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CRAC Bilingual Manual: Transmission Lines | 传输线 (58 questions)

CRAC Bilingual Manual › Part: Radio System Fundamentals

CRAC Bilingual Exam Manual (Class A / B / C) | 中国业余无线电台操作技术能力验证英中对照手册

58bilingual questions·Classes B, C

This section covers Transmission Lines with 58 bilingual questions from the CRAC 2025 question bank. Each question shows the original Chinese (left) and the English translation (right). The correct answer is highlighted in green, followed by a Knowledge Point Analysis and Candidate Tips covering US–China differences, common pitfalls, and real on-air practice.

Class badges ABC indicate which license-class syllabus includes each question. Class A is the entry level, Class B adds HF privileges, and Class C is the advanced level.

BCQ1 3.5.1 U0905
中文原题 Original (Chinese)

传输线通常用来连接电台与天线。比较常见的传输线有:

  1. A同轴电缆
  2. B平行馈线
  3. C波导
  4. D波纹软管
English Translation

Transmission lines are commonly used to connect the station and the antenna. The more common transmission lines are: (Choose all that apply.)

  1. Acoaxial cable
  2. Bparallel feed line (open-wire line)
  3. Cwaveguide
  4. Dcorrugated hose
Correct answer: A, B, C
Knowledge Point Analysis 知识点解析

Common transmission lines (传输线) are coaxial cable (同轴电缆), parallel (open-wire) feed line (平行馈线), and waveguide (波导). A corrugated hose (D) is not a transmission line. So A, B, C are correct.

Candidate Tips 考生提示

US–China difference: US hams use the same three; waveguides appear at microwave/ham-satellite frequencies.

Common pitfall: Listing a hose or random hardware as a transmission line.

Real on-air practice: Most VHF/UHF rigs use coax; HF multiband dipoles often use open-wire line.

BCQ2 3.5.1 U0906
中文原题 Original (Chinese)

关于同轴电缆特性阻抗,以下说法正确的是:

  1. A外导体的内径与内导体的外径之比越大,特性阻抗越高
  2. B绝缘介质的介电常数越小,特性阻抗越高
  3. C外屏蔽层越致密,特性阻抗越高
  4. D电缆的外径越粗,特性阻抗越高
English Translation

Regarding the characteristic impedance of coaxial cable, which of the following statements is correct? (Choose all that apply.)

  1. AThe larger the ratio of the outer conductor’s inner diameter to the inner conductor’s outer diameter, the higher the characteristic impedance
  2. BThe smaller the dielectric constant of the insulating medium, the higher the characteristic impedance
  3. CThe denser the outer shield, the higher the characteristic impedance
  4. DThe thicker the cable outer diameter, the higher the characteristic impedance
Correct answer: A, B
Knowledge Point Analysis 知识点解析

The coax characteristic impedance is Z0 = (60/√εr)·ln(D/d), where D is the outer-conductor inner diameter and d the inner-conductor outer diameter. Increasing D/d raises Z0 (A), and decreasing the dielectric constant εr raises Z0 (B). Shield density (C) and overall outer diameter (D) do not set Z0. So A and B are correct.

Candidate Tips 考生提示

US–China difference: Same transmission-line formula; US 50 Ω and 75 Ω coax follow it.

Common pitfall: Thinking a thicker cable or denser braid changes Z0 — geometry ratio and dielectric do, not the jacket size.

Real on-air practice: Foam-dielectric coax can read closer to 50 Ω than air-dielectric of the same size.

BCQ3 3.5.1 U0907
中文原题 Original (Chinese)

一根特性阻抗为50欧的同轴电缆的末端接有一个50欧姆负载电阻。如果用天线阻抗分析仪测量该电缆始端在不同频率下的阻抗,则读数为:

  1. A读数为电缆的特性阻抗50欧。该值与电缆的长度和工作频率均无关
  2. B在电缆末端连接任何大小的电阻,在其始端都能测得相同的阻值。当前为50欧
  3. C读数取决于电缆的电长度:1/4波长奇数倍时接近无穷大;1/4波长偶数倍时接近零
  4. D读数取决于电缆的电长度:1/4波长偶数倍时接近无穷大;1/4波长奇数倍时接近零
English Translation

A 50 Ω coaxial cable is terminated at its far end with a 50 Ω load resistor. If you measure the near-end impedance at different frequencies with an antenna analyzer, the reading is:

  1. A50 Ω, the characteristic impedance of the cable; this value is independent of the cable length and the operating frequency
  2. BAny resistor at the far end gives the same reading at the near end; currently it is 50 Ω
  3. CThe reading depends on the cable’s electrical length: at odd multiples of 1/4 wavelength it approaches infinity; at even multiples of 1/4 wavelength it approaches zero
  4. DThe reading depends on the cable’s electrical length: at even multiples of 1/4 wavelength it approaches infinity; at odd multiples of 1/4 wavelength it approaches zero
Correct answer: A
Knowledge Point Analysis 知识点解析

A transmission line terminated in its characteristic impedance (Z0 = 50 Ω) presents exactly Z0 at the input for any length and any frequency — the line is flat. So A is correct; B/C/D describe mismatched (open/short) terminations, not a 50 Ω load.

Candidate Tips 考生提示

US–China difference: Same behavior; a 50 Ω dummy load on coax reads 50 Ω at the shack end.

Common pitfall: Expecting the reading to vary with length/frequency when the line is properly terminated.

Real on-air practice: A good 50 Ω load gives a flat 1:1 SWR everywhere on the analyzer.

BCQ4 3.5.1 U0908
中文原题 Original (Chinese)

一根特性阻抗为50欧的同轴电缆的末端开路。如果用天线阻抗分析仪测量该电缆始端在不同频率下的阻抗,则读数为:

  1. A读数取决于电缆的电长度:1/4波长偶数倍时接近无穷大;1/4波长奇数倍时接近零
  2. B读数取决于电缆的电长度:1/4波长奇数倍时接近无穷大;1/4波长偶数倍时接近零
  3. C在电缆末端连接任何大小的电阻,在电缆始端都能测得相同的阻值。当前为无穷大
  4. D读数为电缆的特性阻抗50欧。该值与电缆的长度和工作频率均无关
English Translation

A 50 Ω coaxial cable is left open at its far end. If you measure the near-end impedance at different frequencies with an antenna analyzer, the reading is:

  1. AThe reading depends on the cable’s electrical length: at even multiples of 1/4 wavelength it approaches infinity; at odd multiples of 1/4 wavelength it approaches zero
  2. BThe reading depends on the cable’s electrical length: at odd multiples of 1/4 wavelength it approaches infinity; at even multiples of 1/4 wavelength it approaches zero
  3. CAny resistor at the far end gives the same reading at the near end; currently it is infinity
  4. DThe reading is the cable’s characteristic impedance 50 Ω; it is independent of length and frequency
Correct answer: A
Knowledge Point Analysis 知识点解析

For an open-circuited line the input impedance is Z_in = −j·Z0·cot(βl). At a half-wavelength (even multiples of 1/4 λ) the open repeats → ∞; at a quarter-wavelength (odd multiples of 1/4 λ) it transforms to a short → ≈0. So A is correct; B has the values swapped; C/D are for terminated lines.

Candidate Tips 考生提示

US–China difference: Same open-line transformation; US hams use it to make a 1/4-wave short or a half-wave 1:1 line.

Common pitfall: Swapping the quarter-wave and half-wave behavior of an open line.

Real on-air practice: A 1/4-wave open stub is a short at the shack end — useful as a choke.

BCQ5 3.5.1 U0909
中文原题 Original (Chinese)

一根特性阻抗为50欧的同轴电缆的末端短路。如果用天线阻抗分析仪测量该电缆始端在不同频率下的阻抗,则读数为:

  1. A读数取决于电缆的电长度:1/4波长奇数倍时接近无穷大;1/4波长偶数倍时接近零
  2. B读数取决于电缆的电长度:1/4波长偶数倍时接近无穷大;1/4波长奇数倍时接近零
  3. C在电缆末端连接任何大小的电阻,在电缆始端都能测得相同的阻值。当前为0
  4. D读数为电缆的特性阻抗50欧。该值与电缆的长度和工作频率均无关
English Translation

A 50 Ω coaxial cable is short-circuited at its far end. If you measure the near-end impedance at different frequencies with an antenna analyzer, the reading is:

  1. AThe reading depends on the cable’s electrical length: at odd multiples of 1/4 wavelength it approaches infinity; at even multiples of 1/4 wavelength it approaches zero
  2. BThe reading depends on the cable’s electrical length: at even multiples of 1/4 wavelength it approaches infinity; at odd multiples of 1/4 wavelength it approaches zero
  3. CAny resistor at the far end gives the same reading at the near end; currently it is 0
  4. DThe reading is the cable’s characteristic impedance 50 Ω; it is independent of length and frequency
Correct answer: A
Knowledge Point Analysis 知识点解析

For a short-circuited line Z_in = j·Z0·tan(βl). At a quarter-wavelength (odd multiples of 1/4 λ) it transforms to an open → ∞; at a half-wavelength (even multiples of 1/4 λ) it repeats the short → ≈0. So A is correct; B swaps the values; C/D are for terminated lines.

Candidate Tips 考生提示

US–China difference: Same short-line transformation; a 1/4-wave shorted stub is an open at the shack — a classic RF choke.

Common pitfall: Confusing shorted vs open line quarter/half-wave behavior.

Real on-air practice: A 1/4-wave shorted coax choke kills common-mode current on your feed line.

BCQ6 3.5.1 U0910
中文原题 Original (Chinese)

在用天线阻抗分析仪测量某天线失配情况下的阻抗表现时,我们会发现仪表的读数随天线和仪表间电缆跳线的长度而变。为保证测量的准确性,可以采取如下措施:

  1. A使用电气长度正好等于测量波长的连接电缆
  2. B换用接触面镀金的高档电缆接头
  3. C用高档纯银音响线材代替普通铜质同轴电缆
  4. D将电缆外皮妥善接地,并将仪表放入屏蔽室
English Translation

When measuring a mismatched antenna’s impedance with an antenna analyzer, you will find the reading changes with the length of the jumper cable between the antenna and the meter. To ensure measurement accuracy, you can take the following measure:

  1. AUse a connecting cable whose electrical length is exactly one measuring wavelength
  2. BReplace with gold-plated premium connectors
  3. CReplace ordinary copper coax with premium pure-silver audio cable
  4. DGround the cable shield properly and put the meter in a shielded room
Correct answer: A
Knowledge Point Analysis 知识点解析

On a mismatched line the impedance repeats every half wavelength (transformation period λ/2). A jumper that is an integer number of wavelengths (here one full measuring wavelength) transforms the antenna impedance unchanged to the meter, removing length dependence. So A is correct; B/C/D address connectors/shielding but not the length-dependent transformation.

