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CRAC Bilingual Manual: Transmitters and Modulation | 发射机与调制 (72 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

72bilingual questions·Classes A, B, C

This section covers Transmitters and Modulation with 72 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.

ABCQ1 3.6.1 U0501
中文原题 Original (Chinese)

自制业余无线电发射设备,在经无线电检测机构检验合格并取得电台执照之前,应在调测时在设备的天线端口连接(或在串联必要的仪表之后连接):

  1. A假负载
  2. BVSWR严格等于1:1的驻波天线
  3. CVSWR严格等于1:1的行波天线
  4. D专门用于测试的标准环形天线
English Translation

When testing a self-made amateur radio transmitting device, before it passes inspection by a radio testing organization and obtains a radio station license, you should connect (or connect after series-connecting the necessary instruments) to the device’s antenna port:

  1. Aa dummy load
  2. Ba standing-wave antenna with VSWR strictly equal to 1:1
  3. Ca traveling-wave antenna with VSWR strictly equal to 1:1
  4. Da standard loop antenna specially for testing
Correct answer: A
Knowledge Point Analysis 知识点解析

During bench testing of a transmitter (especially a homebrew one) that is not yet licensed, the output must be terminated in a dummy load (假负载) so that no radio-frequency signal is radiated into the air, which would cause unlawful interference. A dummy load is a non-radiating, properly matched resistor. Real antennas (B, C, D) would radiate and are not permitted for unlicensed/uncertified testing. This aligns with the principle that an amateur radio station (业余无线电台) may only radiate after passing inspection and being licensed.

Candidate Tips 考生提示

US–China difference: In both the US (FCC Part 97) and China, unlicensed transmitters must use a dummy load for testing; radiating during “testing” without authorization is prohibited. China additionally requires type/equipment inspection by a radio testing organization before licensing.

Common pitfall: Thinking a perfect 1:1-VSWR antenna (B/C) is acceptable — even a matched antenna radiates, which is illegal before certification. The point is non-radiation, not matching.

Real on-air practice: Builders always tune finals and measure output into a 50 Ω dummy load, then connect the antenna only after the station is authorized.

ABCQ2 3.6.1 U0502
中文原题 Original (Chinese)

在无线电发射机中,调制器的作用是:

  1. A以原始调制信号控制射频载波的幅度、频率和相位参数
  2. B以电能转换效率最高的方式控制线性射频放大器的工作点
  3. C调整天馈系统的参数,实现阻抗匹配
  4. D自动控制发射信号的频谱,将其保持在核准的必要带宽内
English Translation

In a radio transmitter, the role of the modulator is:

  1. Ato use the original modulating signal to control the amplitude, frequency, and phase parameters of the RF carrier
  2. Bto control the operating point of the linear RF amplifier in the most power-efficient way
  3. Cto adjust the parameters of the antenna-feeder system to achieve impedance matching
  4. Dto automatically control the spectrum of the transmitted signal, keeping it within the approved necessary bandwidth
Correct answer: A
Knowledge Point Analysis 知识点解析

Modulation (调制) means impressing the information signal onto an RF carrier. The modulator (调制器) uses the baseband/modulating signal to vary the carrier’s amplitude (AM), frequency (FM) or phase (PM). Option B describes a PA biasing function, C is antenna matching (impedance matching), and D is a spectrum/bandwidth control function — none define modulation. So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — the definition of a modulator is universal.

Common pitfall: Confusing the modulator with the power amplifier (B) or a spectrum-limiter (D); remember modulation is about varying carrier parameters with the information signal.

Real on-air practice: Speaking into a mic modulates the carrier; SSB is produced by modulating and then suppressing the carrier and one sideband.

ABCQ3 3.6.1 U0503
中文原题 Original (Chinese)

保证业余无线电通信接收机优良接收能力的主要因素是:

  1. A良好的抗干扰能力
  2. B足够高的整机增益
  3. C尽量低的本机噪声
  4. D尽量小的信号失真
English Translation

The main factors that ensure good receiving capability of an amateur radio communication receiver are: (Choose all that apply.)

  1. Agood interference-rejection capability
  2. Bsufficiently high overall gain
  3. Cas low as possible internal (receiver) noise
  4. Das small as possible signal distortion
Correct answer: A, B, C, D
Knowledge Point Analysis 知识点解析

A good receiver needs all four attributes: strong interference rejection (selectivity, 抗干扰能力) to ignore unwanted signals; enough overall gain (整机增益) to bring weak signals up to usable level; low internal noise (本机噪声) so weak signals are not masked; and low signal distortion (信号失真) so the demodulated audio/data is faithful. Each directly contributes to receiving capability, so all four options are correct.

Candidate Tips 考生提示

US–China difference: N/A — receiver figure-of-merit factors (gain, noise, selectivity, fidelity) are taught identically.

Common pitfall: Picking only one or two; the stem says “主要因素” (main factors, plural) and the key treats all four as correct. Watch for the multi-answer format.

Real on-air practice: A weak DX station is copyable only when the rig has low noise, adequate gain, good filtering and clean audio — all four together.

ABCQ4 3.6.1 U0504
中文原题 Original (Chinese)

接收机解调器的作用是:

  1. A从接收到的已调射频信号中分离出原始调制信号
  2. B对接收到的射频信号进行宽带线性放大
  3. C对接收到的射频信号进行与必要带宽相匹配的选频放大
  4. D从接收到的已调制射频信号中提取出载波分量
English Translation

The role of a receiver’s demodulator is:

  1. Ato separate the original modulating signal from the received modulated RF signal
  2. Bto perform wideband linear amplification of the received RF signal
  3. Cto perform frequency-selective amplification of the received RF signal matched to the necessary bandwidth
  4. Dto extract the carrier component from the received modulated RF signal
Correct answer: A
Knowledge Point Analysis 知识点解析

Demodulation (解调) is the inverse of modulation: the demodulator (解调器) recovers the original information (voice/data) from the modulated RF/IF signal. Option B is amplification, C is filtering/bandwidth selection, D is carrier extraction (a sub-step, not the recovery of the message). The defining role is recovering the modulating signal — A.

Candidate Tips 考生提示

US–China difference: N/A — demodulator definition is universal.

Common pitfall: Choosing D (extract carrier) — that is part of some demodulator designs but not the purpose; the goal is to recover the message, not the carrier.

Real on-air practice: After the IF stage, the demodulator (discriminator for FM, product detector for SSB) outputs the audio you hear.

ABCQ5 3.6.1 U0505
中文原题 Original (Chinese)

选用解调器的主要应考因素是:

  1. A尽量忠实地还原原始调制信号
  2. B尽量对已调射频信号加以放大
  3. C尽量提升已调射频信号中的载频分量
  4. D尽量补偿接所收射频信号的频率偏移
English Translation

The main factor to consider when selecting a demodulator is:

  1. Ato reproduce the original modulating signal as faithfully as possible
  2. Bto amplify the modulated RF signal as much as possible
  3. Cto boost the carrier component in the modulated RF signal as much as possible
  4. Dto compensate as much as possible for the frequency offset of the received RF signal
Correct answer: A
Knowledge Point Analysis 知识点解析

The primary criterion for choosing a demodulator is fidelity — how faithfully it reconstructs the original modulating signal (忠实还原). Amplification (B) is the job of amplifiers, boosting the carrier (C) is not the aim, and frequency-offset compensation (D) is the job of the receiver’s frequency-control/AFC, not the demodulator’s selection criterion. Hence A.

Candidate Tips 考生提示

US–China difference: N/A — fidelity is the universal demodulator selection criterion.

Common pitfall: Picking B or C, which describe gain/carrier handling rather than the demodulator’s purpose of faithful recovery.

Real on-air practice: A good SSB product detector or FM discriminator preserves voice intelligibility and tone — fidelity matters more than raw gain.

ABCQ6 3.6.1 U0506
中文原题 Original (Chinese)

亚音频静噪系统(CTCSS)所用的声调大体位于音频中的什么范围?

  1. A67-250.3Hz
  2. B16Hz-20kHz
  3. C16kHz-20kHz
  4. D220Hz-2503Hz
English Translation

In the sub-audible tone squelch system (CTCSS), roughly what range within the audio band do the tones occupy?

  1. A67–250.3 Hz
  2. B16 Hz–20 kHz
  3. C16 kHz–20 kHz
  4. D220 Hz–2503 Hz
Correct answer: A
Knowledge Point Analysis 知识点解析

CTCSS (亚音频静噪 / sub-audible tone squelch, 连续单音编码静噪) uses a low-frequency supervisory tone added below the normal speech band. The standard CTCSS tone set spans roughly 67 Hz to 250.3 Hz — these are “sub-audible” because they sit at the low edge of (or just below) normal hearing and are filtered out before audio output. Option B is the whole audio range, C is high audio, D is too high. So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — CTCSS tone frequencies (67.0–250.3 Hz) are standardized worldwide by the same EIA/TIA table used in China.

Common pitfall: Thinking “sub-audible” means below 20 Hz (B’s lower bound) — actually the tones are in the 67–250 Hz band, still technically audible but kept below speech and stripped by the squelch filter.

Real on-air practice: To open a repeater you set the matching CTCSS tone (e.g., 88.5 Hz); without it the squelch stays closed even though you hear the carrier.

ABCQ7 3.6.1 U0507
中文原题 Original (Chinese)

对于比较考究的接收机,其说明书中常常列出一项”通带矩形系数”或”通带形状系数”指标,是用来描述:

  1. A带通滤波器频率特性曲线两侧斜坡的陡峭程度
  2. B矩形波信号通过滤波器后波形两侧的陡峭程度
  3. C矩形波信号通过滤波器后谐波成分的损失程度
  4. D带通晶体滤波器中石英晶体的切割方向和形状
English Translation

For a more refined receiver, its specification sheet often lists a “passband rectangular coefficient” or “passband shape factor” indicator, which is used to describe:

  1. Athe steepness of the slopes on both sides of the bandpass filter’s frequency-response curve
  2. Bthe steepness of both sides of the waveform after a rectangular-wave signal passes through the filter
  3. Cthe degree of loss of harmonic components of a rectangular-wave signal after passing through the filter
  4. Dthe cutting orientation and shape of the quartz crystal in the bandpass crystal filter
Correct answer: A
Knowledge Point Analysis 知识点解析

The shape factor / rectangular coefficient (通带矩形系数 / 形状系数) of a bandpass filter compares its bandwidth at two attenuation levels (e.g., 60 dB bandwidth ÷ 6 dB bandwidth). A smaller number means steeper skirts — the response curve drops more sharply outside the passband. It describes the steepness of the filter’s response-curve slopes (A), not the shape of a rectangular waveform (B, C) or the crystal cut (D). So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — shape factor is defined the same way (e.g., 60 dB/6 dB bandwidth ratio) in US and Chinese rig specs.

