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EE 499 · 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.

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

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

Primary source: the original lecture notes, Topic 4 — Chapter 4, Radio Wave Propagation, sections 4.1 to 4.3 (pp. 1–14). The decibel link budget and the system-loss entries go beyond the notes and are marked. Presentation View — study slide by slide →

01 — Where we are

From components to the whole link

In Week 8 we learned how RF power behaves through individual components and interfaces: how much is transmitted, reflected or dissipated, and how \(S_{11}\) and \(S_{21}\) describe and measure it. 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?

The answer is a chain, and every link in it is a number we can calculate:

  1. transmit power
  2. transmit antenna gain
  3. spreading through space
  4. receiving aperture
  5. extra propagation losses
  6. received power

02 — The propagation path

Two parts, studied separately

A signal travelling between an earth station and a satellite must pass through the Earth's atmosphere with all its upper layers. The notes divide the propagation path into two parts: the part in the atmosphere, and the part in free space. Each is 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. \(EL\) is the elevation angle of the earth-station antenna, the same angle that Week 3 calculated.

Almost all of the distance is free space; almost all of the weather is in the short red segment. That is why the two are treated so differently: the free-space part is set by geometric spreading over distance (with wavelength entering through the antennas, as section 7 explains), while the atmospheric part depends on frequency, elevation and weather.

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

03 — Is free space lossy?

The surprising answer is no

The notes begin section 4.1, free space loss, with an emphasis: free space is not lossy by any means. It is a perfect dielectric with zero conductivity and zero loss tangent. A uniform plane wave propagates in free space with absolute zero attenuation. From the electromagnetics course, a uniform plane wave travelling along \(+z\) has the field

\[\mathbf E=\hat a_x\,E_0\,e^{j(\omega t-\beta z)},\qquad \beta=\omega\sqrt{\mu_0\varepsilon_0}=\frac{\omega}{c}=\frac{2\pi f}{c},\quad c=3\times10^{8}\ \text{m/s},\quad \lambda=\frac{2\pi}{\beta}\]

The magnitude \(E_0\) remains as it is, with no attenuation, even after an infinite number of kilometres. The power density associated with the field is

\[P=\frac{1}{2}\frac{E_0^{2}}{\eta}\quad[\text{W/m}^{2}]\]

and it does not decrease with distance.

If that is the case, what do we mean by free-space loss? The answer is in the word source.

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

04 — Spherical spreading

A point radiator makes a spherical wave

A point radiator produces a spherical wave, not a plane wave. The field associated with a spherical wave can be written, for example, as

\[\mathbf E=\hat a_\theta\,E_0\,\frac{e^{j(\omega t-\mathbf k\cdot\mathbf r)}}{|\mathbf r|},\qquad |\mathbf k|=\beta=\omega\sqrt{\mu_0\varepsilon_0}=\frac{\omega}{c}\ \ [\text{rad/m}]\]

where \(|\mathbf r|=r\) is the distance from the point source. The magnitude of the field is \(E_0/r\), which decreases as \(r\) increases. Accordingly, the power density associated with the wave decreases with \(r^{2}\), and we can write

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

\(P_t\) is the transmitted power density, \(W_t\) is the total power radiated by the point radiator, and \(4\pi r^{2}\) is the surface area of a sphere of radius \(r\) centred at the point source.

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 the radius quadruples the area, so the density falls to one quarter: \(10\log_{10}4=6\ \text{dB}\) less.

This is the whole content of the \(1/r^{2}\) law. It is geometry: power spread over a surface whose area grows as \(r^{2}\). No energy has gone anywhere.

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

05 — Antenna gain and EIRP

Real antennas are not point radiators

In reality there is no perfect point source. Any antenna produces a group of spherical waves instead of one. Those waves add up in certain directions and cancel out in others, giving the shape of the antenna radiation pattern. Compare a real antenna with a point radiator radiating the same power; the notes define the gain as the ratio of the power densities:

\[\text{Gain}=\frac{\text{real power density}}{\text{isotropic power density}}\qquad(\text{same radiated power})\]

Usually the direction of maximum gain of a given antenna is directed towards the receiver. Thus the power density at a receiver at a distance \(r\) from the transmitting antenna, in the direction of its maximum gain, is

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

where \(G_{Tx}\) is the maximum gain of the transmitting antenna. The product \(G_{Tx}W_t\) appears again in the next section and is given a name there: the effective isotropic radiated power, EIRP.

