University:
Boston University
Course:
CAS PY 521 | Electromagnetic Theory I
Academic year:

2022
Views:

33

Pages:

11
Author:

Carina Watson

Hemispherical Lens Based Imaging Receiver for MIMO Optical Wireless Communications Outline Introduction System description and analysis Simulation results Summary Introduction MIMO Optical Wireless System with Imaging Receiver  MIMO has potential to increase the data rate and the robustness of optical wireless systems  Non imaging receiver -- evenly distributed power --very little diversity  Imaging receiver using conventional lens -- significant spatial diversity -- small field of view (FoV)  Can we increase the FoV with a different lens?  Given a different lens, can we still have spatial diversity? Introduction Imaging Receiver Using Hemispherical Lens  Hemispherical lens -- gives wide field of view, used for cloud recording as early as 1920s -- forms distorted images -- not a problem for IM/DD  Contribution of this work -- Study the MIMO channel gain with hemispherical lens based receiver -- Calculate total received power as a function of angle of incidence and show the wide FoV of the receiver -- Demonstrate spatial diversity by observing the images of the LEDs and calculating the channel matrix. System Description and Analysis System Description  Nt Generalized Lambertian LEDs installed on the celling, pointing down -- LED is placed at S : l sin  cos , l sin  sin , l cos  Therefore  is the angle of incidence  l -- emitting un-polarized light -- Radiation pattern: Ro     m  1 cosm  where m   ln 2 ln  cos 1 2  2  The receiver put on the floor, pointing up -- Lens is of radius R and refraction index n -- Nr photodetectors -- A Nr X Nt Channel Matrix  System Description and Analysis Analysis  Ray tracing -- reflection and refraction on the surface of the lens -- two refractions: change the direction the ray travels and are governed by Snell’s Law n1 sin 1  n2 sin  2 n1 n2 : refractve index of media 1 and media 2, respectively. 1 : angle of incidence  2 : angle of refraction -- two reflections: results in the loss of optical power governed by Fresnel equations Rp 1 , 2   n1 cos  2  n2 cos 1 n1 cos 2  n2 cos 1 Rs 1 , 2   n1 cos1  n2 cos  2 n1 cos1  n2 cos  2 --For un-polarized light, the power transmission coefficient is  Channel gain m Po  m  1 cos  cosTair-lens 1   Tlens-air  3 , 4  rdrd   Pt 2 l 2 3 arcsin  n1 n2  receiver T 1 1 2 Rs  Rp2   2 Simulation Simulation Results Calculated total received power as a function of angle of incidence  Settings -- 5 m X 5 m X 2.5 m room -- One LED on the ceiling pointing down with semi-angle 1 2 -- Receiver put on the floor pointing up with 5 mm lens and a photodetector. -- Therefore the maximum angle of incidence available is 70.5 degrees -5 10 -6 Channel gain 10 -7 10 -8 10 o  1/2=15  -9 10  =30o 1/2 =45o 1/2 o  1/2=60 -10 10 0 10 20 30 40 50 Angle of incidence (degree) 60 Channel gains versus the angle of incidence for Lambertian emitters with varying half power semi-angles 70 2.5 m 5m 5m  Channel gain drops at different rates  Adequate gain provided by some of LEDs at large angel of incidence  Field of view depends on half power semi angles of transmitters: -- Large half power semi-angle = Greater field of view  Large half power semi-angle -- Adequate gain out to 70 degrees angle of incidence Simulation Results Power Density on Imaging Plane  Settings -- 5 m X 5 m X 2.5 m room 2 -- four Lambertian LEDs, with semi-angles 60 degrees, on the ceiling making 30 degrees of angle with the receiver -- Receiver put at the center of the floor pointing up with 5 mm lens and four photodetectors. Each covers one quadrant. 3 1 4 2.5 m 5m 5m 10 0.025 8 6 0.02 4 4 3 y (mm) 2 0.015 0 -2 1 0.01 2 -4 -6 0.005 -8 -10 -10 -5 0 x (mm) 5 10 0 Four LEDs with 30 degrees of angle of incidence  Images of four LEDs are clearly separated -- System with four photodiode receivers would have significant diversity 0.009 0.151 1.124 0.151 0.152 0.01 0.157 1.13    106 H 1.136 0.158 0.011 0.158   0.152 1.13 0.157 0.01   Little correlation between rows or columns -- Good diversity Simulation Results Effect of more widely spaced transmitters -3 x 10 30 2.5 20 2 y (mm) 10 1.5 0 -10 1 -20 0.5 -30 -30 -20 -10 0 x (mm) 10 20 30 0 Four LEDs with 45 degrees of angle of incidence  With more widely spaced transmitters -- Angle of incidence increases -- Overall received power decreases -- Completely separated images 0 0 H 1  0 0 0 0 1 1 0 0 0 0 1  0  0  No correlation between rows and columns -- Full diversity Simulation Results Imaging with Hemispherical lens vs Non Imaging Imaging Non-imaging Optical power distributes unevenly in each image  Various LEDs form separated images h1,2  h1,1  h2,1 h2,2 H     hNr ,1 hNr ,2 h1, Nt   h2, Nt    hNr , Nt  The channel matrix is of full rank Therefore provides full diversity order  The optical power distributes evenly among the photodetectors  h1 h2  h1 h2 H     h1 h2 hNt   hNt    hNt  Unit  ranked  Therefore no diversity provided Summary In this work, we have answered the following questions: Can we increase the FoV with a different lens? The imaging receiver has large field of view with a hemispherical lens– as large as 70 degrees for a Lambertian LED Given a different lens, can we still have spatial diversity? Spatial diversity is also provided by the lens– full ranked channel matrix

Hemispherical Lens Based Imaging Receiver for MIMO Optical Wireless Communications

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