Link budget — path loss, EIRP, G - T, Eb - N0: Exercises

Everything here builds on the parent link-budget note. If a term feels unfamiliar, the wikilinks below each section point to where it was built.

The toolkit (every formula you will need)

Before we start, here is the complete set of equations, each with its symbols named in plain words. Nothing below uses a symbol not listed here.

Definition The three raw quantities behind the ratios

Two symbols do the heavy lifting on this page, so let us name them before any formula uses them:

$C$ = the received carrier power — the actual signal power (in watts, or dBW) that arrives at the receiver after all gains and losses. Think of it as "what is left in the tank when the signal reaches Earth."
$N_{0}$ = the noise power spectral density — the noise power packed into each 1 Hz slice of bandwidth (in watts per hertz). It is set entirely by temperature: $N_{0} = k T_{sys}$ , where $k$ is Boltzmann's constant and $T_{sys}$ is the system noise temperature. This is the single most important line to remember — noise per hertz is just $k$ times temperature.

Every ratio below ( $C / N_{0}$ , $E_{b} / N_{0}$ ) is one of these divided by the other. In decibels, "divide" becomes "subtract", which is why the whole budget is a sum of + gains and - losses.

Formula The link-budget cheat sheet (all in decibels)

EIRP — effective isotropic radiated power (how much power we seem to send in the best direction): $EIRP = P_{t} + G_{t}$ $P_{t}$ = transmitter power in dBW (decibels relative to 1 watt), $G_{t}$ = transmit antenna gain in dBi (decibels relative to an isotropic radiator).

Free-space path loss (Free-Space Path Loss Derivation): $L_{p} = 32.45 + 20 lo g_{10} (r_{km}) + 20 lo g_{10} (f_{MHz})$ $r_{km}$ = distance in kilometres, $f_{MHz}$ = frequency in megahertz.

Figure of merit of the receiver: $\frac{G}{T} = G_{r} - 10 lo g_{10} (T_{sys})$ $G_{r}$ = receive gain (dBi), $T_{sys}$ = system noise temperature in kelvin (Noise Temperature and Noise Figure).

Carrier-to-noise-density ratio (dB-Hz). Since $C / N_{0} = C / (k T_{sys})$ , in decibels this splits into a chain of adds and subtracts: $\frac{C}{N _{0}} = EIRP - L_{p} + \frac{G}{T} - 10 lo g_{10} (k)$ where $k = 1.38 \times 1 0^{- 23}$ J/K is Boltzmann's constant, and $10 lo g_{10} (k) = - 228.6$ dBW/K/Hz.

Energy-per-bit to noise density: $\frac{E _{b}}{N _{0}} = \frac{C}{N _{0}} - 10 lo g_{10} (R_{b})$ $R_{b}$ = data rate in bits per second.

Link margin (the safety cushion): $M = (\frac{E _{b}}{N _{0}})_{received} - (\frac{E _{b}}{N _{0}})_{required}$

Figure — Link budget — path loss, EIRP, G - T, Eb - N0

Read the picture: on the left, a factor of $\times 2$ becomes a $+ 3.01$ dB step; $\times 10$ becomes $+ 20$ dB; a $\div 100$ loss becomes a $- 20$ dB drop. Straight-line addition on the dB (right) axis corresponds to repeated multiplication on the linear (left) axis. This is the entire reason we work in decibels.

Convert P (watts)

P_{dBW} = 10 lo g_{10} (P)

Undo a dB value back to a ratio

ratio

= 1 0^{(dB /10)}

What is

N_{0}

in one line?

N_{0} = k T_{sys}

— noise power in each 1 Hz slice

What does

C

stand for?

the received carrier (signal) power reaching the antenna

Level 1 — Recognition

L1.1 — Power to dBW

A spacecraft transmitter outputs 40 W. Express this in dBW.

Recall Solution

WHAT: convert a linear power to decibels. WHY: the whole budget is done in dB so we can add. $P_{t} = 10 lo g_{10} (40) = 10 \times 1.602 = 16.02 dBW$ Answer: $\approx 16.02$ dBW.

L1.2 — EIRP by addition

The same 40 W transmitter feeds an antenna of gain 18 dBi. Find the EIRP.

Recall Solution

In dB, EIRP is just a sum: $EIRP = P_{t} + G_{t} = 16.02 + 18 = 34.02 dBW$ Answer: $\approx 34.02$ dBW. (No multiplication — that is the whole point of decibels.)

L1.3 — Reading G/T

A receiver has gain $G_{r} = 55$ dBi and system noise temperature $T_{sys} = 100$ K. Compute $G / T$ .

Recall Solution

$\frac{G}{T} = 55 - 10 lo g_{10} (100) = 55 - 10 \times 2 = 55 - 20 = 35 dB/K$ Answer: 35 dB/K.