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

06 — Receiving aperture

From power density to received power

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

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

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

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

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

This is the equation that brings frequency into the picture. For a given gain, a longer wavelength means a larger effective area; for a given area, a shorter wavelength means a larger gain, which is the Week 5 rule \(D\propto A/\lambda^{2}\) read backwards. A high-gain receiving antenna recovers more power from the same incident density because its effective area is larger.

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

07 — Received power and free-space loss

Putting the pieces together

Substitute the power density at the receiver, \(G_{Tx}W_t/4\pi r^{2}\), and the effective area, \(\lambda^{2}G_{Rx}/4\pi\), into \(W_r=P_rA_{\text{eff}}\). The notes write:

\[W_r=G_{Tx}\,\frac{W_t}{4\pi r^{2}}\cdot\frac{\lambda^{2}}{4\pi}\,G_{Rx}=\left(\frac{\lambda}{4\pi r}\right)^{2}G_{Rx}\,G_{Tx}\,W_t\quad[\text{W}]\]

This is the free-space transmission equation, usually met as the Friis formula. It assumes a free-space path, far-field distances, and antennas pointed at each other.

The quantity \(G_{Tx}W_t\) is usually called the effective isotropic radiated power (EIRP): the power an isotropic radiator would need to radiate to give the same density in the direction of the receiver. The free-space loss is usually defined as the factor that remains:

\[\text{FSL}=\left(\frac{4\pi r}{\lambda}\right)^{2}=\left(\frac{4\pi r}{c}\,f\right)^{2},\qquad \text{FSL}_{\text{dB}}=20\log_{10}\!\left(\frac{4\pi r}{\lambda}\right)\ \text{dB}\]

So \(\left(4\pi r/\lambda\right)^{2}\) and \(20\log_{10}(4\pi r/\lambda)\) remain the standard link-budget quantities, and we keep using them. Read the \(\lambda\) in them as a bookkeeping consequence of the aperture relation, not as a dissipative loss that grows with \(f^{2}\).

What happens at a fixed frequency change (beyond the notes)

The warning can be made concrete. If the gains are held fixed, doubling the frequency does raise FSL by \(20\log_{10}2=6\ \text{dB}\). But if instead the antenna areas are held fixed, then \(G\propto A/\lambda^{2}\) and the gains rise with frequency, and the received power becomes

\[W_r=\frac{W_t\,A_{t}\,A_{r}}{\lambda^{2}r^{2}}\]

which increases as the wavelength falls. Same dishes, higher frequency, more received power in free space: the opposite of the naive reading. Which statement is right depends entirely on what is held fixed, and the notes' instruction is to keep \(G\) and \(\lambda\) together.

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

08 — 20 dB per decade

The distance law, drawn

The notes illustrate the distance dependence graphically: received power in decibels against distance, with the power reference usually taken at the beginning of the far-field zone.

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: \(10\log_{10}(10^{2})=20\ \text{dB}\). This rate comes from geometric spreading, so it is the same for any free-space channel whatever the frequency.
Distance factor\(\times2\)\(\times3\)\(\times10\)\(\times100\)
Power density and received power\(\div4\)\(\div9\)\(\div100\)\(\div10^{4}\)
Change in dB\(-6.0\ \text{dB}\)\(-9.5\ \text{dB}\)\(-20\ \text{dB}\)\(-40\ \text{dB}\)

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

09 — Atmospheric loss

The red segment

Section 4.2 of the notes treats the loss that appears in the part of the path that passes through the atmosphere, the red segment of the figure. The atmospheric loss is given by

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

where \(EL\) is the elevation angle of the earth-station antenna, and \(AA_{90}\) is the value of the loss for \(EL=90^{\circ}\), given in the form of a graph against frequency.

