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

This is a conceptual question bank for the link budget topic. No heavy arithmetic lives here — that is what the worked-example pages are for. Instead, each item pokes at a misconception or a boundary case that the maths quietly invites. Read the reveal only after you have committed to an answer out loud.

Every reveal below gives you reasoning, never a bare "yes" or "no". If you can reconstruct the why, you own the concept.

True or false — justify

Each statement is either true or false. The reveal tells you which, and the physical reason.

Doubling the transmitter power

P_{t}

doubles the EIRP in watts.

True. EIRP

= P_{t} \cdot G_{t}

in linear units, so scaling

P_{t}

by 2 scales EIRP by 2 — but in dB that is only a

+ 3

dB change, since

10 lo g_{10} (2) \approx 3

Doubling the distance

r

quadruples the path loss in decibels.

False. Doubling

r

quadruples the linear power ratio (inverse-square), but in dB that is only

20 lo g_{10} (2) \approx 6

dB added, not multiplied — dB values add, they never multiply.

Path loss depends on the antennas you choose at each end.

False. Free-space path loss

L_{p} = (4 π r / λ)^{2}

contains only distance and wavelength; antenna gains are separate terms in the budget. See Free-Space Path Loss Derivation.

A higher frequency link always suffers more free-space path loss for the same distance.

True.

L_{p} \propto (r / λ)^{2}

and

λ = c / f

, so larger

f

means smaller

λ

means larger

L_{p}

— higher frequency, more spreading loss for a fixed dish size unchanged.

Increasing system noise temperature

T_{sys}

raises the G/T figure of merit.

False.

G / T = G_{r} - 10 lo g_{10} (T_{sys})

; a larger

T_{sys}

subtracts more, lowering G/T. Lower noise temperature is better. See Noise Temperature and Noise Figure.

For a fixed received carrier power

C

, sending data faster lowers

E_{b} / N_{0}

True.

E_{b} / N_{0} = (C / N_{0}) - 10 lo g_{10} (R_{b})

, so raising the bit rate

R_{b}

subtracts more, shrinking the energy each bit receives.

EIRP is a property of the transmitter alone, measured before the signal leaves the antenna.

True. EIRP combines transmit power and transmit antenna gain only; it says nothing about the channel or the receiver.

C / N_{0}

and

E_{b} / N_{0}

are the same quantity expressed in different units.

False.

C / N_{0}

(dB-Hz) is a carrier-to-noise-density ratio independent of data rate;

E_{b} / N_{0}

is dimensionless and depends on

R_{b}

. They differ by the

10 lo g_{10} (R_{b})

term.

Spot the error

Each line contains a flawed statement or reasoning. The reveal names the mistake.

"The received power is EIRP minus path loss, so bigger receive antennas cannot help."

The receive antenna gain

G_{r}

is a separate additive term;

C = EIRP - L_{p} + G_{r}

. A bigger dish raises

G_{r}

and directly boosts received power. See Antenna Gain and Effective Aperture.

"Since path loss is 279 dB and EIRP is only 34 dBW, the link is impossible."

You cannot compare EIRP against

L_{p}

alone — you must also add the receive

G / T

term and subtract Boltzmann's constant (

- 228.6

dBW/K/Hz). Those large positive contributions rescue the link.

"G/T uses gain in dBi and temperature in Kelvin, so I subtract 25 directly for a 25 K system."

You subtract

10 lo g_{10} (T_{sys}) = 10 lo g_{10} (25) \approx 14

dB, not the raw number 25. Temperature enters logarithmically.

"To halve the bit error rate, just double the transmit power."

BER falls with

E_{b} / N_{0}

but steeply and non-linearly (roughly exponentially near threshold), so a

+ 3

dB power bump can drop BER by orders of magnitude — the relationship is not proportional. See Modulation Schemes (BPSK, QPSK).

"Using QPSK instead of BPSK doubles the data rate for free with the same

E_{b} / N_{0}

QPSK packs 2 bits per symbol, but for the same bit error rate BPSK and QPSK need the same

E_{b} / N_{0}

— the gain is spectral efficiency, not free performance. The extra bits are not "free" in energy terms.