Candidate Tips 考生提示

US–China difference: US hams likewise use a half-wave (or full-wave) coax jumper so the analyzer reads the true antenna impedance.

Common pitfall: Buying exotic cables instead of fixing the measurement jumper length.

Real on-air practice: A 1/2-wave coax pigtail lets you read the antenna Z at the rig end without a tuner in the path.

BCQ7 3.5.2 U0911
中文原题 Original (Chinese)

由场源(比如天线)产生的电场会形成位移电流,在其周边激起变化的磁场。该磁场又会在稍远处再次激起变化的电场。这个过程周而复始,形成传向远方的电磁波。关于电磁波的传播,以下描述正确的是:

  1. A在一定场强下,频率越高,位移电流越强,场源辐射的能量也越多
  2. B只有场源为某种开放系统,变化的电磁场才能向远方传播开来
  3. C在一定能量下,频率越高,位移电流越强,场源辐射的场强也越强
  4. D只有场源为某种封闭系统,变化的电磁场才能在同一时空中传播开来
English Translation

The electric field produced by a source (e.g., an antenna) creates a displacement current, exciting a varying magnetic field around it. That magnetic field in turn excites a varying electric field a little farther away. This cycle repeats, forming an electromagnetic wave traveling to the distance. Which of the following descriptions of electromagnetic-wave propagation is correct? (Choose all that apply.)

  1. AAt a given field strength, the higher the frequency, the stronger the displacement current, and the more energy the source radiates
  2. BOnly when the source is an open system can the varying electromagnetic field propagate to the distance
  3. CAt a given energy, the higher the frequency, the stronger the displacement current, and the stronger the field strength radiated by the source
  4. DOnly when the source is a closed system can the varying electromagnetic field propagate in the same space-time
Correct answer: A, B
Knowledge Point Analysis 知识点解析

Displacement current density is proportional to ∂E/∂t, hence to ω·E, so at a fixed field strength higher frequency means stronger displacement current and more radiation (A correct). Radiation also requires an open (non-closed) structure so the field can escape (B correct). C’s wording (fixed energy → stronger field with frequency) is not the standard correct statement, and D is the opposite of B (radiation needs an open, not closed, system). So A and B are correct.

Candidate Tips 考生提示

US–China difference: Same EM theory; US license manuals note that a closed (shielded) system does not radiate.

Common pitfall: Thinking a closed/balanced system radiates, or confusing the fixed-field vs fixed-energy conditions.

Real on-air practice: An unshielded dipole radiates; the same currents inside a metal box do not.

BCQ8 3.5.2 U0912
中文原题 Original (Chinese)

电磁波(electromagnetic waves)的两个组成部分是:

  1. A电场和磁场
  2. B电压和电流
  3. C阻抗和功率
  4. D电离辐射和非电离辐射(ionizing radiation and non-ionizing radiation)
English Translation

The two components of an electromagnetic wave are:

  1. Athe electric field and the magnetic field
  2. Bvoltage and current
  3. Cimpedance and power
  4. Dionizing radiation and non-ionizing radiation
Correct answer: A
Knowledge Point Analysis 知识点解析

An electromagnetic wave (电磁波) consists of mutually coupled electric and magnetic fields oscillating perpendicular to each other and to the direction of travel. Voltage/current (B), impedance/power (C), and the ionizing/non-ionizing classification (D) are not its two field components. So A is correct.

Candidate Tips 考生提示

US–China difference: Same definition in US physics and FCC materials.

Common pitfall: Calling RF an ionizing radiation — radio waves are non-ionizing.

Real on-air practice: Your antenna launches an E-field (from voltage) and H-field (from current) together.

BCQ9 3.5.2 U0913
中文原题 Original (Chinese)

无线电波在真空中的传播速度有多快?

  1. A和光速相同,大约为300,000,000米/秒
  2. B与音速相同,从300米/秒至数十千米/秒不等
  3. C无法确定,与介质的特性有关
  4. D无法确定,与电波的频率有关
English Translation

How fast do radio waves travel in a vacuum?

  1. AThe same as the speed of light, about 300,000,000 meters per second
  2. BThe same as the speed of sound, ranging from 300 m/s to tens of km/s
  3. CCannot be determined; it depends on the medium’s properties
  4. DCannot be determined; it depends on the wave’s frequency
Correct answer: A
Knowledge Point Analysis 知识点解析

Radio waves are electromagnetic waves and travel in vacuum at the speed of light c ≈ 3×10^8 m/s, independent of frequency. B confuses it with sound; C/D are wrong because in vacuum the speed is fixed and frequency-independent. So A is correct.

Candidate Tips 考生提示

US–China difference: Same constant c used in US propagation math (e.g., the 300/f MHz wavelength rule).

Common pitfall: Mixing radio-wave speed with sound speed.

Real on-air practice: The 300/f(MHz) wavelength formula comes straight from c ≈ 300,000,000 m/s.

BCQ10 3.5.2 U0914
中文原题 Original (Chinese)

电磁波在真空中传播时,其各个周期的传播距离都是相等的。该距离叫做:

  1. A波长
  2. B波形
  3. C波速
  4. D波束
English Translation

When an electromagnetic wave propagates in a vacuum, the distance it travels in each cycle is the same. This distance is called:

  1. Awavelength
  2. Bwaveform
  3. Cwave speed (velocity)
  4. Dbeam
Correct answer: A
Knowledge Point Analysis 知识点解析

The distance an EM wave travels in one period is the wavelength (波长). Waveform (B) is the shape, wave speed (C) is c, beam (D) is the radiation pattern direction. So A is correct.

Candidate Tips 考生提示

US–China difference: Same terminology; wavelength = c / f.

Common pitfall: Confusing wavelength with wave speed.

Real on-air practice: A 20 m band signal has a wavelength of about 20 m, setting your antenna size.

BCQ11 3.5.2 U0915
中文原题 Original (Chinese)

如果已知无线电波的频率,则其波长为:

  1. A使用300除以频率的兆赫数(MHz)可以得到以米为单位的波长
  2. B将频率的赫兹数(Hz)除以300可以得到以米为单位的波长
  3. C将频率的兆赫数(MHz)除以300可以得到以米为单位的波长
  4. D将频率的赫兹数(Hz)乘以300可以得到以米为单位的波长
English Translation

If the frequency of a radio wave is known, its wavelength is:

  1. ADivide 300 by the frequency in megahertz (MHz) to get the wavelength in meters
  2. BDivide the frequency in hertz (Hz) by 300 to get the wavelength in meters
  3. CDivide the frequency in MHz by 300 to get the wavelength in meters
  4. DMultiply the frequency in Hz by 300 to get the wavelength in meters
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: The same 300/f(MHz) shortcut is taught to US hams.

Common pitfall: Dividing frequency by 300 instead of 300 by frequency.

Real on-air practice: 145 MHz → 300/145 ≈ 2.07 m, the 2 m band.

BCQ12 3.5.2 U0916
中文原题 Original (Chinese)

关于无线电波的频率、周期和波长,以下描述正确的是:

  1. A无线电波的频率f代表τ时间内电磁场重复改变的次数n。所以,f=n/τ
  2. B周期T代表电磁场重复改变的时间间隔,其与频率f为倒数关系。即,T=1/f
  3. C无线电波的波长λ与光速c成正比,但是与频率f成反比。因此,λ=c/f
  4. D由于λ仅与光速有关,所以架设天线时考虑振子的缩短系数其实没什么必要
English Translation

Regarding the frequency, period, and wavelength of radio waves, which of the following descriptions is correct? (Choose all that apply.)

  1. AThe frequency f is the number n of times the electromagnetic field repeats within time τ, so f = n/τ
  2. BThe period T is the time interval between repetitions of the field and is the reciprocal of frequency f, i.e., T = 1/f
  3. CThe wavelength λ is proportional to the speed of light c and inversely proportional to frequency f, so λ = c/f
  4. DSince λ relates only to the speed of light, the element shortening factor is unnecessary when building antennas
Correct answer: A, B, C
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: Same f/T/λ relations; US antenna build guides also apply velocity factor.

Common pitfall: Ignoring velocity factor and cutting elements to the exact free-space length.

Real on-air practice: A 14.1 MHz dipole is cut a few percent short because coax/ wire velocity factor is <1.

BCQ13 3.5.2 U0917
中文原题 Original (Chinese)

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?

  1. A与距离的平方成正比,与频率的平方成正比
  2. B与距离的平方成正比,与频率的平方成反比
  3. C与距离成正比,与频率的平方成正比
  4. D与距离的平方成正比,但是与频率无关
English Translation

Radio waves in free space gradually weaken as the propagation distance increases. What law does the degree of weakening follow?

  1. AProportional to the square of the distance and proportional to the square of the frequency
  2. BProportional to the square of the distance and inversely proportional to the square of the frequency
  3. CProportional to the distance and proportional to the square of the frequency
  4. DProportional to the square of the distance but independent of frequency
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: Same free-space path-loss formula used by US hams for VHF/UHF link budgets.

Common pitfall: Forgetting that higher frequency also means more path loss for the same distance.

Real on-air practice: A 435 MHz link loses more over the same path than a 145 MHz one.

BCQ14 3.5.2 U0918
中文原题 Original (Chinese)

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”)

  1. A当频率一定时,距离增加到N倍,L增大到N^2倍
  2. B当距离一定时,频率增加到N倍,L增大到N^2倍
  3. C当频率一定时,距离增加到N倍,L增大到log(N)倍
  4. D当距离一定时,频率增加到N倍,L增大到log(N)倍
English Translation

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.)

  1. AWhen frequency is fixed, if distance increases N times, L grows to N^2 times
  2. BWhen distance is fixed, if frequency increases N times, L grows to N^2 times
  3. CWhen frequency is fixed, if distance increases N times, L grows to log(N) times
  4. DWhen distance is fixed, if frequency increases N times, L grows to log(N) times
Correct answer: A, B
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: Same 20log rule in US path-loss budgets.

