CRAC Class A (2025) — Questions 541–550

Sensitivity is expressed as a minimum discernible signal (e.g., μV or dBm). A smaller number means the receiver can detect weaker signals — so smaller = better weak-signal capability (灵 敏度越高). Options B/C/D invert or misapply the meaning. So A is correct.

Power-level sensitivity = the minimum input power (at the antenna port) needed for a specified output quality, expressed in dBm or dBμW (功率电平单位). It is not a ratio (B/C/D are ratios/gains). So A is correct.

Voltage-level sensitivity is defined as the minimum electromotive force (电动势, open-circuit EMF of the signal source) at the input needed for a specified output quality, in μV, dBμV or dBmV. The exam key distinguishes "电动势" (EMF, A) from mere "电压" (B); the standard definition uses the source EMF into the specified input impedance. So A is correct (and C is wrong because it is minimum, not maximum).

HF bands carry much higher natural/atmospheric and man- made background noise than VHF/UHF. Because the receiver's noise floor is higher on HF, the smallest *useful/perceivable* signal is about 20 dB stronger there, so the S-meter's S1 reference is set higher (−121 dBm on HF vs −141 dBm on VHF/UHF). Option A captures this. B is a misstatement (HF receivers aren't inherently 20 dB less sensitive); C and D are irrelevant. Keep VHF/UHF, dBm, μV, Ω, HF, DX.

A smaller μV number normally means better sensitivity, but only if measured to the same output-quality criterion (e.g., same SINAD). Since the specifications do not state the quality standard, you cannot validly compare the two. So A is correct. B and C jump to conclusions; D confuses sensitivity with strong-signal handling (dynamic range). Keep μV.

When sensitivity is given directly in "voltage level μV", it is simply the minimum input voltage itself — 1 μV. The 50 Ω impedance and the "voltage level μV" unit mean no conversion is needed; the answer is B (1 μV). Keep Ω, μV.

Using the Chinese exam's voltage-level convention, sensitivity in dBm is computed from the voltage across 50 Ω as P = V² / R = (1×10⁻⁶)² / 50 = 2×10⁻¹⁴ W; dBm = 10·log₁₀(2×10⁻¹⁴ / 10⁻³) = 10·log₁₀(2×10⁻¹¹) ≈ −107 dBm. So A is correct. (Note: this uses V²/R, the voltage-level convention; the alternative EMF formula V²/4R would give −113 dBm, which is option D — but the exam key uses the V²/R form, giving −107 dBm.) Keep Ω, μV, dBm.

The official answer key for this item marks option A (6 dBμV). Note: by the standard formula dBμV = 20·log₁₀(V/1 μV), a 1 μV input equals 0 dBμV (option D). The Chinese exam key here designates 6 dBμV; this translation preserves the official key as required. Candidates should be aware the two conventions can differ and that the published answer key for this particular item is A. Keep Ω, μV, dBμV.

Expressed directly in voltage level μV, sensitivity equals the measured minimum input voltage — 0.5 μV. No conversion needed (the 50 Ω is irrelevant for the plain-μV unit). So B is correct. Keep Ω, μV.

Using the same V²/R convention as U0546: P = (0.5×10⁻⁶)² / 50 = 5×10⁻¹⁵ W; dBm = 10·log₁₀(5×10⁻¹⁵ / 10⁻³) = 10·log₁₀(5×10⁻¹²) ≈ −113 dBm. This is 6 dB lower than the 1 μV case (−107 dBm), consistent with halving the voltage (20·log₁₀(0.5) = −6 dB). So A is correct. Keep Ω, μV, dBm.