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Vibration Isolation of a
LEGO® plate
Myphotonics
The myphotonics project deals with the construction of
opto-mechanical components and optical experiment
implementation using modular systems such as LEGO®.
Michelson-Interferometer
Optische Pinzette
Zweistrahl-Interferometer
The components are low cost and the instructions
that originated them are free to use
OpenAdaptonik and myphotonics can work together
sharing the same purpose.
OpenAdaptronik on myphotonics
The purpose of my project in OpenAdaptronik is to isolate from
environmental vibrations the platform on which the myphotonics
experiments are mounted.
This must be done with low cost instrumentation and a guide line to
make it repeatable by your own.
Simple Vibration Problem
Equation of motion:
𝑚
𝑑2
𝑥
𝑑𝑡2
+ 𝑟
𝑑𝑥
𝑑𝑡
+ 𝑘𝑥 = 𝑟
𝑑𝑥 𝑝
𝑑𝑡
+ 𝑘𝑥 𝑝
𝑚𝑥 + 𝑟𝑥 + 𝑘𝑥 = 𝑟𝑥 𝑝 + 𝑘𝑥 𝑝
Where: 𝑚𝑒 = 𝑚𝑎𝑠𝑠 𝑡𝑜 𝑏𝑒 𝑖𝑠𝑜𝑙𝑎𝑡𝑒𝑑 ; 𝑟 = 𝑑𝑎𝑚𝑝𝑖𝑛𝑔 ; 𝑘 = 𝑠𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠
It is possible to have different behavior
of the system depending on the
damping with fixed mass and stiffness
Effects of damping increasing:
1. Amplitude attenuation at the
resonance frequency
(𝜔0 = 𝑘 𝑚 𝑒 )
2. Amplitude increase at
frequencies higher then the
resonance one
• Passing in the Laplace domain is possible to
find the transfer function of the system
𝐺𝑠𝑦𝑠 =
𝑟𝑆 + 𝑘
𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘
=
𝑋
𝑋 𝑝
𝑟 ↑
Skyhook Solution (feedback control)
Skyhook control is a feedback velocity
control widely use in vibration isolation
for a vehicle.
It simply introduce into the system a
force proportional to the speed of the
mass to be isolated
Equation of motion:
𝑚𝑥 + 𝑟𝑥 + 𝑘𝑥 = 𝒇 + 𝑟𝑥 𝑝 + 𝑘𝑥 𝑝 𝑚𝑥 + 𝑟𝑥 + 𝑘𝑥 = −𝑲𝑥 + 𝑟𝑥 𝑝 + 𝑘𝑥 𝑝
Choosing: 𝐾 = 2𝑚 𝑒 𝜔0 − 𝑟
the best attenuation at the
resonance frequency is obtained
𝐺 𝑠𝑦𝑠𝐶 =
𝑟𝑆 + 𝑘
𝑚 𝑒 𝑆2 + (𝑟 + 𝑲)𝑆 + 𝑘
1. Amplitude attenuation at
the resonance frequency
2. Amplitude attenuation at
frequencies higher then
the resonance one
Skyhook Solution equivalent version
It is possible to see the controlled system above mentioned as the product of two transfer functions.
𝐺𝑠𝑦𝑠𝐶 = 𝐺 𝐶 ∗ 𝐺𝑠𝑦𝑠
Where : 𝐺 𝐶 =
𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘
(𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘) + 𝑲𝑆
𝐺𝑠𝑦𝑠 =
𝑟𝑆 + 𝑘
(𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘)
∗ 𝐺𝑠𝑦𝑠𝐶 =
𝑟𝑆 + 𝑘
(𝑚 𝑒 𝑆2 + (𝑟 + 𝑲)𝑆 + 𝑘)
=
+
Introduction of the inertial mass actuator
The mechanical behavior of the inertial mass
actuator is exactly the same as the system
discussed before plus:
• The electrical behavior
• The mechanical electrical coupling factor
Equations of the system:
𝐿
𝑑𝑖
𝑑𝑡
+ 𝑅𝑖 + 𝐶 𝑖𝑑𝑡 + 𝛹 𝑥 − 𝑥 𝑝 = 𝑉
𝑚𝑥 + 𝑟𝑥 + 𝑘𝑥 = 𝑓 + 𝑟𝑥 𝑝 + 𝑘𝑥 𝑝
𝑓 = 𝛹𝑖
Where : 𝑉 = 𝑖𝑛𝑝𝑢𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
𝑖 = 𝑜𝑢𝑡𝑝𝑢𝑡 𝑐𝑢𝑟𝑟𝑒𝑛𝑡
𝑥 = 𝑜𝑢𝑡𝑝𝑢𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
In the Laplace domain the electrical system has the following expression :
𝐼 =
𝑆
𝐿𝑆2 + 𝑅𝑆 + 𝐶
𝑉 +
𝛹𝑆2
𝐿𝑆2 + 𝑅𝑆 + 𝐶
𝑋 𝑝 − 𝑋
Controller Selection (Proportional Integral Derivative Control)
As for the ideal case (without actuator) it is chosen the velocity of the mass as feedback control variable.