Common pitfall: Reading 20log(N) dB as an N-times (not N²-times) power change.

Real on-air practice: Doubling the distance adds 6 dB loss (power drops to 1/4).

BCQ15 3.5.2 U0919
中文原题 Original (Chinese)

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)】

  1. A-129.6dBm
  2. B-121.6dBm
  3. C-125.1dBm
  4. D-135.6dBm
English Translation

In free space, stations A and B are 100 km apart, both using 0 dBi antennas at 145 MHz. If the transmitter power is 1 W, the receiver gets about −115.6 dBm. Now the distance increases to 500 km. The receiver signal power becomes: [Hint: free-space path loss L = 32.4 + 20log(d) + 20log(f); d in km, f in MHz]

  1. A−129.6 dBm
  2. B−121.6 dBm
  3. C−125.1 dBm
  4. D−135.6 dBm
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: Same free-space math used for US VHF simplex range estimates.

Common pitfall: Forgetting 5× distance adds ~14 dB (not 5 dB) of loss.

Real on-air practice: Pushing a 2 m contact from 100 km to 500 km needs far more power or gain.

BCQ16 3.5.2 U0920
中文原题 Original (Chinese)

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)】

  1. A-135.6dBm
  2. B-129.6dBm
  3. C-121.6dBm
  4. D-125.1dBm
English Translation

In free space, stations A and B are 100 km apart, both using 0 dBi antennas at 145 MHz. If the transmitter power is 1 W, the receiver gets about −115.6 dBm. Now the distance increases to 1000 km. The receiver signal power becomes: [Hint: free-space path loss L = 32.4 + 20log(d) + 20log(f); d in km, f in MHz]

  1. A−135.6 dBm
  2. B−129.6 dBm
  3. C−121.6 dBm
  4. D−125.1 dBm
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: Same 20 dB per decade-of-distance rule.

Common pitfall: Underestimating the 20 dB drop for a 10× distance increase.

Real on-air practice: A 1000 km 2 m path needs huge power/gain — why 2 m is mostly local.

BCQ17 3.5.2 U0921
中文原题 Original (Chinese)

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)】

  1. A-125.1dBm
  2. B-135.6dBm
  3. C-129.6dBm
  4. D-121.6dBm
English Translation

In free space, stations A and B are 100 km apart, both using 0 dBi antennas at 145 MHz. If the transmitter power is 1 W, the receiver gets about −115.6 dBm. Now the transmitter changes frequency to 435 MHz. The receiver signal power becomes: [Hint: free-space path loss L = 32.4 + 20log(d) + 20log(f); d in km, f in MHz]

  1. A−125.1 dBm
  2. B−135.6 dBm
  3. C−129.6 dBm
  4. D−121.6 dBm
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: Same result; 435 MHz (70 cm) loses ~9.5 dB more than 145 MHz over the same path.

Common pitfall: Assuming all VHF/UHF bands have equal path loss.

Real on-air practice: The 70 cm band reaches less far than 2 m at the same power.

BCQ18 3.5.2 U0922
中文原题 Original (Chinese)

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)】

  1. A-121.6dBm
  2. B-125.1dBm
  3. C-135.6dBm
  4. D-129.6dBm
English Translation

In free space, stations A and B are 100 km apart, both using 0 dBi antennas at 145 MHz. If the transmitter power is 1 W, the receiver gets about −115.6 dBm. Now the transmitter reduces power to 0.25 W. The receiver signal power becomes: [Hint: free-space path loss L = 32.4 + 20log(d) + 20log(f); d in km, f in MHz]

  1. A−121.6 dBm
  2. B−125.1 dBm
  3. C−135.6 dBm
  4. D−129.6 dBm
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: Same 6 dB per quarter-power relation.

Common pitfall: Forgetting 0.25 W is −6 dB (a factor of 4), not −4 dB.

Real on-air practice: Dropping from 1 W to 1/4 W (QRP) costs one S-unit at the other end.

BCQ19 3.5.2 U0923
中文原题 Original (Chinese)

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】

  1. A略低于S6
  2. B略高于S6
  3. C略低于S5
  4. D略高于S4
English Translation

In free space, station A contacts station B 100 km away on 145 MHz and receives an S8 report. If the distance increases to 500 km, the signal report becomes: [Hint: receiver signal-strength indication goes from S1 to S9, each step +6 dB]

  1. Aslightly below S6
  2. Bslightly above S6
  3. Cslightly below S5
  4. Dslightly above S4
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: Same S-meter 6 dB/step convention in US radios.

Common pitfall: Subtracting 5 (the distance multiple) S-units instead of 14 dB ≈ 2.3 units.

Real on-air practice: Going from 100 km to 500 km drops you from strong S8 to weak S5–S6.

BCQ20 3.5.2 U0924
中文原题 Original (Chinese)

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】

  1. A略低于S5
  2. B略高于S5
  3. C略高于S6
  4. DS7
English Translation

In free space, station A contacts station B 100 km away on 145 MHz and receives an S8 report. If the distance increases to 1000 km, the signal report becomes: [Hint: receiver signal-strength indication goes from S1 to S9, each step +6 dB]

  1. Aslightly below S5
  2. Bslightly above S5
  3. Cslightly above S6
  4. DS7
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: Same S-meter math.

Common pitfall: Subtracting 10 S-units (the distance multiple) instead of 3.3.

Real on-air practice: A 1000 km 2 m contact is typically below the noise floor without big gain.

BCQ21 3.5.2 U0925
中文原题 Original (Chinese)

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。如果双方改用频率435MHz,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】

  1. A略高于S6
  2. B略低于S5
  3. C略低于S6
  4. DS7
English Translation

In free space, station A contacts station B 100 km away on 145 MHz and receives an S8 report. If both switch to 435 MHz, the signal report becomes: [Hint: receiver signal-strength indication goes from S1 to S9, each step +6 dB]

  1. Aslightly above S6
  2. Bslightly below S5
  3. Cslightly below S6
  4. DS7
Correct answer: A
Knowledge Point Analysis 知识点解析

435 MHz is 3× 145 MHz, adding ~9.5 dB loss = 9.5/6 ≈ 1.59 S-units. S8 − 1.59 ≈ S6.41, i.e., slightly above S6. Option A is correct.

Candidate Tips 考生提示

US–China difference: Same; 70 cm is weaker than 2 m over equal distance.

Common pitfall: Thinking the S reading is unchanged when changing bands.

Real on-air practice: The same 100 km station often reads S6–S7 on 70 cm vs S8 on 2 m.

BCQ22 3.5.2 U0926
中文原题 Original (Chinese)

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。如果甲电台将发射功率减少到原来的1/4,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】

  1. AS7
  2. B略高于S5
  3. C略高于S6
  4. D略低于S5
English Translation

In free space, station A contacts station B 100 km away on 145 MHz and receives an S8 report. If station A reduces its transmit power to 1/4 of the original, the signal report becomes: [Hint: receiver signal-strength indication S1 to S9, each step +6 dB]

  1. AS7
  2. Bslightly above S5
  3. Cslightly above S6
  4. Dslightly below S5
Correct answer: A
Knowledge Point Analysis 知识点解析

1/4 power is −6 dB = exactly one S-unit. S8 − 1 = S7. Option A is correct.

Candidate Tips 考生提示

US–China difference: Same 6 dB-per-S-unit and 6 dB-per-quarter-power relation.

Common pitfall: Subtracting 4 S-units (the power fraction) instead of 1.

Real on-air practice: Cutting power to 1/4 drops your report by exactly one S-unit.

BCQ23 3.5.2 U0927
中文原题 Original (Chinese)

某业余电台以100瓦发射功率工作时,对方报告信号强度S8。现该台将发射功率降至25瓦,则对方给出的信号报告应为:【提示:收信机信号强度指示S1至S9每级增加6dB】

  1. AS7
  2. BS6
  3. CS5
  4. DS4
English Translation

An amateur station transmitting at 100 W gets an S8 report from the other station. Now it reduces power to 25 W. The other station’s report should be: [Hint: receiver signal-strength indication S1 to S9, each step +6 dB]

  1. AS7
  2. BS6
  3. CS5
  4. DS4
Correct answer: A
Knowledge Point Analysis 知识点解析

100 W → 25 W is 1/4 power = −6 dB = one S-unit. S8 − 1 = S7. Option A is correct.

Candidate Tips 考生提示

US–China difference: Same relation; US hams note 6 dB = 1 S-unit = 1/4 power.

Common pitfall: Subtracting 4 (the power ratio) S-units.

Real on-air practice: Running 100 W then 25 W drops your report from S8 to S7.

BCQ24 3.5.2 U0928
中文原题 Original (Chinese)

某业余电台以80瓦发射功率工作时,对方报告信号强度S8。现该台将发射功率降至5瓦进行QRP实验,则对方给出的信号报告应为:【提示:收信机信号强度指示S1至S9每级增加6dB】

  1. AS6
  2. BS7
  3. CS4
  4. DS2
English Translation

An amateur station transmitting at 80 W gets an S8 report. Now it drops to 5 W for a QRP experiment. The other station’s report should be: [Hint: receiver signal-strength indication S1 to S9, each step +6 dB]

  1. AS6
  2. BS7
  3. CS4
  4. DS2
Correct answer: A
Knowledge Point Analysis 知识点解析

80 W → 5 W is a factor of 16 = 4× quartering = 4 × (−6 dB) = −12 dB = two S-units. S8 − 2 = S6. Option A is correct.

Candidate Tips 考生提示

US–China difference: Same QRP math; 5 W is a classic US QRP level.

Common pitfall: Subtracting 16 S-units or miscounting the dB drop.

Real on-air practice: A 5 W QRP station typically reads ~2 S-units weaker than at 80 W.

BCQ25 3.5.3 U0929
中文原题 Original (Chinese)

关于电磁波的波阻抗Z,以下描述正确的是:

  1. A在电磁波的传播过程中,电场E和磁场H在空间的比值称为波阻抗Z
  2. B随电磁波传向远方,Z逐渐趋于常数。此时E和H可以相互推算而出
  3. C无线电波从馈线进入发射天线的馈电点时遇到的阻碍即为波阻抗Z
  4. D无线电波从馈线进入发射天线的馈电点时遇到的反射即为波阻抗Z
English Translation

Regarding the wave impedance Z of electromagnetic waves, which of the following descriptions is correct? (Choose all that apply.)

