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EE 499 · Satellite Technologies

Week 9

Will Enough Power Reach the Receiver?

Week 8 followed RF power through components and interfaces. Now we step outside the hardware and follow it across space, from the transmitting antenna to the receiving antenna — and add up everything that helps and hurts along the way.

By the end: you should be able to explain why free space is not lossy yet the received power falls as \(1/r^{2}\), use gain, EIRP and effective aperture to find received power, calculate free-space loss in decibels, add atmospheric and rain loss, and assemble a decibel link budget.

Primary source: the original lecture notes, Topic 4 — Chapter 4, Radio Wave Propagation (pp. 1–14). The decibel link budget and the system-loss entries go beyond the notes and are marked.

01 — Where we are

From components to the whole link

In Week 8 we learned how RF power behaves through individual components and interfaces. We now step outside the hardware and ask what happens to that power as the electromagnetic wave travels from the transmitting antenna to the receiving antenna.

Will enough RF power reach the receiver for the satellite link to work?

  1. spreading
  2. antenna gain
  3. effective aperture
  4. received power
  5. free-space loss
  6. atmosphere and rain
  7. link budget

Part 1 — free space

Is free space lossy?

The notes open with a surprising answer: no. So what is the “loss” that the rest of the chapter calculates?

02 — Spreading

Two parts, studied separately

The signal path between an earth station and a satellite An earth-station dish on the ground at the left. A band of atmosphere sits above the ground. The signal path leaves the dish at elevation angle EL, crosses the atmosphere as a short red segment, then continues as a long blue segment through free space to a satellite at the upper right. atmosphere earth station local horizon EL satellite in the atmosphere in free space each part of the path is studied separately
Fig. 1 of the notes, redrawn: the red segment is the path through the atmosphere, the long blue segment is free space.

Almost all of the distance is free space; almost all of the weather is in the short red segment. The notes treat them separately.

From the original lecture notes · Topic 4, p. 1 →

02 — Spreading

A plane wave never weakens

Free space is a perfect dielectric: zero conductivity, zero loss tangent. A uniform plane wave along \(+z\) has

\[\mathbf E=\hat a_x E_0\,e^{j(\omega t-\beta z)},\qquad \beta=\frac{\omega}{c}=\frac{2\pi f}{c},\quad \lambda=\frac{2\pi}{\beta},\quad c=3\times10^{8}\ \text{m/s}\]
\[P=\frac{1}{2}\frac{E_0^{2}}{\eta}\ \ [\text{W/m}^{2}]\]

\(E_0\) and \(P\) do not decrease with distance — not after a million kilometres.

From the original lecture notes · Topic 4, pp. 2–3 →

02 — Spreading

A point radiator makes a spherical wave

A point radiator produces a spherical wave. Its field magnitude is \(E_0/r\), so the power density falls as \(r^{2}\):

\[\mathbf E=\hat a_\theta E_0\,\frac{e^{j(\omega t-\mathbf k\cdot\mathbf r)}}{|\mathbf r|},\qquad |\mathbf k|=\beta=\frac{\omega}{c}\]
\[P_t=\frac{W_t}{4\pi r^{2}}\quad[\text{W/m}^{2}]\]

\(W_t\) is the total power radiated; \(4\pi r^{2}\) is the area of a sphere of radius \(r\) centred on the source.

This is geometry, not absorption: the same power crosses every sphere.

From the original lecture notes · Topic 4, pp. 3–4 →

02 — Spreading

Doubling the radius quadruples the area

Power from a point radiator spreading over a sphere A point radiator at the centre of two concentric spheres of radius r and 2r. The same power W_t crosses both. The larger sphere has four times the area, so the power density is one quarter. r 2r a point radiator sends power Wt equally in every direction sphere of radius r sphere of radius 2r area 4πr² power density P = Wt / 4πr² area 4π(2r)² = 16πr², four times larger the same Wt, so the density is one quarter nothing is absorbed: the same power crosses every sphere
The same power \(W_t\) crosses every sphere. Doubling \(r\) quarters the density: \(10\log_{10}4=6\ \text{dB}\) less.

02 — Spreading

Check yourself

1 · Doubling the range

In free space the distance doubles and everything else is unchanged. What happens to the received power?

