EE536 Lab 4

7 pt 1 pin 03 10 mw 3mw 2 friis equation

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: cm S11=- 1.5 dB At Half max power : d’= 2.3 cm Freq= 12.5 GHz IV. Estimations: 1) s11 = 10 log(Γ) ⇔ Γ = 10 s11 10 = 0.7 Pt = (1 − Γ) ⋅ Pin = 0.3 × 10 mW = 3mW 2) Friis Equation : Pr = Pt ⋅ G ² ⋅ ( With path loss L = 10 log( Gain G = Pr λ Pt ⋅ ( )² 4πD λ )² 4πD Pt ) = 23dB Pr = 3.89 = 5.9dB 3) 1 Aeff = G ⋅ λ² = 1.78 ⋅ 10 −4 m² 4π 4) Let the Beamwidth ϴ² d' = 0.23 D → θ = 0.45rad tan(θ / 2) = Directivity : D = V. 4π = 63 = 17.9dB θ² Conclusion From the data measured, we were able to compute the gain of the Rx antenna thanks to Friis transmission equation and to deduce the effective antenna area, the half beam width and the directivity of the antenna. We can note that horn antennas have a high directivity, which means that they are to receive signal from a pretty fixed direction. 2...
View Full Document

This note was uploaded on 02/11/2014 for the course EE 536 taught by Professor Victorpogrebnyak during the Fall '13 term at SUNY Buffalo.

Ask a homework question - tutors are online