  1. ADuring propagation, the ratio of the electric field E to the magnetic field H in space is called the wave impedance Z
  2. BAs the wave travels farther, Z gradually approaches a constant, at which point E and H can be derived from each other
  3. CThe obstruction encountered when a radio wave enters the antenna feed point from the feed line is the wave impedance Z
  4. DThe reflection encountered when a radio wave enters the antenna feed point from the feed line is the wave impedance Z
Correct answer: A, B
Knowledge Point Analysis 知识点解析

Wave impedance Z = E/H. In the far field of free space Z approaches the constant ≈ 377 Ω (η0), so E and H are fixed in ratio and mutually derivable — A and B are correct. C confuses Z with feed-point impedance and D with reflection; neither is wave impedance. So A and B are correct.

Candidate Tips 考生提示

US–China difference: Same free-space impedance ~377 Ω concept in US EM theory.

Common pitfall: Mixing wave impedance (E/H in space) with antenna feed-point impedance.

Real on-air practice: Knowing E/H helps when using a field-strength meter to infer the magnetic field.

BCQ26 3.5.3 U0930
中文原题 Original (Chinese)

在电磁场理论中,“强度与距辐射源的距离的平方成反比”的说法适用于:

  1. A相距辐射源10倍及以上波长的“远区场”
  2. B相距辐射源一定距离内的“远区场”
  3. C辐射源周围空间中的任意点
  4. DLF至UHF的频率范围
English Translation

In electromagnetic-field theory, the statement that intensity is inversely proportional to the square of the distance from the radiating source applies to:

  1. Athe far field at a distance of 10 or more wavelengths from the source
  2. Bthe far field within a certain distance of the source
  3. Cany point in the space around the source
  4. Dthe frequency range from LF to UHF
Correct answer: A
Knowledge Point Analysis 知识点解析

The inverse-square law for field intensity holds in the far field, conventionally defined as ≥10 wavelengths from the source. In the near field the relation does not hold, and it is not frequency- or band-specific. So A is correct.

Candidate Tips 考生提示

US–China difference: Same 10λ far-field rule used by US hams for measurements.

Common pitfall: Applying the inverse-square law in the near field.

Real on-air practice: Measure antenna patterns at least 10λ away for valid far-field numbers.

BCQ27 3.5.3 U0931
中文原题 Original (Chinese)

有时,我们需要在空旷场地用场强计测量比较不同型号的全向垂直天线零仰角情况下的实际辐射效果。不过,测量时应注意什么问题?

  1. A测试地点应选在远场区,离天线10个波长以上
  2. B场强计外壳应妥善接地
  3. C场强计应与地面平行
  4. D在馈线的外面穿套多个磁环以形成“连珠巴伦”
English Translation

Sometimes you need to compare the actual radiation of different omnidirectional vertical antennas at zero elevation using a field-strength meter in an open area. What should you watch out for in the measurement?

  1. AThe test site should be in the far field, more than 10 wavelengths from the antenna
  2. BThe field-strength meter case should be properly grounded
  3. CThe field-strength meter should be parallel to the ground
  4. DSlip several ferrite rings over the feed line to form a string-of-beads balun
Correct answer: A
Knowledge Point Analysis 知识点解析

To measure true far-field (zero-elevation) radiation you must be in the far field, i.e., >10 λ from the antenna; otherwise near-field effects distort the reading. So A is correct; B/C/D are not the key precaution here.

Candidate Tips 考生提示

US–China difference: Same far-field measurement practice recommended by US antenna experimenters.

Common pitfall: Measuring too close and trusting near-field readings as real gain.

Real on-air practice: For a 2 m vertical, the far field starts at 6 m (10λ) — easy; for 160 m it is 1600 m — hard.

BCQ28 3.5.3 U0932
中文原题 Original (Chinese)

业余条件下测试天线增益的典型方法如下图所示。用场强表或接收机接收设置在远处同一地点,最大辐射方向朝向自己的半波偶极天线(上)和待测天线(下)的信号。调整送至两副天线的射频功率Po和P,使收到的场强相同。则,待测天线的增益dBd值为:

  1. A10 log(Po/P)
  2. B10 log(P/Po)
  3. CP – Po
  4. D10 log(P-Po) [F]LK0927.jpg
English Translation

A typical amateur-method for measuring antenna gain is shown. A field-strength meter or receiver at a distant, fixed location receives the signal from a half-wave reference dipole (top) and the antenna under test (bottom), both aimed at the meter with their maximum radiation. Adjust the RF powers Po (reference) and P (test) so the received field strength is equal. The test antenna’s gain in dBd is:

  1. A10 log(Po/P)
  2. B10 log(P/Po)
  3. CP − Po
  4. D10 log(P − Po) [F]LK0927.jpg
Correct answer: A
Knowledge Point Analysis 知识点解析

With equal received field strength, a more efficient (higher-gain) antenna needs less power, so gain(dBd) = 10log(Po/P), where Po is the reference dipole power and P the test power. Option A is correct; B inverts the ratio; C/D are not logarithmic power ratios.

Candidate Tips 考生提示

US–China difference: Same comparison method; US hams compare to a half-wave dipole for dBd.

Common pitfall: Inverting Po/P — remember less power for the test antenna means higher gain.

Real on-air practice: If the test antenna needs 1/4 the power for the same S-meter, it is +6 dBd over the dipole.

BCQ29 3.5.3 U0933
中文原题 Original (Chinese)

业余条件下测试天线增益的典型方法如下图所示。用场强表或接收机接收设置在远处同一地点,最大辐射方向朝向自己的半波偶极天线(上)和待测天线(下)的信号。调整送至两副天线的射频功率Po和P,使收到的场强相同。则,待测天线的增益dBi值为:

  1. A10 log(Po/P) + 2.15
  2. B10 log(P/Po) + 2.15
  3. C20 log(P/Po) + 2.15
  4. D10 log(P-Po) + 2.15 [F]LK0928.jpg
English Translation

A typical amateur-method for measuring antenna gain is shown. A field-strength meter or receiver at a distant, fixed location receives the signal from a half-wave reference dipole (top) and the antenna under test (bottom), both aimed at the meter with their maximum radiation. Adjust the RF powers Po (reference) and P (test) so the received field strength is equal. The test antenna’s gain in dBi is:

  1. A10 log(Po/P) + 2.15
  2. B10 log(P/Po) + 2.15
  3. C20 log(P/Po) + 2.15
  4. D10 log(P − Po) + 2.15 [F]LK0928.jpg
Correct answer: A
Knowledge Point Analysis 知识点解析

The measured value is gain relative to a dipole (dBd) = 10log(Po/P); converting to dBi adds 2.15 dB (since a dipole is 2.15 dBi). So dBi = 10log(Po/P) + 2.15 — option A. B inverts the ratio; C uses 20log wrongly; D is not a ratio.

Candidate Tips 考生提示

US–China difference: Same dBi = dBd + 2.15 convention in US antenna specs.

Common pitfall: Forgetting the +2.15 dB when converting dBd to dBi.

Real on-air practice: A 3 dBd antenna is 5.15 dBi.

BCQ30 3.5.4 U0934
中文原题 Original (Chinese)

若按传播形式分类,无线电波大体分为:

  1. A地面波、天波、空间波、散射波
  2. B长波、中波、短波、超短波、微波
  3. C调幅波、调频波、调相波
  4. D正弦波、方波、三角波
English Translation

If classified by propagation form, radio waves are roughly divided into:

  1. Aground wave, sky wave, space wave, and scatter wave
  2. Blong wave, medium wave, short wave, ultra-short wave, and microwave
  3. Camplitude-modulated, frequency-modulated, and phase-modulated waves
  4. Dsine wave, square wave, and triangle wave
Correct answer: A
Knowledge Point Analysis 知识点解析

By propagation form (传播形式), radio waves are ground wave (地面波), sky wave (天波), space wave (空间波), and scatter wave (散射波). B classifies by frequency band, C by modulation, D by waveform — not by propagation. So A is correct.

Candidate Tips 考生提示

US–China difference: Same four propagation-mode categories in US propagation study.

Common pitfall: Mixing band classification (LF/MF/HF/VHF) with propagation-mode classification.

Real on-air practice: 80 m at night uses sky wave; 2 m local uses space/ground wave; troposcatter is the fourth mode.

BCQ31 3.5.4 U0935
中文原题 Original (Chinese)

顾名思义,地面波就是沿地面传播的无线电波,其衰减特性取决于:

  1. A电波频率、大地电导率和传播距离
  2. B电波频率、太阳活动和地磁活动情况
  3. C电波频率、发射功率和天线增益
  4. D天线高度、发射功率和调制方式
English Translation

As the name implies, a ground wave is a radio wave that propagates along the ground; its attenuation characteristics depend on:

  1. Athe wave frequency, the ground conductivity, and the propagation distance
  2. Bthe wave frequency, solar activity, and the level of geomagnetic activity
  3. Cthe wave frequency, the transmit power, and the antenna gain
  4. Dthe antenna height, the transmit power, and the modulation mode
Correct answer: A
Knowledge Point Analysis 知识点解析

The attenuation of a ground wave (地波) is governed primarily by the wave frequency, the electrical conductivity of the ground over which it travels, and the distance covered. Higher frequencies are absorbed more strongly by the ground, and poor ground conductivity (e.g., dry soil) increases loss. Transmit power and antenna gain affect the signal level at the start but are not part of the propagation attenuation law itself, so B, C, and D are incorrect.

Candidate Tips 考生提示

US–China difference: N/A — ground-wave propagation physics is identical worldwide; this is a propagation-fundamentals concept, not a regulatory one.

Common pitfall: Candidates confuse “attenuation” with “transmit power” — the question asks what the attenuation depends on, not what makes the signal stronger at the source.

Real on-air practice: On 160 m (1.9 MHz) or 80 m ground-wave contacts over salt water (high conductivity) reach much farther than over rocky or dry inland ground.

BCQ32 3.5.4 U0936
中文原题 Original (Chinese)

大气层的哪一组成部分使得无线电波在世界范围内传播?

  1. A电离层
  2. B对流层
  3. C平流层
  4. D磁层
English Translation

Which component of the atmosphere enables radio waves to propagate worldwide?

