Mumbai University > EXTC > Sem 8 > Satellite Communication and Networks

Marks: 8M

A 12 GHz receiver consists of an RF stage with gain G1 =30dB and noise temperature T1 = 20 K, a down converter with gain G2 = 10dB and noise temperature T2 = 360 K and an IF amplifier stage with gain G3 =15 dB and noise temperature T3 = 1000 K. Calculate the effective noise temperature and noise figure of the system. Take reference temperature as 290 K. Compute the noise figure specifications of the three stages and then compute an overall noise figure from the individual noise figure specifications.

The three stages are connected in cascade.

G1 = 30 dB = 103 =1000

G2 = 10 dB = 101 =100

G3 = 15 dB = 101.5 =31.6227

The equivalent noise temperature Teq

$T_{eq}=T_{1}+\frac{T_2}{G_1} +\frac{T_3}{G_1 G_2 }$

$T_{eq}=20+\frac{360}{1000}+\frac{1000}{(1000)(100)}=20.37 K$

Equivalent noise factor of the system $F_{eq}$

$T_{eq} = (F_{eq} -1)T_o$

20.37 =$ (F_{eq} -1) 290$

Equivalent noise factor of the system $NF_{eq}$

$NF_{eq}$ = 10 log 1.0701 = 0.2942 dB

To compute noise figure of the three stages

$T_1 = (F_1 -1)T_o 20 = (F_1 -1) 290 F1 = 1.0689 Noise figure_1 = 10 log 1.0689 = 0.2893 dB$

$T_2 = (F_2 -1)To 360 = (F_2 -1)290 F2 =2.2413 Noise figure_2 = 10 log 2.2413 = 3.505 dB$

$T_3 = (F_3 -1)To 1000 = (F_3 -1)290 F3 =4.4482 Noise figure_3 = 10 log 4.4482 = 6.4818 dB$

The equivalent noise factor Feq

$F_{eq}=F_{1}+\frac{(F_2-1)}{G_1} +\frac{(F_3-1)}{G_1 G_2}$

$F_{eq}=1.0689+\frac{2.2413-1}{1000}+\frac{4.4482-1}{(1000)(100)}=1.0701$

The equivalent noise figure $NF_eq = 10log (F_{eq} =10log(1.0701) =0.2942 dB$