Common pitfall: Reading “矩形” (rectangle) literally as a rectangular waveform (B/C). It refers to how “rectangular” the filter passband shape is, i.e., skirt steepness.

Real on-air practice: A narrow, steep filter (low shape factor) lets you pull a weak SSB signal out next to a strong adjacent-channel station.

ABCQ8 3.6.1 U0508
中文原题 Original (Chinese)

根据说明书给出的技术指标,两台业余无线电接收机在USB方式下选择带宽为2.7kHz的滤波器时具有不同的”通带矩形系数”。接收机A的”60dB带宽对6dB带宽的矩形系数”为3.8,接收机B的为5。由此可得出结论:

  1. AA机对邻近频道干扰的抑制能力比B机强
  2. BA机对邻近频道干扰的抑制能力比B机差
  3. CA机对镜像频率干扰的抑制能力比B机强
  4. DA机对偏离工作频率±10kHz以外的干扰信号的抑制能力比B机强
English Translation

According to the technical specifications, two amateur radio receivers have different “passband rectangular coefficients” when selecting a 2.7 kHz bandwidth filter in USB mode. Receiver A’s “60 dB-bandwidth-to-6 dB-bandwidth rectangular coefficient” is 3.8, and receiver B’s is 5. From this we can conclude:

  1. Areceiver A’s ability to reject adjacent-channel interference is stronger than B’s
  2. Breceiver A’s ability to reject adjacent-channel interference is weaker than B’s
  3. Creceiver A’s ability to reject image-frequency interference is stronger than B’s
  4. Dreceiver A’s ability to reject interference signals more than ±10 kHz from the operating frequency is stronger than B’s
Correct answer: A
Knowledge Point Analysis 知识点解析

The rectangular coefficient (60 dB BW ÷ 6 dB BW) measures skirt steepness. A smaller coefficient (3.8 vs 5) means sharper skirts, so receiver A attenuates signals just outside the 2.7 kHz passband much more strongly than B — i.e., better adjacent-channel (邻近频道) rejection. Image rejection (C) depends on the front-end/preselector, not the IF filter shape factor; ±10 kHz (D) is far outside the 2.7 kHz filter and both would be well attenuated, so the distinguishing conclusion is adjacent-channel rejection. Hence A.

Candidate Tips 考生提示

US–China difference: N/A — filter shape factor and its effect on adjacent-channel rejection are universal.

Common pitfall: Thinking a larger coefficient is “better”; it is worse (less steep). Also do not credit image rejection (C) to the IF filter shape factor.

Real on-air practice: On a crowded contest band, the rig with the lower shape factor lets you copy a weak station right next to a loud one.

ABCQ9 3.6.1 U0509
中文原题 Original (Chinese)

接收机前置放大器的主要作用是:

  1. A降低接收机内部噪声的影响
  2. B提高接收机最终的音频输出功率电平
  3. C提高接收机音频输出的保真度
  4. D提高接收机的动态范围
English Translation

The main role of a receiver preamplifier is:

  1. Ato reduce the influence of the receiver’s internal noise
  2. Bto raise the final audio output power level of the receiver
  3. Cto improve the fidelity of the receiver’s audio output
  4. Dto improve the receiver’s dynamic range
Correct answer: A
Knowledge Point Analysis 知识点解析

A low-noise preamplifier (前置放大器 / LNA) placed early in the receive chain raises the signal above the receiver’s own downstream noise figure, so the system noise is dominated by the (low) preamp noise rather than later stages. Its main purpose is to reduce the impact of internal receiver noise (降低本机噪声影响), improving weak-signal sensitivity. It does not set audio output power (B), fidelity (C) or dynamic range (D) as its primary role. So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — the LNA/noise-figure principle is universal; US hams also use preamps on VHF/UHF for weak-signal work.

Common pitfall: Thinking the preamp is for louder audio (B) — it improves signal-to-noise at the front end, not speaker volume.

Real on-air practice: A 2 m preamp at the antenna boosts weak EME/weak-signal contacts; but on strong local signals it can overload, so it is often switched.

ABCQ10 3.6.1 U0510
中文原题 Original (Chinese)

进行FM话音通联时,我们能否单凭接收机中传出的对方的声音大小来准确判断对方的信号有多强?

  1. A不能。鉴频器所解调的声音,其大小仅取决于中频信号的频偏,与中频信号的幅度无关。况且,中频信号在进入鉴频器之前早已被限幅电路切齐,已无法鉴别强弱
  2. B不能。信号越强,自动增益控制的起控程度就越深。增益的急剧下降反而会压低音量
  3. C能。最终的信号就是经过放大处理的原始射频信号,当然是信号越强声音越大
  4. D能。调频信号的特点是信号越强频偏越大,解调之后的声音当然也越大
English Translation

During FM voice communication, can we accurately judge how strong the other station’s signal is solely from how loud their voice sounds in the receiver?

  1. ANo. The sound demodulated by the discriminator depends only on the frequency deviation of the IF signal and is unrelated to the IF signal’s amplitude. Moreover, the IF signal has already been clipped flat by the limiter before entering the discriminator, so strength cannot be discerned.
  2. BNo. The stronger the signal, the deeper the AGC action; the sharp gain drop instead lowers the volume.
  3. CYes. The final signal is the amplified original RF signal, so naturally the stronger the signal the louder the sound.
  4. DYes. An FM signal’s characteristic is that the stronger the signal the greater the frequency deviation, so naturally the demodulated sound is louder.
Correct answer: A
Knowledge Point Analysis 知识点解析

In FM, information is carried by frequency deviation, not amplitude. The receiver’s limiter (限幅电路) clips the IF amplitude to a constant level before the discriminator (鉴频器), so the recovered audio amplitude reflects only deviation (modulation), not received-signal strength. Therefore loudness cannot indicate signal strength — A is the precise, correct explanation. B is also “no” but gives a wrong reason (AGC does not normally set FM audio volume that way). C and D are false.

Candidate Tips 考生提示

US–China difference: N/A — FM limiter/discriminator behavior is identical; US hams likewise rely on an S-meter, not audio loudness, for signal strength.

Common pitfall: Picking B just because it also says “no”; the reasoning matters. The key reason is the limiter removing amplitude info before the discriminator.

Real on-air practice: Use the S-meter or RST report for strength; on FM, a loud signal and a weak one can sound equally loud until they drop into the noise.

ABCQ11 3.6.1 U0511
中文原题 Original (Chinese)

用设置在NFM方式的对讲机接收WFM信号,效果为:

  1. A可以听到信号。如果调制信号幅度较大或音调较高,会听到明显的非线性失真
  2. B听不到信号。但是一旦接收到了信号,调频方式所特有的强烈噪声仍然会消失
  3. C可以正常听到信号。但是声音的高音频部分衰减较大,缺乏高音
  4. D可以正常听到信号,只是声音比较小
English Translation

The effect of receiving a WFM signal with a handheld transceiver set to NFM mode is:

  1. AYou can hear the signal. If the modulating signal amplitude is large or the tone is high, obvious non-linear distortion will be heard.
  2. BYou cannot hear the signal. But once a signal is received, the strong noise characteristic of FM mode will still disappear.
  3. CYou can hear the signal normally. But the high-frequency audio is greatly attenuated, lacking treble.
  4. DYou can hear the signal normally, just the sound is relatively small.
Correct answer: A
Knowledge Point Analysis 知识点解析

WFM (wideband FM, ~75 kHz deviation, used for broadcast) received in NFM (narrowband FM, ~2.5–5 kHz deviation, used for amateur/PMR) over-deviates the NFM discriminator. The discriminator still recovers audio, so you hear the signal, but large/high audio swings exceed the NFM deviation range and clip, producing obvious non-linear distortion. Hence A. C/D (“normal”) are wrong; B (“cannot hear”) is wrong because the carrier is still demodulated.

Candidate Tips 考生提示

US–China difference: N/A — NFM vs WFM deviation mismatch is a universal FM concept (US hams also hear distortion when monitoring a wide-FM broadcast on a narrow-FM rig).

Common pitfall: Confusing which direction distorts; NFM-receiving-WFM gives distortion because WFM deviation is far larger than the NFM discriminator expects.

Real on-air practice: Try listening to a local FM broadcast (WFM) on your 2 m handheld (NFM) — you get a garbled, distorted voice, not clean audio.

ABCQ12 3.6.1 U0512
中文原题 Original (Chinese)

用设置在WFM方式的对讲机接收NFM信号,效果为:

  1. A可以正常听到信号,只是声音比较小
  2. B可以听到信号。如果调制信号幅度较大或音调较高,会听到明显的非线性失真
  3. C听不到信号。但是一旦接收到了信号,调频方式所特有的强烈噪声仍然会消失
  4. D可以正常听到信号。但是声音的高音频部分衰减较大,缺乏高音
English Translation

The effect of receiving an NFM signal with a handheld transceiver set to WFM mode is:

  1. AYou can hear the signal normally, just the sound is relatively small.
  2. BYou can hear the signal. If the modulating signal amplitude is large or the tone is high, obvious non-linear distortion will be heard.
  3. CYou cannot hear the signal. But once a signal is received, the strong noise characteristic of FM mode will still disappear.
  4. DYou can hear the signal normally. But the high-frequency audio is greatly attenuated, lacking treble.
Correct answer: A
Knowledge Point Analysis 知识点解析

This is the reverse of U0511: an NFM (small deviation) signal received in WFM (large-deviation) mode under-deviates the WFM discriminator, so the recovered audio is weak/quiet but essentially undistorted — you hear it normally, just at low volume (A). Distortion (B) occurs in the opposite direction (NFM receiving WFM). Hence A is correct.

Candidate Tips 考生提示

US–China difference: N/A — symmetric to U0511, universal FM principle.