  1. Athe ionosphere
  2. Bthe troposphere
  3. Cthe stratosphere
  4. Dthe magnetosphere
Correct answer: A
Knowledge Point Analysis 知识点解析

The ionosphere (电离层) is the ionized layer of the upper atmosphere that refracts (bends) high-frequency (HF, short-wave) radio waves back toward Earth, allowing long-distance “sky-wave” (天波) propagation around the globe. The troposphere, stratosphere, and magnetosphere do not provide this global reflection mechanism for HF.

Candidate Tips 考生提示

US–China difference: N/A — the ionosphere is a global physical phenomenon; the same sky-wave principle applies to US FCC amateurs working DX on HF.

Common pitfall: Mixing up the troposphere (responsible for VHF/UHF ducting and rain scatter, not worldwide HF) with the ionosphere.

Real on-air practice: HF DX contacts on 20 m or 15 m across continents rely entirely on ionospheric (F-layer) refraction.

BCQ33 3.5.4 U0937
中文原题 Original (Chinese)

HF通信术语”静寂区”和”越距”是指:

  1. A“静寂区”是指天波、地波和空间波都未能覆盖的区域。在此区域中我们说传播”越距”了
  2. B“静寂区”是指超出视距,导致空间波传播不到的区域。在此区域中我们说传播”越距”了
  3. C“静寂区”是指因障碍物遮挡,空间波无法覆盖的区域。在此区域中我们说传播”越距”了
  4. D“静寂区”是指短波通信卫星下行信号无法覆盖的区域。在此区域中我们说传播”越距”了
English Translation

The HF-communication terms “silent zone” (skip zone) and “skip distance” mean:

  1. Athe “silent zone” is an area where neither the sky wave, the ground wave, nor the space wave reaches; we say propagation has “skipped” this area
  2. Bthe “silent zone” is an area beyond line-of-sight where the space wave cannot reach; we say propagation has “skipped” this area
  3. Cthe “silent zone” is an area where the space wave cannot reach because of obstacle shadowing; we say propagation has “skipped” this area
  4. Dthe “silent zone” is an area not covered by the downlink signal of a short-wave communication satellite; we say propagation has “skipped” this area
Correct answer: A
Knowledge Point Analysis 知识点解析

The skip zone (静寂区 / silent zone) is the ring-shaped region around a transmitter where neither the ground wave (地波) nor the refracted sky wave (天波) reaches — the sky wave is refracted back to Earth beyond a certain distance (the skip distance, 越距). It is a direct consequence of HF ionospheric reflection geometry, not satellite coverage or obstacle shadowing.

Candidate Tips 考生提示

US–China difference: N/A — skip zone and skip distance are universal HF propagation concepts used identically by amateurs everywhere.

Common pitfall: Conflating the skip zone with a line-of-sight obstruction (B, C) or with satellite coverage (D). The silent zone is specifically the gap between ground-wave and sky-wave coverage.

Real on-air practice: You may hear a European station 8000 km away but not a station 500 km away — the closer station lies inside your skip zone for that band and time.

BCQ34 3.5.4 U0938
中文原题 Original (Chinese)

“衰减”和”衰落”都是无线电通信领域中的常见名词。它们的含义分别为:

  1. A衰减是指信号通过信道或电路后功率减少;衰落是指信号通过信道或电路后发生幅度随时间而变的起伏
  2. B衰减是指信号通过信道或电路后发生幅度随时间而变的起伏;衰落是指信号通过信道或电路后功率减少
  3. C衰减和衰落是一回事,都是指信号通过信道或电路后功率减少
  4. D衰减和衰落是一回事,都是指信号通过信道或电路后发生幅度随时间而变的起伏
English Translation

“Attenuation” and “fading” are both common terms in radio communication. Their meanings are respectively:

  1. Aattenuation means the signal power decreases after passing through a channel or circuit; fading means the signal amplitude fluctuates over time after passing through a channel or circuit
  2. Battenuation means the signal amplitude fluctuates over time after passing through a channel or circuit; fading means the signal power decreases after passing through a channel or circuit
  3. Cattenuation and fading are the same thing, both meaning the signal power decreases after passing through a channel or circuit
  4. Dattenuation and fading are the same thing, both meaning the signal amplitude fluctuates over time after passing through a channel or circuit
Correct answer: A
Knowledge Point Analysis 知识点解析

Attenuation (衰减) is a steady, average reduction in signal power as it travels through a medium or circuit. Fading (衰落) is a time-varying fluctuation of the received amplitude caused by changing propagation paths (e.g., multipath or ionospheric changes). They are distinct: one is a constant loss, the other a random variation. Option A defines both correctly.

Candidate Tips 考生提示

US–China difference: N/A — these are standard IEEE/ITU definitions used identically in US amateur and commercial radio.

Common pitfall: Swapping the two definitions (option B) or assuming they are the same (C, D). Remember: attenuation = power loss; fading = time-varying amplitude.

Real on-air practice: On 10 m during an opening you may see the S-meter swing up and down by 2 S-units — that is fading, on top of the band’s overall path attenuation.

BCQ35 3.5.4 U0939
中文原题 Original (Chinese)

无线电信号经地面和电离层交相反射之后会有什么改变?

  1. A信号的极化特性会伴随时间随机改变
  2. B信号的不同频率成分会伴随时间随机衰落
  3. C信号的上下边带会伴随时间随机反转
  4. D信号中随时都会夹杂强力广播电台的播音
English Translation

After a radio signal is repeatedly reflected between the ground and the ionosphere, what changes occur? (Choose all that apply.)

  1. Athe polarization of the signal changes randomly with time
  2. Bdifferent frequency components of the signal fade randomly with time
  3. Cthe upper and lower sidebands of the signal reverse randomly with time
  4. Dstrong broadcast-station audio is intermittently mixed into the signal
Correct answer: A, B
Knowledge Point Analysis 知识点解析

Multi-hop sky-wave propagation makes the ionosphere act like a rotating, scattering mirror: the polarization (极化) of the returning wave is randomized (A), and different frequency components fade independently — this is selective fading (选择性衰落), matching B. Sidebands do not “reverse” (C), and broadcast audio intrusion (D) is ordinary interference, not a propagation effect. Hence only A and B are correct.

Candidate Tips 考生提示

US–China difference: N/A — polarization rotation and selective fading are inherent to ionospheric multi-hop propagation everywhere.

Common pitfall: Selecting C or D by intuition; the phenomenon described is purely about polarization and frequency-selective fading, not sideband reversal or broadcasting.

Real on-air practice: On a long-path 20 m contact you may notice your received audio “warbling” and your antenna polarization seemingly irrelevant — classic multi-hop fading.

BCQ36 3.5.4 U0940
中文原题 Original (Chinese)

用SDR接收机的频谱显示器观察短波RTTY等2FSK调制的数据通信信号,我们应当看到两个幅度相等的谱峰。但是实际观察结果是这两个谱峰的高度在随机变化。造成这种现象的原因是:

  1. A电离层的选择性衰落
  2. B接收机的工作点漂移
  3. C发射机的ALC不稳定
  4. D接收机的AGC不稳定
English Translation

When observing a short-wave 2FSK data signal such as RTTY with an SDR receiver’s spectrum display, we should see two equal-amplitude spectral peaks. In practice, however, the heights of the two peaks vary randomly. The cause of this phenomenon is:

  1. Aselective fading of the ionosphere
  2. Bdrift of the receiver’s operating point
  3. Cinstability of the transmitter’s ALC
  4. Dinstability of the receiver’s AGC
Correct answer: A
Knowledge Point Analysis 知识点解析

A 2FSK signal (such as RTTY, 无线电传) has two equal tones; in the ionosphere the two frequencies experience selective fading (选择性衰落) independently, so their received amplitudes vary randomly. Receiver AGC acts on the overall signal and would not make the two peaks differ; transmitter ALC and operating-point drift would affect both tones together, not split them. Hence A is correct.

Candidate Tips 考生提示

US–China difference: N/A — RTTY selective fading is observed by US and Chinese HF digital operators alike.

Common pitfall: Blaming the receiver (AGC/operating point) when the cause is propagation; the key clue is the two peaks changing relative to each other.

Real on-air practice: While copying RTTY on 14.080 MHz you may see the mark and space peaks rise and fall out of step on your SDR waterfall — that is ionospheric selective fading, and shifting frequency often helps.

BCQ37 3.5.4 U0941
中文原题 Original (Chinese)

假设接收和发射天线均使用半波长偶极天线,则在依靠电离层反射的远距离通联中,接收和发射天线的最佳极化方式应当安排为:

  1. A不确定。天波反射的特点是信号强度、频率成分和极化会随机改变
  2. B接收和发射天线均位于垂直于两台站连线的平面内,极化保持一致
  3. C接收和发射天线均位于垂直于两台站连线的平面内,极化彼此正交
  4. D发射天线垂直极化,接收天线的极化应当平行于两台站之间的连线
English Translation

Assuming both the receiving and transmitting antennas are half-wave dipoles, in long-distance contacts relying on ionospheric reflection, how should the optimum polarization of the receive and transmit antennas be arranged?

  1. Auncertain — the characteristic of sky-wave reflection is that the signal strength, frequency components, and polarization change randomly
  2. Bboth receive and transmit antennas lie in the plane perpendicular to the line joining the two stations, with matching polarization
  3. Cboth receive and transmit antennas lie in the plane perpendicular to the line joining the two stations, with orthogonal polarization
  4. Dthe transmit antenna is vertically polarized and the receive antenna’s polarization should be parallel to the line joining the two stations
Correct answer: A
Knowledge Point Analysis 知识点解析

Because ionospheric (sky-wave) reflection randomizes the polarization (极化) of the returned wave, there is no fixed optimum polarization to “arrange” between stations — it is unpredictable (A). This contrasts with line-of-sight (space-wave) links where matching polarization matters. Options B, C, D assume a deterministic polarization relationship that does not hold for sky-wave paths.

Candidate Tips 考生提示

US–China difference: N/A — sky-wave polarization randomness is a global propagation fact.

Common pitfall: Applying line-of-sight polarization-matching logic (B/C/D) to an ionospheric path. On HF the reflection scrambles polarization, so fixed matching is pointless.

Real on-air practice: HF operators rarely worry about horizontal vs vertical polarization for sky-wave DX; the ionosphere will rotate it anyway.

BCQ38 3.5.4 U0942
中文原题 Original (Chinese)

甲、乙业余电台相距2000千米,均使用1/2波长水平偶极天线进行HF通联。现其中一方改用1/2波长垂直偶极天线,则改变前后的通信效果有什么不同?

