Calculate the power density at a ground station for which the range is 40,000 km and the power delivered to matched load at a ground station receiver if the antenna gain is 50 dB the downlink frequency is 4 GHZ. - Marks: 10 M

Year: Dec 2014

Noise factor (F)

Noise factor is alternative way to represent amplifier noise. In defining the noise factor of an amplifier the source is taken to be at room temperature denoted by $T_0$ usually taken at $290K$.

The input noise from such a source is kT_0 and the output noise from the amplifier is

$N_{0,out}=FGkT_0$

Here G is the available power gain of the amplifier and F is noise factor.

Relationship between noise temperature and noise factor

$T_e=(F-1) T_0$

Noise figure is simply F expressed in decibels

$Noise figure=[F]=10logF$

Given:

$[EIRP]=49.4dBW$

$Range(r)=40000 Km$

$G_Rx=50dB$

Downlink frequency $(f_D )=4GHz$

To find:

• Power density=?
• Power delivered i.e power received at the earth station

Solution:

Power flux density is given as: $ψ_M=\frac{EIRP}{4πr^2 }$

So in dB

$[ψ_M ]=[EIRP]-10 log⁡(\frac{1}{4πr^2 })$

$=49.4-10 log⁡(\frac{1}{4π(40000×10^3)^2 })$

$ψ_M=\frac{212.433dBW}{m^2}$

Power received is calculated as $P_R=P_t G_t G_r.(\frac{λ}{4πr})^2$

In dB

$[P_R ]=[EIRP]+[G_R ]-10log⁡(\frac{4πr}{λ})^2$

$[P_R ]=49.4+50-10log⁡(\frac{4π.40000×10^3.4×10^9}{3×10^8})^2$

$=49.4+50-196.524$

$P_R=-97.124dBW$