Question

3.4-17 An aperture antenna with a true area of 12 m2 is excited by a 100-kW transmitter at 4.6 GHz. Radiated power is measured to be 96.3 kW when the antenna's power gain is 18,440 (or 42.66 dB). Find (a) radiation efficiency and loss, (b) directivity, (c) aperture efficiency, and (d) effective area.

Answer 1

For this aperture antenna, A_true = 12 m^2 f = 4.6 GHZ P_in = 100 KW P_rad = 96.3 KW G_p = 18440 (= 42.66 dB) a) radiation efficiency, eta = P_rad/P_in = 96.3/100 = 0.963 or, eta = 96.3% radiation loss, P_loss = P_in - P_rad rightarrow P_loss = 100 - 96.3 rightarrow P_loss = 3.7KW b) eta = G_p/G_alpha = Power gain/directivity rightarrow G_d = G_p/eta or, G_d = 18440/0.963 rightarrow G_d = 19148.5 (42.82 dB) c) Gain and effective area are related as - G_p = 4 pi Ae/lambda^2 \r\n or, Ae = lambda^2/4 pi G_p = (3 times 10^8/4.6 times 10^9)^2/4 pi times 18440 or, Ae = 6.24 m^2 d) Aperture efficiency is - eta_ap = Ae/A_true = 6.24/12 or, eta_ap = 0.52 or, eta_ap = 52%