Link Budget with Interference
Simulation Parameters |
FR1 |
FR3 |
|---|---|---|
Satellite Altitude |
LEO1200 |
LEO1200 |
Terrestrial environment |
Rural |
Rural |
LOS probability |
1 |
1 |
Antenna Aperture (m) |
1 |
1 |
Beam Radius (Km) |
110.26 |
110.26 |
EIRP density (dBW/MHz) |
40 |
40 |
Guard Band (KHz) |
0 |
0 |
Channel Bandwidth B (MHz) |
30 |
10 |
EIRP (dBW) |
54.77 |
50 |
Elevation angle |
85.26 |
85.26 |
Slant range d (Km) |
1203.46 |
1203.46 |
Free space pathloss (dB), \(PL_{FS}\) |
160.85 |
160.85 |
Shadow loss (dB), \(PL_{SM}\) |
0.42 |
0.96 |
Additional Loss(dB), \(PL_{AD}\) |
2 |
2 |
Total Pathloss, L(dB) |
163.27 |
168.069 |
Tx Angular Antenna Gain (dBm) |
-17.82 |
-17.82 |
Rx Antenna Gain |
0 |
0 |
Noise figure \(N_{f}\) |
7 |
7 |
Rx Antenna Temp Ta (K) |
290 |
290 |
RX equivalent antenna Temp, \(T\ \lbrack dBK\rbrack\) |
31.62 |
31.62 |
Receiver G/T |
-31.62 |
-31.62 |
Boltzmann constant \(k\ \left\lbrack \frac{dBW}{\frac{K}{Hz}} \right\rbrack\) |
-228.6 |
-228.6 |
CNR (dB) |
-4.12 |
-4.66 |
Noise (dBm) |
-99.06 |
-103.83 |
Received Power (dBm) |
-103.18 |
-108.49 |
Interference Power (dBm) (based on geometric interference model) |
-106.37 |
-116.51 |
CINR (dB) |
-4.86 |
-4.89 |
Throughput (Mbps) |
6.57 |
2.15 |
Table-1: Simulation Parameters and Performance comparison for FR1 and FR3 bands.
Calculations for FRF 1
We consider the satellite co-ordinates as \((0,\ 0,\ 1200)\) km, and the UE co-ordinates as \((77.05,\ 63.0,\ 0)\ \)km. The elevation angle is calculated as arctan of the ratio of the satellite altitude to the distance between the satellite and the UE in the XY place.
The slant height used in NetSim is per 38.811, equation 6.6-3, i.e.
For a link between a ground station and a LEO satellite operating at 1200 km with elevation angle \(\alpha = 86.59{^\circ}\).
The free space pathloss \(PL_{FS}\ \)for a channel with \(f_{c} = 2\ GHz\), or \(\lambda = \frac{c}{f} = 0.15m,\ \)and \(d = 1203.46\ km\) is
The receiver antenna \(G/T\ \)is given by the expression
We know \(T_{0} = 290\), and when we apply \(T_{a} = 290\), we obtain
We know that bandwidth \(B\ \lbrack dBHz\rbrack = 10 \times \log_{10}\left( B\lbrack Hz\rbrack \right) = 10 \times \log_{10}{\left( 30 \times 10^{6} \right) = 74.77\ \ dBHz}\)
And finally, we obtain \(CNR\) as
\(= \ 54.77 + ( - 17.82) + ( - 31.62) - ( - 228.6) - 160.85 - 0.42 - 2 - 74.77 = \ - 4.12\) dB
Interference power\(,\ I,\) obtained based on the geometric interference model is \(- 101.95\ dBm\)
CINR calculated is \(- 2.77\ dB\)
Calculations for FRF 3
We consider the satellite co-ordinates as \((0,\ 0,\ 1200)\) km, and the UE co-ordinates as \((77.05,\ 63.0,\ 0)\ \)km. The elevation angle is calculated as arctan of the ratio of the satellite altitude to the distance between the satellite and the UE in the XY place.
The slant height used in NetSim is per 38.811, equation 6.6-3, i.e.
For a link between a ground station and a LEO satellite operating at 1200 km with elevation angle \(\alpha = 86.59{^\circ}\).
The free space pathloss \(PL_{FS}\ \)for a channel with \(f_{c} = 2\ GHz\), or \(\lambda = \frac{c}{f} = 0.15m,\ \)and \(d = 1203.46\ km\) is
The receiver antenna \(G/T\ \)is given by the expression
We know \(T_{0} = 290\), and when we apply \(T_{a} = 290\), we obtain
We know that bandwidth \(B\ \lbrack dBHz\rbrack = 10 \times \log_{10}\left( B\lbrack Hz\rbrack \right) = 10 \times \log_{10}{\left( 10 \times 10^{6} \right) = 70\ dBHz}\)
And finally, we obtain \(CNR\) as
\(= \ 50 + ( - 17.82) + ( - 31.62) - ( - 228.6) - 160.85 - 0.96 - 2 - 70 = \ - 4.66\ dB\)
Interference power\(,\ I,\) obtained based on the geometric interference model is \(- 116.51\ dBm\)
CINR calculated is \(- 4.89\ dB\)