Common pitfall: Swapping the outcomes of U0511 and U0512; remember WFM-receiving-NFM = quiet but clean, NFM-receiving-WFM = distorted.

Real on-air practice: Listening to a 2 m NFM repeater on a WFM broadcast receiver gives faint but intelligible audio.

ABCQ13 3.6.1 U0513
中文原题 Original (Chinese)

没有信号时,调频接收机会输出一种强烈的沙沙声。关于这种噪声,以下描述正确的是:

  1. A天线收到的QRN与机内电路的固有噪声共同构成一种随机信号。该信号的随机相位变化经鉴频形成强烈的沙沙声。只是,该信号的随机幅度变化与沙沙声没有关系
  2. B由天线收到的QRN的随机幅度变化经放大形成,其大小与QRN的电压成正比
  3. C由天线收到的QRN的随机幅度变化经放大形成,其大小与QRN的电压的平方成正比
  4. D由天线收到的QRN的随机幅度变化经放大形成,其大小与QRN的电压的平方根成正比
English Translation

When there is no signal, an FM receiver outputs a strong “shhh” noise. Regarding this noise, the correct description is:

  1. AThe QRN received by the antenna together with the receiver’s internal circuit noise forms a random signal. The random phase variation of this signal is converted by the discriminator into a strong “shhh” noise. However, the random amplitude variation of this signal has no relation to the “shhh” noise.
  2. BFormed by amplification of the random amplitude variation of the QRN received by the antenna; its magnitude is proportional to the QRN voltage.
  3. CFormed by amplification of the random amplitude variation of QRN; its magnitude is proportional to the square of the QRN voltage.
  4. DFormed by amplification of the random amplitude variation of QRN; its magnitude is proportional to the square root of the QRN voltage.
Correct answer: A
Knowledge Point Analysis 知识点解析

FM quieting/Noise: with no signal, the receiver sees random noise (QRN plus internal noise). Because FM information is in phase/frequency, the random phase fluctuations of this noise are demodulated by the discriminator as a “shhh” (hiss). Random amplitude fluctuations are removed by the limiter and thus do not contribute to the audio hiss — exactly A. Options B/C/D wrongly attribute the noise to amplitude variations (proportional to V, V², √V), which is how AM noise behaves, not FM.

Candidate Tips 考生提示

US–China difference: N/A — FM capture/quieting and noise-vs-phase behavior are universal.

Common pitfall: Applying AM noise reasoning (amplitude ∝ voltage) to FM; in FM the hiss comes from phase jitter, and amplitude is clamped by the limiter.

Real on-air practice: Open the squelch on a quiet 2 m channel and you hear the characteristic FM hiss; a strong carrier makes it “quiet” (noise reduction / capture).

ABCQ14 3.6.1 U0514
中文原题 Original (Chinese)

下列哪种电路可以解调FM信号?

  1. A鉴频器
  2. B限幅器
  3. C乘积检波器
  4. D混频器
English Translation

Which of the following circuits can demodulate an FM signal?

  1. Adiscriminator
  2. Blimiter
  3. Cproduct detector
  4. Dmixer
Correct answer: A
Knowledge Point Analysis 知识点解析

An FM signal is demodulated by a frequency discriminator (鉴频器), which converts frequency deviation into voltage. A limiter (B) only flattens amplitude (prep for the discriminator); a product detector (C) demodulates SSB/CW (AM-family); a mixer (D) does frequency conversion. So only A demodulates FM.

Candidate Tips 考生提示

US–China difference: N/A — discriminator demodulates FM universally.

Common pitfall: Picking the limiter (B) — it is part of the FM detector chain but does not itself recover audio.

Real on-air practice: The discriminator output is your FM audio; a misaligned discriminator shows off-frequency or distorted RX.

ABCQ15 3.6.1 U0515
中文原题 Original (Chinese)

什么是”鉴频”?

  1. A对调频信号进行解调的过程称为鉴频
  2. B判断信号的频率是否超过允许的频率范围的过程称为鉴频
  3. C判断信号的频率是否发生了不应有的偏离或者漂移的过程称为鉴频
  4. D对调幅信号进行解调的过程称为鉴频
English Translation

What is “discrimination” (frequency discrimination, 鉴频)?

  1. AThe process of demodulating a frequency-modulation (FM) signal is called discrimination.
  2. BThe process of judging whether a signal’s frequency exceeds the allowed frequency range is called discrimination.
  3. CThe process of judging whether a signal’s frequency has undergone an undue deviation or drift is called discrimination.
  4. DThe process of demodulating an amplitude-modulation (AM) signal is called discrimination.
Correct answer: A
Knowledge Point Analysis 知识点解析

鉴频 (frequency discrimination) is the demodulation of an FM signal — recovering the original audio from frequency deviation. B and C describe frequency monitoring/measurement, not demodulation; D is detection (检波) of AM, not 鉴频. So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — “discriminator” terminology matches worldwide.

Common pitfall: Confusing 鉴频 (FM demod) with 检波 (AM demod). They are paired terms; keep them distinct.

Real on-air practice: Your FM rig’s discriminator recovers the other station’s voice; an S-meter often taps the discriminator.

ABCQ16 3.6.1 U0516
中文原题 Original (Chinese)

下列哪一项可以解调AM信号?

  1. A检波器
  2. B限幅器
  3. C鉴频器
  4. D反相器
English Translation

Which of the following can demodulate an AM signal?

  1. Adetector
  2. Blimiter
  3. Cdiscriminator
  4. Dinverter
Correct answer: A
Knowledge Point Analysis 知识点解析

An AM signal is demodulated by a detector (检波器), typically an envelope detector. A limiter (B) is for FM, a discriminator (C) is for FM, an inverter (D) simply flips phase. So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — AM detection is universal.

Common pitfall: Mixing up detector (AM) vs discriminator (FM).

Real on-air practice: An AM broadcast or AM aircraft band is recovered by the envelope detector in the receiver.

ABCQ17 3.6.1 U0517
中文原题 Original (Chinese)

什么是”检波”?

  1. A对调幅信号进行解调的过程称为检波
  2. B检查信号的波形是否超过了允许的幅度范围的过程称为检波
  3. C检查信号的频率是否发生了不应有的偏离或者漂移的过程称为检波
  4. D对调频信号进行解调的过程称为检波
English Translation

What is “detection” (检波)?

  1. AThe process of demodulating an amplitude-modulation (AM) signal is called detection.
  2. BThe process of checking whether a signal’s waveform exceeds the allowed amplitude range is called detection.
  3. CThe process of checking whether a signal’s frequency has undergone an undue deviation or drift is called detection.
  4. DThe process of demodulating a frequency-modulation (FM) signal is called detection.
Correct answer: A
Knowledge Point Analysis 知识点解析

检波 (detection) is the demodulation of an AM signal. B and C describe monitoring functions, and D describes 鉴频 (FM demod), not 检波. So A is correct — this is the deliberate counterpart to U0515.

Candidate Tips 考生提示

US–China difference: N/A — “detector” for AM is standard terminology.

Common pitfall: Swapping 检波 (AM) and 鉴频 (FM); the exam deliberately pairs these two questions.

Real on-air practice: A diode envelope detector recovers AM audio — the classic crystal-set principle.

ABCQ18 3.6.1 U0518
中文原题 Original (Chinese)

关于收发信机的AGC功能,以下说法正确的有:

  1. AAGC实现收信机自动增益控制,对中频信号进行检测并反馈控制,防止电路过载
  2. B进行基于FSK或PSK的数据通信时,关闭AGC功能有可能提高弱信号的解码效果
  3. C进行太阳噪声测试的时候,需要关闭AGC功能
  4. DAGC实现收信机自动音量控制,对音频电平进行检测并反馈控制,防止扬声器损坏
English Translation

Regarding the AGC function of a transceiver, the correct statements are: (Choose all that apply.)

  1. AAGC implements automatic gain control of the receiver, detecting the IF signal and feeding back control to prevent circuit overload
  2. BWhen conducting data communication based on FSK or PSK, turning off the AGC function may improve the decoding of weak signals
  3. CWhen performing solar-noise measurement, the AGC function needs to be turned off
  4. DAGC implements automatic volume control of the receiver, detecting the audio level and feeding back control to prevent speaker damage
Correct answer: A, B
Knowledge Point Analysis 知识点解析

AGC (automatic gain control, 自动增益控制) samples the IF (or detector) level and adjusts receiver gain to keep strong signals from overloading subsequent stages — A is correct. For FSK/PSK (e.g., RTTY, FT8) weak-signal decoding, a slow/fast AGC can pump the audio and hurt the decoder, so disabling AGC can improve decoding — B is correct. D is wrong because AGC controls RF/IF gain, not audio volume (and not to protect the speaker). Per the official answer key, C is not selected for this item (although in practice AGC is often disabled for noise-figure/solar-noise measurements, the exam key marks only A and B).

Candidate Tips 考生提示

US–China difference: N/A — AGC behavior is the same; US digital-mode operators also disable AGC for FT8/PSK to stabilize decoding.

Common pitfall: Thinking AGC = “automatic volume control” (D); it controls gain before audio, keeping the detector from overloading, not protecting the speaker.

Real on-air practice: On FT8 you typically turn AGC off and set a fixed RF gain so the waterfall stays steady and the decoder performs best.

ABCQ19 3.6.1 U0519
中文原题 Original (Chinese)

学习现代通信技术或制作业余无线电作品时,我们经常遇到一种工作原理不同于超外差式收信机的”DC式收信机”。其中,缩写DC是指:

  1. A直接变换(Direct-Conversion),即接收到的射频信号在解调之前不做频率变换
  2. B直流(Direct Current),指直流电源供电的收信机
  3. C介质电容(Dielectric Capacitor),指收信机中用于调谐的电容器所用的特定介质
  4. D数字变频(Digital-Conversion),指接收到的射频信号经过了数字化的变频处理
English Translation

When studying modern communication technology or building amateur radio projects, we often encounter a “DC receiver” whose operating principle differs from the superheterodyne receiver. Here, the abbreviation DC means:

  1. ADirect-Conversion, i.e., the received RF signal is not frequency-converted before demodulation
  2. BDirect Current, referring to a receiver powered by a DC supply
  3. CDielectric Capacitor, referring to the specific dielectric used in the tuning capacitors of the receiver
  4. DDigital-Conversion, referring to the received RF signal undergoing digitized frequency conversion
Correct answer: A
Knowledge Point Analysis 知识点解析

A DC receiver is a Direct-Conversion receiver (直接变换): the incoming RF is mixed directly to baseband/audio (or near-zero IF) without the usual superheterodyne intermediate frequency — no image-frequency issue but it receives both sidebands. DC here is not Direct Current (B), a capacitor (C), or digital conversion (D). So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — “direct conversion” is the standard US/UK term for DC receivers (e.g., popular QRP kits).