Level 2 — Application

L2.1 — Path loss at the Moon

Compute the free-space path loss for a link to a lunar orbiter: distance $r = 3.84 \times 1 0^{5}$ km, frequency $f = 2.2$ GHz (S-band).

Recall Solution

Step 1 — units. WHY: the constant 32.45 was derived assuming $r$ is in km and $f$ in MHz; feed it anything else and the constant is wrong. $r = 3.84 \times 1 0^{5}$ km (already km); $f = 2.2$ GHz $= 2200$ MHz. Step 2 — plug in. WHY each term: the $20 lo g_{10} (r)$ term is the inverse-square spreading loss written in dB (the signal spreads over a sphere of area $4 π r^{2}$ , see figure s02); the $20 lo g_{10} (f)$ term is there because a higher-frequency wave has a smaller catching aperture for a fixed dish. $L_{p} = 32.45 + 20 lo g_{10} (3.84 \times 1 0^{5}) + 20 lo g_{10} (2200)$

$20 lo g_{10} (3.84 \times 1 0^{5}) = 20 \times 5.584 = 111.69$ dB
$20 lo g_{10} (2200) = 20 \times 3.342 = 66.85$ dB $L_{p} = 32.45 + 111.69 + 66.85 = 210.99 dB$ Answer: $\approx 211.0$ dB.

The picture below shows why distance costs you dB: the same power spread over a bigger sphere means less power per square metre.

L2.2 — Same distance, higher frequency

Repeat L2.1 but at X-band, $f = 8.4$ GHz. By how many dB does the path loss increase?

Recall Solution

Only the frequency term changes. Going from 2200 MHz to 8400 MHz is a factor of $8400/2200 = 3.818$ . $Δ L_{p} = 20 lo g_{10} (3.818) = 20 \times 0.5818 = 11.64 dB$ So $L_{p} = 210.99 + 11.64 = 222.63$ dB. Answer: path loss rises by $\approx 11.6$ dB (to $\approx 222.6$ dB). WHY it rises: higher frequency = shorter wavelength = smaller effective aperture for a fixed physical dish, so free-space loss (which scales as $f^{2}$ ) grows.

L2.3 — Eb/N0 from C/N0

A link delivers $C / N_{0} = 68$ dB-Hz and runs at $R_{b} = 1$ Mbps. Find $E_{b} / N_{0}$ .

Recall Solution

WHY subtract the rate: each second of signal carries $R_{b}$ bits, so the received power is shared among $R_{b}$ bits per second — more bits means less energy per bit, hence we subtract $10 lo g_{10} (R_{b})$ . $R_{b} = 1 \times 1 0^{6}$ so $10 lo g_{10} (1 0^{6}) = 60$ dB-Hz. $\frac{E _{b}}{N _{0}} = 68 - 60 = 8 dB$ Answer: 8 dB.

Level 3 — Analysis

L3.1 — How far can we double the range?

A link currently closes (works) with exactly 3 dB of margin. Ignoring pointing and atmosphere, how much extra path loss can we absorb, and what range multiplier does that correspond to?

Recall Solution

WHAT: margin is spare loss we can afford. WHY: every term except $L_{p}$ is fixed, so 3 dB of margin = 3 dB more $L_{p}$ before the link breaks. Path loss scales as $20 lo g_{10} (r)$ . Setting $20 lo g_{10} (k_{r}) = 3$ dB for range multiplier $k_{r}$ : $k_{r} = 1 0^{3/20} = 1 0^{0.15} = 1.413$ Answer: we can go $\approx 1.41 \times$ farther before the margin vanishes. (Not $2 \times$ ! To double range you need $20 lo g_{10} (2) = 6.02$ dB.)

L3.2 — Trading data rate for margin

A downlink has $E_{b} / N_{0} = 6$ dB at $R_{b} = 4$ Mbps, but the modem needs $E_{b} / N_{0} = 9.6$ dB for reliable BPSK decoding. We cannot change power or antennas. To what data rate must we drop to hit exactly 9.6 dB?

Recall Solution

Key idea: $C / N_{0}$ is fixed (power, loss, G/T unchanged). Only $R_{b}$ moves $E_{b} / N_{0}$ , through $E_{b} / N_{0} = C / N_{0} - 10 lo g_{10} (R_{b})$ . We are short by $9.6 - 6 = 3.6$ dB. We recover exactly that by lowering the rate: $10 lo g_{10} (\frac{4 Mbps}{R _{b}}) = 3.6 \Rightarrow \frac{4 Mbps}{R _{b}} = 1 0^{0.36} = 2.291$ $R_{b} = \frac{4}{2.291} = 1.746 Mbps$ Answer: $\approx 1.75$ Mbps. Slowing down buys margin because each bit gets a larger share of the same received power.