"Boltzmann's constant is a loss, so we subtract

+ 228.6

dB in the budget."

We add

228.6

(i.e. subtract the negative

- 228.6

dBW/K/Hz). Because

k

is tiny,

10 lo g_{10} (k)

is a large negative number, and removing it helps the

C / N_{0}

Why questions

Explain the reason, not just the fact.

Why do we work the whole link budget in decibels instead of linear watts?

Because gains and losses multiply in linear units but add in dB; addition of a dozen terms is far easier and less error-prone than multiplying huge and tiny numbers spanning 30 orders of magnitude.

Why does path loss grow with

1/ r^{2}

even though nothing is absorbing the wave?

The same total power spreads over an ever-larger spherical shell of area

4 π r^{2}

; the energy is not lost, just diluted — power density falls as

1/ r^{2}

purely from geometry.

Why is

E_{b} / N_{0}

the preferred metric for digital links rather than plain signal-to-noise ratio?

E_{b} / N_{0}

normalises out the data rate, so it directly predicts bit error rate and ties cleanly to the Shannon-Hartley Theorem information-theory limit, letting us compare links running at different speeds fairly.

Why does forward error correction let a link close at a lower

E_{b} / N_{0}

than an uncoded one?

FEC adds redundant bits that let the decoder correct errors, so it tolerates a noisier channel — trading raw bit rate for a "coding gain" of several dB in required

E_{b} / N_{0}

. See Forward Error Correction.

Why do deep-space ground stations chase very low noise temperatures with cryogenic amplifiers?

Because

N_{0} = k T_{sys}

sets the noise floor, and for a fixed weak signal from a distant probe, halving

T_{sys}

directly raises

G / T

and

C / N_{0}

— the only lever left once the signal itself is fixed. See Deep Space Network (DSN).

Edge cases

Boundary and degenerate scenarios the formulas quietly assume.

What does the path-loss formula predict as

r \to 0

, and why is that unphysical?

L_{p} = (4 π r / λ)^{2} \to 0

, implying zero loss (or even "gain") at zero range — but the formula assumes a far-field spherical wave, so it simply does not apply in the near field close to the antenna.

If the receive antenna gain

G_{r}

is 0 dBi (isotropic receiver), does the link still work?

It can — 0 dBi means gain factor 1 (no help, no hurt); the term just contributes

+ 0

to the budget. The link survives only if EIRP,

G / T

from a low

T_{sys}

, and low data rate compensate for path loss.

What happens to

E_{b} / N_{0}

as the data rate

R_{b} \to 0

E_{b} / N_{0} = (C / N_{0}) - 10 lo g_{10} (R_{b})

grows without bound as

R_{b} \to 0

, since each vanishingly rare bit hoards all the carrier energy — reflecting why very slow links can close over enormous distances.

At the Shannon limit, what is the minimum

E_{b} / N_{0}

a link can theoretically tolerate?

About

- 1.6

dB (

ln 2 \approx 0.693

in linear terms); below this no coding scheme, however clever, can achieve arbitrarily low error rates. See Shannon-Hartley Theorem.

If two spacecraft are equidistant but one uses half the wavelength, which has higher path loss, and by how much in dB?

The half-wavelength (double-frequency) link has higher loss, by

20 lo g_{10} (2) \approx 6

dB, since

L_{p} \propto (1/ λ)^{2}

and halving

λ

quadruples the linear loss.

Recall Self-test: the three "quiet subtractions"

Name the three terms in a link budget that are subtracted (or feel like subtractions) and one trap each. Answer ::: (1) Path loss $L_{p}$ — trap: it is huge but antenna gains and $G / T$ offset it. (2) $10 lo g_{10} (R_{b})$ in $E_{b} / N_{0}$ — trap: faster data lowers $E_{b} / N_{0}$ . (3) $10 lo g_{10} (k)$ Boltzmann — trap: it is a large negative number you effectively add back.