Using Decibels in Communication Systems

Correct answer: (b). The decibel expresses a ratio on a logarithmic scale — it always compares two quantities and is therefore a relative, dimensionless unit.

Correct answer: (d). By definition, dB of power uses the factor 10: G = 10 log₁₀(P_out/P_in). The factor 20 is reserved for voltage/current ratios (amplitude quantities).

Correct answer: (c). "Deci" means one-tenth, so 1 bel = 10 decibels. The bel (after Alexander Graham Bell) proved too large for practical use, giving rise to the dB.

Correct answer: (a). dBm is an absolute power unit referenced to 1 mW: P(dBm) = 10 log₁₀(P/1 mW). 0 dBm = 1 mW; 30 dBm = 1 W.

Correct answer: (b). dBi = gain relative to an isotropic antenna (radiates equally in all directions, theoretical). Gain relative to a half-wave dipole is expressed in dBd, where 0 dBd ≈ 2.15 dBi.

Correct answer: (b) 3 dB. 10 log₁₀(2) ≈ 3.01 dB. Note the contrast: doubling voltage (across the same impedance) gives 20 log₁₀(2) ≈ 6.02 dB. Rule of thumb: +3 dB doubles power, +6 dB doubles voltage. voltage (across the same impedance) gives 20 log₁₀(2) ≈ 6.02 dB. Remember the rule of thumb: +3 dB doubles power, +6 dB doubles voltage.

Correct answer: (c). By definition 0 dBm = 10 log₁₀(1 mW / 1 mW) = 0, so 0 dBm = 1 mW. Useful anchors: 10 dBm = 10 mW, 20 dBm = 100 mW, 30 dBm = 1 W.

Correct answer: (b). This is the great practical advantage: a cascaded link P₁×G₁×G₂×L₃… becomes a sum of dB terms. Gains of 100×, 1000× and a loss of ½ become +20, +30 and −3 dB — easily added on a link budget. (Impedance still matters when converting between dBm and dBµV.)

Correct answer: (a). Voltage gain in dB uses the 20-factor: G_v = 20 log₁₀(V_out/V_in) = 30 dB → ratio = 10^30/20 = 10^1.5 ≈ 31.6. (1000× would be 60 dB of voltage, or 30 dB of power.)

Correct answer: (c). A 10 dB loss means −10 dB = 10 log₁₀(P_out/P_in) → ratio = 10⁻¹ = 0.1. Remember the rule of thumb: +10 dB = ×10 power; −10 dB = ÷10 power.

Correct answer: (d). 30 dBm = 10 log₁₀(P/1 mW) → P = 1 mW × 10³ = 1000 mW = 1 W. Quick method: every 10 dB = ×10 in power, so 0 dBm (1 mW) → 10 dBm (10 mW) → 20 dBm (100 mW) → 30 dBm (1 W).

Correct answer: (b). S and N are powers, so the 10-factor applies. Conveniently, SNR(dB) = S(dBm) − N(dBm) — subtracting two absolute dB quantities. E.g., signal −70 dBm, noise −95 dBm → SNR = 25 dB.

Correct answer: (c). Noise figure NF = 10 log₁₀(F) where F is the noise factor. NF = 3 dB → F = 10^0.3 ≈ 2 — the SNR at the output is half the SNR at the input (in linear terms). It is a ratio of SNRs, not an absolute noise power.

Correct answer: (a). EIRP = P_tx + G_ant − L_feeder (all in dB units). E.g., 30 dBm transmitter, 15 dBi dish, 2 dB cable loss → EIRP = 43 dBm (≈ 20 W effective toward the antenna's peak direction).

Correct answer: (a) 13 dBm. EIRP = P_tx − L_cable + G_ant = 10 − 3 + 6 = 13 dBm. Watch the signs: cable loss subtracts, antenna gain adds.

Correct answer: (a) 21 dBm ≈ 126 mW. Total gain = 20 − 6 + 10 − 3 = +21 dB; output = 0 dBm + 21 dB = 21 dBm = 10^2.1 mW ≈ 126 mW. Option (b) is the classic trap: 21 dB is a ratio, not an absolute power. In linear terms the chain gives 100 × 0.25 × 10 × 0.5 = 125× → same result.

Correct answer: (a) 16 dBm. 10 log₁₀(40 mW / 1 mW) = 10 × 1.602 = 16 dBm. Shortcut: 40 mW = 4 × 10 mW → 6 dB (factor of 4) + 10 dB (factor of 10) = 16 dBm.

Correct answer: (a) 7 dB. Required output = −85 dBm, available input = −92 dBm, so minimum gain = −85 − (−92) = 7 dB. Working in dB turns this "sensitivity budget" into simple subtraction of absolute levels.

Correct answer: (b) −77 dBm. Receive power = EIRP − L_path + G_rx − L_feeder = 30 − 110 + 5 − 2 = −77 dBm. This is the standard link-budget equation used in every radio design.

Correct answer: (c) 224 mV. P = 1 mW = V²/R → V = √(0.001 × 50) = √0.05 ≈ 0.2236 V = 224 mV RMS. Memorize the 50 Ω anchors: 0 dBm = 224 mV, 0 dBµV = 1 µV, and dBµV = dBm + 107 dB (in 50 Ω).