Part 2 — the transmitter

Real antennas are not point radiators

Gain concentrates the same radiated power into the direction we want. It does not create power.

03 — Antenna gain

Gain compares a real antenna with an isotropic one

Any real antenna produces a group of spherical waves that add in some directions and cancel in others — the radiation pattern. The notes define gain by comparing power densities for the same radiated power:

\[\text{Gain}=\frac{\text{real power density}}{\text{isotropic power density}}\]

The direction of maximum gain is usually pointed at the receiver, so at distance \(r\):

\[P_r=G_{Tx}P_t=\frac{G_{Tx}\,W_t}{4\pi r^{2}}\quad[\text{W/m}^{2}]\]

From the original lecture notes · Topic 4, pp. 5–6 →

03 — Antenna gain

What the product G·W really means

\[\text{EIRP}=G_{Tx}\,W_t\]

The power an isotropic radiator would need to radiate to give the same density in the direction of the receiver.

Part 3 — the receiver

From power density to received power

A density is power per square metre. The receiving antenna collects it over an area.

04 — Receiving aperture

The receiver collects over an effective area

\[W_r=P_r\,A_{\text{eff}}\]

\(A_{\text{eff}}\) is the effective area of the receiving antenna.

Recalling from the antenna course, the effective area is tied to gain and wavelength:

\[A_{\text{eff}}=\frac{\lambda^{2}}{4\pi}\,G\]

This is the equation that brings frequency in. For fixed gain, a longer wavelength means a larger area; for a fixed area, a shorter wavelength means more gain — Week 5’s \(D\propto A/\lambda^{2}\) read backwards. A high-gain receiving antenna recovers more power because its effective area is larger.

From the original lecture notes · Topic 4, p. 6 →

04 — Receiving aperture

Check yourself

2 · A bigger receiving dish

Two receiving antennas see the same power density at the same frequency. The first has twice the gain of the second. Why does it recover more power?

Part 4 — putting the pieces together

The free-space transmission equation

Density at the receiver, times effective area, gives the received power — and a factor we call free-space loss.

05 — Received power

Substitute and collect

  1. Density at the receiver: \(P_r=\dfrac{G_{Tx}W_t}{4\pi r^{2}}\).
  2. Effective area: \(A_{\text{eff}}=\dfrac{\lambda^{2}}{4\pi}G_{Rx}\).
  3. Multiply: \(W_r=P_rA_{\text{eff}}\).
\[W_r=\left(\frac{\lambda}{4\pi r}\right)^{2}G_{Rx}\,G_{Tx}\,W_t\quad[\text{W}]\]

The free-space transmission (Friis) equation: free-space path, far field, antennas pointed at each other.

From the original lecture notes · Topic 4, pp. 7–8 →

05 — Received power

The factor that is left over

With \(\text{EIRP}=G_{Tx}W_t\), the factor that remains is the free-space loss:

\[\text{FSL}=\left(\frac{4\pi r}{\lambda}\right)^{2}=\left(\frac{4\pi r f}{c}\right)^{2},\qquad \text{FSL}_{\text{dB}}=20\log_{10}\frac{4\pi r}{\lambda}\]
Distancegrows as \(r^{2}\): spreading over a sphere
Wavelengthenters through the receiving aperture \(\lambda^{2}/4\pi\)
Notabsorption — empty space dissipates nothing; spreading is frequency-independent

So \(W_r=\text{EIRP}\cdot G_{Rx}/\text{FSL}\).

05 — Received power

A trap the notes warn about

Geometric spreadingideal free space is lossless; spreading over \(4\pi r^{2}\) is frequency-independent
The frequency termenters the FSPL/Friis expression through the gain and effective-aperture relation:
\[A_{\text{eff}}=\frac{\lambda^{2}G}{4\pi}\]

Change the frequency and \(G_{Tx}\), \(G_{Rx}\) must change too. \(\left(4\pi r/\lambda\right)^{2}\) stays a standard link-budget quantity; its \(\lambda\) is not dissipative absorption. For real, localized antennas, received power falls with the square of the distance — 20 dB per decade.