The \(1/\sin(EL)\) factor says that a slanted path crosses more atmosphere than a vertical one: at \(30^{\circ}\) the path through the atmosphere is twice as long, so the loss is twice as large in decibels.

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 of atmospheric loss for the whole atmosphere against frequency, on a logarithmic scale. The loss is small at the lower end, about \(0.03\ \text{dB}\), and has two peaks: water vapour near \(22.3\ \text{GHz}\) at about \(1\ \text{dB}\), and oxygen near \(60\ \text{GHz}\) at about \(200\ \text{dB}\).

The graph in the notes is drawn on a logarithmic axis with only those values labelled, so a numerical value of \(AA_{90}\) at a particular frequency is read from the graph, or stated in the problem. Two things follow without any numbers: the oxygen peak makes frequencies near \(60\ \text{GHz}\) useless for a link to space, and the water-vapour peak makes \(22\ \text{GHz}\) a poor choice for the same reason.

Beyond the notes. Unlike free-space spreading, atmospheric loss is genuine absorption: the oxygen and water-vapour molecules take energy from the wave. It is the first loss of this chapter that is dissipative in the Week 8 sense.

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

10 — Rain attenuation

Weather as a loss in decibels

Section 4.3: weather conditions affect the quality of the signal passing through the atmosphere. The effect can be modelled as an attenuation factor in decibels:

\[RA=\alpha L\ \ [\text{dB}],\qquad \alpha=a\,R_p^{\,b}\ \ [\text{dB/km}],\qquad L=L_s\,r_p\ \ [\text{km}]\]

\(R_p\) is the rain rate in mm/hr. \(a\) and \(b\) are constants given in tables for both vertical and horizontal polarizations against frequency. \(L_s\) is the total distance through the rain layer, shown in the figure, and \(r_p\) is a modification factor that depends on the percentage of time \(R_p\) is exceeded.

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
Illustration for the \(RA\) calculation. The earth station sits at height \(R_0\) and the rain layer reaches height \(R_R\), under a layer of ice. \(L_s\) is the slant path inside the rain; \(L_G\) is its horizontal projection.

The modification factor \(r_p\) is given in the notes for three percentages of time, with \(L_G\) in kilometres:

\[r_p=\begin{cases}\dfrac{10}{10+L_G}, & p=0.001\,\%\\[2mm]\dfrac{90}{90+4L_G}, & p=0.01\,\%\\[2mm]\dfrac{180}{180+L_G}, & p=0.1\,\%\end{cases}\qquad\Longrightarrow\qquad RA=\alpha L=a\,R_p^{\,b}\,L_s\,r_p\ \ [\text{dB}]\]

The value depends on frequency, because \(a\) and \(b\) are functions of frequency. No unit conversion is needed: the empirical formulas include constants measured with specific units, so you only substitute into the right equation.

Beyond the notes. Reading the geometry off the figure: \(L_s=(R_R-R_0)/\sin(EL)\) and \(L_G=L_s\cos(EL)\). The percentage \(p\) is the fraction of time that the rain rate is exceeded: \(p=0.01\,\%\) means rain this heavy occurs for about 53 minutes in a year, so a rain allowance is a statistical choice and not a worst case.

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

11 — The link budget

Gains added, losses subtracted

Beyond the notes. The notes end with the individual losses. This section assembles their results into the form an engineer actually uses.

If dBW, dBi and adding levels still feel slippery, the auxiliary dB Gym drills exactly this.

Decibels turn multiplication into addition, and Week 8 introduced them for exactly this reason. Take the logarithm of the received-power equation, add the two propagation losses, and include the losses in the hardware at each end. Working in dBW (decibels relative to one watt) with gains in dBi (decibels relative to an isotropic antenna):

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

With \(\text{EIRP}=W_t+G_{Tx}-L_{Tx}\) in dBW, this is \(W_r=\text{EIRP}-\text{FSL}-AA-RA+G_{Rx}-L_{Rx}\). Gains are added; losses are subtracted.

