NetSim v14.4 Help

Contents:

  • Introduction
    • Earth fixed spot beams and cells
    • Bands
    • Band Frequency Information
    • Simulation GUI
      • Configure Non-Terrestrial Networks
      • Set device properties
      • Configure reports
      • GUI Parameters
  • Link budget
    • Link Budget Calculations: Example 1 LEO 600
    • Link Budget Calculations: Example 2 LEO 1200
  • Satellite Antenna Pattern
    • Link Budget Calculations: Example 3 LEO 600 and LEO 1200
  • Interference Models
    • CIR based Interference
    • Link Budget Calculations: Example 4 LEO 600 and LEO 1200
  • Radio Measurements Log
  • Frequency Reuse
  • Link Budget with Interference
    • Calculations for FRF 1
    • Calculations for FRF 3
  • Beam Radius Calculations
    • Hexagonal tessellation based on beam radius
  • Featured Examples
    • Impact of LEO Altitude Variation on SNR and Path Loss
    • SNR Variation Across Outdoor Scenarios with Varying Transmit Power
    • SNR and Pathloss variation with varying elevation angles
    • 3GPP 38.821 Set 1 system level simulation
      • Introduction
      • Objective
      • Part1: Network Scenario
      • Simulation Setup
      • Parameter configuration
      • Results and Discussion
      • Part-2: Peak Throughput and Spectral Efficiency Evaluation
  • Limitations and assumptions
NetSim v14.4 Help
  • Link Budget with Interference

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.

\[\theta = \tan^{- 1}\left( \frac{Z_{sat}}{\sqrt{\left( X_{UE} - X_{sat} \right)^{2} + \left( Y_{UE} - Y_{Sat} \right)^{2}}} \right)\ = \tan^{- 1}\left( \frac{1200}{99.53} \right) = 85.26{^\circ}\]

The slant height used in NetSim is per 38.811, equation 6.6-3, i.e.

\[d = \sqrt{R_{E}^{2}\sin^{2}(\alpha) + h_{o}^{2} + 2h_{o}R_{E}} - R_{E}\sin(\alpha)\]

For a link between a ground station and a LEO satellite operating at 1200 km with elevation angle \(\alpha = 86.59{^\circ}\).

\[d = \sqrt{\left( 6.371 \cdot 10^{6} \right)^{2} \cdot \sin^{2}(85.26{^\circ}) + \left( 12 \cdot 10^{5} \right)^{2} + 2 \cdot \left( 12 \cdot 10^{5} \right)\left( 6.371 \cdot 10^{6} \right)} - 6.371 \cdot 10^{6}\sin(85.26{^\circ})\]
\[d = 1203.46\ km\]

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

\[PL_{FS} = 20\log_{10}{\left( \frac{4\pi d}{\lambda} \right) =}20\log_{10}\left( \frac{4\pi\left( 1203.46 \times 10^{3} \right)\left( 2 \times 10^{9} \right)}{3 \times 10^{8}} \right) = 160.85\ dB\]

The receiver antenna \(G/T\ \)is given by the expression

\[Rx\frac{G}{T} = G_{rx} - N_{f} - 10\log_{10}{\left( T_{0} + \left( T_{a} - T_{0} \right) \times 10^{- 0.1 \times N_{f}} \right),}\]

We know \(T_{0} = 290\), and when we apply \(T_{a} = 290\), we obtain

\[Rx\frac{G_{rx}}{T} = 0 - 7\ - \ 10\log_{10}(290) = \ - 31.62\ dB/K\]

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

\[CNR = EIRP + G(\theta) + Rx\frac{G}{T} - k - PL_{FS} - PL_{SM} - PL_{AD} - B\]

\(= \ 54.77 + ( - 17.82) + ( - 31.62) - ( - 228.6) - 160.85 - 0.42 - 2 - 74.77 = \ - 4.12\) dB

\[Rx\ Power(dBm) = CNR + Noise = \ - - 4.12\ \ + \ ( - 99.06) = \ - 103.18\ dBm\]

