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

This is the hands-on companion to the link-budget topic note. The parent note built the four pillars — EIRP, path loss, $G / T$ , and $E_{b} / N_{0}$ . Here we use them, and we deliberately hunt down every awkward corner: the tiny numbers, the huge numbers, the "it doesn't close" failure case, and the exam trick.

Before anything else, let us pin down the one piece of grammar the whole page speaks in.

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

Figure 1 — Why we live in decibels. Two bars carry the same number. The tall red bar on the left is the raw linear ratio "one billion" (

1 0^{9}

), a tower so tall it needs a log axis just to draw. The short blue bar on the right is that same ratio expressed in decibels: a friendly

90

. The yellow arrow labelled "10 log10" is the one operation that shrinks the left bar into the right. Every example below lives on the tame right-hand bar.

The scenario matrix

Here is every class of case a link-budget problem can throw at you. Each of the worked examples that follow is tagged with the cell(s) it covers, so by the end no cell is left unvisited.

#	Cell class	What makes it tricky	Covered by
A	Basic forward pass	plug numbers straight through, no traps	Ex 1
B	Very large input (deep-space range)	huge path loss, exponent bookkeeping	Ex 2
C	Very small / near input (LEO, short range)	small loss, does the same formula still hold?	Ex 3
D	Zero / degenerate input	$R_{b} \to$ tiny, $r \to$ tiny, $G = 0$ dBi isotropic	Ex 4
E	Sign flip — negative margin	the link that fails; interpret the minus	Ex 5
F	Limiting behaviour	double the range, double the frequency → how many dB?	Ex 6
G	Real-world word problem	pick the data rate that just closes the link	Ex 7
H	Exam-style twist	units mismatch (Hz vs dB-Hz), Shannon comparison	Ex 8

Constants and one more symbol we reuse everywhere:

Here $C$ is the received carrier power, $N_{0}$ the noise power per hertz (equal to $k T$ ), $C / N_{0}$ their ratio (units dB-Hz), $R_{b}$ the bit rate in bits per second, and $E_{b} / N_{0}$ the energy-per-bit to noise-density ratio that decides the error rate.

Example 1 — Basic forward pass (Cell A)

Forecast: Guess before computing — will $C$ be positive or negative in dBW? (A negative dBW just means "less than a watt".)

Convert power to dBW. $P_{t} = 10 lo g_{10} (20) = 13.01 dBW$ . Why this step? The whole budget is a sum, so every quantity must first become a dB number.
EIRP is transmit power plus transmit gain. $EIRP = 13.01 + 15 = 28.01 dBW$ . Why this step? From the parent note, $EIRP = P_{t} \cdot G_{t}$ in linear units, which is addition in dB.
Received carrier $= EIRP - L_{p} + G_{r} = 28.01 - 180 + 40 = - 111.99 dBW$ . Why this step? Path loss is the only minus here; the receive antenna gain claws some back.

Verify: $- 111.99$ dBW $= 1 0^{- 11.199} W \approx 6.3 \times 1 0^{- 12} W$ — sub-picowatt, exactly the "faint whisper" scale we expect for a real radio link. Sign is negative → far below 1 W, as forecast. ✓

Example 2 — Very large input: deep-space path loss (Cell B)

Forecast: The constant $32.45$ is fixed; will the range term or the frequency term dominate the total?

Put $f$ in MHz. $8.4 GHz = 8400 MHz$ . Why this step? The $32.45$ constant is calibrated for kilometres and megahertz; feed it anything else and the number is nonsense.
Range term. $20 lo g_{10} (2.5 \times 1 0^{8}) = 20 (8.3979) = 167.96 dB$ . Why this step? $lo g_{10}$ of a big number is just its "number of digits", so a hundred-million-km distance is only ~168 dB.
Frequency term. $20 lo g_{10} (8400) = 20 (3.9243) = 78.49 dB$ . Why this step? Higher frequency means a shorter wavelength, and a receiving antenna of fixed gain has a smaller effective collecting area at short wavelengths — so a larger fraction of the wave is missed. That penalty rises with frequency, and it enters the formula through this same $20 lo g_{10}$ term.
Sum. $L_{p} = 32.45 + 167.96 + 78.49 = 278.90 dB$ . Why this step? Each contribution is independent, so we add.

Verify: The range term (168 dB) dominates the frequency term (78 dB), as forecast — distance is the killer in deep space. In linear terms $1 0^{27.89}$ is roughly the ratio of the observable universe's volume to a grain of sand — space really is that unforgiving. Matches parent's $278.89$ dB (rounding). ✓

Figure 2 — The path-loss tower for Mars. Three stacked bars build up the total

L_{p}

: a fixed yellow base (

32.45

dB, the unit-conversion constant), a tall red slab (the

167.96

dB range term), and a shorter blue slab (the

78.49

dB frequency term). The red range slab is more than twice the blue frequency slab — a picture of why distance, not frequency, is the deep-space enemy. See Free-Space Path Loss Derivation for where the

32.45

comes from.