  1. A通信效果的变化不确定,取决于天波反射过程中电波极化的随机变化
  2. B通信效果变差
  3. C通信效果变好
  4. D通信效果不变
English Translation

Two amateur stations A and B, 2000 km apart, both use half-wave horizontal dipoles for HF contacts. If one station switches to a half-wave vertical dipole, how does the communication effect change compared with before?

  1. Athe change in communication effect is uncertain, depending on the random variation of polarization during sky-wave reflection
  2. Bthe communication effect becomes worse
  3. Cthe communication effect becomes better
  4. Dthe communication effect stays the same
Correct answer: A
Knowledge Point Analysis 知识点解析

On a 2000 km sky-wave (天波) path the ionosphere randomizes polarization, so changing one antenna from horizontal to vertical does not give a predictable improvement or degradation (A). Because the returned wave’s polarization is scrambled, “better,” “worse,” and “unchanged” are all unreliable predictions. Only A correctly expresses this uncertainty.

Candidate Tips 考生提示

US–China difference: N/A — same sky-wave polarization randomness for all HF operators.

Common pitfall: Picking B, C, or D thinking vertical vs horizontal has a definite effect; on ionospheric paths it is essentially random.

Real on-air practice: An HF station can use a vertical and still work a station using a horizontal dipole with no penalty — the ionosphere handles the polarization mismatch.

BCQ39 3.5.4 U0943
中文原题 Original (Chinese)

影响电离层的短波传播特性的主要因素有:

  1. A太阳黑子活动、太阳耀斑和地磁活动
  2. B季节和昼夜变化
  3. C工作频率和通信距离
  4. D高空云量和气温变化
English Translation

The main factors affecting the short-wave propagation characteristics of the ionosphere are: (Choose all that apply.)

  1. Asunspot activity, solar flares, and geomagnetic activity
  2. Bseasonal and day-night variations
  3. Cthe operating frequency and the communication distance
  4. Dhigh-altitude cloud cover and air-temperature variations
Correct answer: A, B, C
Knowledge Point Analysis 知识点解析

Ionospheric short-wave propagation is governed by solar and geomagnetic conditions (sunspots 太阳黑子, solar flares 太阳耀斑, geomagnetic activity 地磁活动 — A), by time of day and season (B), and by the chosen frequency and path length (C). Weather such as clouds and air temperature (D) does not significantly affect ionospheric refraction. Thus A, B, C are correct.

Candidate Tips 考生提示

US–China difference: N/A — the ionospheric drivers are identical worldwide; US operators use the same VOACAP-style predictors.

Common pitfall: Mistakenly including weather (D); ionospheric propagation depends on solar/geomagnetic and temporal factors, not tropospheric weather.

Real on-air practice: Checking a propagation forecast (solar flux, K-index) before a scheduled 40 m contact is exactly using factors A and B.

BCQ40 3.5.4 U0944
中文原题 Original (Chinese)

喜爱HF通联的爱好者大都了解术语”最高可用频率(MUF)”。其涵义为:

  1. A在地球上的两点间通过天波建立HF联络时可以使用的最高频率
  2. B通常,MUF在夜间降低,在白天显著升高
  3. C当MUF偶尔达到数百兆赫兹时,VHF/UHF可能出现超视距传播
  4. D使用较低仰角的天线进行DX通联时,需将MUF估计得低一些
English Translation

HF enthusiasts mostly know the term “Maximum Usable Frequency (MUF).” Its meaning is: (Choose all that apply.)

  1. Athe highest frequency that can be used to establish an HF contact between two points on Earth via the sky wave
  2. Bgenerally, the MUF decreases at night and rises significantly during the day
  3. Cwhen the MUF occasionally reaches several hundred megahertz, VHF/UHF beyond-line-of-sight propagation may appear
  4. Dwhen using a lower-elevation-angle antenna for DX contacts, the MUF should be estimated lower
Correct answer: A, B
Knowledge Point Analysis 知识点解析

The Maximum Usable Frequency (最高可用频率 / MUF) is the highest frequency that will still be refracted back to Earth between two points via the ionosphere (A). Because ionization is driven by sunlight, the MUF is typically low at night and high during the day (B). Per the official key, only A and B are marked correct; C and D are not part of the marked answer set.

Candidate Tips 考生提示

US–China difference: N/A — MUF is a global ionospheric term used identically by US and Chinese HF operators.

Common pitfall: On multi-answer questions, only mark what the official key lists (A, B). Do not add C or D even if they seem plausible; the marked answer is fixed.

Real on-air practice: If the 20 m band closes at night, the MUF has dropped below 14 MHz — switch to 40 m or 80 m, which are below the nighttime MUF.

BCQ41 3.5.4 U0945
中文原题 Original (Chinese)

若已知最高可用频率(MUF)为20MHz,则DX通联成功率最高的业余频段为:

  1. A18MHz
  2. B14MHz
  3. C21MHz
  4. D24MHz
English Translation

If the Maximum Usable Frequency (MUF) is known to be 20 MHz, the amateur band with the highest success rate for DX contacts is:

  1. A18 MHz
  2. B14 MHz
  3. C21 MHz
  4. D24 MHz
Correct answer: A
Knowledge Point Analysis 知识点解析

The optimum working frequency for reliable DX is typically about 0.85 × MUF — just below the MUF so the signal is still refracted but not marginal. With MUF = 20 MHz, the best amateur band is 18 MHz (the 17 m band), which lies safely below 20 MHz. 21 MHz and 24 MHz exceed the MUF and will not be refracted; 14 MHz works but is well below optimum, giving a lower rate than 18 MHz.

Candidate Tips 考生提示

US–China difference: N/A — the “work just below the MUF” rule is universal for HF DX.

Common pitfall: Choosing the band closest to but above the MUF (21 MHz), forgetting that frequencies above the MUF are not reflected and escape to space.

Real on-air practice: When a propagation map shows MUF 20 MHz for a path, tune 17 m (18.1 MHz) first for the best chance of a two-way contact.

BCQ42 3.5.4 U0946
中文原题 Original (Chinese)

如果按距离地表的高度从高到低排列,对短波传播有主要影响的电离层有:

  1. AF2、F1、E、D
  2. BC1、C2、D1、D2
  3. CF、E2、E1、D
  4. DE1、E2、F1、F2
English Translation

If arranged from highest to lowest altitude above the Earth’s surface, the ionospheric layers that mainly affect short-wave propagation are:

  1. AF2, F1, E, D
  2. BC1, C2, D1, D2
  3. CF, E2, E1, D
  4. DE1, E2, F1, F2
Correct answer: A
Knowledge Point Analysis 知识点解析

The principal ionospheric layers affecting HF propagation, from highest to lowest altitude, are F2 (highest), F1, E, and D (lowest). The D layer (lowest) mostly absorbs, while F2 (highest) is the main reflector for long-distance HF. Options listing non-standard layer names (C1, C2, E2) or wrong ordering are incorrect.

Candidate Tips 考生提示

US–China difference: N/A — the D/E/F1/F2 layer model is universal in HF propagation texts.

Common pitfall: Forgetting that F2 is the highest layer and D the lowest; some candidates reverse the order.

Real on-air practice: F2-layer refraction is what brings you 20 m and 15 m DX; the D layer is why 10 m dies during the day at low solar activity (excess absorption).

BCQ43 3.5.4 U0947
中文原题 Original (Chinese)

如果按离子密度从高到低排列,对短波传播有主要影响的电离层有:

  1. AF2、F1、E、D
  2. BC1、C2、D1、D2
  3. CF、E2、E1、D
  4. DE1、E2、F1、F2
English Translation

If arranged from highest to lowest electron (ion) density, the ionospheric layers that mainly affect short-wave propagation are:

  1. AF2, F1, E, D
  2. BC1, C2, D1, D2
  3. CF, E2, E1, D
  4. DE1, E2, F1, F2
Correct answer: A
Knowledge Point Analysis 知识点解析

By electron density, the same principal layers rank F2 (highest density, the strongest refractor), F1, E, and D (lowest). The F2 layer’s high density is what allows it to refract the highest frequencies. The ordering by density coincides with the altitude ordering here, so F2, F1, E, D is correct.

Candidate Tips 考生提示

US–China difference: N/A — ionospheric layer density ordering is a global standard.

Common pitfall: Assuming D (lowest altitude) has highest density; in fact the highest layer F2 has the greatest electron density.

Real on-air practice: Higher electron density (high solar flux) raises the MUF, opening higher bands like 10 m — directly tied to F2 density.

BCQ44 3.5.4 U0948
中文原题 Original (Chinese)

电离层对短波传播的影响主要体现为:

  1. AF2、F1和E层反射电波
  2. BD层不反射电波,但是吸收电波
  3. CD2和D1层不反射电波,但是吸收电波
  4. DC2、C1和F层白天反射电波,夜晚吸收电波
English Translation

The influence of the ionosphere on short-wave propagation is mainly reflected in: (Choose all that apply.)

  1. Athe F2, F1, and E layers reflect radio waves
  2. Bthe D layer does not reflect radio waves but absorbs them
  3. Cthe D2 and D1 layers do not reflect radio waves but absorb them
  4. Dthe C2, C1, and F layers reflect waves by day and absorb waves by night
Correct answer: A, B
Knowledge Point Analysis 知识点解析

The F2, F1, and E layers reflect (refract) HF sky waves (A), while the lowest D layer does not reflect but absorbs energy — especially at lower frequencies and by day (B). “D1/D2” and “C1/C2” are not the standard layer nomenclature for this effect, so C and D are not part of the marked answer. The official key marks A and B.

Candidate Tips 考生提示

US–China difference: N/A — the reflecting/ absorbing roles of E/F vs D layers are universal HF facts.

Common pitfall: Thinking the D layer reflects (it only absorbs); also do not be misled by fictional layer names like D1/D2 in C.

Real on-air practice: 80 m signals are strongly absorbed by the daytime D layer, which is why 80 m is mainly a nighttime band.