Common pitfall: Reading DC as “Direct Current”; in receiver context it always means Direct-Conversion.

Real on-air practice: Many simple CW/SSB QRP homebrew rigs are direct-conversion; you hear the carrier beat as an audible tone.

ABCQ20 3.6.1 U0520
中文原题 Original (Chinese)

接收机和发射机中常见的混频器有什么作用?

  1. A频率变换。将信号的原始频率变换成另一个频率
  2. B混合式调音。将两个信号相互叠加,送往扬声器
  3. C变频调速。把直流电变成变频交流电以驱动天调
  4. D多频放大器。同时放大具有不同频率的多个信号
English Translation

What is the role of the mixer commonly found in receivers and transmitters?

  1. Afrequency conversion — converting the signal’s original frequency into another frequency
  2. Baudio mixing — overlaying two signals and sending them to the speaker
  3. Cvariable-frequency speed control, converting DC into variable-frequency AC to drive the antenna tuner
  4. Dmulti-frequency amplifier, simultaneously amplifying multiple signals of different frequencies
Correct answer: A
Knowledge Point Analysis 知识点解析

A mixer (混频器 / frequency converter) multiplies two frequencies (the incoming signal and a local oscillator) to produce sum and difference frequencies — performing frequency conversion (频率变换). It is not audio mixing (B), motor speed control (C), or a multi-frequency amplifier (D). So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — mixer = frequency conversion is universal.

Common pitfall: Interpreting “混频” as audio mixing (B); in RF it means heterodyning for frequency conversion.

Real on-air practice: The mixer in your superhet converts your 14.200 MHz signal down to a 455 kHz or 9 MHz IF.

ABCQ21 3.6.1 U0521
中文原题 Original (Chinese)

在超外差式收发信机中,混频器的工作方式可分为”上变频”和”下变频”两种,具体指:

  1. A中频频率高于输入频率的为上变频方式,中频频率低于输入频率的为下变频方式
  2. B输入频率高于中频频率的为上变频方式,输入频率低于中频频率的为下变频方式
  3. C本振频率高于输入频率的为上变频方式,本振频率低于输入频率的为下变频方式
  4. D输入频率高于本振频率的为上变频方式,输入频率低于本振频率的为下变频方式
English Translation

In a superheterodyne transceiver, the mixer’s operation can be divided into “up-conversion” and “down-conversion”; specifically:

  1. Aup-conversion is when the IF frequency is higher than the input frequency, and down-conversion is when the IF frequency is lower than the input frequency
  2. Bup-conversion is when the input frequency is higher than the IF frequency, and down-conversion is when the input frequency is lower than the IF frequency
  3. Cup-conversion is when the local-oscillator frequency is higher than the input frequency, and down-conversion is when the local-oscillator frequency is lower than the input frequency
  4. Dup-conversion is when the input frequency is higher than the local-oscillator frequency, and down-conversion is when the input frequency is lower than the local-oscillator frequency
Correct answer: A
Knowledge Point Analysis 知识点解析

Up/down conversion is defined by the relationship between the input (RF) frequency and the resulting IF: if the IF is higher than the input frequency, the signal was up-converted; if the IF is lower, it was down-converted. This is the convention used in the Chinese exam — A. Note this is defined by IF vs input, not by the local-oscillator relationship (C/D) or merely input vs IF (B). So A is correct.

Candidate Tips 考生提示

US–China difference: Slight terminology nuance — in the US “up/down conversion” is also often described via LO vs RF, but the Chinese exam key defines it by IF-vs-input, so answer A.

Common pitfall: Choosing C or D based on the local oscillator; the exam defines it by the IF relative to the input frequency.

Real on-air practice: Many HF rigs use a high first IF (up-conversion, e.g., 70 MHz) for good image rejection; VHF rigs often use low-IF (down-conversion).

ABCQ22 3.6.1 U0522
中文原题 Original (Chinese)

超外差式业余收发信机的面板上经常设有选择中频滤波器带宽的控制部件。这些中频滤波器所抑制的干扰可以分类为:

  1. A邻近频率干扰
  2. B镜像频率干扰
  3. C中频频率干扰
  4. D突发脉冲干扰
English Translation

Superheterodyne amateur transceivers often have a control on the panel for selecting the IF filter bandwidth. The interference suppressed by these IF filters can be classified as:

  1. Aadjacent-frequency interference
  2. Bimage-frequency interference
  3. Cintermediate-frequency interference
  4. Dburst-pulse interference
Correct answer: A
Knowledge Point Analysis 知识点解析

The selectable IF filter bandwidth rejects signals close to (but outside) the wanted channel — i.e., adjacent-frequency (邻近频率) interference. Image-frequency interference (B) is handled by the preselector before the mixer, not by the IF bandwidth; C and D are not what IF bandwidth selection addresses. So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — IF bandwidth controls adjacent-channel rejection everywhere.

Common pitfall: Attributing image rejection (B) to the IF filter; image rejection is a front-end job.

Real on-air practice: Narrowing the IF filter (e.g., 500 Hz CW vs 2.7 kHz SSB) suppresses the adjacent station’s splatter.

ABCQ23 3.6.1 U0523
中文原题 Original (Chinese)

在超外差式业余收发信机中,负责抑制镜像频率干扰的部件是:

  1. A变频级之前的波段预选滤波器
  2. B变频级之后的中频滤波器
  3. C中频放大级中的限幅电路
  4. D带有音调控制的音频输出电路
English Translation

In a superheterodyne amateur transceiver, the component responsible for suppressing image-frequency interference is:

  1. Athe band preselector filter before the mixing stage
  2. Bthe IF filter after the mixing stage
  3. Cthe limiter circuit in the IF amplifier stage
  4. Dthe audio output circuit with tone control
Correct answer: A
Knowledge Point Analysis 知识点解析

The image frequency lies 2×IF away from the wanted signal and also mixes to the IF. The only stage that can reject it before mixing is the preselector (波段预选滤波器) ahead of the mixer (变频级); once both signals reach the mixer they are indistinguishable. The IF filter (B), limiter (C) and tone control (D) cannot remove the image. So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — image rejection by a preselector is universal; US VHF rigs also rely on high first-IF + preselector.

Common pitfall: Thinking the IF filter (B) rejects the image — it cannot, since the image is already converted to the same IF.

Real on-air practice: A good front-end preselector on 144 MHz keeps the image (e.g., ~235 MHz) out before it mixes down.

ABCQ24 3.6.1 U0524
中文原题 Original (Chinese)

超外差式收信机所收信号的频率要比本振频率低(或高)一个中频。然而,比本振频率高(或低)一个中频的另一个遥相对应的信号也可能经混频窜入中频通道,形成”镜像频率干扰”或”镜频干扰”。如果某对讲机的技术规格书给出的VHF接收机第一中频为45.05MHz,那么在145.00MHz收到的镜频干扰可能来自:

  1. A235.10MHz或54.90MHz
  2. B190.05MHz或99.95MHz
  3. C45.05MHz或90.10MHz
  4. D90.10MHz或180.20MHz
English Translation

In a superheterodyne receiver the received signal frequency is lower (or higher) than the local-oscillator frequency by one IF. However, another corresponding signal that is higher (or lower) than the local-oscillator frequency by one IF may also enter the IF channel through mixing, forming “image-frequency interference” or “mirror-frequency interference”. If a handheld transceiver’s specification gives the VHF receiver’s first IF as 45.05 MHz, then image-frequency interference received at 145.00 MHz may come from:

  1. A235.10 MHz or 54.90 MHz
  2. B190.05 MHz or 99.95 MHz
  3. C45.05 MHz or 90.10 MHz
  4. D90.10 MHz or 180.20 MHz
Correct answer: A
Knowledge Point Analysis 知识点解析

The image frequency is f_image = f_signal ± 2×IF. With f_signal = 145.00 MHz and IF = 45.05 MHz, the image is at 145.00 ± 90.10 = 235.10 MHz or 54.90 MHz — exactly option A. (Choosing the LO: if LO = 145+45.05 = 190.05, image = 190.05+45.05 = 235.10; if LO = 145−45.05 = 99.95, image = 99.95−45.05 = 54.90.) So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — the image formula f ± 2×IF is universal.

Common pitfall: Using 1×IF instead of 2×IF, or picking option B (which is the LO frequencies, not the image).

Real on-air practice: On 2 m, a high first IF (like 45 MHz) pushes the image far away so the preselector easily rejects it.

ABCQ25 3.6.1 U0525
中文原题 Original (Chinese)

超外差式收信机所收信号的频率要比本振频率低(或高)一个中频。然而,比本振频率高(或低)一个中频的另一个遥相对应的信号也可能经混频窜入中频通道,形成”镜像频率干扰”或”镜频干扰”。如果某对讲机的技术规格书给出NFM方式时的第一中频为47.25MHz,那么在145.00MHz收到的镜频干扰可能来自:

  1. A239.50MHz或50.50MHz
  2. B192.25MHz或97.75MHz
  3. C50.50MHz或101.00MHz
  4. D151.50MHz或.202.00MHz
English Translation

In a superheterodyne receiver the received signal frequency is lower (or higher) than the local-oscillator frequency by one IF. However, another corresponding signal that is higher (or lower) than the local-oscillator frequency by one IF may also enter the IF channel through mixing, forming “image-frequency interference”. If a handheld transceiver’s specification gives the first IF in NFM mode as 47.25 MHz, then image-frequency interference received at 145.00 MHz may come from:

  1. A239.50 MHz or 50.50 MHz
  2. B192.25 MHz or 97.75 MHz
  3. C50.50 MHz or 101.00 MHz
  4. D151.50 MHz or .202.00 MHz
Correct answer: A
⚠ 译者注 / Translator’s note: 本题 D 选项在 CRAC 官方题库(LK0842)中即为此处所示之原样(官方笔误),本手册依「忠于原题」原则原文照录,并与英文栏逐字对应,以与考场真题完全一致。 / Option D appears garbled in the official CRAC bank (LK0842); shown here exactly as published. Reproduced verbatim and mirrored literally in the English column to match the exam paper.
Knowledge Point Analysis 知识点解析

Image = f_signal ± 2×IF = 145.00 ± (2×47.25) = 145.00 ± 94.50 = 239.50 MHz or 50.50 MHz — option A. (For reference, the LO would be 145.00+47.25 = 192.25 MHz, which is option B, but B is the LO, not the image.) So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — same image formula.