TermGain or lossUnitWhere it comes from
Transmitter power \(W_t\)sourcedBWthe amplifier
Tx-side loss \(L_{Tx}\)lossdBcable, guide, connectors and filters between amplifier and antenna: Week 8's \(-S_{21}\)
Tx antenna gain \(G_{Tx}\)gaindBithe antenna pattern (section 5)
Free-space losslossdB\(20\log_{10}(4\pi r/\lambda)\) (section 7)
Atmospheric loss \(AA\)lossdB\(AA_{90}/\sin(EL)\) (section 9)
Rain loss \(RA\)lossdB\(\alpha L\) (section 10)
Rx antenna gain \(G_{Rx}\)gaindBisets \(A_{\text{eff}}\) (section 6)
Rx-side loss \(L_{Rx}\)lossdBfeed, cable and connectors before the receiver
Received power \(W_r\)resultdBWthe sum of the rows above

A budget gives the received power. Deciding whether that power is enough needs a model of the receiver, which depends on noise and is beyond the notes and this course so far. The worked example therefore uses an illustrative required received-power level, not a receiver-sensitivity model, and the margin is simple decibel bookkeeping.

12 — Worked example

A Ku-band downlink from geostationary orbit

Added course example. The notes give no complete numerical link, so this one is built only from the equations above and from earlier weeks. Use \(c=3\times10^{8}\ \text{m/s}\).

GivenDownlink at \(f=12\ \text{GHz}\) from a geostationary satellite to an earth station at elevation \(EL=30^{\circ}\). The satellite transmits \(W_t=20\ \text{W}\) with \(G_{Tx}=30\ \text{dBi}\) and a Tx-side loss of \(1.0\ \text{dB}\). The dish has \(G_{Rx}=41\ \text{dBi}\) and an Rx-side loss of \(0.5\ \text{dB}\). Read from the graph for this frequency: \(AA_{90}=0.1\ \text{dB}\). Rain, exceeded \(0.01\,\%\) of the time: \(R_p=40\ \text{mm/hr}\), \(R_R-R_0=3\ \text{km}\), with illustrative constants \(a=0.02\), \(b=1.2\). For bookkeeping only, take an illustrative required received-power level of \(-135\ \text{dBW}\) — not a receiver-sensitivity model.
  1. Distance. With Week 3's \(R_E=6378\ \text{km}\) and \(a_{GEO}=42{,}164\ \text{km}\), the slant range at \(30^{\circ}\) is \[r=-R_E\sin EL+\sqrt{(R_E\sin EL)^{2}+a_{GEO}^{2}-R_E^{2}}=38{,}612\ \text{km}\]
  2. Wavelength and free-space loss. \(\lambda=c/f=25\ \text{mm}\); \[\text{FSL}_{\text{dB}}=20\log_{10}\dfrac{4\pi\times38.612\times10^{6}}{0.025}=205.76\ \text{dB}\]
  3. EIRP. \(W_t=10\log_{10}20=13.01\ \text{dBW}\), so \(\text{EIRP}=13.01+30-1.0=42.01\ \text{dBW}\).
  4. Atmosphere. \(AA=0.1/\sin30^{\circ}=0.20\ \text{dB}\).
  5. Rain. \(\alpha=0.02\times40^{1.2}=1.673\ \text{dB/km}\); \(L_s=3/\sin30^{\circ}=6.00\ \text{km}\); \(L_G=6\cos30^{\circ}=5.196\ \text{km}\); \(r_{0.01}=\dfrac{90}{90+4\times5.196}=0.812\); \(L=L_sr_p=4.874\ \text{km}\); \(RA=1.673\times4.874=8.15\ \text{dB}\).
  6. Receive effective area. \(G_{Rx}=10^{4.1}=12{,}589\), so \(A_{\text{eff}}=\lambda^{2}G_{Rx}/4\pi=0.626\ \text{m}^{2}\).
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
The signal level in dBW through the link, including the rain allowance. The enormous free-space step is recovered, almost entirely, by the two antenna gains.
TermGain / lossClear skyRain, \(0.01\,\%\)
Transmitter powersource\(+13.01\ \text{dBW}\)\(+13.01\ \text{dBW}\)
Tx-side lossloss\(-1.00\ \text{dB}\)\(-1.00\ \text{dB}\)
Tx antenna gaingain\(+30.00\ \text{dBi}\)\(+30.00\ \text{dBi}\)
Free-space lossloss\(-205.76\ \text{dB}\)\(-205.76\ \text{dB}\)
Atmospheric lossloss\(-0.20\ \text{dB}\)\(-0.20\ \text{dB}\)
Rain lossloss\(0\)\(-8.15\ \text{dB}\)
Rx antenna gaingain\(+41.00\ \text{dBi}\)\(+41.00\ \text{dBi}\)
Rx-side lossloss\(-0.50\ \text{dB}\)\(-0.50\ \text{dB}\)
Received powerresult\(-123.45\ \text{dBW}\)\(-131.60\ \text{dBW}\)
Over the illustrative \(-135\ \text{dBW}\) leveldB bookkeeping\(+11.55\ \text{dB}\)\(+3.40\ \text{dB}\)

