This document discusses the design and testing of an optomechanical fiber Bragg grating (FBG) sensor for measuring acceleration. It presents two approaches to modeling the sensor as a multi-degree of freedom system and analyzing its dynamic response. The first approach models it as two uncoupled single-degree of freedom systems, determining the natural frequencies and mode shapes. The second approach models it as a rigid body with one rotational and one translational degree of freedom. Both approaches show the sensor has a linear response and is dynamically stable over time.

2. Abdallah
Mohamed
Mario
Gamal
Mostafa El
Bahrawy
Mohamed EhabAbdelrahman
Nagy

5. Problem
Need
light propagation is changed (modulated)
by a mechanical variable
Opto-mechanical
sensors
Optomechanical advantages
• Non-contact sensors
• High precision
• Limited sensitivity/ bandwidth due to thermal and
electrical noise
• Immunity to electromagnetic interference
• Wide dynamic range
• Reliable operation

9. Strain is used to measure
Acceleration ε= (
𝑴
𝑬𝑨
) × 𝒂Linear Relation
K=
𝑬𝑨
𝑳
Stiffness of FBG
𝑴
𝒙
𝒚
λ 𝐵 = 2𝑛 𝑒𝑓𝑓Λ
∆λ 𝐵=λ 𝐵 1 − 𝜌 𝑎 ∆𝜀 + (𝛼 + ζ)∆𝑇
T.F =
𝑍(𝑠)
𝐴(𝑠)
= −
𝑚
𝑚𝑠2+𝑘
𝑚 𝑧 + 𝑘𝑧 = −𝑚𝑎
No
damping!!
2nd Order system
FBG
Small light range
(150 nm -248 nm
No effect of ∆𝑇
on ∆λ
Constant
𝑇 𝑛 𝑒𝑓𝑓
Input
Acceleration
(𝑎)
Output
Change in
wavelength (∆λ)

10. Schematic
Block
Diagram

12. Sensitivity = 𝟎. 𝟐
𝒏𝒎
𝒎
𝒔 𝟐
compared to 10
v/g which resolution of electric accelerometers!
No hysteresis due to absence of
damping.
Dynamic Error:
Bandwidth
Analysis:
y = 2E-10x + 1E-11
0
5E-10
1E-09
1.5E-09
2E-09
2.5E-09
3E-09
3.5E-09
4E-09
4.5E-09
5E-09
0 5 10 15 20 25 30
deltalamda(nm)
-Acceleration(m/s^2)
Calibration Curve 𝝎 𝒏 =
𝒌
𝒎
= 𝟐𝟗𝟎. 𝟗 𝒓𝒂𝒅𝒔−𝟏
1.85
𝒎
𝒔 𝟐
𝒏𝒎
Resolution:
𝑑𝑒 =
𝑦
𝑘𝑥 𝑠
− 1
Linearity

13. t(s)
A(𝑚𝑠−2
)
Harmonic
input
Ramp
Input
Step
input
Calibration curve Under effect of different
inputs

14. Harmonic inputRamp inputStep input
A(𝑚𝑠−2
)
t(s)t(s)t(s)
A(𝑚𝑠−2
) A(𝑚𝑠−2
)

16. 1st approach

17. 𝐿 = 𝑇 + 𝑉
𝑑
𝑑𝑡
𝜕𝑇
𝜕 𝑞 𝑗
−
𝜕𝑇
𝜕𝑞 𝑗
+
𝜕𝐷
𝜕 𝑞 𝑗
+
𝜕𝑉
𝜕𝑞 𝑗
= 𝑄𝑗
𝑑
𝑑𝑡
𝜕𝐿
𝜕 𝑥𝑗
−
𝜕𝐿
𝜕𝑥𝑗
= 0
𝑇 =
1
2
m1z1
. 2
+
1
2
m2z2
. 2
𝑉 =
1
2
[z1
2(k1+k2)−2k2z1z2+z2
2(k2+k3)]
𝑚 𝑞 + 𝑘 𝑞 = 𝑄
𝑚
2
0
0
𝑚
2
𝑧1
..
𝑧2
.. +
5𝑘
12
−
𝑘
6
−
𝑘
6
5𝑘
12
𝑧1
𝑧2
=
1
2
𝑚
𝑦
..
𝑦
..
𝑴 𝟏 𝑴 𝟐
𝒙 𝟏 𝒙 𝟐
𝒌 𝟑 =
𝟏
𝟐
𝑲𝒌 𝟏 =
𝟏
𝟐
𝑲 𝒌 𝟐=1000 K
𝒌 & 𝒎 are
Real
Positive definite
Using
System is
Dynamically
uncoupled
System is
Statically
coupled
Stable
𝒚
Using Single Degree
system as Multi degree
system