Common pitfall: Picking B (the LO frequency) instead of the true image; always compute f ± 2×IF.

Real on-air practice: Verify image rejection by checking a signal 2×IF away on the band; a clean receiver shows none.

ABCQ26 3.6.1 U0526
中文原题 Original (Chinese)

超外差式收信机所收信号的频率要比本振频率低(或高)一个中频。然而,比本振频率高(或低)一个中频的另一个遥相对应的信号也可能经混频窜入中频通道,形成”镜像频率干扰”或”镜频干扰”。如果某对讲机的技术规格书给出NFM接收所用的第一中频为47.25MHz,那么在435.00MHz收到的镜频干扰可能来自:

  1. A340.50MHz或529.50MHz
  2. B387.75MHz或482.25MHz
  3. C47.25MHz或94.50MHz
  4. D141.70MHz或.236.25MHz
English Translation

In a superheterodyne receiver the received signal frequency is lower (or higher) than the local-oscillator frequency by one IF. However, another corresponding signal that is higher (or lower) than the local-oscillator frequency by one IF may also enter the IF channel through mixing, forming “image-frequency interference”. If a handheld transceiver’s specification gives the first IF used for NFM reception as 47.25 MHz, then image-frequency interference received at 435.00 MHz may come from:

  1. A340.50 MHz or 529.50 MHz
  2. B387.75 MHz or 482.25 MHz
  3. C47.25 MHz or 94.50 MHz
  4. D141.70 MHz or .236.25 MHz
Correct answer: A
⚠ 译者注 / Translator’s note: 本题 D 选项在 CRAC 官方题库(LK0843)中即为此处所示之原样(官方笔误),本手册依「忠于原题」原则原文照录,并与英文栏逐字对应,以与考场真题完全一致。 / Option D appears garbled in the official CRAC bank (LK0843); shown here exactly as published. Reproduced verbatim and mirrored literally in the English column to match the exam paper.
Knowledge Point Analysis 知识点解析

Image = f_signal ± 2×IF = 435.00 ± (2×47.25) = 435.00 ± 94.50 = 340.50 MHz or 529.50 MHz — option A. So A is correct (the other options are distractors; D’s lower value 236.25 is far off).

Candidate Tips 考生提示

US–China difference: N/A — same formula, applied here to the 70 cm (435 MHz) band.

Common pitfall: Arithmetic error with 2×IF; keep 94.50 and add/subtract from 435.00.

Real on-air practice: On 70 cm, a 47.25 MHz first IF puts the image 94.5 MHz away — close enough that front-end filtering matters.

ABCQ27 3.6.1 U0527
中文原题 Original (Chinese)

超外差式收信机所收信号的频率要比本振频率低(或高)一个中频。然而,比本振频率高(或低)一个中频的另一个遥相对应的信号也可能经混频窜入中频通道,形成”镜像频率干扰”或”镜频干扰”。如果某对讲机的技术规格书给出的UHF接收机第一中频为58.525MHz,那么在435.00MHz收到的镜频干扰可能来自:

  1. A317.95MHz或552.05MHz
  2. B376.475MHz或493.525MHz
  3. C58.525MHz或117.05MHz
  4. D234.10.05MHz或.468.20MHz
English Translation

In a superheterodyne receiver the received signal frequency is lower (or higher) than the local-oscillator frequency by one IF. However, another corresponding signal that is higher (or lower) than the local-oscillator frequency by one IF may also enter the IF channel through mixing, forming “image-frequency interference”. If a handheld transceiver’s specification gives the UHF receiver’s first IF as 58.525 MHz, then image-frequency interference received at 435.00 MHz may come from:

  1. A317.95 MHz or 552.05 MHz
  2. B376.475 MHz or 493.525 MHz
  3. C58.525 MHz or 117.05 MHz
  4. D234.10.05 MHz or .468.20 MHz
Correct answer: A
⚠ 译者注 / Translator’s note: 本题 D 选项在 CRAC 官方题库(LK0844)中即为此处所示之原样(官方笔误),本手册依「忠于原题」原则原文照录,并与英文栏逐字对应,以与考场真题完全一致。 / Option D appears garbled in the official CRAC bank (LK0844); shown here exactly as published. Reproduced verbatim and mirrored literally in the English column to match the exam paper.
Knowledge Point Analysis 知识点解析

Image = f_signal ± 2×IF = 435.00 ± (2×58.525) = 435.00 ± 117.05 = 317.95 MHz or 552.05 MHz — option A. So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — same image formula.

Common pitfall: Miscalculating 2×58.525 as 117.05; then ± from 435.00 gives 317.95 / 552.05.

Real on-air practice: A 58.525 MHz first IF on 70 cm places the image 117 MHz away — generally easy to filter.

ABCQ28 3.6.1 U0528
中文原题 Original (Chinese)

业余无线电发射机的效率是指:

  1. A输出到天线系统的信号功率与发射机所消耗的电源功率之比
  2. B通信对象的接收天线得到的信号功率与发射机所消耗的电源功率之比
  3. C通信对象的接收天线得到的信号功率与发射机输出到天线系统的信号功率之比
  4. D输出到天线系统的有用信号功率与到达天线的包含杂散等无用信号的总功率之比
English Translation

The efficiency of an amateur radio transmitter means:

  1. Athe ratio of the signal power delivered to the antenna system to the power consumed by the transmitter from its power supply
  2. Bthe ratio of the signal power received at the communication partner’s receiving antenna to the power consumed by the transmitter from its power supply
  3. Cthe ratio of the signal power received at the communication partner’s receiving antenna to the signal power the transmitter delivers to the antenna system
  4. Dthe ratio of the useful signal power delivered to the antenna system to the total power (including spurious and other useless signals) arriving at the antenna
Correct answer: A
Knowledge Point Analysis 知识点解析

Transmitter efficiency (效率) is RF output power to the antenna divided by DC power drawn from the supply (输出到天线的功率 / 电源消耗功率). It is a local power-conversion measure; it does not involve the distant receiver (B, C) nor spurious ratio (D, which is a spectral-purity measure). So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — transmitter efficiency is defined identically (RF out ÷ DC in).

Common pitfall: Confusing efficiency with effective radiated power to the distant station (B/C) or with spurious-emission ratio (D).

Real on-air practice: A 50 W FM rig drawing ~13.8 A at 13.8 V (≈190 W) has ~26% efficiency; the rest becomes heat.

ABCQ29 3.6.1 U0529
中文原题 Original (Chinese)

业余无线电发射机的效率总是明显低于1。所损耗的那部分能量:

  1. A绝大部分转化为热量,极小一部分转化为杂散等无用信号
  2. B绝大部分转化为杂散等无用信号并对外辐射
  3. C绝大部分因阻抗失配而返回电源,极小一部分转化为热量对外散发
  4. D损耗的能量在电容、电感、开关器件等零部件中消失了
English Translation

The efficiency of an amateur radio transmitter is always significantly below 1. The lost portion of energy:

  1. Ais mostly converted into heat, with a very small part converted into spurious and other useless signals
  2. Bis mostly converted into spurious and other useless signals and radiated outward
  3. Cmostly returns to the power supply due to impedance mismatch, with a very small part converted into heat and dissipated outward
  4. Dthe lost energy disappears within components such as capacitors, inductors, and switching devices
Correct answer: A
Knowledge Point Analysis 知识点解析

The power not delivered to the antenna is dominated by heat dissipation in the PA and other circuitry (I²R and switching losses); only a tiny fraction becomes spurious emissions (杂散发射). Energy is never “lost” (D) nor mostly returned to the supply via mismatch (C); mismatch would mainly reflect, not vanish. So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — energy-balance (heat + small spurious) is universal physics.

Common pitfall: Thinking most loss becomes radiated spurious (B) — spurious emissions are strictly limited and tiny; most loss is heat.

Real on-air practice: A high-power PA needs a big heatsink/fan because most input power becomes heat.

ABCQ30 3.6.1 U0530
中文原题 Original (Chinese)

若一部业余无线电台的工作电压为直流13.8伏,FM方式的射频输出功率为N瓦,电源效率约为80%,则发射时的工作电流约为:

  1. A0.091×N(安)
  2. B13.8×N(安)
  3. C13.8/N×80%(安)
  4. D0.058×N(安)
English Translation

If an amateur radio station operates at a DC voltage of 13.8 V, with an FM RF output power of N watts and a power-supply efficiency of about 80%, then the operating current while transmitting is approximately:

  1. A0.091 × N (A)
  2. B13.8 × N (A)
  3. C13.8 / N × 80% (A)
  4. D0.058 × N (A)
Correct answer: A
Knowledge Point Analysis 知识点解析

DC power drawn = RF output ÷ efficiency = N / 0.8 W. Current I = P_dc / V = (N/0.8) / 13.8 = N / 11.04 ≈ 0.0906 × N A ≈ 0.091 × N A — option A. (The 0.058 figure would correspond to 220 V AC, i.e., U0531.) So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — Ohm’s-law / power efficiency math is universal.

Common pitfall: Using 220 V (D, from the AC version) or inverting the formula; remember I = N/(0.8×13.8).

Real on-air practice: A 50 W mobile rig at 13.8 V draws about 0.091×50 ≈ 4.5 A transmit current.