Check the arithmetic the other way. In linear terms \(W_r=W_t\,G_{Tx}G_{Rx}(\lambda/4\pi r)^{2}\) with all the losses applied as factors gives \(4.52\times10^{-13}\ \text{W}=-123.45\ \text{dBW}\) in clear sky, the same answer. The received power is a few tenths of a picowatt, and it stays above the illustrative \(-135\ \text{dBW}\) level even through heavy rain, because \(42\ \text{dBW}\) of EIRP and a \(41\ \text{dBi}\) dish together recover most of the \(206\ \text{dB}\) lost to spreading.

13 — Real systems

When the spacecraft is very far away

A spacecraft millions or billions of kilometres away faces the same equation with a much larger \(r\). Every factor of \(100\) in distance beyond geostationary range costs another \(40\ \text{dB}\), and every factor of \(1000\) another \(60\ \text{dB}\). Which terms in the budget can make communication possible despite that?

Transmit power \(W_t\)limited by the spacecraft's power supply
Transmit gain \(G_{Tx}\)a larger spacecraft antenna concentrates the same power into a narrower beam
Receive gain \(G_{Rx}\)a very large ground antenna has a very large effective area \(A_{\text{eff}}\)
Distance \(r\)fixed by the mission: it is the term we cannot change

That is why the large ground antennas in NASA's Deep Space Network matter: their size is a gain term in the budget. The deep-space resources in Discussions let you look at which dishes are active and how far away the spacecraft they track really are.

14 — Week summary

Will enough power reach the receiver?

Week 8 asked how RF power passes through components and interfaces. Week 9 followed it across space and added up the whole link.

Free space is not lossy

A plane wave has zero attenuation. The \(1/r^{2}\) fall is spreading over a sphere of area \(4\pi r^{2}\), not absorption, and it does not depend on frequency.

Spreading

\(P_t=\dfrac{W_t}{4\pi r^{2}}\) for a point radiator; doubling \(r\) gives \(-6\ \text{dB}\); ten times \(r\) gives \(-20\ \text{dB}\).

Gain and EIRP

\(P_r=\dfrac{G_{Tx}W_t}{4\pi r^{2}}\); \(\text{EIRP}=G_{Tx}W_t\). Gain redistributes power, it does not create it.

Aperture

\(W_r=P_rA_{\text{eff}}\), \(A_{\text{eff}}=\dfrac{\lambda^{2}}{4\pi}G\).

Friis and FSL

\(W_r=\left(\dfrac{\lambda}{4\pi r}\right)^{2}G_{Rx}G_{Tx}W_t\); \(\text{FSL}_{\text{dB}}=20\log_{10}\dfrac{4\pi r}{\lambda}\). Keep \(G\) and \(\lambda\) together.

Atmosphere

\(AA=AA_{90}/\sin(EL)\), with \(AA_{90}\) read from the graph against frequency.

Rain

\(RA=\alpha L=aR_p^{\,b}L_sr_p\) dB; \(a\) and \(b\) from tables.