18. 1st Approach

19. 𝑘 − 𝜔2
𝑚 𝑋 = 0
𝑘 − 𝜔2 𝑚 = 0 Natural Frequencies
Mode shapes
𝜔1 = 290.991 𝑟𝑎𝑑𝑠−1
𝜔2 = 18406.191𝑟𝑎𝑑𝑠−1
Fundamental
𝑋1 =
1
1
𝑋2 =
1
−1
𝑋 1
𝑇
𝑚 𝑋 2 = 0 𝑋 1
𝑇
𝑘 𝑋 2 = 0
𝑋 1
𝑇
𝑚 𝑋 1 ≠ 0 𝑋 1
𝑇
𝑘 𝑋 1 ≠ 0
𝑋 2
𝑇
𝑚 𝑋 2 ≠ 0 𝑋 2
𝑇
𝑘 𝑋 2 ≠ 0
Orthogonal  Stable all
over time !
Mode 1
Mode 2
𝑈 =
1 1
1 −1 𝑢1 =
x1 + 𝑥2
2
, 𝑢2 =
x1 − 𝑥2
2
Principal Coordinates

20. Approach 1.A
F=
𝟏
𝟐
𝒎
𝒚
..
𝒚
..
Bode plot of X1 and X2
𝜔1
A(𝑚𝑠−2)
𝜔(
rad
s
)

21. Approach 1.B
F= 𝒎𝒚
..
𝟎
Bode plot of X1
𝜔1A(𝑚𝑠−2
)
𝜔(
rad
s
)

22. Approach 1.B
Cont’d
𝛺 = 0 𝑡𝑜 𝛺 = 300 𝛺 = 300 𝑡𝑜 𝛺 = 20000 𝛺 = 20000 𝑡𝑜 𝛺 = 30000
Bode plot of X2
t(s)
A(𝑚𝑠−2
)

23. Auto Correlation = 𝟎. 𝟎𝟎𝟓𝟔𝟐𝟔𝟒𝟗𝟑𝟖𝟑𝟕𝟖𝟐𝟓𝟕𝟏𝟔𝟔 Auto Correlation = 𝟎. 𝟎𝟏𝟐𝟕𝟒𝟔𝟏𝟒𝟖𝟖𝟔𝟑𝟖𝟖𝟎𝟓𝟓𝟑
Approach 1
Using Mathematica
𝑞1 (m)
t(s)
𝑞2(𝑚)
t(s)

25. M
R
r 𝑚1 0
0 𝐽1
𝑥
𝜃
+
𝑘1 −𝑘1 𝑅
−𝑘1 𝑅 𝑘1 𝑅2 + 𝑘3 𝑟2
𝑥
𝜃
=
𝐹(𝑡)
𝑇(𝑡)
𝜔1 = 1.717 𝑟𝑎𝑑𝑠−1
𝜔2 = 290.991𝑟𝑎𝑑𝑠−1
Fundamental
𝑋1 =
1
1333.43
𝑋2 =
1
0.00976
Real
Positive definite
System is
Dynamically
uncoupled
System is
Statically
coupled
Stable
Orthogonal
𝑈 =
1 1
1 −1
𝑢1 = 2y −
𝜃
1333.42
,
𝑢2 =
θ
1333.42
− 𝑦
Using single degree of freedom
as a building block
Principal Coordinates

26. Approach 2.A
F=
𝟏𝟎
𝟎
(N)
Bode plot of X1
t(s)
A(𝑚𝑠−2
)

27. Approach 2.B
Cont’d
Bode plot of X2
F=
𝟏𝟎
𝟎
(N)
t(s)
A(𝑚𝑠−2)

28. Auto Correlation = 𝟎. 𝟎𝟎𝟏𝟖 Auto Correlation = 𝟎. 𝟎𝟎𝟏𝟖
Approach 2
t(s)
t(s)
A(𝑚𝑠−2) A(𝑚𝑠−2)

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31. Any Questions?