Link Budget Calculation

This appendix shows the link budget calculation of all the satellite links (including the radio links of the payloads). The method used was taken from [35] (section 13.3).

Distance to Satellite at Horizon

The distance to the satellite at the horizon (the maximum theoretical distance between the satellite and a ground station) can be calculated using Equation (6).

(6)\[d = \sqrt{2\cdot R_{e}\cdot h + h^{2}}\]

Where:

\(R_{e}\) = Earth radius = 6378 km
\(h\) = Satellite altitude = 500 km
\(d\) = Distance to the satellite at the horizon

So, the distance to the satellite at the horizon is:

(7)\[d = \sqrt{2\cdot 6378\cdot 500 + 500^{2}} \cong \mathbf{2574\ km}\]

Free-Space Path Loss

The free-space path loss (\(FSPL\)) can be calculated using Equation (8).

(8)\[FSPL = \left( \frac{4\pi d f}{c} \right)^{2}\]

Where:

\(d\) = Distance between the satellite and the ground station
\(f\) = Radiofrequency
\(c\) = Speed of light

The FSPL value in decibels can be calculated with Equation (9).

(9)\[\begin{split}\begin{split} FSPL^{dB} & = 20\log\left(\frac{4\pi}{c}\right) + 20\log\left(d\right) + 20\log\left(f\right) \\ & = 32,45 + 20\log\left(\frac{d}{1\ km}\right) + 20\log\left(\frac{f}{1\ MHz}\right) \\ \end{split}\end{split}\]

The minimum distance between the satellite and a ground station is the satellite altitude, in this case: 500 km. The maximum distance is the distance at the horizon, defined by Equation (7).

Downlink

Considering the frequency of the downlink as 468 MHz, the minimum and maximum FSBL is:

\[FSPL^{dB}_{min} = 32,45 + 20\log\left(\frac{500}{1\ km}\right) + 20\log\left(\frac{468}{1\ MHz}\right) = \mathbf{139,8\ dB}\]

\[FSPL^{dB}_{max} = 32,45 + 20\log\left(\frac{2574}{1\ km}\right) + 20\log\left(\frac{468}{1\ MHz}\right) = \mathbf{154,1\ dB}\]

\[\mathbf{139,8 \leq FSPL^{dB} \leq 154,1\ dB}\]

Uplink

Considering the frequency of the uplink as 401 MHz, the minimum and maximum FSBL is:

\[FSPL^{dB}_{min} = 32,45 + 20\log\left(\frac{500}{1\ km}\right) + 20\log\left(\frac{401}{1\ MHz}\right) = \mathbf{138,5\ dB}\]

\[FSPL^{dB}_{max} = 32,45 + 20\log\left(\frac{2574}{1\ km}\right) + 20\log\left(\frac{401}{1\ MHz}\right) = \mathbf{152,7\ dB}\]

\[\mathbf{138,5 \leq FSPL^{dB} \leq 152,7\ dB}\]

Uplink (Payload)

Considering the frequency of the main payload’s uplink is 401,635 MHz, the minimum and maximum FSBL is:

\[FSPL^{dB}_{min} = 32,45 + 20\log\left(\frac{500}{1\ km}\right) + 20\log\left(\frac{401,635}{1\ MHz}\right) = \mathbf{138,5\ dB}\]

\[FSPL^{dB}_{max} = 32,45 + 20\log\left(\frac{2574}{1\ km}\right) + 20\log\left(\frac{401,635}{1\ MHz}\right) = \mathbf{152,7\ dB}\]

\[\mathbf{138,5 \leq FSPL^{dB} \leq 152,7\ dB}\]

Power at Receiver

The power of the signal at the receiver can be estimated using Equation (10).

(10)\[P_{r} = P_{t} + G_{t} + G_{r} - L_{p} - L_{s}\]

Where:

\(P_{r}\) = Power at the receiver
\(P_{t}\) = Transmitter power
\(G_{t}\) = Antenna gain of the transmitter
\(G_{r}\) = Antenna gain of the receiver
\(L_{p}\) = FSPL (Free-Space Path Loss)
\(L_{s}\) = Other losses in the system

Considering the worst scenario with the maximum possible distance between the satellite and a ground station, the power at the receiver for each link is calculated below.

Downlink

\[P_{r} = 30 + 0 + 12 - 154,1 - 5 = -117,1\ dBm\]

\[\mathbf{P_{r} \geq -117,1\ dBm}\]

Uplink

\[P_{r} = 44 + 12 + 0 - 152,7 - 5 = -101,7\ dBm\]

\[\mathbf{P_{r} \geq -101,7\ dBm}\]

Uplink (Payload)

\[P_{r} = 30 + 3 + 0 - 152,7 - 5 = -124,7\ dBm\]

\[\mathbf{P_{r} \geq -124,7\ dBm}\]

Signal-to-Noise-Ratio

The Signal-to-Noise-Ratio (SNR) of a transmitted signal at the receiver can be expressed using Equation (11):

(11)\[SNR = \frac{E_{b}}{N_{0}} = \frac{P_{t}G_{t}G_{r}}{kT_{s}RL_{p}}\]

Where:

\(P_{t}\) = Transmitter power
\(G_{t}\) = Antenna gain of the transmitter
\(G_{r}\) = Receiver gain
\(k\) = Boltzmann’s constant (\(\cong 1,3806 \times 10^{-23}\ J/K\))
\(T_{s}\) = System noise temperature
\(R\) = Data rate in bits per second (bps)
\(L_{p}\) = Free-Space Path Loss (FSPL)

The system noise temperature (\(T_{s}\)) can be defined using Equation (12).