+
The controller must be easy to implement 𝑃𝐼𝐷(𝑡) = 𝐾 𝑝 𝑥 + 𝐾𝐼 𝑥 𝑑𝑡 + 𝐾 𝑑
𝑑𝑥
𝑑𝑡
In the Laplace domain 𝑃𝐼𝐷 𝑆 = 𝐾 𝑝 + 𝐾𝐼
1
𝑆
+ 𝐾 𝑑 𝑆 𝑋
Unfortunately is not possible to implement the PID controller with this type of configuration but must be
write in a different way due to the derivative part.
𝑃𝐼𝐷 = 𝐾𝑝 + 𝐾𝐼
1
𝑆
+ 𝐾 𝑑
𝑁
1 + 𝑁
1
𝑆
𝑃𝐼𝐷 =
𝑆2
𝐾1 + 𝑆𝐾2 + 𝐾3
𝑆(𝑆 + 𝑁)
𝐾 𝑎
Choice of the Gains
The choice of the gains was done by trying to neglect in the control transfer function the
contribution of the electrical behavior in order to not affect the electrical part of the system but
influencing only the mechanical part.
𝐾1 = 1 ; 𝐾2 = 𝑅
𝐿 ; 𝐾3 = 𝐶
𝐿 ; 𝐾 𝑎 = 𝑁
𝛼𝛹 𝑁 = 𝑲 = 2𝑚 𝑒 𝜔0 − 𝑟 ; 𝛼 = 5
If the inductance L is sufficiently
small (generally true for an inertial
mass actuator) the control transfer
function con be simplified as:
Where :
𝐺 𝐶 =
𝛼𝐿(𝑆 + 𝑁)(𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘)
𝛼𝐿(𝑆 + 𝑁)(𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘) + 𝑁𝑆
𝐺 𝐶𝑖𝑑𝑒𝑎𝑙 =
𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘
(𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘) + 𝑲𝑆
That is really close to:
Results on 1 d.o.f. system
Frequency response Time response for simulation of 10 seconds
2 degrees of freedom Problem
Preliminary hypothesis and assumptions: • Rigid body
• Same actuators on both sides
• Co-located control
-
-
Each actuator is controlled by the same PID control and each one of them has the same gains as the 1
d.o.f. problem excluding the N parameter that for now on will be expressed as:
𝑁 = 2𝑚 𝑒 𝜔01 − 𝑛𝑟 ; 𝜔01 = 𝑛𝑘 𝑚 𝑒
Where n is the number of the springs present in the system and 𝜔01 is
the first resonance frequency due to the translating mode
𝑚 𝑒 ; 𝐽𝑒
1 2
Results on 2 d.o.f. system
Due to the symmetry of the problem the response from the system from the input disturbance in 1 and
the output velocity in 1 is the same as the behavior of the system from 2 and 2
The choice of the regulators is working well also for the 2 d.o.f. system making each point (1 or 2)
behaves similar to a single d.o.f. system.
Finally the Plate – the 3 degrees of freedom Problem
ϕ
𝑚 𝑒 ; 𝐽φ ; 𝐽ϑ
1
4
3
2
Preliminary hypothesis
and assumptions:
• Rigid body
• 4 equal actuators
• Co-located control
• 4 equal PID control
We should see 3 resonance picks in the transfer function
but, being the two mass moment of inertia really close to
each other for the LEGO® plate, they are confused.
Conclusions and future developments
• For the PID control tuned in the way shown no modification of the control gains (a part from N) is
necessary to pass from a 1 d.o.f. problem to a 3 d.o.f. problem as long as the hypothesis are valid.