L3.3 — Which upgrade wins?

Your link is 2 dB short. You can either (A) cool the receiver, cutting $T_{sys}$ from 120 K to 60 K, or (B) add a 3 dB transmit power amplifier. Which gains more, and by how much does the winner exceed the need?

Recall Solution

Option A changes $G / T$ via the $- 10 lo g_{10} (T_{sys})$ term. Halving $T$ from 120 K to 60 K: $Δ (G / T) = - 10 lo g_{10} (60) + 10 lo g_{10} (120) = 10 lo g_{10} (120/60) = 10 lo g_{10} (2) = 3.01 dB$ Option B adds exactly 3.00 dB to EIRP. Both flow one-for-one into $C / N_{0}$ (and thus $E_{b} / N_{0}$ ). A wins by a hair: 3.01 dB vs 3.00 dB. Either clears the 2 dB shortfall — A leaves $3.01 - 2 = 1.01$ dB of new margin. WHY halving temperature ≈ 3 dB: noise power is proportional to $T_{sys}$ , and halving a power ratio is always $10 lo g_{10} (2) = 3.01$ dB. This is the same 3 dB you meet everywhere in this subject.

Level 4 — Synthesis

L4.1 — Build a full Mars downlink budget

Given:

$P_{t} = 12$ W
$G_{t} = 25$ dBi
$r = 2.0 \times 1 0^{8}$ km, $f = 8.4$ GHz
Ground station $G / T = 50$ dB/K (a DSN 34-m dish)
$R_{b} = 500$ kbps

Find EIRP, $L_{p}$ , $C / N_{0}$ , and $E_{b} / N_{0}$ . If BPSK with FEC needs $E_{b} / N_{0} = 2.5$ dB, what is the margin?

Recall Solution

Step 1 — EIRP. $P_{t} = 10 lo g_{10} (12) = 10.79$ dBW. $EIRP = 10.79 + 25 = 35.79 dBW$ Step 2 — Path loss. $f = 8400$ MHz, $r = 2.0 \times 1 0^{8}$ km.

$20 lo g_{10} (2.0 \times 1 0^{8}) = 20 \times 8.301 = 166.02$ dB
$20 lo g_{10} (8400) = 20 \times 3.924 = 78.49$ dB $L_{p} = 32.45 + 166.02 + 78.49 = 276.96 dB$ Step 3 — C/N0. WHY $+ 228.6$ : the term is $- 10 lo g_{10} (k) = - (- 228.6) = + 228.6$ — we are dividing by the tiny noise density $N_{0} = k T_{sys}$ , and dividing by a tiny number makes the ratio huge, so it adds. $\frac{C}{N _{0}} = 35.79 - 276.96 + 50 - (- 228.6) = 37.43 dB-Hz$ Step 4 — Eb/N0. $R_{b} = 5 \times 1 0^{5}$ , so $10 lo g_{10} (5 \times 1 0^{5}) = 56.99$ dB-Hz. $\frac{E _{b}}{N _{0}} = 37.43 - 56.99 = - 19.56 dB$ Step 5 — Margin. $M = - 19.56 - 2.5 = - 22.06$ dB. The link FAILS by 22 dB. Interpretation: at 500 kbps this budget is hopeless. See L4.2 for the fix.

L4.2 — Make the failing link close

Take the L4.1 system. Keeping everything else fixed, what is the highest data rate that still gives exactly the required $E_{b} / N_{0} = 2.5$ dB?

Recall Solution

WHY this works: $C / N_{0}$ is a fixed property of the hardware and geometry — it does not know or care about the data rate. The rate only enters through the $- 10 lo g_{10} (R_{b})$ term, so we simply ask "what rate makes $E_{b} / N_{0}$ land exactly on the requirement?" Rearranging $E_{b} / N_{0} = C / N_{0} - 10 lo g_{10} (R_{b}) = 2.5$ : $10 lo g_{10} (R_{b}) = 37.43 - 2.5 = 34.93 dB-Hz$ Undo the dB (raise 10 to the power dB/10, because that inverts $10 lo g_{10}$ ): $R_{b} = 1 0^{34.93/10} = 1 0^{3.493} = 3113 bits/s \approx 3.1 kbps$ Answer: $\approx 3.1$ kbps. Deep-space links really are this slow — this is why missions use heavy FEC and small data rates.

Level 5 — Mastery

L5.1 — Shannon sanity check

Our L4 link has $C / N_{0} = 37.43$ dB-Hz. Suppose the channel bandwidth is $B = 1$ MHz. Using the Shannon-Hartley Theorem $C_{cap} = B lo g_{2} (1 + SNR)$ , is the L4.2 rate of 3.1 kbps anywhere near the theoretical ceiling?