From the original lecture notes · Topic 4, p. 8 →

05 — Received power

It depends on what you hold fixed

Beyond the notes. Making the warning concrete:

Gains held fixeddoubling \(f\) raises FSL by \(20\log_{10}2=6\ \text{dB}\); received power falls by 6 dB.
Areas held fixed\(G\propto A/\lambda^{2}\), so gains rise with frequency and the received power rises.
\[W_r=\frac{W_t\,A_t\,A_r}{\lambda^{2}r^{2}}\]

Same dishes, higher frequency, more received power in free space. Keep \(G\) and \(\lambda\) together.

05 — Received power

The distance law, drawn

Received power against distance in free space Received power in decibels against distance on a logarithmic axis at 1, 10, 100 and 1000 distance units. The line falls by 20 dB for every factor of ten in distance, a slope of 20 dB per decade. Wr [dB] r 20 dB 20 dB power at the reference distance (start of the far field) slope = 20 dB / decade fixed for any free-space channel and independent of frequency 1101001000
Every factor of ten in distance costs 20 dB — geometric spreading, whatever the frequency.
Distance factor\(\times2\)\(\times3\)\(\times10\)\(\times100\)
Change in dB\(-6.0\)\(-9.5\)\(-20\)\(-40\)

From the original lecture notes · Topic 4, p. 9 →

05 — Received power

Check yourself

3 · Raising the frequency

The frequency is doubled while \(G_{Tx}\), \(G_{Rx}\) and the distance are fixed. What happens to the free-space loss \((4\pi r/\lambda)^{2}\)?

05 — Received power

Check yourself

4 · Does gain create power?

A transmitting antenna has a gain of 30 dBi. Does that mean it radiates 1000 times the power the transmitter delivers to it?

Part 5 — the red segment

The part of the path that is lossy

Unlike free space, the atmosphere genuinely absorbs. The notes give one formula for the air and one for rain.

06 — Atmosphere and rain

Atmospheric loss and elevation

\[AA=AA_{90}\,\operatorname{cosec}(EL)=\frac{AA_{90}}{\sin(EL)}\]

\(EL\) is the earth-station elevation angle; \(AA_{90}\) is the loss at \(EL=90^{\circ}\), given as a graph against frequency.

A slanted path crosses more atmosphere: at \(30^{\circ}\) the loss in decibels is twice the zenith value.

Beyond the notes. Unlike spreading, this is genuine absorption by oxygen and water vapour — the first dissipative loss of the chapter.

From the original lecture notes · Topic 4, pp. 10–11 →

06 — Atmosphere and rain

Two absorption peaks

Zenith atmospheric loss against frequency, schematic Atmospheric loss at zenith in decibels on a logarithmic scale against frequency. It starts near 0.03 dB above the critical frequency, rises to a water-vapour peak of about 1 dB near 22.3 gigahertz, falls, then rises to an oxygen peak of about 200 dB near 60 gigahertz. AA₉₀ [dB] 200 dB 1 dB 0.03 dB log scale water vapour H₂O oxygen O₂ fcritical 22.3 60 f (GHz) schematic of the notes' graph: only the labelled values are given; read AA₉₀ from the graph
The notes’ graph on a logarithmic scale: about \(0.03\ \text{dB}\) at the low end, a water-vapour peak near \(22.3\ \text{GHz}\) (about \(1\ \text{dB}\)) and an oxygen peak near \(60\ \text{GHz}\) (about \(200\ \text{dB}\)).

A value of \(AA_{90}\) at a given frequency is read from the graph or given in the problem.

06 — Atmosphere and rain

Weather as a loss in decibels

\[RA=\alpha L,\quad \alpha=aR_p^{\,b}\ [\text{dB/km}],\quad L=L_s\,r_p\ [\text{km}]\]

\(R_p\): rain rate in mm/hr. \(a,b\): table constants by frequency and polarization. \(L_s\): path through the rain layer. \(r_p\): modification factor for the percentage of time \(p\).

\[r_p=\begin{cases}\dfrac{10}{10+L_G}, & p=0.001\%\\[2mm]\dfrac{90}{90+4L_G}, & p=0.01\%\\[2mm]\dfrac{180}{180+L_G}, & p=0.1\%\end{cases}\]