• The solution found sims to have positive effect on the problem.
Conclusions
Future developments
• A more precise theory should be carried out in order to demonstrate appropriately the effectiveness
of the control found.
• Introduction of non rigid body in the system.
• Experimental test on the platform must be done to find the vibration modes of the structure.
• Find a sensor for the purpose of the control.
• Identify the real reliability of the control with low cost control board  study the effects of the
digitalization of the signal on the control and find if it is possible to implement in a low cost board.
• Finally build the experiment and demonstrate if it works.
Thanks for the
attention
Vibration Isolation of a LEGO® plate
Vibration Isolation of a LEGO® plate
𝐼 =
𝑆
𝐿𝑆2 + 𝑅𝑆 + 𝐶
𝑉 +
𝛹𝑆2
𝐿𝑆2 + 𝑅𝑆 + 𝐶
𝑋 𝑝 − 𝑋
𝑉 = −
𝑁
𝛼𝛹
∗
𝑆2
+ 𝑅
𝐿 𝑆 + 𝐶
𝐿
𝑆 𝑆 + 𝑁
∗ 𝑋 = −
𝑁
𝛼𝛹
∗
𝑆2
+ 𝑅
𝐿 𝑆 + 𝐶
𝐿
𝑆 + 𝑁
∗ 𝑋
𝑆
𝐿𝑆2 + 𝑅𝑆 + 𝐶
𝑉 =
𝑆
𝐿 𝑆2 + 𝑅
𝐿 𝑆 + 𝐶
𝐿
∗
𝑁
𝛼𝛹
∗
𝑆2
+ 𝑅
𝐿 𝑆 + 𝐶
𝐿
𝑆 + 𝑁
=
𝑁𝑆
𝛼𝛹𝐿(𝑆 + 𝑁)
𝐼 =
𝛹𝑆2
𝐿𝑆2 + 𝑅𝑆 + 𝐶
𝑋 𝑝 −
𝛹𝑆2
𝐿𝑆2 + 𝑅𝑆 + 𝐶
+
𝑁𝑆
𝛼𝛹𝐿(𝑆 + 𝑁)
𝑋
𝛹𝑆2
𝐿𝑆2 + 𝑅𝑆 + 𝐶
+
𝑁𝑆
𝛼𝛹𝐿(𝑆 + 𝑁)
=
𝑆 𝛼𝛹2
𝐿𝑆 𝑆 + 𝑁 + 𝑁 𝐿𝑆2
+ 𝑅𝑆 + 𝐶
𝛼𝛹𝐿(𝑆 + 𝑁) 𝐿𝑆2 + 𝑅𝑆 + 𝐶
=
𝑆𝑁 𝐿𝑆2
+ 𝑅𝑆 + 𝐶
𝛼𝛹𝐿(𝑆 + 𝑁) 𝐿𝑆2 + 𝑅𝑆 + 𝐶
=
𝑁𝑆
𝛼𝛹𝐿(𝑆 + 𝑁)
𝑋 =
1
𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘
𝛹𝐼 +
𝑆𝑟 + 𝑘
𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘
𝑋 𝑝 = 𝐺 𝑥 𝛹𝐼 + 𝐺 𝑥𝑝 𝑋 𝑝
𝑋 = 𝐺 𝑥 𝛹
𝛹𝑆2
𝐿𝑆2 + 𝑅𝑆 + 𝐶
𝑋 𝑝 − 𝐺 𝑥 𝛹
𝑁𝑆
𝛼𝛹𝐿 𝑆 + 𝑁
𝑋 + 𝐺 𝑥𝑝 𝑋 𝑝
1 + 𝐺 𝑥 𝛹
𝑁𝑆
𝛼𝛹𝐿 𝑆 + 𝑁
𝑋 = 𝐺 𝑥 𝛹
𝛹𝑆2
𝐿𝑆2 + 𝑅𝑆 + 𝐶
+ 𝐺 𝑥𝑝 𝑋 𝑝
1 + 𝐺 𝑥 𝛹
𝑁𝑆
𝛼𝛹𝐿 𝑆 + 𝑁
= 1 +
𝑆𝑁
𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘 (𝑆 + 𝑁)𝛼𝛹𝐿
𝛹 =
𝛼𝐿(𝑆 + 𝑁)(𝑚 𝑒 𝑆2
+ 𝑟𝑆 + 𝑘)
𝛼𝐿(𝑆 + 𝑁)(𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘) + 𝑁𝑆

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Vibration Isolation of a LEGO® plate

  • 1. Vibration Isolation of a LEGO® plate
  • 2. Myphotonics The myphotonics project deals with the construction of opto-mechanical components and optical experiment implementation using modular systems such as LEGO®. Michelson-Interferometer Optische Pinzette Zweistrahl-Interferometer The components are low cost and the instructions that originated them are free to use OpenAdaptonik and myphotonics can work together sharing the same purpose.
