Question 10.1: An X-band amplifier has a gain of 20 dB and a 1 GHz bandwidt...
An X-band amplifier has a gain of 20 dB and a 1 GHz bandwidth. Its equivalent noise temperature is to be measured via the Y-factor method. The following data are obtained:
For T_{1} = 290 K, N_{1} = −62.0 dBm.
For T_{2} = 77 K , N_{2} = −64.7 dBm.
Determine the equivalent noise temperature of the amplifier. If the amplifier is used with a source having an equivalent noise temperature of T_{s} = 450 K, what is the output noise power from the amplifier, in dBm?
From (10.8)
Y=\frac{N_{1}}{N_{2}}=\frac{T_{1}+T_{e}}{T_{2}+T_{e}}>1,, the Y-factor in dB is
Y = (N_{1} − N_{2}) dB = (−62.0) − (−64.7) = 2.7 dB,
which is a numeric value of Y = 1.86. Using (10.9) gives the equivalent noise temperature as
T_{e}=\frac{T_{1} – YT_{2}}{Y – 1}=\frac{290 – (1.86)(77)}{1.86 – 1} =170 KIf a source with an equivalent noise temperature of T_{s} = 450 K drives the amplifier, the noise power into the amplifier will be kT_{s} B. The total noise power out of the amplifier will be
N_{0}=GkT_{s}B + GkT_{e}B =100(1.38 \times 10^{-23})(10^{9})(450 + 170)= 8.56 \times 10^{-10} W=-60.7 dBm