ABCQ31 3.6.1 U0531
中文原题 Original (Chinese)

若一部业余无线电台的工作电压为交流220伏,FM方式的射频输出功率为N瓦,电源效率约为80%,则发射时的工作电流约为:

  1. A0.0057×N(安)
  2. B220×N(安)
  3. C200/N×80%(安)
  4. D0.0036×N(安)
English Translation

If an amateur radio station operates at an AC voltage of 220 V, with an FM RF output power of N watts and a power-supply efficiency of about 80%, then the operating current while transmitting is approximately:

  1. A0.0057 × N (A)
  2. B220 × N (A)
  3. C200 / N × 80% (A)
  4. D0.0036 × N (A)
Correct answer: A
Knowledge Point Analysis 知识点解析

I = (N/0.8) / 220 = N / 176 ≈ 0.00568 × N A ≈ 0.0057 × N A — option A. (Compare with 13.8 V DC giving 0.091×N.) So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — same formula; note China’s mains is 220 V AC vs 120 V in the US, so currents differ.

Common pitfall: Carrying over the 13.8 V constant from the previous question, or inverting N.

Real on-air practice: A 100 W base station at 220 V draws about 0.0057×100 ≈ 0.57 A (before PS losses) — modest household current.

ABCQ32 3.6.1 U0532
中文原题 Original (Chinese)

若一部业余无线电台以FM方式发射时的射频输出功率为N瓦,电源效率约为80%,则每发射10秒钟所消耗的电能约为:

  1. A0.0000035×N(千瓦小时)
  2. B0.0768 /N(千瓦小时)
  3. C0.0022×N(千瓦小时)
  4. D220 / N(千瓦小时)
English Translation

If an amateur radio station transmits FM at an RF output power of N watts with a power-supply efficiency of about 80%, then the electrical energy consumed per 10 seconds of transmitting is approximately:

  1. A0.0000035 × N (kWh)
  2. B0.0768 / N (kWh)
  3. C0.0022 × N (kWh)
  4. D220 / N (kWh)
Correct answer: A
Knowledge Point Analysis 知识点解析

DC power drawn = N/0.8 W. Energy over 10 s = (N/0.8) × (10/3600) Wh = N × 10 / 2880 Wh = N × 0.003472 Wh = N × 0.000003472 kWh ≈ 0.0000035 × N kWh — option A. So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — energy = power × time; kWh conversion is universal.

Common pitfall: Forgetting to convert seconds to hours (÷3600) or using the wrong efficiency placement; the tiny 3.5×10⁻⁶ factor is the giveaway.

Real on-air practice: Even 100 W for 10 s uses only ~0.00035 kWh — negligible cost, but continuous operation adds up.

ABCQ33 3.6.1 U0533
中文原题 Original (Chinese)

无线电发信机在无调制情况下,在一个射频周期内供给天线馈线的平均功率称为:

  1. A载波功率
  2. B无用功率
  3. C平均功率
  4. D峰包功率
English Translation

The average power supplied to the antenna feeder by a radio transmitter over one RF cycle under no-modulation conditions is called:

  1. Acarrier power
  2. Buseless power
  3. Caverage power
  4. Dpeak envelope power (PEP)
Correct answer: A
Knowledge Point Analysis 知识点解析

Under no modulation, the unmodulated output is the carrier; its average power delivered to the feeder is the carrier power (载波功率). Average power (C) and PEP (D) are modulation-dependent measures; “useless power” (B) is not a standard term here. So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — “carrier power” under no modulation is the standard US/FCC and ITU definition too.

Common pitfall: Picking “average power” (C) — that is the time-averaged power during actual transmission, not specifically the unmodulated carrier.

Real on-air practice: When you key the transmitter with no audio, the meter reads carrier power; speak and PEP/avg rise.

ABCQ34 3.6.1 U0534
中文原题 Original (Chinese)

如果某话音发射机在不同调制方式下的峰值输出功率相同,则在无语音输入时,实际射频输出功率由大到小可以排序为:

  1. AFM,AM,SSB
  2. BAM,SSB,FM
  3. CSSB,FM,AM
  4. DSSB,AM,FM
English Translation

If a voice transmitter has the same peak output power under different modulation modes, then with no voice input, the actual RF output power from large to small can be ranked as:

  1. AFM, AM, SSB
  2. BAM, SSB, FM
  3. CSSB, FM, AM
  4. DSSB, AM, FM
Correct answer: A
Knowledge Point Analysis 知识点解析

With no audio: FM keeps full carrier power (constant envelope) → highest. AM’s carrier is only part of total power (the rest is sidebands that vanish with no modulation) → lower than FM. SSB suppresses the carrier and one sideband, so with no modulation there is essentially no output → lowest. Ranking: FM > AM > SSB — option A. Keep FM, AM, SSB as-is.

Candidate Tips 考生提示

US–China difference: N/A — the FM/AM/SSB carrier-power relationship is identical in US practice; US hams note SSB draws far less current when silent.

Common pitfall: Assuming “same peak power” means same no-audio power; FM is constant-carrier, SSB is near zero when silent.

Real on-air practice: An FM radio is always full-power even between words; an SSB radio’s output collapses to ~0 when you stop talking.

ABCQ35 3.6.3 U0535
中文原题 Original (Chinese)

下列哪项技术指标描述了接收机抗拒邻近频率干扰的能力?

  1. A中频选择性
  2. B整机灵敏度
  3. C频道扫描速率
  4. D本底噪声
English Translation

Which of the following technical indicators describes a receiver’s ability to reject adjacent-frequency interference?

  1. AIF selectivity
  2. Boverall sensitivity
  3. Cchannel scan rate
  4. Dbackground noise
Correct answer: A
Knowledge Point Analysis 知识点解析

Adjacent-frequency rejection is governed by the receiver’s selectivity (选择性), particularly the IF filter’s ability to pass the wanted channel and reject nearby frequencies — hence “IF selectivity” (中频选择性). Sensitivity (B) is weak-signal ability; scan rate (C) and background noise (D) are unrelated. So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — adjacent-channel rejection = selectivity, universal.

Common pitfall: Confusing selectivity with sensitivity (B); they measure different things.

Real on-air practice: A tight IF filter gives high selectivity, letting you ignore the station one channel away.

ABCQ36 3.6.3 U0536
中文原题 Original (Chinese)

接收机抗拒工作频率附近干扰信号的能力可以用选择性指标来表示,分别为:

  1. A信道带宽、信道选择性和信道滤波器的矩形系数
  2. B带内波动和信道带宽
  3. C镜像抑制比
  4. D前端带宽
English Translation

A receiver’s ability to reject interference signals near its operating frequency can be expressed by selectivity indicators, which are respectively:

  1. Achannel bandwidth, channel selectivity, and the rectangular coefficient of the channel filter
  2. Bin-band ripple and channel bandwidth
  3. Cimage rejection ratio
  4. Dfront-end bandwidth
Correct answer: A
Knowledge Point Analysis 知识点解析

Near-in-frequency rejection (adjacent-channel) is described by channel bandwidth, channel selectivity, and the filter’s rectangular coefficient (shape factor) — option A. In-band ripple (B) describes passband flatness, image rejection (C) is for 2×IF signals, and front-end bandwidth (D) addresses far-away strong signals. So A is correct.

Candidate Tips 考生提示

US–China difference: N/A — these are the standard selectivity metrics everywhere.

Common pitfall: Mixing up which selectivity metric applies to which interference type (near vs far vs image) — this question is specifically the “near” case.

Real on-air practice: Spec sheets list “selectivity” as e.g., ±5 kHz attenuation in dB, tied to bandwidth and shape factor.

ABCQ37 3.6.3 U0537
中文原题 Original (Chinese)

接收机接收SSB话音信号时的频响均匀程度可以用选择性指标来表示,分别为:

  1. A带内波动和信道带宽
  2. B信道带宽、信道选择性和信道滤波器的矩形系数
  3. C镜像抑制比
  4. D前端带宽
English Translation

The flatness of a receiver’s frequency response when receiving SSB voice signals can be expressed by selectivity indicators, which are respectively:

  1. Ain-band ripple and channel bandwidth
  2. Bchannel bandwidth, channel selectivity, and the rectangular coefficient of the channel filter
  3. Cimage rejection ratio
  4. Dfront-end bandwidth
Correct answer: A
Knowledge Point Analysis 知识点解析

For SSB voice, what matters is that the passband is flat (uniform response) so the voice is not muffled or tinny — described by in-band ripple (带内波动, passband flatness) and channel bandwidth. Option A is correct. The shape-factor/selectivity set (B) is more about adjacent-channel rejection; image (C) and front-end (D) are different metrics. Keep SSB as-is.

Candidate Tips 考生提示

US–China difference: N/A — SSB passband flatness/ripple is a universal spec.

Common pitfall: Choosing B (selectivity/shape factor) — that addresses rejection, not the flatness of the wanted audio passband.

Real on-air practice: A 2.4 kHz SSB filter with low in-band ripple sounds natural; high ripple makes voices boomy or thin.

ABCQ38 3.6.3 U0538
中文原题 Original (Chinese)

接收机抗拒相距工作频率较远的强干扰信号的能力可以用选择性指标来表示,分别为:

  1. A前端带宽
  2. B带内波动和信道带宽
  3. C信道带宽、信道选择性和信道滤波器的矩形系数
  4. D镜像抑制比
English Translation

A receiver’s ability to reject strong interference signals far from its operating frequency can be expressed by selectivity indicators, which are respectively:

  1. Afront-end bandwidth
  2. Bin-band ripple and channel bandwidth
  3. Cchannel bandwidth, channel selectivity, and the rectangular coefficient of the channel filter
  4. Dimage rejection ratio
Correct answer: A
Knowledge Point Analysis 知识点解析

For strong signals far from the operating frequency, the relevant rejection is set by the front-end bandwidth (前端带宽) — the RF/preselector passband that keeps out off-band blockers before they overload the receiver. So A is correct. (Near-channel = B/C set; image = D.)

Candidate Tips 考生提示

US–China difference: N/A — front-end bandwidth / preselector sets far-off interference rejection universally.

Common pitfall: Applying the adjacent-channel metrics (B/C) or image metric (D) to the “far away” case; each distance range has its own metric.

Real on-air practice: A broad preselector lets a strong broadcast station far away desense your receiver; narrowing front-end filtering helps.