  • 3. OpenAdaptronik on myphotonics The purpose of my project in OpenAdaptronik is to isolate from environmental vibrations the platform on which the myphotonics experiments are mounted. This must be done with low cost instrumentation and a guide line to make it repeatable by your own.
  • 4. Simple Vibration Problem Equation of motion: 𝑚 𝑑2 𝑥 𝑑𝑡2 + 𝑟 𝑑𝑥 𝑑𝑡 + 𝑘𝑥 = 𝑟 𝑑𝑥 𝑝 𝑑𝑡 + 𝑘𝑥 𝑝 𝑚𝑥 + 𝑟𝑥 + 𝑘𝑥 = 𝑟𝑥 𝑝 + 𝑘𝑥 𝑝 Where: 𝑚𝑒 = 𝑚𝑎𝑠𝑠 𝑡𝑜 𝑏𝑒 𝑖𝑠𝑜𝑙𝑎𝑡𝑒𝑑 ; 𝑟 = 𝑑𝑎𝑚𝑝𝑖𝑛𝑔 ; 𝑘 = 𝑠𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 It is possible to have different behavior of the system depending on the damping with fixed mass and stiffness Effects of damping increasing: 1. Amplitude attenuation at the resonance frequency (𝜔0 = 𝑘 𝑚 𝑒 ) 2. Amplitude increase at frequencies higher then the resonance one • Passing in the Laplace domain is possible to find the transfer function of the system 𝐺𝑠𝑦𝑠 = 𝑟𝑆 + 𝑘 𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘 = 𝑋 𝑋 𝑝 𝑟 ↑
  • 5. Skyhook Solution (feedback control) Skyhook control is a feedback velocity control widely use in vibration isolation for a vehicle. It simply introduce into the system a force proportional to the speed of the mass to be isolated Equation of motion: 𝑚𝑥 + 𝑟𝑥 + 𝑘𝑥 = 𝒇 + 𝑟𝑥 𝑝 + 𝑘𝑥 𝑝 𝑚𝑥 + 𝑟𝑥 + 𝑘𝑥 = −𝑲𝑥 + 𝑟𝑥 𝑝 + 𝑘𝑥 𝑝 Choosing: 𝐾 = 2𝑚 𝑒 𝜔0 − 𝑟 the best attenuation at the resonance frequency is obtained 𝐺 𝑠𝑦𝑠𝐶 = 𝑟𝑆 + 𝑘 𝑚 𝑒 𝑆2 + (𝑟 + 𝑲)𝑆 + 𝑘 1. Amplitude attenuation at the resonance frequency 2. Amplitude attenuation at frequencies higher then the resonance one
  • 6. Skyhook Solution equivalent version It is possible to see the controlled system above mentioned as the product of two transfer functions. 𝐺𝑠𝑦𝑠𝐶 = 𝐺 𝐶 ∗ 𝐺𝑠𝑦𝑠 Where : 𝐺 𝐶 = 𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘 (𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘) + 𝑲𝑆 𝐺𝑠𝑦𝑠 = 𝑟𝑆 + 𝑘 (𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘) ∗ 𝐺𝑠𝑦𝑠𝐶 = 𝑟𝑆 + 𝑘 (𝑚 𝑒 𝑆2 + (𝑟 + 𝑲)𝑆 + 𝑘) =
  • 7. + Introduction of the inertial mass actuator The mechanical behavior of the inertial mass actuator is exactly the same as the system discussed before plus: • The electrical behavior • The mechanical electrical coupling factor Equations of the system: 𝐿 𝑑𝑖 𝑑𝑡 + 𝑅𝑖 + 𝐶 𝑖𝑑𝑡 + 𝛹 𝑥 − 𝑥 𝑝 = 𝑉 𝑚𝑥 + 𝑟𝑥 + 𝑘𝑥 = 𝑓 + 𝑟𝑥 𝑝 + 𝑘𝑥 𝑝 𝑓 = 𝛹𝑖 Where : 𝑉 = 𝑖𝑛𝑝𝑢𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑖 = 𝑜𝑢𝑡𝑝𝑢𝑡 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑥 = 𝑜𝑢𝑡𝑝𝑢𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 In the Laplace domain the electrical system has the following expression : 𝐼 = 𝑆 𝐿𝑆2 + 𝑅𝑆 + 𝐶 𝑉 + 𝛹𝑆2 𝐿𝑆2 + 𝑅𝑆 + 𝐶 𝑋 𝑝 − 𝑋
