CRAC Class C (2025) — Questions 671–680
U0915. If the frequency of a radio wave is known, its wavelength is:
如果已知无线电波的频率，则其波长为：
Exam tips · 考试提示
Wavelength λ(m) = c / f = 300,000,000 / (f_Hz) = 300 / (f_MHz). So dividing 300 by the frequency in MHz gives meters — option A. B/C/D have the arithmetic inverted or use the wrong unit. So A is correct.
U0916. Regarding the frequency, period, and wavelength of radio waves, which of the following descriptions is correct? (Choose all that apply.)
关于无线电波的频率、周期和波长，以下描述正确的是：
Exam tips · 考试提示
A, B and C are the standard definitions: f = n/τ, T = 1/f, and λ = c/f. D is false — real conductors have a velocity factor below 1, so a physical element must be shortened relative to the free- space length; the shortening factor matters. So A, B, C are correct.
U0917. Radio waves in free space gradually weaken as the propagation distance increases. What law does the degree of weakening follow?
自由空间中的无线电波随传播距离的增加而逐渐减弱。其减弱 程度遵循什么规律？
Exam tips · 考试提示
Free-space path loss L = 32.4 + 20log(d) + 20log(f) dB. In power terms the loss grows with d² (spreading) and with f² (shorter wavelength spreads faster), so the weakening is proportional to the square of distance and the square of frequency — option A. B has frequency inverted; C/D omit or misstate the frequency term.
U0918. Radio waves in free space spread and weaken with distance, forming path loss. The loss can be described as L = 32.4 + 20log(d) + 20log(f) (L in dB, distance d in km, frequency f in MHz). Which of the following descriptions of L is correct? (X＾ M means X to the power M.) (Choose all that apply.)
自由空间中的无线电波随传播距离的增加逐渐发散减弱，形成 路径损耗。该损耗可以描述为L=32.4 + 20log(d) + 20log(f) （其中，损耗L的单位为dB，距离d的单位是km而频率f的单位 为MHz）。关于L，以下描述正确的是：（“X＾M”表示“X 的M次方”）
Exam tips · 考试提示
L increases by 20log(N) dB when either d or f is multiplied by N; in power ratio that is N². So with frequency fixed, N× distance → loss ×N² (A); with distance fixed, N× frequency → loss ×N² (B). C and D wrongly say log(N) for the power multiple. So A and B are correct.
U0919. In free space, stations
自由空间中甲乙两电台相距100km，均使用增益为0dBi的天线 工作于145MHz。如果发射方的功率为1W，则接收方可获得 约-115.6dBm的信号功率。现将两电台的距离增至500km， 则接收方的信号功率变为：【提示：自由空间中无线电波的路 径损耗L =32.4 + 20log(d) + 20log(f)；其中，d为发射方与 接收方之间的距离（km），f为工作频率（MHz）】
Exam tips · 考试提示
Distance 100 → 500 km is 5×. Path loss rises by 20log(5) ≈ 13.98 dB ≈ 14 dB. So the signal drops by ~14 dB: −115.6 − 14 = −129.6 dBm. Option A is correct.
U0920. In free space, stations
自由空间中甲乙两电台相距100km，均使用增益为0dBi的天线 工作于145MHz。如果发射方的功率为1W，则接收方可获得 约-115.6dBm的信号功率。现将两电台的距离增至1000km， 则接收方的信号功率变为：【提示：自由空间中无线电波的路 径损耗L =32.4 + 20log(d) + 20log(f)；其中，d为发射方与 接收方之间的距离（km），f为工作频率（MHz）】
Exam tips · 考试提示
Distance 100 → 1000 km is 10×. Path loss rises by 20log(10) = 20 dB. Signal drops by 20 dB: −115.6 − 20 = −135.6 dBm. Option A is correct.
U0921. In free space, stations
自由空间中甲乙两电台相距100km，均使用增益为0dBi的天线 工作于145MHz。如果发射方的功率为1W，则接收方可获得 约-115.6dBm的信号功率。现发射方将频率变为435MHz，则 接收方的信号功率变为：【提示：自由空间中无线电波的路径 损耗L =32.4 + 20log(d) + 20log(f)；其中，d为发射方与接 收方之间的距离（km），f为工作频率（MHz）】
Exam tips · 考试提示
Frequency 145 → 435 MHz is 3×. Path loss rises by 20log(3) ≈ 9.54 dB ≈ 9.5 dB. Signal drops: −115.6 − 9.5 = −125.1 dBm. Option A is correct.
U0922. In free space, stations
自由空间中甲乙两电台相距100km，均使用增益为0dBi的天线 工作于145MHz。如果发射方的功率为1W，则接收方可获得 约-115.6dBm的信号功率。现发射方将功率降低为0.25W，则 接收方的信号功率变为：【提示：自由空间中无线电波的路径 损耗L =32.4 + 20log(d) + 20log(f)；其中，d为发射方与接 收方之间的距离（km），f为工作频率（MHz）】
Exam tips · 考试提示
Power 1 W → 0.25 W is a factor of 1/4, i.e., −6 dB. Signal drops 6 dB: −115.6 − 6 = −121.6 dBm. Option A is correct.
U0923. In free space, station
自由空间中的甲电台在145MHz联络相距100km的乙电台并获 得信号报告S8。如果两电台的距离增至500km，则信号报告变 为：【提示：收信机信号强度指示从S1至S9每级增加6dB】
Exam tips · 考试提示
500 km is 5× the 100 km distance, adding ~14 dB path loss. Each S-unit is 6 dB, so the signal falls 14/6 ≈ 2.33 S-units: S8 − 2.33 ≈ S5.67, i.e., slightly below S6. Option A is correct.
U0924. In free space, station
自由空间中的甲电台在145MHz联络相距100km的乙电台并获 得信号报告S8。如果两电台的距离增至1000km，则信号报告 变为：【提示：收信机信号强度指示从S1至S9每级增加6dB】
Exam tips · 考试提示
1000 km is 10× the distance, adding 20 dB loss = 20/6 ≈ 3.33 S- units. S8 − 3.33 ≈ S4.67, i.e., slightly below S5. Option A is correct.