BCQ45 3.5.4 U0949
中文原题 Original (Chinese)

讨论1.9MHz或3.5MHz等短波低频段DX通信时,业余无线电爱好者常会谈及术语”灰线”。这是指:

  1. A地球上白昼与黑夜交汇的区域
  2. B地球上有极光活动的区域
  3. C连接地球上具有相同最高可用频率的地点形成的线
  4. D地球上通讯双方所在地点的大圆连线
English Translation

When discussing DX communication on low short-wave bands such as 1.9 MHz or 3.5 MHz, amateur operators often mention the term “gray line.” This refers to:

  1. Athe region on Earth where day and night meet
  2. Bthe region on Earth with auroral activity
  3. Cthe line connecting points on Earth that have the same Maximum Usable Frequency
  4. Dthe great-circle line connecting the two communicating stations
Correct answer: A
Knowledge Point Analysis 知识点解析

The gray line (灰线) is the terminator — the moving boundary on Earth between the sunlit hemisphere and the dark hemisphere. Along this line the D-layer absorption is minimal on both ends while the F-layer is still active, creating excellent low-band DX conditions. It is not the auroral zone (B), an MUF contour (C), or a great-circle path (D).

Candidate Tips 考生提示

US–China difference: N/A — “gray-line propagation” is a term used identically by US and other HF amateurs.

Common pitfall: Confusing the gray line with the auroral oval (B) or with the great-circle bearing to a DX station (D).

Real on-air practice: 80 m and 160 m DX hunters aim their CQ at the gray line to work stations at the opposite terminator at sunrise/sunset.

BCQ46 3.5.4 U0950
中文原题 Original (Chinese)

业余无线电爱好者经常利用”灰线”来建立1.9MH或3.5MHz等短波低频段DX联络。这是因为:

  1. A通信双方同时位于灰线时,最有可能利用天波的多跳反射路径建立联络
  2. B通信双方位于灰线两侧4000千米以外的对称点时,传播效果最佳
  3. C通信双方位于灰线同一侧4000千米以外的两点时,传播效果最佳
  4. D通信双方应避免同时处于灰线才能获得更稳定的传播
English Translation

Amateur operators often use the “gray line” to establish DX contacts on low short-wave bands such as 1.9 MHz or 3.5 MHz. This is because:

  1. Awhen both stations are simultaneously on the gray line, it is most likely that a multi-hop sky-wave reflection path can be established
  2. Bwhen both stations are on opposite sides of the gray line, more than 4000 km from it symmetrically, propagation is best
  3. Cwhen both stations are on the same side of the gray line, more than 4000 km apart, propagation is best
  4. Dboth stations should avoid being on the gray line at the same time to obtain more stable propagation
Correct answer: A
Knowledge Point Analysis 知识点解析

At the gray line (terminator) the D-layer absorption is low at both ends while the F-layer remains reflective, so a two-station path straddling the terminator can support reliable multi-hop sky-wave (天波) propagation on low bands (A). Options B, C, D describe geometries or advice contrary to the well-known gray-line effect.

Candidate Tips 考生提示

US–China difference: N/A — gray-line low-band DX is exploited by amateurs worldwide.

Common pitfall: Thinking both stations must be far from the gray line (B/C) or avoid it (D); the advantage comes precisely from both being near the terminator.

Real on-air practice: A 160 m contact between Europe and Japan often succeeds only in the brief gray-line window at their mutual sunrise/sunset.

BCQ47 3.5.4 U0951
中文原题 Original (Chinese)

如果你尝试用”长路径”联络某个业余电台,你的定向天线应当指向:

  1. A该台短路径方向的反方向
  2. B垂直于灰线,但是背向灰线
  3. C可能出现北极光的方向
  4. D可能出现南极光的方向
English Translation

If you try to contact a certain amateur station via the “long path,” your directional antenna should point:

  1. Ain the direction opposite to the short-path bearing to that station
  2. Bperpendicular to the gray line, but away from the gray line
  3. Ctoward the direction where aurora borealis may appear
  4. Dtoward the direction where aurora australis may appear
Correct answer: A
Knowledge Point Analysis 知识点解析

Every great-circle path between two stations has a short path and a long path (the complementary bearing, 180° opposite). To work the long path (长路径) you aim your beam 180° away from the short-path bearing (A). The gray line and aurora directions (B, C, D) are unrelated to the long-path aiming rule.

Candidate Tips 考生提示

US–China difference: N/A — short-path/long-path beamheading is identical for all HF operators.

Common pitfall: Pointing the antenna at the short-path bearing when trying long path; the correct aim is directly opposite (±180°).

Real on-air practice: To work a European station from the US on the long path, you point your beam due east (toward Europe’s short path is west, so long path is east across Asia).

BCQ48 3.5.5 U0952
中文原题 Original (Chinese)

进行短波电离层传播预测所必需的参数为:

  1. A太阳黑子平均数
  2. B地磁活动指数
  3. C通信双方的位置
  4. D通信时间
English Translation

The parameters necessary for short-wave ionospheric propagation prediction are: (Choose all that apply.)

  1. Athe sunspot number (average)
  2. Bthe geomagnetic-activity index
  3. Cthe locations of the two communicating stations
  4. Dthe communication time
Correct answer: A, B, C, D
Knowledge Point Analysis 知识点解析

Ionospheric propagation prediction requires the solar conditions (sunspot number 太阳黑子平均数, and a flux/geomagnetic index such as the A/K index — A and B), the great-circle geometry between the two stations (their locations — C), and the time of day/month (D), since ionization varies diurnally and seasonally. All four are mandatory inputs, so A, B, C, D are correct.

Candidate Tips 考生提示

US–China difference: N/A — propagation predictors (e.g., VOACAP) use the same four inputs worldwide.

Common pitfall: Missing the time parameter (D); many candidates think only solar/geomagnetic data and location matter, but time-of-day is essential.

Real on-air practice: Before a scheduled 40 m sked you run a prediction using SSN, K-index, both grid locators, and the UTC time of the sked.

BCQ49 3.5.5 U0953
中文原题 Original (Chinese)

尝试进行传播预测时,太阳通量的最低值可以取为50左右。其最高值可以取为:

  1. A300
  2. B280
  3. C250
  4. D200
English Translation

When attempting a propagation prediction, the minimum value of the solar flux can be taken as about 50. Its maximum value can be taken as:

  1. A300
  2. B280
  3. C250
  4. D200
Correct answer: A
Knowledge Point Analysis 知识点解析

The 10.7 cm solar flux (F107, 太阳通量) normally ranges from about 65 (solar minimum) to roughly 300 (solar maximum). For prediction purposes the low end is taken near 50 and the high end near 300, so 300 is the correct maximum value.

Candidate Tips 考生提示

US–China difference: N/A — the F10.7 cm flux scale (≈65–300) is the standard solar index used by US and international propagation models.

Common pitfall: Picking a mid-range number (200–280); the question asks the practical upper bound used in predictions, which is about 300.

Real on-air practice: During a solar maximum you will see F107 values approaching 200–250, opening 10 m and 6 m for worldwide DX.

BCQ50 3.5.5 U0954
中文原题 Original (Chinese)

反映地磁活动程度的常见指标是A指数和K指数。K指数的取值范围是:

  1. A0-9
  2. B0-8
  3. C0-7
  4. D0-6
English Translation

Common indicators of geomagnetic activity are the A-index and the K-index. The range of the K-index is:

  1. A0–9
  2. B0–8
  3. C0–7
  4. D0–6
Correct answer: A
Knowledge Point Analysis 知识点解析

The K-index (K指数) is a three-hourly geomagnetic-activity measure on a quasi-logarithmic scale from 0 (very quiet) to 9 (extreme storm). The A-index is the daily average converted to a linear scale. Thus the K-index range is 0–9.

Candidate Tips 考生提示

US–China difference: N/A — the Kp/K-index 0–9 scale is the global standard used by NOAA and amateurs everywhere.

Common pitfall: Confusing the K-index (0–9) with the linear A-index; the K scale caps at 9, not 6–8.

Real on-air practice: When K > 4–5, HF propagation degrades and auroral openings may appear on VHF; many DXers post “K-index” in their propagation notices.

BCQ51 3.5.5 U0955
中文原题 Original (Chinese)

太阳黑子活动的平均周期约为:

  1. A11.2年,每个周期中的太阳活跃程度有所差别
  2. B11.2年,每个周期中的太阳活跃程度完全相同
  3. C38年,每个周期中的太阳活跃程度有所差别
  4. D38年,每个周期中的太阳活跃程度完全相同
English Translation

The average period of sunspot activity is about:

  1. A11.2 years, with the level of solar activity differing in each cycle
  2. B11.2 years, with the level of solar activity identical in each cycle
  3. C38 years, with the level of solar activity differing in each cycle
  4. D38 years, with the level of solar activity identical in each cycle
Correct answer: A
Knowledge Point Analysis 知识点解析

The solar (sunspot) cycle (太阳黑子活动周期) averages about 11.2 years. Each successive cycle differs in intensity (some are strong, some weak), so the correct statement is A: ~11.2 years with varying activity. The 38-year figure refers to the longer Gleissberg modulation, not the basic cycle.

Candidate Tips 考生提示

US–China difference: N/A — the ~11-year sunspot cycle is a universal astronomical fact.

Common pitfall: Choosing B (identical activity) or the 38-year value (C/D); cycles are ~11 years and are never identical.

Real on-air practice: We are in Solar Cycle 25; its strength determines how often 10 m and 6 m open for DX compared with the previous weak Cycle 24.

BCQ52 3.5.5 U0956
中文原题 Original (Chinese)

太阳黑子活动的强弱是用”太阳黑子平均数(SSN)”来描述的。其一般规律为:

  1. A较大的SSN利于短波DX通信
  2. B较小的SSN利于短波DX通信
  3. C在地磁活动剧烈的年份,SSN与短波DX通信关系不大
  4. D只在发生太阳耀斑时,SSN才影响短波DX通信
English Translation

The strength of sunspot activity is described by the “sunspot number (SSN).” The general rule is:

  1. Aa larger SSN favors short-wave DX communication
  2. Ba smaller SSN favors short-wave DX communication
  3. Cin years of intense geomagnetic activity, the SSN has little relation to short-wave DX communication
  4. Dthe SSN affects short-wave DX communication only when a solar flare occurs
Correct answer: A
Knowledge Point Analysis 知识点解析

A higher sunspot number (SSN, 太阳黑子平均数) means a more strongly ionized ionosphere and a higher MUF, which opens the higher HF bands (10 m, 6 m) for long-distance (DX) work. So larger SSN favors short-wave DX (A). Smaller SSN (B) suppresses it; C and D overstate the exceptions.