  • 8. Controller Selection (Proportional Integral Derivative Control) As for the ideal case (without actuator) it is chosen the velocity of the mass as feedback control variable. + The controller must be easy to implement 𝑃𝐼𝐷(𝑡) = 𝐾 𝑝 𝑥 + 𝐾𝐼 𝑥 𝑑𝑡 + 𝐾 𝑑 𝑑𝑥 𝑑𝑡 In the Laplace domain 𝑃𝐼𝐷 𝑆 = 𝐾 𝑝 + 𝐾𝐼 1 𝑆 + 𝐾 𝑑 𝑆 𝑋 Unfortunately is not possible to implement the PID controller with this type of configuration but must be write in a different way due to the derivative part. 𝑃𝐼𝐷 = 𝐾𝑝 + 𝐾𝐼 1 𝑆 + 𝐾 𝑑 𝑁 1 + 𝑁 1 𝑆 𝑃𝐼𝐷 = 𝑆2 𝐾1 + 𝑆𝐾2 + 𝐾3 𝑆(𝑆 + 𝑁) 𝐾 𝑎
  • 9. Choice of the Gains The choice of the gains was done by trying to neglect in the control transfer function the contribution of the electrical behavior in order to not affect the electrical part of the system but influencing only the mechanical part. 𝐾1 = 1 ; 𝐾2 = 𝑅 𝐿 ; 𝐾3 = 𝐶 𝐿 ; 𝐾 𝑎 = 𝑁 𝛼𝛹 𝑁 = 𝑲 = 2𝑚 𝑒 𝜔0 − 𝑟 ; 𝛼 = 5 If the inductance L is sufficiently small (generally true for an inertial mass actuator) the control transfer function con be simplified as: Where : 𝐺 𝐶 = 𝛼𝐿(𝑆 + 𝑁)(𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘) 𝛼𝐿(𝑆 + 𝑁)(𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘) + 𝑁𝑆 𝐺 𝐶𝑖𝑑𝑒𝑎𝑙 = 𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘 (𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘) + 𝑲𝑆 That is really close to:
  • 10. Results on 1 d.o.f. system Frequency response Time response for simulation of 10 seconds
  • 11. 2 degrees of freedom Problem Preliminary hypothesis and assumptions: • Rigid body • Same actuators on both sides • Co-located control - - Each actuator is controlled by the same PID control and each one of them has the same gains as the 1 d.o.f. problem excluding the N parameter that for now on will be expressed as: 𝑁 = 2𝑚 𝑒 𝜔01 − 𝑛𝑟 ; 𝜔01 = 𝑛𝑘 𝑚 𝑒 Where n is the number of the springs present in the system and 𝜔01 is the first resonance frequency due to the translating mode 𝑚 𝑒 ; 𝐽𝑒 1 2
  • 12. Results on 2 d.o.f. system Due to the symmetry of the problem the response from the system from the input disturbance in 1 and the output velocity in 1 is the same as the behavior of the system from 2 and 2 The choice of the regulators is working well also for the 2 d.o.f. system making each point (1 or 2) behaves similar to a single d.o.f. system.
  • 13. Finally the Plate – the 3 degrees of freedom Problem ϕ 𝑚 𝑒 ; 𝐽φ ; 𝐽ϑ 1 4 3 2 Preliminary hypothesis and assumptions: • Rigid body • 4 equal actuators • Co-located control • 4 equal PID control We should see 3 resonance picks in the transfer function but, being the two mass moment of inertia really close to each other for the LEGO® plate, they are confused.
