Question 14.8: Complete the link budget for a satellite system with the fol......
Complete the link budget for a satellite system with the following parameters.
| Uplink | |
| 1. Earth station transmitter output power at saturation, 2000 W | 33 dBW |
| 2. Earth station back-off loss | 3 dB |
| 3. Earth station branching and feeder losses | 4 dB |
| 4. Earth station transmit antenna gain (from Figure 27, 15 m at 14 GHz) | 64 dB |
| 5. Additional uplink atmospheric losses | 0.6 dB |
| 6. Free-space path loss (from Figure 28, at 14 GHz) | 206.5 dB |
| 7. Satellite receiver G/T_e ratio | -5.3 dBK{}^{-1} |
| 8. Satellite branching and feeder losses | 0 dB |
| 9. Bit rate | 120 Mbps |
| 10. Modulation scheme | 8-PSK |
| Downlink | |
| 1. Satellite transmitter output power at saturation, 10 W | 10 dBW |
| 2. Satellite back-off loss | 0.1 dB |
| 3. Satellite branching and feeder losses | 0.5 dB |
| 4. Satellite transmit antenna gain (from Figure 27, 0.37 m at 12 GHz) | 30.8 dB |
| 5. Additional downlink atmospheric losses | 0.4 dB |
| 6. Free-space path loss (from Figure 28, at 14 GHz) | 205.6 dB |
| 7. Earth station receive antenna gain (15 m, 12 GHz) | 62 dB |
| 8. Earth station branching and feeder losses | 0 dB |
| 9. Earth station equivalent noise temperature | 270 K |
| 10. Earth station G/T_e ratio | 37.7 dBK{}}^{-1} |
| 11. Bit rate | 120 Mbps |
| 12. Modulation scheme | 8 -PSK |
Uplink budget: Expressed as a log,
EIRP (earth station) = P_{t}+A_{t}-L_{bo}-L_{bf}
= 33 dBW + 64 dB − 3 dB − 4 dB = 90 dBW
Carrier power density at the satellite antenna:
C′ = EIRP (earth station) − L_{p} – L_{u}
= 90 dBW − 206.5 dB − 0.6 dB = −117.1 dBW
C/N_{0} at the satellite:
\frac{C}{N_{0}} = \frac{C}{KT_{e}}=\frac{C}{T_{e}}\,×\,\frac{1}{K} where\frac{C}{T_{e}} = C′ × \frac{G}{T_{e}}
Thus, \frac{C}{N_{0}} = C′ × \frac{G}{T_{e}}\,×\,\frac{1}{K}
Expressed as a log,
\frac{C}{N_{0}}= C′ + \frac{G}{T_{e}}\, -\, 10\,log(1.38\,×\,10^{-23})
= −117.1 dBW + (−5.3 dBK^{-1}) − (− 228.6 dBWK) = 106.2 dB
Thus, \frac{E_{b}}{N_{0}} = \frac{C/f_{b}}{N_{0}} = \frac{C}{N_{0}} – \,10\,log f_{b}
= 106.2 dB − 10 (log 120 × 10^{6}) = 25.4 dB
and for a minimum bandwidth system,
\frac{C}{N}=\frac{E_{b}}{N_{0}}-\frac{B}{f_{b}}=25.4 -\, 10\, log\frac{40\,×\,10^{6}}{120\,×\,10^{6}}=30.2\, dB
Downlink budget: Expressed as a log,
EIRP (satellite transponder) = P_{t}+A_{t}-L_{bo}-L_{bf}
= 10 dBW + 30.8 dB − 0.1 dB − 0.5 dB
= 40.2 dBW
Carrier power density at earth station antenna:
C′ = EIRP − L_{p} \,-\, L_{d}
= 40.2 dBW − 205.6 dB − 0.4 dB = −165.8 dBW
C/N_{0} at the earth station receiver:
\frac{C}{N_{0}} = \frac{C}{KT_{e}}=\frac{C}{T_{e}}\,×\,\frac{1}{K} where\frac{C}{T_{e}} = C′ × \frac{G}{T_{e}}
Thus, \frac{C}{N_{0}} = C′ × \frac{G}{T_{e}}\,×\,\frac{1}{K}
Expressed as a log,
\frac{C}{N_{0}}= C′ + \frac{G}{T_{e}}\, -\, 10\,log(1.38\,×\,10^{-23})
= −165.8 dBW + (37.7 dBK^{-1}) − (− 228.6 dBWK) = 100.5 dB
An alternative method of solving for C/N_{0} is
\frac{C}{N_{0}}= C′ + A_{r}-T_{e}-K
= −165.8 dBW + 62 dB − 10 log 270 − (−228.6 dBWK)
= −165.8 dBW + 62 dB − 24.3 dBK^{-1} + 228.6 dBWK = 100.5 dB
\frac{E_{b}}{N_{0}} = \frac{C}{N_{0}} – \,10\,log f_{b}
= 100.5 dB − 10 log( 120 × 10^{6})
= 100.5 dB − 80.8 dB = 19.7 dB
and for a minimum bandwidth system,
\frac{C}{N}=\frac{E_{b}}{N_{0}}-\frac{B}{f_{b}}= 19.7 -\, 10\, log\frac{40\,×\,10^{6}}{120\,×\,10^{6}}=24.5\, dB
With careful analysis and a little algebra, it can be shown that the overall energy of bit-to-noise density ratio (E_{b}/N_{0}), which includes the combined effects of the uplink ratio (E_{b}/N_{0})_{u} and the downlink ratio (E_{b}/N_{0})_{d}, is a standard product over the sum relationship and is expressed mathematically as
\frac{E_{b}}{N_{0}}(overall) = \frac{(E_{b}/N_{0})_{u}(E_{b}/N_{0})_{d}}{(E_{b}/N_{0})_{u}\,+\,(E_{b}/N_{0})_{d}} (28)
where all E_{b}/N_{0} ratios are in absolute values. For Example 25, the overall E_{b}/N_{0} ratio is
\frac{E_{b}}{N_{0}}(overall) = \frac{(346.7)(93.3)}{346.7 \,+\, 93.3}=73.5
= 10 log 73.5 = 18.7 dB