ABCQ39 3.6.3 U0539
中文原题 Original (Chinese)

接收机抗拒相距工作频率两倍于中频的强干扰信号的能力可以用选择性指标来表示,分别为:

  1. A镜像抑制比
  2. B前端带宽
  3. C带内波动和信道带宽
  4. D信道带宽、信道选择性和信道滤波器的矩形系数
English Translation

A receiver’s ability to reject strong interference signals at a frequency twice the IF away from its operating frequency can be expressed by selectivity indicators, which are respectively:

  1. Aimage rejection ratio
  2. Bfront-end bandwidth
  3. Cin-band ripple and channel bandwidth
  4. Dchannel bandwidth, channel selectivity, and the rectangular coefficient of the channel filter
Correct answer: A
Knowledge Point Analysis 知识点解析

A signal at 2×IF from the operating frequency is the image frequency (镜像频率 = f ± 2×IF), so its rejection is the image rejection ratio (镜像抑制比). Option A is correct. This closes the four-part series (near = B/C, far = front-end, image = this).

Candidate Tips 考生提示

US–China difference: N/A — image rejection ratio is the universal metric for 2×IF spurious responses.

Common pitfall: Picking front-end bandwidth (B) — that handles “far” but not specifically the 2×IF image, which is the textbook image case.

Real on-air practice: Spec sheets quote image rejection in dB; a high first IF improves it mechanically.

ABCQ40 3.6.3 U0540
中文原题 Original (Chinese)

接收机灵敏度指标的数值大小具有什么意义?

  1. A灵敏度指标的数值越小,接收微弱信号的能力越强
  2. B灵敏度指标的数值越大,接收微弱信号的能力越强
  3. C灵敏度指标的数值越小,对与有用信号同时出现的干扰信号的响应越灵敏
  4. D灵敏度指标的数值越大,对与有用信号同时出现的干扰信号的响应越灵敏
English Translation

What is the significance of the magnitude of a receiver’s sensitivity indicator?

  1. AThe smaller the sensitivity value, the stronger the ability to receive weak signals
  2. BThe larger the sensitivity value, the stronger the ability to receive weak signals
  3. CThe smaller the sensitivity value, the more sensitive the response to interference appearing together with the useful signal
  4. DThe larger the sensitivity value, the more sensitive the response to interference appearing together with the useful signal
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: N/A — “lower number = more sensitive” holds for both μV/dBm sensitivity specs worldwide.

Common pitfall: Thinking bigger is better (B); sensitivity is an inverse measure.

Real on-air practice: A rig rated 0.1 μV is more sensitive than one rated 0.5 μV (all else equal).

ABCQ41 3.6.3 U0541
中文原题 Original (Chinese)

用功率电平表示接收机的灵敏度具有什么意义?常用单位是什么?

  1. A输出信号达到一定质量标准时输入信号的最小功率电平。单位是dBm或dBμW
  2. B输出信号达到一定质量标准时输出信号与输入信号的功率电平之比。单位是dB
  3. C输出信号达到一定质量标准时输入信号与输出信号的功率电平之比。单位是dB
  4. D输出信号维持一定质量标准时输入信号的最小和最大功率电平之比。单位是dB
English Translation

What is the significance of expressing a receiver’s sensitivity in terms of power level, and what are the common units?

  1. Athe minimum input-signal power level when the output signal meets a specified quality standard; the units are dBm or dBμW
  2. Bthe ratio of output-signal to input-signal power level when the output meets a specified quality standard; the unit is dB
  3. Cthe ratio of input-signal to output-signal power level when the output meets a specified quality standard; the unit is dB
  4. Dthe ratio of the minimum to maximum input-signal power level when the output maintains a specified quality standard; the unit is dB
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: N/A — dBm/dBμW sensitivity is standard; US specs also use dBm.

Common pitfall: Treating sensitivity as a gain ratio (B/C/D); sensitivity is an absolute minimum input level, not a ratio.

Real on-air practice: A typical VHF receiver sensitivity might be −120 dBm for 12 dB SINAD.

ABCQ42 3.6.3 U0542
中文原题 Original (Chinese)

用电压电平表示接收机的灵敏度具有什么意义?常用单位是什么?

  1. A输出信号达到一定质量标准时输入信号的最小电动势。单位是μV、dBμV或dBmV
  2. B输出信号达到一定质量标准时输入信号的最小电压。单位是μV、dBμV或dBmV
  3. C输出信号达到一定质量标准时输入信号的最大电动势。单位是μV或mV
  4. D输出信号达到一定质量标准时输出信号与输入信号的电压之比。单位是dB
English Translation

What is the significance of expressing a receiver’s sensitivity in terms of voltage level, and what are the common units?

  1. Athe minimum electromotive force of the input signal when the output signal meets a specified quality standard; the units are μV, dBμV, or dBmV
  2. Bthe minimum voltage of the input signal when the output signal meets a specified quality standard; the units are μV, dBμV, or dBmV
  3. Cthe maximum electromotive force of the input signal when the output signal meets a specified quality standard; the units are μV or mV
  4. Dthe ratio of the output-signal to input-signal voltage when the output signal meets a specified quality standard; the unit is dB
Correct answer: A
Knowledge Point Analysis 知识点解析

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).

Candidate Tips 考生提示

US–China difference: US specs usually quote sensitivity in μV (often meaning the applied signal voltage); the Chinese exam specifically uses “电动势” (EMF) terminology, so choose A.

Common pitfall: Picking B (“最小电压”) because it looks equivalent — the official key uses “电动势” (EMF), making A the marked answer.

Real on-air practice: A spec like “0.2 μV sensitivity” means the generator EMF needed for a given SINAD.

ABCQ43 3.6.3 U0543
中文原题 Original (Chinese)

业余无线电接收机大多具备信号强度指示功能。在VHF/UHF频段,信号强度的最小刻度S1与输入信号功率电平-141dBm(50Ω上的电压电平0.02μV)相一致;而在HF频段,S1则与-121dBm的输入信号(50Ω上的0.2μV)相对应。这是因为:

  1. AHF频段的背景噪声较VHF/UHF频段为高,使得可感知最小信号电平相应高出约20dB
  2. B由于技术原因,HF接收机的灵敏度都比VHF/UHF频段的低大约20dB
  3. CHF业余电台功率大一些,而VHF/UHF的相对较小,这导致信号强度的刻度标准差异
  4. DHF业余电台主要用于DX通信,而VHF/UHF的面向本地通信,刻度可以更随意一些
English Translation

Most amateur radio receivers have a signal-strength indication function. In the VHF/UHF bands, the minimum scale S1 corresponds to an input-signal power level of −141 dBm (a voltage level of 0.02 μV across 50 Ω); whereas in the HF band, S1 corresponds to an input signal of −121 dBm (0.2 μV across 50 Ω). This is because:

  1. Athe background noise in the HF band is higher than in the VHF/UHF bands, so the perceivable minimum signal level is correspondingly about 20 dB higher
  2. Bfor technical reasons, HF receivers all have sensitivity about 20 dB lower than VHF/UHF receivers
  3. CHF amateur stations have larger power while VHF/UHF ones are relatively smaller, causing the difference in signal-strength scale standards
  4. DHF amateur stations are mainly for DX communication while VHF/UHF are for local communication, so the scale can be more arbitrary
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: N/A — the S-meter/S1 and HF-noise-floor reasoning are the same; US hams also see weaker S1 references on VHF than HF.

Common pitfall: Attributing the difference to receiver sensitivity (B) rather than the external noise floor; the S-meter is referenced to usable signal above noise.

Real on-air practice: On 20 m at night the band noise itself may be S3–S5; on 2 m a quiet location shows S0/S1 with no signals.

ABCQ44 3.6.3 U0544
中文原题 Original (Chinese)

甲乙两种型号的业余无线电收发信机在技术规格中给出的接收机灵敏度指标分别为0.1 μV和0.15μV。关于该指标,正确的推论是:

  1. A无法比较二者接收微弱信号的能力,测量灵敏度时所用的输出信号质量标准未知
  2. B可以推断甲机接收微弱信号的能力比乙机的差,因为灵敏度指标的数值较小
  3. C可以推断甲机接收微弱信号的能力比乙机的高,因为可以接收的信号更微弱
  4. D可以推断甲机承受强信号的能力比乙机的低,因为其灵敏度数值比较小
English Translation

Two models of amateur transceivers, A and B, give receiver sensitivity indicators of 0.1 μV and 0.15 μV respectively in their specifications. The correct inference about this indicator is:

  1. Ait is impossible to compare their ability to receive weak signals, because the output-signal quality standard used when measuring sensitivity is unknown
  2. Bwe can infer that A’s ability to receive weak signals is worse than B’s, because the sensitivity value is smaller
  3. Cwe can infer that A’s ability to receive weak signals is better than B’s, because it can receive a weaker signal
  4. Dwe can infer that A’s ability to withstand strong signals is lower than B’s, because its sensitivity value is smaller
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: N/A — comparing sensitivities requires the same reference (e.g., 10 dB S/N or 12 dB SINAD); US buyers check the fine print too.

Common pitfall: Automatically saying “0.1 μV is better than 0.15 μV” (C) — without the common measurement standard the comparison is invalid.

Real on-air practice: Always read the footnotes: “0.1 μV at 10 dB S/N” vs “0.15 μV at 12 dB SINAD” are not directly comparable.

ABCQ45 3.6.3 U0545
中文原题 Original (Chinese)

当输出信号满足规定质量标准时,在输入阻抗为50欧的某接收机天线输入端口处测得的输入信号最小电压为1μV。如果用电压电平μV表示该机的灵敏度,则为:

  1. A2μV
  2. B1μV
  3. C50μV
  4. D0.5μV
English Translation

When the output signal meets the specified quality standard, the minimum input-signal voltage measured at the antenna input port of a receiver with 50 Ω input impedance is 1 μV. If the sensitivity is expressed in voltage level μV, it is:

  1. A2 μV
  2. B1 μV
  3. C50 μV
  4. D0.5 μV
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: N/A — quoting sensitivity as a μV figure is identical practice.

Common pitfall: Trying to convert with the 50 Ω (which would be needed only for dBm/dBμV); here the unit is plain μV, so the number stays 1 μV.

Real on-air practice: A spec “sensitivity 1 μV” means that input voltage yields the rated output quality.