From the original lecture notes · Topic 4, pp. 12–14 →

06 — Atmosphere and rain

The slant path through the rain

Geometry of the slant path through the rain layer An ice layer sits above a rain layer. An earth station on a hill of height R0 looks up at elevation EL. The slant path inside the rain layer has length Ls and rises R_R minus R0, the rain height above the station. Its horizontal projection is L_G. ice rain ES Ls LG EL RR R₀ RR − R₀
Earth station at height \(R_0\); rain layer up to \(R_R\). \(L_s\) is the slant path in rain; \(L_G\) its horizontal projection.

Beyond the notes. From the figure, \(L_s=(R_R-R_0)/\sin EL\) and \(L_G=L_s\cos EL\). The empirical formulas carry their own units, so no conversion is needed; and \(p=0.01\%\) is about 53 minutes a year, so a rain allowance is a statistical choice, not a worst case.

Part 6 — adding it all up

Gains added, losses subtracted

The notes give the individual losses. Decibels turn the whole chain into a column of numbers.

07 — The link budget

Taking logarithms

Beyond the notes. Working in dBW with gains in dBi:

\[W_r=W_t+G_{Tx}-L_{Tx}-\text{FSL}-AA-RA+G_{Rx}-L_{Rx}\]

With \(\text{EIRP}=W_t+G_{Tx}-L_{Tx}\): \(\ W_r=\text{EIRP}-\text{FSL}-AA-RA+G_{Rx}-L_{Rx}\) — all in dBW or dB.

Keep units consistent: powers in dBW (or dBm), gains in dBi, losses in dB. Never mix dBW with dBm without converting.

07 — The link budget

Every row is a gain or a loss

TermTypeWhere it comes from
Transmitter power \(W_t\)source, dBWthe amplifier
Tx-side loss \(L_{Tx}\)losscable, guide, connectors: Week 8’s \(-S_{21}\)
Tx gain \(G_{Tx}\)gainantenna pattern
Free-space lossloss\(20\log_{10}(4\pi r/\lambda)\)
Atmospheric \(AA\), rain \(RA\)loss\(AA_{90}/\sin EL\); \(\alpha L\)
Rx gain \(G_{Rx}\)gainsets \(A_{\text{eff}}\)
Rx-side loss \(L_{Rx}\)lossfeed, cable, connectors

07 — The link budget

Keep the loss words apart

Free-space lossgeometric spreading — no energy absorbed
Atmospheric and rain lossabsorption by the atmosphere
Tx- and Rx-side lossinsertion loss of hardware (Week 8)
Mismatchreflection inside that hardware — already part of its insertion loss
Evanescent decayfield falling in a guide below cutoff (Week 7) — none of the above

A budget gives the received power. Whether that is enough needs a receiver model, which depends on noise and is beyond the notes; the worked example uses an illustrative required level, not a receiver-sensitivity model.

07 — The link budget

A Ku-band downlink from geostationary orbit

Added course example. The notes contain no complete numerical link.

Given. \(f=12\ \text{GHz}\), geostationary satellite, earth station at \(EL=30^{\circ}\). \(W_t=20\ \text{W}\), \(G_{Tx}=30\ \text{dBi}\), Tx-side loss \(1.0\ \text{dB}\). \(G_{Rx}=41\ \text{dBi}\), Rx-side loss \(0.5\ \text{dB}\). \(AA_{90}=0.1\ \text{dB}\). Rain exceeded \(0.01\%\) of the time: \(R_p=40\ \text{mm/hr}\), \(R_R-R_0=3\ \text{km}\), \(a=0.02\), \(b=1.2\). Illustrative required received-power level for dB bookkeeping only: \(-135\ \text{dBW}\) — not a receiver-sensitivity model.

  1. Find the slant range, wavelength and free-space loss.
  2. Find the EIRP, and the atmospheric loss.
  3. Find the rain attenuation.
  4. Assemble the budget, clear sky and rain, and the dB difference from the illustrative level.

Use Week 3’s \(R_E=6378\ \text{km}\) and \(a_{GEO}=42{,}164\ \text{km}\).