(12)\[T_{s} = T_{ant} + T_{r}\]

with:

(13)\[T_{r} = \frac{T_{0}}{L_{r}} (F - L_{r})\]

and:

(14)\[F = 1 + \frac{T_{r}}{T_{0}}\]

Combining Equations (12), (13) and (14):

(15)\[T_{s} = T_{ant} + \left( \frac{T_{0}(1 - L_{r})}{L_{r}} \right) + \left( \frac{T_{0} (F - 1)}{L_{r}} \right)\]

Where:

\(T_{ant}\) = Antenna noise temperature
\(T_{0}\) = Reference temperature (usually 290 K)
\(L_{r}\) = Line loss between the antenna and the receiver
\(F\) = Noise figure of the receiver
\(T_{r}\) = Noise temperature of the receiver

The SNR value in decibels can be calculated using the Equation (16):

(16)\[\begin{split}\begin{split} SNR^{dB} & = 10\log_{10}\left( \frac{E_{b}}{N_{0}} \right) = 10\log_{10} \left( \frac{P_{t}G_{t}G_{r}}{kT_{s}RL_{p}} \right) \\ & = P_{t}^{dBm} - 30 + G_{t}^{dB} + G_{r}^{dB} - L_{p}^{dB} - 10\log k - 10\log T_{s} - 10\log R \end{split}\end{split}\]

Considering other losses in the system (\(L_{s}\)) (cable and connection losses as an example), the Equation (16) can be corrected as presented in Equation (17).

(17)\[SNR^{dB} = P_{t}^{dBm} - 30 + G_{t}^{dB} + G_{r}^{dB} - L_{p}^{dB} - L_{s}^{dB} - 10\log k - 10\log T_{s} - 10\log R\]

Downlink

Using Equations (17) and (12), with:

\(P_{t} = 30\ dBm\)
\(G_{t} = 0\ dBi\)
\(G_{r} = 12\ dBi\)
\(L_{p} = 154,1\ dB\)
\(L_{s} = 5\ dB\)
\(R = 4800\ bps\)
\(T_{0} = 290\ K\)
\(T_{r} = 290\ K\)
\(T_{ant} = 300\ K\)
\(F = 2\ dB\)
\(L_{r} = 0,89\ (0,5\ dB)\)

\[SNR^{dB} = 30 - 30 + 0 + 12 - 154,1 - 5 + 228,6 - 28,21 - 36,81 = 16,48\ dB\]

\[\mathbf{SNR^{dB} \geq 16,48\ dB}\]

Uplink

Using Equations (17) and (12), with:

\(P_{t} = 44\ dBm\)
\(G_{t} = 12\ dBi\)
\(G_{r} = 0\ dBi\)
\(L_{p} = 152,7\ dB\)
\(L_{s} = 5\ dB\)
\(R = 4800\ bps\)
\(T_{0} = 290\ K\)
\(T_{r} = 290\ K\)
\(T_{ant} = 300\ K\)
\(F = 2\ dB\)
\(L_{r} = 0,89\ (0,5\ dB)\)

\[SNR^{dB} = 44 - 30 + 12 + 0 - 152,7 - 5 + 228,6 - 28,21 - 36,81 = 31,88\ dB\]

\[\mathbf{SNR^{dB} \geq 31,88\ dB}\]

Uplink (Payload)

Using Equations (17) and (12), with:

\(P_{t} = 30\ dBm\)
\(G_{t} = 3\ dBi\)
\(G_{r} = 0\ dBi\)
\(L_{p} = 152,7\ dB\)
\(L_{s} = 5\ dB\)
\(R = 400\ bps\)
\(T_{0} = 290\ K\)
\(T_{r} = 290\ K\)
\(T_{ant} = 300\ K\)
\(F = 2\ dB\)
\(L_{r} = 0,89\ (0,5\ dB)\)

\[SNR^{dB} = 30 - 30 + 3 + 0 - 152,7 - 5 + 228,6 - 28,21 - 26,02 = 19,67\ dB\]

\[\mathbf{SNR^{dB} \geq 19,67\ dB}\]

Link Margin

From [35], the minimum SNR value at the received considering a \(10^{-5}\) bit error rate is:

Downlink: \(SNR^{dB} \geq 9,6\ dB\)
Uplink: \(SNR^{dB} \geq 9,6\ dB\)
Uplink (payload): \(SNR^{dB} \geq 9,6\ dB\)

And considering the link margin as the SNR of the link minus the SNR threshold for a given bit error, the link margin of the radio links of the satellite are:

Downlink: \(16,48 - 9,6 = \mathbf{6,88\ dB}\)
Uplink: \(31,88 - 9,6 = \mathbf{22,28\ dB}\)
Uplink (payload): \(19,67 - 9,6 = \mathbf{10,07\ dB}\)