Which expression suits best when the solid angle obtained by the noise source is less than antenna solid angle?(a) PA ΩA=PB ΩBand ΔTA=\(\frac{\Omega_B}{\Omega_A} T_B\)(b) PA ΩB=PB ΩA and ΔTA=\(\frac{\Omega_B}{\Omega_A} T_B\)(c) ΔTA=\(\frac{\Omega_A}{\Omega_B} T_B\) and PA ΩB=PB ΩA(d) ΔTA=\(\frac{\Omega_A}{\Omega_B} T_B\) and PA ΩA=PB ΩBI have been asked this question in class test.This intriguing question comes from Antenna Noise Temperature topic in portion Antenna Parameters – II of Antennas

Which expression suits best when the solid angle obtained by the noise

1.	Which expression suits best when the solid angle obtained by the noise source is less than antenna solid angle?(a) PA ΩA=PB ΩBand ΔTA=\(\frac{\Omega_B}{\Omega_A} T_B\)(b) PA ΩB=PB ΩA and ΔTA=\(\frac{\Omega_B}{\Omega_A} T_B\)(c) ΔTA=\(\frac{\Omega_A}{\Omega_B} T_B\) and PA ΩB=PB ΩA(d) ΔTA=\(\frac{\Omega_A}{\Omega_B} T_B\) and PA ΩA=PB ΩBI have been asked this question in class test.This intriguing question comes from Antenna Noise Temperature topic in portion Antenna Parameters – II of Antennas
Answer» Right answer is (a) PA ΩA=PB ΩBand ΔTA=\(\frac{\Omega_B}{\Omega_A} T_B\) To elaborate: When the solid ANGLE OBTAINED by the noise sourceΩB is greater than antenna solid angle ΩA, then RELATION between noise TEMPERATURE introduced by beam TB and the antenna temperatureTAis given by TA= TB (If Lossless antenna). For radio astronomy, ΩB<ΩA and TA≠ TB; ΔTA=\(\frac{\Omega_B}{\Omega_A} T_B\) and PA ΩA=PB ΩB

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