07 — The link budget

Distance, free-space loss, EIRP

  1. Range. \(r=-R_E\sin EL+\sqrt{(R_E\sin EL)^{2}+a_{GEO}^{2}-R_E^{2}}=38{,}612\ \text{km}\).
  2. Wavelength and FSL. \(\lambda=c/f=25\ \text{mm}\); \(\text{FSL}=20\log_{10}\dfrac{4\pi\times38.612\times10^{6}}{0.025}=205.76\ \text{dB}\).
  3. EIRP. \(10\log_{10}20=13.01\ \text{dBW}\), so \(\text{EIRP}=13.01+30-1.0=42.01\ \text{dBW}\).
  4. Atmosphere. \(AA=0.1/\sin30^{\circ}=0.20\ \text{dB}\).
  5. Receive area. \(G_{Rx}=10^{4.1}=12{,}589\); \(A_{\text{eff}}=\lambda^{2}G_{Rx}/4\pi=0.626\ \text{m}^{2}\).

07 — The link budget

Rain attenuation

  1. \(\alpha=0.02\times40^{1.2}=1.673\ \text{dB/km}\).
  2. \(L_s=3/\sin30^{\circ}=6.00\ \text{km}\); \(L_G=6\cos30^{\circ}=5.196\ \text{km}\).
  3. \(r_{0.01}=\dfrac{90}{90+4\times5.196}=0.812\); \(L=L_sr_p=4.874\ \text{km}\).
  4. \(RA=\alpha L=1.673\times4.874=8.15\ \text{dB}\).

Constants \(a,b\) and \(AA_{90}\) are illustrative values given with the problem; in practice they are read from the notes’ graph and tables.

07 — The link budget

The signal level through the link

Level diagram of the worked link budget A staircase of signal level in dBW for the worked example. It starts at 13.0 dBW, loses 1.0 dB, gains 30 dB, drops by 205.8 dB of free-space loss, loses 0.2 dB in the atmosphere and 8.2 dB in rain, gains 41 dB at the receiving antenna, loses 0.5 dB, and ends at about minus 131.6 dBW. 0 dBW 13.01Wt−1Tx loss+30Tx gain−205.8FSL−0.2atmos.−8.15rain+41Rx gain−0.5Rx loss -131.6 dBW
Signal level in dBW, including the rain allowance. The large free-space step is recovered, almost entirely, by the two antenna gains.

07 — The link budget

The link-budget table

TermClear skyRain, 0.01 %
Transmitter power\(+13.01\ \text{dBW}\)\(+13.01\)
Tx-side loss\(-1.00\ \text{dB}\)\(-1.00\)
Tx antenna gain\(+30.00\ \text{dBi}\)\(+30.00\)
Free-space loss\(-205.76\ \text{dB}\)\(-205.76\)
Atmospheric loss\(-0.20\ \text{dB}\)\(-0.20\)
Rain loss\(0\)\(-8.15\)
Rx antenna gain\(+41.00\ \text{dBi}\)\(+41.00\)
Rx-side loss\(-0.50\ \text{dB}\)\(-0.50\)
Received power\(-123.45\ \text{dBW}\)\(-131.60\ \text{dBW}\)
Over illustrative \(-135\ \text{dBW}\)\(+11.55\ \text{dB}\)\(+3.40\ \text{dB}\)

Linear check: \(4.52\times10^{-13}\ \text{W}=-123.45\ \text{dBW}\) in clear sky — the same answer.

07 — The link budget

Check yourself

5 · Gains and losses

Which of these is a gain in a link budget, to be added?

08 — Real systems

Which terms can make a very long link work?

Beyond geostationary range, every factor of \(100\) in distance costs another \(40\ \text{dB}\), and every factor of \(1000\) another \(60\ \text{dB}\).

Transmit powerlimited by the spacecraft’s supply
Transmit gaina larger antenna puts the same power into a narrower beam
Receive gaina very large ground dish has a very large \(A_{\text{eff}}\)
Distancefixed by the mission

That is why the large dishes of NASA’s Deep Space Network matter: size is a gain term in the budget. See the deep-space resources in Discussions (NASA DSN Now, Eyes on the Solar System).

