Rectangular Planar Array wher e, Rectangular Planar Array an - - PowerPoint PPT Presentation

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Rectangular Planar Array wher e, Rectangular Planar Array an - - PowerPoint PPT Presentation

Rectangular Planar Array wher e, Rectangular Planar Array an where k = 2 / d The principal maximum(m = n = 0) and grating lobes can be located by: m = 0, 1, 2,. n = 0, 1, 2,. Radiation Pattern of 5x5 Planar Array


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Rectangular Planar Array

wher e,

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SLIDE 2

Rectangular Planar Array

where k = 2π/λ

The principal maximum(m = n = 0) and grating lobes can be located by:

an d m = 0, 1, 2,…. n = 0, 1, 2,….

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SLIDE 3

Radiation Pattern of 5x5 Planar Array

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Directivity of Planar Array

Directivity of Rectangular Array For Broadside Array: Directivity of Circular Array

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Example: Calculate the array factor of a 7-elements hexagonal array (2 elements in first and third rows, 3 elements in the second row).

Group 2- 2x2 array Group 1- 3x1 array λ/2 λ/2

Hexagonal Array – 7 Elements

x y

Total Array Factor = Array Factor of (Group 1 + Group 2)

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SLIDE 6

Array factor of Group 2: M = 2, N = 2 Array factor of Group 2: M = 3, N = 1

AF of Hexagonal Array – 7 Elements

Total Array Factor = AF1 + AF2

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Example: Calculate the array factor of a 19-elements hexagonal array (3 elements in first and fifth rows, 4 elements in the second and fourth rows and 5 in the third row) Total Array Factor = Array Factor of (Group 1 + Group 2 + Group 3)

Hexagonal Array – 19 Elements

Group 3- 3x2 array Group 1- 5x1 array Group 2- 4x2 array λ/2 λ/2 x y

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Array Factor of Group 1: M=5, N=1 Array Factor of Group 2: M=4, N=2

AF of Hexagonal Array – 19 Elements

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Array Factor of Group 3: M=3, N=2

AF of Hexagonal Array – 19 Elements (Contd.)

Total Array Factor:

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Circular Vs Hexagonal Array

Planar Circular Array Planar Hexagonal Array

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Microstrip Antennas

  • Prof. Girish Kumar

Electrical Engineering Department, IIT Bombay

gkumar@ee.iitb.ac.in (022) 2576 7436

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Rectangular Microstrip Antenna (RMSA)

Co-axial feed Side View r Ground plane h Top View L W X Y x

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Microwave Integrated Circuits (MIC) vs MSA

Parameters MIC MSA Dielectric Constant (εr) Large Small Thickness (h) Small Large Width (W) Generally Small (impedance dependent) Generally Large Radiation Minimum (small fringing fields) Maximum (large fringing fields) Examples Filters, power dividers, couplers, amplifiers, etc. Antennas

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Substrates for MSA

Substrate Dielectric Constant (εr) Loss tangent (tanδ) Cost Alumina 9.8 0.001 Very High Glass Epoxy 4.4 0.02 Low Duroid / Arlon 2.2 0.0009 Very High Foam 1.05 0.0001 Low/ Medium Air 1 NA

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SLIDE 15

Advantages

  • Light weight, low volume, low profile, planar

configuration, which can be made conformal

  • Low fabrication cost and ease of mass production
  • Linear and circular polarizations are possible
  • Dual frequency antennas can be easily realized
  • Feed lines and matching network can be easily

integrated with antenna structure

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SLIDE 16

Disadvantages

  • Narrow bandwidth (1 to 5%)
  • Low power handling capacity
  • Practical limitation on Gain (around 30 dB)
  • Poor isolation between the feed and

radiating elements

  • Excitation of surface waves
  • Tolerance problem requires good quality

substrate, which are expensive

  • Polarization purity is difficult to achieve
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SLIDE 17

Applications

  • Pagers and mobile phones
  • Doppler and other radars
  • Satellite communication
  • Radio altimeter
  • Command guidance and telemetry in

missiles

  • Feed elements in complex antennas
  • Satellite navigation receiver
  • Biomedical radiator
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SLIDE 18

Various Microstrip Antenna Shapes

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MSA Feeding Techniques

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Coaxial Feed

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Microstrip Line Feed

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Microstrip Feed (contd.)

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Electromagnetically Coupled Feed

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Aperture Coupled Feed

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RMSA: Resonance Frequency

where m and n are orthogonal modes of excitation.

Fundamental mode is TM10 mode, where m =1 and n = 0.

L Le W We ~ x

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RMSA – Characterization

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RMSA: Design Equations

Smaller or larger W can be taken than the W obtained from this expression.

BW α W and Gain α W Choose feed-point x between L/6 to L/4.

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RMSA: Design Example

Design a RMSA for Wi-Fi application (2.400 to 2.483 GHz)

Chose Substrate: εr = 2.32, h = 0.16 cm and tan δ = 0.001

= 3 x 1010 / ( 2 x 2.4415 x 109 x √1.66) = 4.77 cm. W = 4.7 cm is taken = 2.23 Le = 3 x 1010 / ( 2 x 2.4415 x 109 x √2.23) cm = 4.11 cm L = Le – 2 ∆L = 4.11 – 2 x 0.16 / √2.23 = 3.9 cm

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RMSA: Design Example – Simulation using IE3D

L = 3.9 cm, W = 4.7 cm, x = 0.7 cm εr = 2.32, h = 0.16 cm and tan δ = 0.001 Zin = 54Ω at f = 2.414 GHz BW for |S11| < -10 dB is from 2.395 to 2.435 GHz = 40 MHz

Designed f = 2.4415 and Simulated f = 2.414 GHz % error = 1.1%. Also, BW is small.

SOLUTION: Increase h and reduce L