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) | 160.85 | 160.85 |
| Shadow loss (dB) | 0.42 | 0.96 |
| Additional Loss (dB) | 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 | 7 | 7 |
| Rx Antenna Temp \(T_a\) (K) | 290 | 290 |
| RX equivalent antenna Temp | 31.62 | 31.62 |
| Receiver G/T | \(-\)31.62 | \(-\)31.62 |
| Boltzmann constant | \(-\)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) (geometric model) | \(-\)106.37 | \(-\)116.51 |
| CINR (dB) | \(-\)4.86 | \(-\)4.89 |
| Throughput (Mbps) | 6.57 | 2.15 |
Calculations for FRF 1¶
We consider the satellite co-ordinates as \((0, 0, 1200)\) km, and the UE co-ordinates as \(\left(77.05, 63.0, 0\right)\)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 plane.
\[\begin{equation} \theta =85.26^{\circ} \end{equation}\]
The slant height used in NetSim is per 38.811, equation 6.6-3, i.e.
\[\begin{equation} d=\sqrt{R_{E}^{2}\sin^{2}\!\left(\alpha \right)+h_{o}^{2}+2h_{o}R_{E}}-R_{E}\sin\!\left(\alpha \right) \end{equation}\]
For a link between a ground station and a LEO satellite operating at 1200 km with elevation angle \(\alpha =85.26^{\circ}\).
\[\begin{multline} d=\sqrt{\left(6.371\cdot 10^{6}\right)^{2}\cdot \sin^{2}\!\left(85.26^{\circ}\right)+\left(12\cdot 10^{5}\right)^{2}+2\cdot \left(12\cdot 10^{5}\right)\left(6.371\cdot 10^{6}\right)}\\ -6.371\cdot 10^{6}\sin\!\left(85.26^{\circ}\right) \end{multline}\]
\[\begin{equation} d=1203.46 \text{ km} \end{equation}\]
The free space pathloss \(PL_{FS}\) for a channel with \(f_{c}=2\) GHz, or \(\lambda =\frac{c}{f}=0.15\) m, and \(d=1203.46\) km is
\[\begin{equation} PL_{FS}=20\log_{10}\!\left(\frac{4\pi d}{\lambda }\right)=20\log_{10}\!\left(\frac{4\pi \left(1203.46\times 10^{3}\right)\left(2\times 10^{9}\right)}{3\times 10^{8}}\right)=160.85 \text{ dB} \end{equation}\]
The receiver antenna \(G/T\) is given by the expression
\[\begin{equation} Rx\frac{G}{T}=G_{rx}-N_{f}-10\log_{10}\!\left(T_{0}+\left(T_{a}-T_{0}\right)\times 10^{-0.1\times N_{f}}\right) \end{equation}\]
We know \(T_{0}=290\), and when we apply \(T_{a}=290\), we obtain
\[\begin{equation} Rx\frac{G_{rx}}{T}=0-7 - 10\log_{10}\!\left(290\right)= -31.62 \text{ dB/K} \end{equation}\]
We know that bandwidth \(B\;[\text{dBHz}]=10\times \log_{10}\!\left(B\;[\text{Hz}]\right)=10\times \log_{10}\!\left(30\times 10^{6}\right)=74.77\) dBHz
And finally, we obtain \(CNR\) as
\[\begin{equation} CNR=EIRP+G\!\left(\theta \right)+Rx\frac{G}{T}-k-PL_{FS}-PL_{SM}-PL_{AD}-B \end{equation}\]
\(= 54.77+(-17.82)+(-31.62)-(-228.6)-160.85-0.42-2-74.77= -4.12\) dB
\[\begin{equation} Rx\;\text{Power}\;(\text{dBm})=CNR+\text{Noise}= -4.12 + \left(-99.06\right)= -103.18 \text{ dBm} \end{equation}\]
Interference power, \(I,\) obtained based on the geometric interference model is \(-101.95\) dBm
\[\begin{equation} CINR \;(\text{Linear})=\frac{Rx\;\text{Power} }{\text{Noise}+\text{Interference Power}}=\frac{10^{\frac{-103.18}{10}}}{10^{\frac{-99.06}{10}}+ 10^{\frac{-106.37}{10}}} \end{equation}\]
\[\begin{equation} =\frac{4.82\times 10^{-11}}{1.24\times 10^{-10}+ 2.31\times 10^{-11}}=\frac{4.82\times 10^{-11}}{ 1.471\times 10^{-10}}=0.327 \end{equation}\]
\[\begin{equation} CINR\;(\text{dBm})=10\times \log_{10}\left(0.327\right)= -4.86 \text{ dB} \end{equation}\]
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 \(\left(77.05, 63.0, 0\right)\)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 plane.