08 — Summary

The chain that carries power across space

Spreading\(P_t=W_t/4\pi r^{2}\); free space is not lossy
Gain, EIRP\(P_r=G_{Tx}W_t/4\pi r^{2}\); gain redistributes power
Aperture\(W_r=P_rA_{\text{eff}}\), \(A_{\text{eff}}=\lambda^{2}G/4\pi\)
Friis, FSL\(W_r=(\lambda/4\pi r)^{2}G_{Rx}G_{Tx}W_t\); \(20\log_{10}(4\pi r/\lambda)\)
Atmosphere, rain\(AA=AA_{90}/\sin EL\); \(RA=aR_p^{\,b}L_sr_p\)
Budgetadd gains, subtract losses, in dBW and dB

The received power is set by the transmitted power, both antenna gains, the wavelength and the distance, less the losses in the atmosphere and the hardware.

After the lecture

Week 9 exercises — practice problems

The lecture ends here. What follows is a set of five problems to work through afterwards, with a worked solution after each one. Attempt the problem before advancing to its solution; use \(c=3\times10^{8}\ \text{m/s}\) throughout.

09 — Exercises

Exercise 1 — the 1/r² law

Given. A point radiator radiates \(W_t=50\ \text{W}\) in free space.

  1. Find the power density at \(r=10\ \text{km}\).
  2. Find it at \(20\ \text{km}\), and the ratio to the \(10\ \text{km}\) value in decibels.
  3. Find it at \(100\ \text{km}\), and the change relative to \(10\ \text{km}\).
  4. Has any energy been absorbed? Explain.

09 — Exercises

Exercise 1 — solution

  1. \(P_t=\dfrac{50}{4\pi(10^{4})^{2}}=3.98\times10^{-8}\ \text{W/m}^{2}\).
  2. \(P_t(20\ \text{km})=9.95\times10^{-9}\ \text{W/m}^{2}\): one quarter, \(-6.02\ \text{dB}\).
  3. \(P_t(100\ \text{km})=3.98\times10^{-10}\ \text{W/m}^{2}\): a factor \(10^{-2}\), \(-20\ \text{dB}\).
  4. No. The same \(50\ \text{W}\) crosses every sphere; it is spread over \(100\) times the area at \(100\ \text{km}\).

09 — Exercises

Exercise 2 — effective aperture

Given. A receiving antenna at \(12\ \text{GHz}\) has a gain of \(40\ \text{dB}\). The incident power density is \(2\times10^{-14}\ \text{W/m}^{2}\).

  1. Find the wavelength and the linear gain.
  2. Find \(A_{\text{eff}}\).
  3. Find the received power in watts and dBW.
  4. The gain is raised to \(43\ \text{dB}\). What happens to \(A_{\text{eff}}\) and the received power?

09 — Exercises

Exercise 2 — solution

  1. \(\lambda=25\ \text{mm}\); \(G=10^{4}\).
  2. \(A_{\text{eff}}=\dfrac{(0.025)^{2}\times10^{4}}{4\pi}=0.497\ \text{m}^{2}\).
  3. \(W_r=2\times10^{-14}\times0.497=9.95\times10^{-15}\ \text{W}=-140.0\ \text{dBW}\).
  4. \(43\ \text{dB}\) is about twice the gain, so \(A_{\text{eff}}\approx0.99\ \text{m}^{2}\) and the power rises by \(3\ \text{dB}\), to \(-137.0\ \text{dBW}\).

09 — Exercises

Exercise 3 — free-space loss

Given. A geostationary satellite at the sub-satellite point, \(r=35{,}786\ \text{km}\).

  1. Calculate \(\text{FSL}_{\text{dB}}\) at \(4\ \text{GHz}\).
  2. Calculate it at \(12\ \text{GHz}\), and the difference.
  3. Show the difference equals \(20\log_{10}\) of the frequency ratio.
  4. Does larger FSL mean more energy absorbed? One sentence.

