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:

𝑚𝑑2𝑥

𝑑𝑡2+ 𝑟

𝑑𝑥

𝑑𝑡+ 𝑘𝑥 = 𝑟

𝑑𝑥𝑝𝑑𝑡

+ 𝑘𝑥𝑝 𝑚𝑥 + 𝑟𝑥 + 𝑘𝑥 = 𝑟𝑥 𝑝 + 𝑘𝑥𝑝

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 𝑚𝑒 )

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

𝑚𝑒𝑆2 + 𝑟𝑆 + 𝑘

=𝑋

𝑋𝑝

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 + 𝑘𝑥𝑝 𝑚𝑥 + 𝑟𝑥 + 𝑘𝑥 = −𝑲𝑥 + 𝑟𝑥 𝑝 + 𝑘𝑥𝑝

Choosing: 𝐾 = 2𝑚𝑒𝜔0 − 𝑟 the best attenuation at the resonance frequency is obtained

𝐺𝑠𝑦𝑠𝐶 =𝑟𝑆 + 𝑘

𝑚𝑒𝑆2 + (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 = 𝐺𝐶 ∗ 𝐺𝑠𝑦𝑠

Where : 𝐺𝐶 =𝑚𝑒𝑆2 + 𝑟𝑆 + 𝑘

(𝑚𝑒𝑆2 + 𝑟𝑆 + 𝑘) + 𝑲𝑆 𝐺𝑠𝑦𝑠 =

𝑟𝑆 + 𝑘

(𝑚𝑒𝑆2 + 𝑟𝑆 + 𝑘) ∗ 𝐺𝑠𝑦𝑠𝐶 =

𝑟𝑆 + 𝑘

(𝑚𝑒𝑆2 + (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:

𝐿𝑑𝑖

𝑑𝑡+ 𝑅𝑖 + 𝐶 𝑖𝑑𝑡 + 𝛹 𝑥 − 𝑥 𝑝 = 𝑉

𝑚𝑥 + 𝑟𝑥 + 𝑘𝑥 = 𝑓 + 𝑟𝑥 𝑝 + 𝑘𝑥𝑝𝑓 = 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 + 𝐶𝑉 +

𝛹𝑆2

𝐿𝑆2 + 𝑅𝑆 + 𝐶𝑋𝑝 − 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) = 𝐾𝑝𝑥 + 𝐾𝐼 𝑥 𝑑𝑡 + 𝐾𝑑𝑑𝑥

𝑑𝑡

In the Laplace domain 𝑃𝐼𝐷 𝑆 = 𝐾𝑝 + 𝐾𝐼1

𝑆+ 𝐾𝑑𝑆 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 + 𝐾𝐼1

𝑆+ 𝐾𝑑

𝑁

1 + 𝑁1𝑆

𝑃𝐼𝐷 =𝑆2𝐾1 + 𝑆𝐾2 + 𝐾3

𝑆(𝑆 + 𝑁)𝐾𝑎

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 =𝑅𝐿 ; 𝐾3 =

𝐶𝐿 ; 𝐾𝑎 =

𝑁𝛼𝛹 𝑁 = 𝑲 = 2𝑚𝑒𝜔0 − 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 :

𝐺𝐶 =𝛼𝐿(𝑆 + 𝑁)(𝑚𝑒𝑆2 + 𝑟𝑆 + 𝑘)

𝛼𝐿(𝑆 + 𝑁)(𝑚𝑒𝑆2 + 𝑟𝑆 + 𝑘) + 𝑁𝑆

𝐺𝐶𝑖𝑑𝑒𝑎𝑙 =𝑚𝑒𝑆2 + 𝑟𝑆 + 𝑘

(𝑚𝑒𝑆2 + 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 = 2𝑚𝑒𝜔01 − 𝑛𝑟 ; 𝜔01 = 𝑛𝑘 𝑚𝑒

Where n is the number of the springs present in the system and 𝜔01 is the first resonance frequency due to the translating mode

𝑚𝑒 ; 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

𝑚𝑒 ; 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

𝐼 =𝑆

𝐿𝑆2 + 𝑅𝑆 + 𝐶𝑉 +

𝛹𝑆2

𝐿𝑆2 + 𝑅𝑆 + 𝐶𝑋𝑝 − 𝑋

𝑉 = −𝑁

𝛼𝛹∗𝑆2 + 𝑅

𝐿 𝑆 + 𝐶𝐿

𝑆 𝑆 + 𝑁∗ 𝑋 = −

𝑁

𝛼𝛹∗𝑆2 + 𝑅

𝐿 𝑆 + 𝐶𝐿

𝑆 + 𝑁∗ 𝑋

𝑆

𝐿𝑆2 + 𝑅𝑆 + 𝐶𝑉 =

𝑆

𝐿 𝑆2 + 𝑅𝐿 𝑆 + 𝐶

𝐿 ∗𝑁

𝛼𝛹∗𝑆2 + 𝑅

𝐿 𝑆 + 𝐶𝐿

𝑆 + 𝑁=

𝑁𝑆

𝛼𝛹𝐿(𝑆 + 𝑁)

𝐼 =𝛹𝑆2

𝐿𝑆2 + 𝑅𝑆 + 𝐶𝑋𝑝 −

𝛹𝑆2

𝐿𝑆2 + 𝑅𝑆 + 𝐶+

𝑁𝑆

𝛼𝛹𝐿(𝑆 + 𝑁)𝑋

𝛹𝑆2

𝐿𝑆2 + 𝑅𝑆 + 𝐶+

𝑁𝑆

𝛼𝛹𝐿(𝑆 + 𝑁)=𝑆 𝛼𝛹2𝐿𝑆 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

𝑚𝑒𝑆2 + 𝑟𝑆 + 𝑘

𝛹𝐼 +𝑆𝑟 + 𝑘

𝑚𝑒𝑆2 + 𝑟𝑆 + 𝑘

𝑋𝑝 = 𝐺𝑥𝛹𝐼 + 𝐺𝑥𝑝𝑋𝑝

𝑋 = 𝐺𝑥𝛹𝛹𝑆2

𝐿𝑆2 + 𝑅𝑆 + 𝐶𝑋𝑝 − 𝐺𝑥𝛹

𝑁𝑆

𝛼𝛹𝐿 𝑆 + 𝑁𝑋 + 𝐺𝑥𝑝𝑋𝑝

1 + 𝐺𝑥𝛹𝑁𝑆

𝛼𝛹𝐿 𝑆 + 𝑁𝑋 = 𝐺𝑥𝛹

𝛹𝑆2

𝐿𝑆2 + 𝑅𝑆 + 𝐶+ 𝐺𝑥𝑝 𝑋𝑝

1 + 𝐺𝑥𝛹𝑁𝑆

𝛼𝛹𝐿 𝑆 + 𝑁= 1 +

𝑆𝑁

𝑚𝑒𝑆2 + 𝑟𝑆 + 𝑘 (𝑆 + 𝑁)𝛼𝛹𝐿

𝛹 =𝛼𝐿(𝑆 + 𝑁)(𝑚𝑒𝑆

2 + 𝑟𝑆 + 𝑘)

𝛼𝐿(𝑆 + 𝑁)(𝑚𝑒𝑆2 + 𝑟𝑆 + 𝑘) + 𝑁𝑆