Link budget

In dBW and dB: add gains, subtract losses. Received power \(=\text{EIRP}-\text{FSL}-AA-RA+G_{Rx}-L_{Rx}\).

Study this week as slides

15 — Check yourself

Five quick checks

1 · Doubling the range

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

2 · Raising the frequency

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

3 · 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?

4 · 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?

5 · Gains and losses

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

16 — Week 9 exercises

Practice problems

The lecture ends above. What follows is a separate set of five problems to work through afterwards. Each one hides its worked solution behind a disclosure — attempt the problem first, then open it. Use \(c=3\times10^{8}\ \text{m/s}\) throughout, as the lecture notes do.

Problems 1 to 4 use only the equations in the notes. Problem 5 assembles a budget and uses the added decibel form. The values of \(AA_{90}\), \(a\) and \(b\) are given in the problems; in practice they are read from the graph and the tables.

Exercise 1

Power density and the \(1/r^{2}\) 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 value at \(10\ \text{km}\) in decibels.
  3. Find it at \(100\ \text{km}\), and the change in decibels relative to \(10\ \text{km}\).
  4. Has any energy been absorbed between \(10\) and \(100\ \text{km}\)? Explain.
Show worked solution
  1. \(P_t=\dfrac{50}{4\pi(10^{4})^{2}}=3.98\times10^{-8}\ \text{W/m}^{2}\).
  2. \(P_t(20\ \text{km})=\dfrac{50}{4\pi(2\times10^{4})^{2}}=9.95\times10^{-9}\ \text{W/m}^{2}\), exactly one quarter, \(10\log_{10}(1/4)=-6.02\ \text{dB}\).
  3. \(P_t(100\ \text{km})=3.98\times10^{-10}\ \text{W/m}^{2}\), a factor \(10^{-2}\) below the \(10\ \text{km}\) value: \(-20\ \text{dB}\), one decade of distance.
  4. No. The same \(50\ \text{W}\) crosses every sphere centred on the source; it is spread over an area that is \(100\) times larger at \(100\ \text{km}\) than at \(10\ \text{km}\).

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. Calculate the wavelength and the linear gain.
  2. Calculate the effective area \(A_{\text{eff}}\).
  3. Calculate the received power in watts and in dBW.
  4. The gain is raised to \(43\ \text{dB}\). What happens to \(A_{\text{eff}}\) and to the received power?
Show worked solution
  1. \(\lambda=\dfrac{3\times10^{8}}{12\times10^{9}}=25\ \text{mm}\); \(G=10^{40/10}=10^{4}\).
  2. \(A_{\text{eff}}=\dfrac{\lambda^{2}G}{4\pi}=\dfrac{(0.025)^{2}\times10^{4}}{4\pi}=0.497\ \text{m}^{2}\).
  3. \(W_r=P\,A_{\text{eff}}=2\times10^{-14}\times0.497=9.95\times10^{-15}\ \text{W}=-140.0\ \text{dBW}\).
  4. \(43\ \text{dB}\) is a factor of \(1.995\approx2\) more gain, so \(A_{\text{eff}}\) rises to \(0.992\ \text{m}^{2}\) and the received power rises by \(3\ \text{dB}\), to \(-137.0\ \text{dBW}\).

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 that the difference equals \(20\log_{10}\) of the frequency ratio.
  4. Does a larger free-space loss mean that more energy has been absorbed? Explain in one sentence.
Show worked 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. \(\lambda=25\ \text{mm}\): \(\text{FSL}=20\log_{10}\dfrac{4\pi\times35.786\times10^{6}}{0.025}=205.1\ \text{dB}\). The difference is \(9.54\ \text{dB}\).
  3. \(20\log_{10}(12/4)=20\log_{10}3=9.54\ \text{dB}\) ✓
  4. No. The spreading itself is geometric and frequency-independent; the extra \(9.54\ \text{dB}\) appears because the gains were held fixed, so the receiving aperture \(\lambda^{2}G/4\pi\) shrinks at the higher frequency. No energy is absorbed by empty space.

