762.4
๐ + 1.489 ร 104
Internal inductances of
the motors causes
exponential behavior
Given a sudden jolt of
voltage, the current
spikes, but quickly
dissipates
๐(๐ )
๐(๐ )=
49.14
๐ + 41.17
Performance
Rise time: 0.1s
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๐(๐ )
๐(๐ )=
2.294 ร 106๐
๐ (๐ + 0.3856)
Motor 0 Response
Motor 1 Response
Motor 0 transfer function outputs angular velocity given
some torque
๐(๐ )
๐(๐ )=
12644๐
๐ 2 + 0.00213๐ + 14.11
Oscillation occurs due to the spring behavior
Angular velocity decays to 0 due to friction
Given a impulse torque, the motor 1 angular velocity
decays exponentially due to friction
With a constant torque, motor 1 speeds up to its
maximum angular velocity
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Root Locus
Power amplifier
Motor 1Motor 0
Electrical
The electrical system, amplifier , and
motor 1 are stable according to the
root locus
To be continued in part 2 (PID Control)
Motor 0 has complex root locus, which
explains the damped oscillation