Example 3 — Very small input: a nearby LEO pass (Cell C)

Forecast: Compared with Mars (279 dB), how much smaller — half? A tenth?

Range term. $20 lo g_{10} (500) = 20 (2.6990) = 53.98 dB$ . Why this step? Same formula — nothing about it assumed a large distance, so it works identically for small $r$ .
Frequency term. $20 lo g_{10} (437) = 20 (2.6405) = 52.81 dB$ .
Sum. $L_{p} = 32.45 + 53.98 + 52.81 = 139.24 dB$ .

Verify: $139$ dB versus Mars's $279$ dB — the loss is half in dB, but because dB are logarithmic, in linear power that is $1 0^{(279 - 139) /10} = 1 0^{14}$ : the Mars signal is a hundred trillion times weaker despite "only" double the dB. This is the lesson of Cell C: small dB differences hide enormous linear ones. ✓

Example 4 — Degenerate inputs: isotropic antenna, unit bit rate (Cell D)

Forecast: In (a), does a 0-dBi gain add anything? In (b), does a 1-bit/s rate subtract anything?

(a) Convert power. $5 W = 10 lo g_{10} 5 = 6.99 dBW$ . Why this step? Standard first move.
(a) Add the gain. $EIRP = 6.99 + 0 = 6.99 dBW$ . Why this step? $0$ dBi means gain factor $= 1$ ; the antenna focuses nothing, so EIRP equals raw power. This is the definition of isotropic — the degenerate baseline all gains are measured against.
(b) Rate in dB. $10 lo g_{10} (1) = 0 dB-Hz$ . Why this step? $lo g_{10} 1 = 0$ : a one-bit-per-second link spreads all the power onto a single bit each second, subtracting nothing.
(b) Energy per bit. $E_{b} / N_{0} = 40 - 0 = 40 dB$ .

Verify: Both degenerate inputs collapse their term to zero, exactly as $lo g_{10} (1) = 0$ demands. A slower link always has higher $E_{b} / N_{0}$ — here the slowest possible rate gives the entire $C / N_{0}$ to each bit. Consistent. ✓

Example 5 — The link that fails: negative margin (Cell E)

Forecast: Margin = available minus required. Will it be positive or negative?

Margin definition. $Margin = (E_{b} / N_{0})_{available} - (E_{b} / N_{0})_{required} = 2.3 - 4.0 = - 1.7 dB$ . Why this step? A budget "closes" only when what you have exceeds what you need; the difference is the safety cushion.
Interpret the sign. Negative margin means the signal is $1 0^{1.7/10} = 1.48 \times$ too weak — you are short by 48% of the needed power. Why this step? Converting the dB deficit back to linear shows exactly how far you missed.
Fix options. To recover $1.7$ dB you could (i) drop the bit rate by a factor $1 0^{0.17} = 1.48$ , or (ii) raise EIRP by $1.7$ dB, or (iii) improve $G / T$ by $1.7$ dB. Why this step? Because everything is additive in dB, a deficit anywhere can be repaired anywhere else by the same number of dB.

Verify: A $- 1.7$ dB margin means BER too high to lock — the receiver hears noise. Reduce $R_{b}$ from, say, $2$ Mbps to $2/1.48 = 1.35$ Mbps and the term $- 10 lo g_{10} (R_{b})$ rises by $1.7$ dB, pushing margin to $0$ . Signs and remedies consistent. ✓ See Forward Error Correction for how coding lowers the required threshold instead.

Example 6 — Limiting behaviour: doubling range and frequency (Cell F)

Forecast: Guess "3 dB" or "6 dB" for each before reading on.

The rule of doubling. For any $20 lo g_{10}$ term, doubling the argument adds $20 lo g_{10} (2) = 20 (0.3010) = 6.02 dB$ . Why this step? Because $lo g_{10} (2 x) = lo g_{10} x + lo g_{10} 2$ ; the constant $lo g_{10} 2 \approx 0.301$ is worth memorising.
(a) Double the range. $L_{p}$ rises by $6.02 dB$ . Why this step? $r$ sits inside a $20 lo g_{10}$ , so its doubling costs one "6-dB brick".
(b) Double the frequency. $L_{p}$ rises by $6.02 dB$ as well. Why this step? $f$ also sits inside a $20 lo g_{10}$ — identical geometry.

Verify: Check Example 3 vs a hypothetical $1000$ -km pass: range term would be $20 lo g_{10} (1000) = 60.00$ dB, up from $53.98$ dB — a rise of $6.02$ dB. Exactly one brick. ✓

Figure 3 — Path loss climbs 6 dB per doubling. The blue curve plots

L_{p}

against distance on a logarithmic horizontal axis, which turns the inverse-square law into a straight rising line. The two yellow dots sit at

500

km and

1000

km; stepping between them — one doubling — lifts the line by exactly

6.02

dB (yellow arrow). The lone red dot far to the right is Mars at

\sim 279

dB. This is the Fris Transmission Equation drawn as a picture.