Recall Solution

Step 1 — get SNR (linear) in the bandwidth. WHY divide by $B$ : $C / N_{0}$ is signal-to-noise per hertz; to get the ordinary SNR we must spread that noise across all $B$ hertz the channel occupies, i.e. $SNR = (C / N_{0}) / B$ . In dB: $37.43 - 10 lo g_{10} (1 0^{6}) = 37.43 - 60 = - 22.57$ dB. As a ratio: $SNR = 1 0^{- 22.57/10} = 1 0^{- 2.257} = 5.53 \times 1 0^{- 3}$ . Step 2 — Shannon capacity. WHY the approximation: when $SNR ≪ 1$ (here it is $\sim 0.006$ , far below 1), the curve $lo g_{2} (1 + x)$ is nearly its tangent at the origin, $lo g_{2} (1 + x) \approx x / ln 2$ . Using it avoids a calculator and shows the physics — capacity is linear in SNR when signal is weak. $C_{cap} = 1 0^{6} \times lo g_{2} (1 + 5.53 \times 1 0^{- 3}) = 1 0^{6} \times 7.96 \times 1 0^{- 3} = 7960 bits/s$ Step 3 — compare. Our practical 3113 bps sits at $3113/7960 = 39%$ of the Shannon limit. Answer: yes, we are within a factor $\approx 2.6$ of the ceiling — respectable, and the remaining gap is exactly what better FEC and modulation try to close.

L5.2 — The atmosphere strikes

Re-open L4.1's link but running at the survivable 3.0 kbps ( $E_{b} / N_{0} \approx 2.6$ dB, tiny margin). A rainstorm at the ground station adds 4 dB of extra loss and raises $T_{sys}$ from the value giving $G / T = 50$ to one giving $G / T = 47$ dB/K. Does the link survive? Quantify the new margin against the 2.5 dB requirement.

Recall Solution

Step 1 — new C/N0. Two penalties hit: +4 dB extra path/atmospheric loss (subtract 4) and $G / T$ drops by 3 dB (subtract 3). $\frac{C}{N _{0}}_{new} = 37.43 - 4 - 3 = 30.43 dB-Hz$ Step 2 — new Eb/N0 at 3.0 kbps. $10 lo g_{10} (3000) = 34.77$ dB-Hz. $\frac{E _{b}}{N _{0}} = 30.43 - 34.77 = - 4.34 dB$ Step 3 — margin. $M = - 4.34 - 2.5 = - 6.84$ dB. The link FAILS by 6.84 dB during the storm. Engineer's takeaway: this is why deep-space links are designed with weather margin built in, and why the DSN can switch to a lower rate mid-pass. The rain both attenuates the signal and warms the sky the antenna sees — a double hit that the $G / T$ term captures.

L5.3 — Where does the 3 dB come from, structurally?

Prove from the formulas that halving the data rate and doubling the received carrier power produce exactly the same change in $E_{b} / N_{0}$ . Which one also changes $C / N_{0}$ ?

Recall Solution

Recall from the toolkit that $C$ is the received carrier power (dBW) and $N_{0} = k T_{sys}$ is the noise per hertz, so $\frac{E _{b}}{N _{0}} = \frac{C}{N _{0}} - 10 lo g_{10} (R_{b}), \frac{C}{N _{0}} = C - 10 lo g_{10} (k T_{sys}) .$ Case 1 — halve the data rate ( $R_{b} \to R_{b} /2$ ). The rate appears only in the $- 10 lo g_{10} (R_{b})$ term: $- 10 lo g_{10} (\frac{R _{b}}{2}) = - 10 lo g_{10} (R_{b}) + 10 lo g_{10} (2) = - 10 lo g_{10} (R_{b}) + 3.01.$ So $E_{b} / N_{0}$ rises by +3.01 dB. But $R_{b}$ does not appear in $C / N_{0}$ , so $C / N_{0}$ is unchanged. Case 2 — double the received carrier ( $C \to 2 C$ in linear terms). In dB, $C$ increases by $10 lo g_{10} (2) = 3.01$ dB. Because $C$ sits inside $C / N_{0}$ , that ratio rises by +3.01 dB, and since $E_{b} / N_{0} = C / N_{0} - 10 lo g_{10} (R_{b})$ , $E_{b} / N_{0}$ also rises by +3.01 dB. Conclusion: both changes lift $E_{b} / N_{0}$ by the identical +3.01 dB. The difference is that doubling the power also raises $C / N_{0}$ , while halving the rate leaves $C / N_{0}$ untouched (it only re-slices the same received power among fewer bits). This is the structural reason "3 dB" recurs everywhere — it is simply $10 lo g_{10} (2)$ , the decibel fingerprint of a factor of two.