Exercise 4

Received power, two ways

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

  1. Calculate the EIRP in watts and in dBW.
  2. Calculate the free-space loss in dB.
  3. Find the received power in dBW using decibels.
  4. Find it again in watts using \(W_r=\left(\lambda/4\pi r\right)^{2}G_{Rx}G_{Tx}W_t\), and convert to dBW to check.
Show worked solution
  1. \(G_{Tx}=10^{2.6}=398\), so \(\text{EIRP}=398\times10=3.98\ \text{kW}\); in dBW, \(10+26=36.0\ \text{dBW}\).
  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. \(W_r=\left(\dfrac{0.05}{4\pi\times38\times10^{6}}\right)^{2}\times10^{3.6}\times10^{2.6}\times10=1.74\times10^{-13}\ \text{W}\), and \(10\log_{10}(1.74\times10^{-13})=-127.6\ \text{dBW}\) ✓

Exercise 5

A complete link budget

Given. A \(12\ \text{GHz}\) geostationary downlink, earth-station elevation \(EL=45^{\circ}\), for which the slant range is \(37{,}412\ \text{km}\). The satellite transmits \(100\ \text{W}\) with \(G_{Tx}=33\ \text{dBi}\) and a Tx-side loss of \(0.8\ \text{dB}\); the dish has \(G_{Rx}=45\ \text{dBi}\) and an Rx-side loss of \(0.7\ \text{dB}\). \(AA_{90}=0.1\ \text{dB}\). Rain, exceeded \(0.01\,\%\) of the time: \(R_p=25\ \text{mm/hr}\), \(R_R-R_0=2.5\ \text{km}\), \(a=0.02\), \(b=1.2\).

  1. Calculate \(\text{FSL}\) and the atmospheric loss \(AA\).
  2. Calculate the rain attenuation \(RA\): \(\alpha\), \(L_s\), \(L_G\), \(r_p\), \(L\).
  3. Assemble the budget as a table and find the received power in clear sky and in rain.
  4. Which single term is largest, and which terms recover most of it?

Part 4 is the point: the table is arithmetic; seeing which terms dominate is the engineering.

Show worked solution
  1. \(\lambda=25\ \text{mm}\): \(\text{FSL}=20\log_{10}\dfrac{4\pi\times37.412\times10^{6}}{0.025}=205.49\ \text{dB}\). \(AA=0.1/\sin45^{\circ}=0.14\ \text{dB}\).
  2. \(\alpha=0.02\times25^{1.2}=0.952\ \text{dB/km}\); \(L_s=2.5/\sin45^{\circ}=3.536\ \text{km}\); \(L_G=3.536\cos45^{\circ}=2.50\ \text{km}\); \(r_{0.01}=\dfrac{90}{90+4\times2.5}=0.900\); \(L=3.536\times0.9=3.182\ \text{km}\); \(RA=0.952\times3.182=3.03\ \text{dB}\).
  3. Clear sky: \(20.00-0.8+33-205.49-0.14+45-0.7=-109.13\ \text{dBW}\). With rain: \(-109.13-3.03=-112.16\ \text{dBW}\).
TermGain / lossValue
Transmitter power (100 W)source\(+20.00\ \text{dBW}\)
Tx-side lossloss\(-0.80\ \text{dB}\)
Tx antenna gaingain\(+33.00\ \text{dBi}\)
Free-space lossloss\(-205.49\ \text{dB}\)
Atmospheric lossloss\(-0.14\ \text{dB}\)
Rain lossloss\(-3.03\ \text{dB}\)
Rx antenna gaingain\(+45.00\ \text{dBi}\)
Rx-side lossloss\(-0.70\ \text{dB}\)
Received powerresult\(-112.16\ \text{dBW}\) (clear sky \(-109.13\ \text{dBW}\))

Part 4. The free-space loss, \(205.5\ \text{dB}\), dwarfs every other term, including rain. The two antenna gains, \(78\ \text{dB}\) together, and the transmitter power recover most of it. The hardware losses and the atmosphere together cost under \(2\ \text{dB}\), so they are cheap to ignore and expensive to neglect only at the margin; the rain, at \(3\ \text{dB}\), is the one term that varies with the weather.