\[\begin{equation} \theta =85.26^{\circ} \end{equation}\]
The slant height used in NetSim is per 38.811, equation 6.6-3, i.e.
\[\begin{equation} d=\sqrt{R_{E}^{2}\sin^{2}\!\left(\alpha \right)+h_{o}^{2}+2h_{o}R_{E}}-R_{E}\sin\!\left(\alpha \right) \end{equation}\]
For a link between a ground station and a LEO satellite operating at 1200 km with elevation angle \(\alpha =85.26^{\circ}\).
\[\begin{multline} d=\sqrt{\left(6.371\cdot 10^{6}\right)^{2}\cdot \sin^{2}\!\left(85.26^{\circ}\right)+\left(12\cdot 10^{5}\right)^{2}+2\cdot \left(12\cdot 10^{5}\right)\left(6.371\cdot 10^{6}\right)}\\ -6.371\cdot 10^{6}\sin\!\left(85.26^{\circ}\right) \end{multline}\]
\[\begin{equation} d=1203.46 \text{ km} \end{equation}\]
The free space pathloss \(PL_{FS}\) for a channel with \(f_{c}=2\) GHz, or \(\lambda =\frac{c}{f}=0.15\) m, and \(d=1203.46\) km is
\[\begin{equation} PL_{FS}=20\log_{10}\!\left(\frac{4\pi d}{\lambda }\right)=20\log_{10}\!\left(\frac{4\pi \left(1203.46\times 10^{3}\right)\left(2\times 10^{9}\right)}{3\times 10^{8}}\right)=160.85 \text{ dB} \end{equation}\]
The receiver antenna \(G/T\)is given by the expression
\[\begin{equation} Rx\frac{G}{T}=G_{rx}-N_{f}-10\log_{10}\!\left(T_{0}+\left(T_{a}-T_{0}\right)\times 10^{-0.1\times N_{f}}\right) \end{equation}\]
We know \(T_{0}=290\), and when we apply \(T_{a}=290\), we obtain
\[\begin{equation} Rx\frac{G_{rx}}{T}=0-7 - 10\log_{10}\!\left(290\right)= -31.62 \text{ dB/K} \end{equation}\]
\[\begin{equation} Rx\frac{G}{T}=G_{rx}-N_{f}-10\log_{10}\!\left(T_{0}+\left(T_{a}-T_{0}\right)\times 10^{-0.1\times N_{f}}\right) \end{equation}\]
We know that bandwidth \(B\;[\text{dBHz}]=10\times \log_{10}\!\left(B\;[\text{Hz}]\right)=10\times \log_{10}\!\left(10\times 10^{6}\right)=70\) dBHz
And finally, we obtain \(CNR\) as
\[\begin{equation} CNR=EIRP+G\!\left(\theta \right)+Rx\frac{G}{T}-k-PL_{FS}-PL_{SM}-PL_{AD}-B \end{equation}\]
\(= 50+(-17.82)+(-31.62)-(-228.6)-160.85-0.96-2-70= -4.66\) dB
\[\begin{equation} Rx\;\text{Power}\;(\text{dBm})=CNR+\text{Noise}= -4.66 + \left(-103.83\right)= -108.49 \text{ dBm} \end{equation}\]
Interference power, \(I,\) obtained based on the geometric interference model is \(-116.51\) dBm
\[\begin{equation} CINR \;(\text{Linear})=\frac{Rx\;\text{Power} }{\text{Noise}+\text{Interference Power}}=\frac{10^{\frac{-108.49}{10}}}{10^{\frac{-103.83}{10}}+ 10^{\frac{-116.51}{10}}} \end{equation}\]
\[\begin{equation} =\frac{1.41\times 10^{-11}}{4.13\times 10^{-11}+ 2.33\times 10^{-12}}=\frac{1.41}{4.353}=0.324 \end{equation}\]
\[\begin{equation} CINR\;(\text{dBm})=10\times \log_{10}\left(0.324\right)= -4.89 \text{ dB} \end{equation}\]
CINR calculated is \(-4.89\) dB