  • 14. Conclusions and future developments • For the PID control tuned in the way shown no modification of the control gains (a part from N) is necessary to pass from a 1 d.o.f. problem to a 3 d.o.f. problem as long as the hypothesis are valid. • The solution found sims to have positive effect on the problem. Conclusions Future developments • A more precise theory should be carried out in order to demonstrate appropriately the effectiveness of the control found. • Introduction of non rigid body in the system. • Experimental test on the platform must be done to find the vibration modes of the structure. • Find a sensor for the purpose of the control. • Identify the real reliability of the control with low cost control board  study the effects of the digitalization of the signal on the control and find if it is possible to implement in a low cost board. • Finally build the experiment and demonstrate if it works.
  • 18. 𝐼 = 𝑆 𝐿𝑆2 + 𝑅𝑆 + 𝐶 𝑉 + 𝛹𝑆2 𝐿𝑆2 + 𝑅𝑆 + 𝐶 𝑋 𝑝 − 𝑋 𝑉 = − 𝑁 𝛼𝛹 ∗ 𝑆2 + 𝑅 𝐿 𝑆 + 𝐶 𝐿 𝑆 𝑆 + 𝑁 ∗ 𝑋 = − 𝑁 𝛼𝛹 ∗ 𝑆2 + 𝑅 𝐿 𝑆 + 𝐶 𝐿 𝑆 + 𝑁 ∗ 𝑋 𝑆 𝐿𝑆2 + 𝑅𝑆 + 𝐶 𝑉 = 𝑆 𝐿 𝑆2 + 𝑅 𝐿 𝑆 + 𝐶 𝐿 ∗ 𝑁 𝛼𝛹 ∗ 𝑆2 + 𝑅 𝐿 𝑆 + 𝐶 𝐿 𝑆 + 𝑁 = 𝑁𝑆 𝛼𝛹𝐿(𝑆 + 𝑁) 𝐼 = 𝛹𝑆2 𝐿𝑆2 + 𝑅𝑆 + 𝐶 𝑋 𝑝 − 𝛹𝑆2 𝐿𝑆2 + 𝑅𝑆 + 𝐶 + 𝑁𝑆 𝛼𝛹𝐿(𝑆 + 𝑁) 𝑋 𝛹𝑆2 𝐿𝑆2 + 𝑅𝑆 + 𝐶 + 𝑁𝑆 𝛼𝛹𝐿(𝑆 + 𝑁) = 𝑆 𝛼𝛹2 𝐿𝑆 𝑆 + 𝑁 + 𝑁 𝐿𝑆2 + 𝑅𝑆 + 𝐶 𝛼𝛹𝐿(𝑆 + 𝑁) 𝐿𝑆2 + 𝑅𝑆 + 𝐶 = 𝑆𝑁 𝐿𝑆2 + 𝑅𝑆 + 𝐶 𝛼𝛹𝐿(𝑆 + 𝑁) 𝐿𝑆2 + 𝑅𝑆 + 𝐶 = 𝑁𝑆 𝛼𝛹𝐿(𝑆 + 𝑁)
  • 19. 𝑋 = 1 𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘 𝛹𝐼 + 𝑆𝑟 + 𝑘 𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘 𝑋 𝑝 = 𝐺 𝑥 𝛹𝐼 + 𝐺 𝑥𝑝 𝑋 𝑝 𝑋 = 𝐺 𝑥 𝛹 𝛹𝑆2 𝐿𝑆2 + 𝑅𝑆 + 𝐶 𝑋 𝑝 − 𝐺 𝑥 𝛹 𝑁𝑆 𝛼𝛹𝐿 𝑆 + 𝑁 𝑋 + 𝐺 𝑥𝑝 𝑋 𝑝 1 + 𝐺 𝑥 𝛹 𝑁𝑆 𝛼𝛹𝐿 𝑆 + 𝑁 𝑋 = 𝐺 𝑥 𝛹 𝛹𝑆2 𝐿𝑆2 + 𝑅𝑆 + 𝐶 + 𝐺 𝑥𝑝 𝑋 𝑝 1 + 𝐺 𝑥 𝛹 𝑁𝑆 𝛼𝛹𝐿 𝑆 + 𝑁 = 1 + 𝑆𝑁 𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘 (𝑆 + 𝑁)𝛼𝛹𝐿 𝛹 = 𝛼𝐿(𝑆 + 𝑁)(𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘) 𝛼𝐿(𝑆 + 𝑁)(𝑚 𝑒 𝑆2 + 𝑟𝑆 + 𝑘) + 𝑁𝑆