Candidate Tips 考生提示

US–China difference: N/A — higher SSN → better HF DX is true for amateurs everywhere.

Common pitfall: Inverting the relationship (B) — beginners sometimes think “fewer sunspots = better,” but the opposite holds for HF sky-wave DX.

Real on-air practice: During solar maximum (high SSN) you can work worldwide on 10 m with low power; during minimum you are mostly limited to 80/40 m.

BCQ53 3.5.5 U0957
中文原题 Original (Chinese)

太阳耀斑可以引发电离层扰动(SID)。其对短波通信的影响是:

  1. A低频率通信所受的影响超过高频率的
  2. B高纬度地区传播路径所受的影响超过低纬度的
  3. C卫星通信所受的影响超过地面台站间直射波的
  4. D地球上黑夜区域所受的影响超过白昼区域的
English Translation

A solar flare can trigger a Sudden Ionospheric Disturbance (SID). Its effect on short-wave communication is:

  1. Alower-frequency communication is affected more than higher-frequency communication
  2. Bpropagation paths in high-latitude regions are affected more than those in low-latitude regions
  3. Csatellite communication is affected more than direct waves between ground stations
  4. Dthe night side of Earth is affected more than the day side
Correct answer: A
Knowledge Point Analysis 知识点解析

A Sudden Ionospheric Disturbance (电离层扰动 / SID) dramatically increases D-layer absorption on the sunlit side, and the effect is strongest at lower frequencies (the D layer absorbs low bands like 80 m/160 m most severely, while higher bands such as 15 m/10 m are less affected). So A is correct; the disturbance is a daytime (not night) effect.

Candidate Tips 考生提示

US–China difference: N/A — SID physics (greater low-frequency absorption, daytime only) is global.

Common pitfall: Thinking higher frequencies are hit hardest (B-style), or that the night side is affected (D); SID is a dayside, low-frequency absorption event.

Real on-air practice: During a big flare, 80 m may go dead at noon while 20 m is still usable — a textbook SID.

BCQ54 3.5.5 U0958
中文原题 Original (Chinese)

业余无线电爱好者在预测HF传播时经常用到一个缩写为F107的参数。其意义是:

  1. A太阳10.7cm波长射电辐射通量指数
  2. B107MHz调频广播信号的典型传播距离
  3. C电离层对10.7MHz电波的衰减指数
  4. D最高可用频率与10.7MHz的比值
English Translation

Amateur operators often use a parameter abbreviated F107 when predicting HF propagation. Its meaning is:

  1. Athe solar radio-flux index at the 10.7 cm wavelength
  2. Bthe typical propagation distance of 107 MHz FM broadcast signals
  3. Cthe attenuation index of the ionosphere for 10.7 MHz waves
  4. Dthe ratio of the Maximum Usable Frequency to 10.7 MHz
Correct answer: A
Knowledge Point Analysis 知识点解析

F107 (太阳10.7cm波长射电辐射通量指数) is the solar flux at a wavelength of 10.7 cm (a microwave proxy for solar UV/EUV output that drives ionospheric ionization). It is a primary input to HF propagation models. It has nothing to do with 107 MHz broadcasts, 10.7 MHz attenuation, or a frequency ratio.

Candidate Tips 考生提示

US–China difference: N/A — the F10.7 flux is the standard solar index used by US NOAA and in VOACAP.

Common pitfall: Misreading “10.7” as 10.7 MHz (C) or 107 MHz (B); it is a wavelength in centimeters of solar radio emission.

Real on-air practice: Propagation websites list “SFI” (solar flux index) — that is F107; an SFI above ~150 means good high-band openings.

BCQ55 3.5.5 U0959
中文原题 Original (Chinese)

业余无线电爱好者在预测HF传播时经常使用缩写为F107的参数,其值大体都在50-300的范围内。如果该数值增大,则:

  1. A安静太阳的辐射强度增加,电离层密度变大。这些都有利于F层反射DX信号
  2. B10.7MHz无线电波受到电离层的衰减变大,远距离传播的条件变差
  3. C107MHz调频广播可能超视距传播,收到50-300千米VHF信号的几率将增加
  4. D安静太阳的辐射强度增加,电离层密度变大。这些不利于DX信号的传播
English Translation

Amateur operators often use the parameter abbreviated F107 when predicting HF propagation; its value is generally in the range 50–300. If this value increases, then:

  1. Athe radiation intensity of the quiet Sun increases and the ionospheric density grows; these are favorable for the F layer to reflect DX signals
  2. B10.7 MHz radio waves suffer greater ionospheric attenuation and long-distance propagation conditions worsen
  3. C107 MHz FM broadcasts may propagate beyond line of sight, increasing the chance of receiving 50–300 km VHF signals
  4. Dthe radiation intensity of the quiet Sun increases and the ionospheric density grows; these are unfavorable for DX-signal propagation
Correct answer: A
Knowledge Point Analysis 知识点解析

A higher F107 means stronger solar (quiet-Sun) radio flux, more ionizing UV/EUV, and a denser ionosphere — which raises the MUF and improves F-layer reflection of DX signals (A). Option D states the opposite, and B/C confuse the 10.7 cm index with actual 10.7/107 MHz radio propagation.

Candidate Tips 考生提示

US–China difference: N/A — rising F10.7 improving HF propagation is universal.

Common pitfall: Inverting the cause/effect (D) or tying F107 to 10.7 MHz/107 MHz bands (B/C).

Real on-air practice: When the solar flux climbs toward 200, 6 m E-skip and 10 m worldwide openings become common.

BCQ56 3.5.6 U0960
中文原题 Original (Chinese)

我国《无线电频率划分规定》指出,122.25-123GHz业余频段用于业余业务,但是不包括卫星业余业务。这是因为:

  1. A60、120、183GHz左右存在大气吸收频带,电波衰减大,不适合地面至空间的卫星业余通信
  2. B120GHz附近频段的宇宙射线特别强烈,干扰卫星通信,包括卫星间的各种业务
  3. C业余无线电爱好者尚无力制作工作于120GHz及以上频段的设备
  4. D120GHz这类频段的天线太大,不适合业余性质的卫星通信研究
English Translation

China’s Radio Frequency Allocation Regulations (《无线电频率划分规定》) state that the 122.25–123 GHz amateur band is allocated to the amateur service but excludes the amateur-satellite service. This is because:

  1. Aaround 60, 120, and 183 GHz there are atmospheric-absorption bands where radio waves are strongly attenuated, which is unsuitable for ground-to-space amateur satellite communication
  2. Bnear 120 GHz cosmic rays are especially intense and interfere with satellite communication, including various inter-satellite services
  3. Camateur operators are still unable to build equipment operating at 120 GHz and above
  4. Dantennas for bands such as 120 GHz are too large, unsuitable for amateur satellite-communication research
Correct answer: A
Knowledge Point Analysis 知识点解析

The Radio Frequency Allocation Regulations of the PRC (《无线电频率划分规定》) allocate 122.25–123 GHz to the amateur service (业余业务) but not the amateur-satellite service (卫星业余业务) because the molecular-absorption bands of the atmosphere near 60, 120, and 183 GHz cause severe attenuation that prevents reliable ground-to-space links. The other options cite nonexistent cosmic-ray or equipment limitations.

Candidate Tips 考生提示

US–China difference: The 122 GHz atmospheric-absorption fact is global (ITU/Radio Regulations), but band allocations themselves differ between administrations; China’s allocation excludes the satellite service here.

Common pitfall: Picking a “technology limitation” answer (C/D); the reason is a physical atmospheric-absorption band, not amateur capability.

Real on-air practice: At millimeter-wave (EHF) frequencies even rain and air absorption dominate path loss, which is why these bands are used only for very short terrestrial links.

BCQ57 3.5.6 U0961
中文原题 Original (Chinese)

以下频段受降雨影响最为严重的是:

  1. A极高频EHF(毫米波)
  2. B高频HF(短波)
  3. C低频LF(长波)
  4. D特高频UHF(分米波)
English Translation

Among the following bands, the one most severely affected by rainfall is:

  1. Athe extremely high frequency EHF (millimeter wave)
  2. Bthe high frequency HF (short wave)
  3. Cthe low frequency LF (long wave)
  4. Dthe ultra-high frequency UHF (decimetric wave)
Correct answer: A
Knowledge Point Analysis 知识点解析

Rain attenuation increases sharply with frequency; the EHF (极高频 / millimeter-wave) band around 30–300 GHz is the most rain-affected because raindrop size becomes comparable to the wavelength. HF, LF, and even UHF are far less affected. Hence EHF (millimeter wave) is correct.

Candidate Tips 考生提示

US–China difference: N/A — rain attenuation increasing with frequency is a universal propagation fact; US microwave/ EME operators see the same on 10 GHz and up.

Common pitfall: Picking UHF (D) thinking “higher = worse” but not going far enough; EHF/ millimeter wave is the worst.

Real on-air practice: On 10 GHz (microwave) contests, heavy rain can drop a contact’s signal by 10–20 dB — and 24/47 GHz are even more rain-sensitive.

BCQ58 3.5.6 U0962
中文原题 Original (Chinese)

根据ITU的建议,植被在一定程度上吸收无线电波。大体规律是:

  1. A频率越高,吸收越多
  2. B频率越低,吸收越多
  3. C吸收程度与植物种类有关,与频率无关
  4. D吸收程度与季节有关,与频率无关
English Translation

According to ITU recommendations, vegetation absorbs radio waves to some extent. The general rule is:

  1. Athe higher the frequency, the greater the absorption
  2. Bthe lower the frequency, the greater the absorption
  3. Cthe absorption depends on the plant species but not on frequency
  4. Dthe absorption depends on the season but not on frequency
Correct answer: A
Knowledge Point Analysis 知识点解析

Per the ITU (国际电信联盟) recommendations, foliage/vegetation absorption of radio waves increases with frequency (higher frequencies, especially VHF/UHF and above, are absorbed more by leaves and moisture). So “the higher the frequency, the greater the absorption” (A) is correct; frequency is the governing factor, not species/season alone.

Candidate Tips 考生提示

US–China difference: N/A — the ITU foliage-attenuation model is the same reference used by US operators planning VHF/UHF paths through trees.

Common pitfall: Inverting the relationship (B) or claiming frequency independence (C/D).

Real on-air practice: A 2 m (144 MHz) signal passing through a forest loses more than an HF signal; at 1.2 GHz the same trees can block the path entirely.

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