Example 7 — Real-world word problem: choose the data rate (Cell G)

Forecast: Deep-space rates are famously slow — kilobits? Megabits?

EIRP. $P_{t} = 10 lo g_{10} (15) = 11.76 dBW$ , so $EIRP = 11.76 + 22 = 33.76 dBW$ . Why this step? Same opening move; the transmit side collapses to one number.
Carrier-to-noise density. $\frac{C}{N _{0}} = 33.76 - 278.89 + 49.02 + 228.6 = 32.49 dB-Hz .$ Why this step? This is the master equation — everything the receiver has before we decide how fast to spend it. (The $+ 228.6$ is the $- 10 lo g_{10} k$ term; subtracting a negative adds.)
Solve for the rate. $E_{b} / N_{0} = C / N_{0} - 10 lo g_{10} (R_{b})$ . Setting $E_{b} / N_{0} = 4.0$ : $10 lo g_{10} (R_{b}) = 32.49 - 4.0 = 28.49 dB-Hz .$ Why this step? At the maximum rate we spend exactly the threshold energy per bit — margin zero.
Undo the log. $R_{b} = 1 0^{2.849} = 706 bit/s$ . Why this step? $1 0^{(\cdot) /10}$ is the inverse of $10 lo g_{10}$ , turning dB-Hz back into bits per second.

Verify: $\approx 706$ bit/s — genuinely the ballpark of early Mars downlinks before high-gain antennas and coding. Plug back: $10 lo g_{10} (706) = 28.49$ , so $E_{b} / N_{0} = 32.49 - 28.49 = 4.0$ dB exactly at threshold. ✓ Faster rates need better Modulation Schemes (BPSK, QPSK) or the Deep Space Network (DSN)'s 70-m dishes.

Example 8 — Exam-style twist: units and the Shannon ceiling (Cell H)

Forecast: Can the true bit rate ever exceed the Shannon capacity? (No — that's the point.)

(a) Spot the trap. The student subtracted a raw hertz count ( $1000$ ) from a decibel quantity ( $32.49$ dB-Hz). You may only subtract a dB from a dB, so the bandwidth must first be turned into decibels. Why this step? $C / N_{0}$ is a ratio-per-hertz in dB; the noise it must be divided by is $N = N_{0} B$ , and dividing in dB means subtracting the dB value of $B$ — never the bare Hz number.
(a) Convert the bandwidth. $B = 1000 Hz \to 10 lo g_{10} (1000) = 30 dB-Hz$ . Why this step? Now both quantities live on the same dB ruler and can legally be subtracted.
(a) Correct SNR. $SNR = C / N_{0} - B_{dB} = 32.49 - 30 = 2.49 dB$ . Why this step? SNR is $C / N$ where $N = N_{0} B$ ; in dB that is exactly $C / N_{0}$ minus $B$ -in-dB. (The student's answer of $- 967.5$ was nonsense; the real SNR is a modest $+ 2.49$ dB.)
(b) Linear SNR. $SNR_{lin} = 1 0^{2.49/10} = 1.774$ . Why this step? The Shannon-Hartley Theorem uses a linear ratio inside its logarithm, so we must leave dB first.
(b) Capacity. $C_{shan} = B lo g_{2} (1 + SNR_{lin}) = 1000 lo g_{2} (2.774) = 1000 (1.472) = 1472 bit/s$ . Why this step? Shannon's law sets the absolute ceiling for error-free bits given SNR and bandwidth.

Verify: $706 bit/s < 1472 bit/s$ — our practical rate sits comfortably below the Shannon ceiling, as physics demands. The $\approx 2 \times$ gap is exactly the "coding gap" that Forward Error Correction exists to close. ✓

Recall Self-test: can you reach the bottom line?

A dB is ten times which function of a ratio? ::: $lo g_{10}$ of the ratio. Doubling either range or frequency adds how many dB of path loss? ::: About $6.02$ dB. A margin of $- 1.7$ dB means the signal is how many times too weak (linear)? ::: $1 0^{0.17} \approx 1.48 \times$ too weak. To go from $C / N_{0}$ (dB-Hz) to $E_{b} / N_{0}$ (dB) you subtract what? ::: $10 lo g_{10} (R_{b})$ , the bit rate in dB. Why must you convert bandwidth from Hz to dB-Hz before subtracting it from $C / N_{0}$ ? ::: You can only add/subtract dB with dB — mixing raw Hz and dB is meaningless. A $0$ -dBi antenna contributes how much to EIRP? ::: Nothing — it is the isotropic baseline, gain factor 1. What does the $T$ in $G / T$ stand for, and is bigger better? ::: System noise temperature in kelvins; smaller is better (less receiver hiss).