09 — Exercises

Exercise 3 — solution

  1. \(\lambda=75\ \text{mm}\): \(\text{FSL}=20\log_{10}\dfrac{4\pi\times35.786\times10^{6}}{0.075}=195.6\ \text{dB}\).
  2. \(\lambda=25\ \text{mm}\): \(205.1\ \text{dB}\); difference \(9.54\ \text{dB}\).
  3. \(20\log_{10}(12/4)=20\log_{10}3=9.54\ \text{dB}\) ✓
  4. No: spreading is geometric and frequency-independent; the extra \(9.54\ \text{dB}\) arises because the gains are held fixed, so the aperture \(\lambda^{2}G/4\pi\) shrinks. Nothing is absorbed.

09 — Exercises

Exercise 4 — received power, two ways

Given. A \(6\ \text{GHz}\) uplink: \(W_t=10\ \text{W}\), \(G_{Tx}=26\ \text{dB}\), \(G_{Rx}=36\ \text{dB}\), \(r=38{,}000\ \text{km}\). Free space only.

  1. Find the EIRP in watts and dBW.
  2. Find the free-space loss.
  3. Find the received power in dBW using decibels.
  4. Find it in watts using \(W_r=(\lambda/4\pi r)^{2}G_{Rx}G_{Tx}W_t\) and convert to check.

09 — Exercises

Exercise 4 — solution

  1. \(G_{Tx}=398\); \(\text{EIRP}=3.98\ \text{kW}=36.0\ \text{dBW}\).
  2. \(\lambda=50\ \text{mm}\): \(\text{FSL}=20\log_{10}\dfrac{4\pi\times38\times10^{6}}{0.05}=199.6\ \text{dB}\).
  3. \(W_r=36.0+36-199.6=-127.6\ \text{dBW}\).
  4. \(W_r=1.74\times10^{-13}\ \text{W}\), and \(10\log_{10}(1.74\times10^{-13})=-127.6\ \text{dBW}\) ✓

09 — Exercises

Exercise 5 — a complete link budget

Given. A \(12\ \text{GHz}\) geostationary downlink, \(EL=45^{\circ}\), slant range \(37{,}412\ \text{km}\). \(100\ \text{W}\), \(G_{Tx}=33\ \text{dBi}\), Tx-side loss \(0.8\ \text{dB}\); \(G_{Rx}=45\ \text{dBi}\), Rx-side loss \(0.7\ \text{dB}\); \(AA_{90}=0.1\ \text{dB}\). Rain exceeded \(0.01\%\): \(R_p=25\ \text{mm/hr}\), \(R_R-R_0=2.5\ \text{km}\), \(a=0.02\), \(b=1.2\).

  1. Find FSL and \(AA\).
  2. Find \(RA\): \(\alpha, L_s, L_G, r_p, L\).
  3. Assemble the budget; find the received power in clear sky and in rain.
  4. Which term is largest, and which recover most of it?

09 — Exercises

Exercise 5 — solution

  1. \(\text{FSL}=205.49\ \text{dB}\); \(AA=0.1/\sin45^{\circ}=0.14\ \text{dB}\).
  2. \(\alpha=0.02\times25^{1.2}=0.952\ \text{dB/km}\); \(L_s=3.536\ \text{km}\); \(L_G=2.50\ \text{km}\); \(r_p=\dfrac{90}{90+10}=0.900\); \(L=3.182\ \text{km}\); \(RA=3.03\ \text{dB}\).
  3. Clear sky: \(20.00-0.8+33-205.49-0.14+45-0.7=-109.13\ \text{dBW}\). With rain: \(-112.16\ \text{dBW}\).
  4. Free-space loss (\(205.5\ \text{dB}\)) dwarfs every other term; the two antenna gains (\(78\ \text{dB}\)) and the transmitter power recover most of it. Hardware and atmosphere cost under \(2\ \text{dB}\); rain, \(3\ \text{dB}\), is the term that varies with weather.

EE 499 · Week 9

Week 9 complete

  1. spreading
  2. gain
  3. aperture
  4. received power
  5. atmosphere and rain
  6. link budget

Week 8 asked how RF power passes through components. Week 9 followed it across space and added up the whole link: free-space spreading is geometric and frequency-independent (the \(\lambda\) in FSPL comes from the aperture relation); gain redistributes power; and a decibel budget turns the chain into gains added and losses subtracted.

Read the Week 9 article