Interference power\(,\ I,\) obtained based on the geometric interference model is \(- 101.95\ dBm\)

\[CINR\ (Linear) = \frac{Rx\ Power\ }{Noise + Interference\ Power} = \frac{10^{\frac{- 103.18}{10}}}{10^{\frac{- 99.06}{10}} + \ 10^{\frac{- 106.37}{10}}}\]
\[= \frac{4.82 \times 10^{- 11}}{1.24 \times 10^{- 10} + \ 2.31 \times 10^{- 11}} = \frac{4.82 \times 10^{- 11}}{\ 1.471 \times 10^{- 10}} = 0.327\]
\[CINR\ (dBm) = 10 \times log_{10}(0.327) = \ - 4.86\ dB\]

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.

\[\theta = \tan^{- 1}\left( \frac{Z_{sat}}{\sqrt{\left( X_{UE} - X_{sat} \right)^{2} + \left( Y_{UE} - Y_{Sat} \right)^{2}}} \right)\ = \tan^{- 1}\left( \frac{1200}{99.53} \right) = 85.26{^\circ}\]

The slant height used in NetSim is per 38.811, equation 6.6-3, i.e.

\[d = \sqrt{R_{E}^{2}\sin^{2}(\alpha) + h_{o}^{2} + 2h_{o}R_{E}} - R_{E}\sin(\alpha)\]

For a link between a ground station and a LEO satellite operating at 1200 km with elevation angle \(\alpha = 86.59{^\circ}\).

\[d = \sqrt{\left( 6.371 \cdot 10^{6} \right)^{2} \cdot \sin^{2}(85.26{^\circ}) + \left( 12 \cdot 10^{5} \right)^{2} + 2 \cdot \left( 12 \cdot 10^{5} \right)\left( 6.371 \cdot 10^{6} \right)} - 6.371 \cdot 10^{6}\sin(85.26{^\circ})\]
\[d = 1203.46\ km\]

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

\[PL_{FS} = 20\log_{10}{\left( \frac{4\pi d}{\lambda} \right) =}20\log_{10}\left( \frac{4\pi\left( 1203.46 \times 10^{3} \right)\left( 2 \times 10^{9} \right)}{3 \times 10^{8}} \right) = 160.85\ dB\]

The receiver antenna \(G/T\ \)is given by the expression

\[Rx\frac{G}{T} = G_{rx} - N_{f} - 10\log_{10}{\left( T_{0} + \left( T_{a} - T_{0} \right) \times 10^{- 0.1 \times N_{f}} \right),}\]

We know \(T_{0} = 290\), and when we apply \(T_{a} = 290\), we obtain

\[Rx\frac{G_{rx}}{T} = 0 - 7\ - \ 10\log_{10}(290) = \ - 31.62\ dB/K\]
\[Rx\frac{G}{T} = G_{rx} - N_{f} - 10\log_{10}{\left( T_{0} + \left( T_{a} - T_{0} \right) \times 10^{- 0.1 \times N_{f}} \right),}\]

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

\[CNR = EIRP + G(\theta) + Rx\frac{G}{T} - k - PL_{FS} - PL_{SM} - PL_{AD} - B\]

\(= \ 50 + ( - 17.82) + ( - 31.62) - ( - 228.6) - 160.85 - 0.96 - 2 - 70 = \ - 4.66\ dB\)

\[Rx\ Power(dBm) = CNR + Noise = \ - 4.66\ + \ ( - 103.83) = \ - 108.49\ dBm\]

Interference power\(,\ I,\) obtained based on the geometric interference model is \(- 116.51\ dBm\)

\[CINR\ (Linear) = \frac{Rx\ Power\ }{Noise + Interference\ Power} = \frac{10^{\frac{- 108.49}{10}}}{10^{\frac{- 103.83}{10}} + \ 10^{\frac{- 116.51}{10}}}\]
\[= \frac{1.41 \times 10^{- 11}}{4.13 \times 10^{- 11} + \ 2.33 \times 10^{- 12}} = \frac{1.41}{4.353} = 0.324\]
\[CINR\ (dBm) = 10 \times log_{10}(0.324) = \ - 4.89\ dB\]

CINR calculated is \(- 4.89\ dB\)

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