ABCQ46 3.6.3 U0546
中文原题 Original (Chinese)

当输出信号满足规定质量标准时,在输入阻抗为50欧的某接收机天线输入端口处测得的输入信号最小电压为1μV。如果用功率电平dBm表示该机的灵敏度,则为:

  1. A-107dBm
  2. B-103dBm
  3. C-73dBm
  4. D-113dBm
English Translation

When the output signal meets the specified quality standard, the minimum input-signal voltage measured at the antenna input port of a receiver with 50 Ω input impedance is 1 μV. If the sensitivity is expressed in power level dBm, it is:

  1. A−107 dBm
  2. B−103 dBm
  3. C−73 dBm
  4. D−113 dBm
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: US specs often define sensitivity as the generator EMF (V²/4R), which would give −113 dBm here; the Chinese exam uses the V²/R voltage-level form, yielding −107 dBm. Follow the exam key (A).

Common pitfall: Using the 4R (EMF) formula and picking D (−113 dBm); the exam’s “电压电平” convention uses R, not 4R.

Real on-air practice: 1 μV into 50 Ω ≈ −107 dBm under this convention; handy for comparing μV and dBm specs.

ABCQ47 3.6.3 U0547
中文原题 Original (Chinese)

当输出信号满足规定质量标准时,在输入阻抗为50欧的某接收机天线输入端口处测得的输入信号最小电压为1μV。如果用电压电平dBμV表示该机的灵敏度,则为:

  1. A6dBμV
  2. B-2dBμV
  3. C3dBμV
  4. D0dBμV
English Translation

When the output signal meets the specified quality standard, the minimum input-signal voltage measured at the antenna input port of a receiver with 50 Ω input impedance is 1 μV. If the sensitivity is expressed in voltage level dBμV, it is:

  1. A6 dBμV
  2. B−2 dBμV
  3. C3 dBμV
  4. D0 dBμV
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: In standard US/ITU practice, 1 μV = 0 dBμV (option D); the Chinese exam key for this item is A (6 dBμV) — an inconsistency to note when studying from this bank.

Common pitfall: Assuming the straightforward 0 dBμV (D); the exam answer key is A, so memorize the key as given even though the math suggests D.

Real on-air practice: When converting specs, always double-check which reference the source uses; mismatches like this appear in some question banks.

ABCQ48 3.6.3 U0548
中文原题 Original (Chinese)

当输出信号满足规定质量标准时,在输入阻抗为50欧的某接收机天线输入端口处测得的输入信号最小电压为0.5μV。如果用电压电平μV表示该机的灵敏度,则为:

  1. A1μV
  2. B0.5μV
  3. C50μV
  4. D2.5μV
English Translation

When the output signal meets the specified quality standard, the minimum input-signal voltage measured at the antenna input port of a receiver with 50 Ω input impedance is 0.5 μV. If the sensitivity is expressed in voltage level μV, it is:

  1. A1 μV
  2. B0.5 μV
  3. C50 μV
  4. D2.5 μV
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: N/A — direct μV sensitivity is the same concept.

Common pitfall: Doubling or halving the number; the voltage-level μV figure is the measured voltage verbatim.

Real on-air practice: A 0.5 μV sensitivity figure is simply half of the 1 μV case’s input voltage.

ABCQ49 3.6.3 U0549
中文原题 Original (Chinese)

当输出信号满足规定质量标准时,在输入阻抗为50欧的某接收机天线输入端口处测得的输入信号最小电压为0.5μV。如果用功率电平dBm表示该机的灵敏度,则为:

  1. A-113 dBm
  2. B-107dBm
  3. C-103 dBm
  4. D-73 dBm
English Translation

When the output signal meets the specified quality standard, the minimum input-signal voltage measured at the antenna input port of a receiver with 50 Ω input impedance is 0.5 μV. If the sensitivity is expressed in power level dBm, it is:

  1. A−113 dBm
  2. B−107 dBm
  3. C−103 dBm
  4. D−73 dBm
Correct answer: A
Knowledge Point Analysis 知识点解析

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.

Candidate Tips 考生提示

US–China difference: N/A — under the exam’s V²/R convention the math is consistent; US EMF convention would shift both by +6 dB but the relationship holds.

Common pitfall: Reusing −107 dBm (B) from the 1 μV case; halving voltage drops sensitivity by 6 dB to −113 dBm.

Real on-air practice: Each halving of the μV figure is −6 dBm; a handy shortcut when comparing sensitivity specs.

ABCQ50 3.6.3 U0550
中文原题 Original (Chinese)

当输出信号满足规定质量标准时,在输入阻抗为50欧的某接收机天线输入端口处测得的输入信号最小电压为0.5μV。如果用电压电平dBμV表示该机的灵敏度,则为:

  1. A0dBμV
  2. B-2dBμV
  3. C3dBμV
  4. D6dBμV
English Translation

When the output signal meets the specified quality standard, the minimum input-signal voltage measured at the antenna input port of a receiver with a 50-ohm input impedance is 0.5 μV. If this receiver’s sensitivity is expressed as a voltage level in dBμV, it is:

  1. A0 dBμV
  2. B−2 dBμV
  3. C3 dBμV
  4. D6 dBμV
Correct answer: A
Knowledge Point Analysis 知识点解析

A receiver’s sensitivity is the minimum usable input signal; dBμV is a voltage level referenced to 1 μV (dBμV = 20·log10(V/1 μV)). The official answer key marks A. Note for study: 0.5 μV computes to 20·log10(0.5) ≈ −6 dBμV, which is not among the listed options; the marking follows the source key, so memorize the key’s choice rather than the raw arithmetic here.

Candidate Tips 考生提示

US–China difference: N/A — the dBμV sensitivity convention is the same internationally.

Common pitfall: Confusing dBμV (referenced to 1 μV) with dBm (referenced to 1 mW); they are different units.

Real on-air practice: A VHF receiver spec of “0.2 μV for 12 dB SINAD” is a typical sensitivity figure you will see.

ABCQ51 3.6.3 U0551
中文原题 Original (Chinese)

已知某接收机的天线输入阻抗为50Ω,灵敏度指标为2μV,则功率电平相应为:

  1. A-107dBm
  2. B0.02dBm
  3. C-103dBm
  4. D-113dBm
English Translation

Given a receiver whose antenna input impedance is 50 Ω and whose sensitivity specification is 2 μV, the corresponding power level is:

  1. A−107 dBm
  2. B0.02 dBm
  3. C−103 dBm
  4. D−113 dBm
Correct answer: A
Knowledge Point Analysis 知识点解析

Converting a voltage sensitivity to a power level uses P = V²/R and dBm = 10·log10(P/1 mW). The official answer key marks A. Study note: 1 μV into 50 Ω equals −107 dBm; 2 μV would be about −101 dBm, which is not among the options — the source key assigns −107 dBm, so learn the keyed answer.

Candidate Tips 考生提示

US–China difference: N/A — the dBm power-level convention is identical worldwide.

Common pitfall: Forgetting that a voltage must be squared and divided by impedance before converting to dBm.

Real on-air practice: Receiver sensitivity is often quoted both as μV and as dBm; knowing the −107 dBm = 1 μV @50 Ω anchor is handy.

ABCQ52 3.6.3 U0552
中文原题 Original (Chinese)

已知某接收机的天线输入阻抗为50Ω,灵敏度指标为1μV,则功率电平相应为:

  1. A-113dBm
  2. B-107dBm
  3. C0.01dBm
  4. D-103dBm
English Translation

Given a receiver whose antenna input impedance is 50 Ω and whose sensitivity specification is 1 μV, the corresponding power level is:

  1. A−113 dBm
  2. B−107 dBm
  3. C0.01 dBm
  4. D−103 dBm
Correct answer: A
Knowledge Point Analysis 知识点解析

The official answer key marks A. Study note: by the standard physics, 1 μV into 50 Ω gives P = (1 μV)²/50 Ω = 2×10⁻¹⁴ W, i.e. 10·log10(2×10⁻¹¹) ≈ −107 dBm — which is option B. The source key nonetheless marks A, so for the exam memorize the keyed answer; be aware the physically correct value is −107 dBm (B).

Candidate Tips 考生提示

US–China difference: N/A — the dBm reference is the same internationally.

Common pitfall: Mixing up the μV-to-dBm conversion; anchor on “1 μV @ 50 Ω = −107 dBm”.

Real on-air practice: When comparing receiver specs, convert everything to dBm so different manufacturers’ numbers are comparable.

ABCQ53 3.6.3 U0553
中文原题 Original (Chinese)

制约现代无线电接收机灵敏度的主要因素是:

  1. A机内噪声
  2. B放大电路的增益
  3. C放大电路的稳定性
  4. D电源噪声
English Translation

The main factor limiting the sensitivity of a modern radio receiver is:

  1. Ainternal (receiver) noise
  2. Bthe gain of the amplifier circuit
  3. Cthe stability of the amplifier circuit
  4. Dpower-supply noise
Correct answer: A
Knowledge Point Analysis 知识点解析

A modern receiver’s sensitivity is fundamentally limited by its own internal noise (机内噪声) — the thermal and device noise generated in the front end. No amount of gain (B) or stability (C) can recover a signal buried below the receiver’s noise floor; power-supply noise (D) is a lesser, separable issue.

Candidate Tips 考生提示

US–China difference: N/A — receiver noise floor limits sensitivity identically worldwide.

Common pitfall: Thinking more gain improves weak-signal reception; extra gain also amplifies noise.

Real on-air practice: On EME you add a low-noise preamp at the antenna to beat the receiver’s internal noise floor.

ABCQ54 3.6.3 U0554
中文原题 Original (Chinese)

术语”信噪比(SNR)”在业余无线电领域广为使用。它是指:

  1. A有用信号功率对噪声功率的比值
  2. B有用信号峰值电压对噪声峰值电压的比值
  3. C有用信号功率对有用信号功率及噪声功率之和的比值
  4. D有用信号峰值电压对有用信号峰值电压及噪声峰值电压之和的比值
English Translation

The term “signal-to-noise ratio (SNR)” is widely used in amateur radio. It refers to:

  1. Athe ratio of useful-signal power to noise power
  2. Bthe ratio of useful-signal peak voltage to noise peak voltage
  3. Cthe ratio of useful-signal power to the sum of useful-signal power and noise power
  4. Dthe ratio of useful-signal peak voltage to the sum of useful-signal peak voltage and noise peak voltage
Correct answer: A
Knowledge Point Analysis 知识点解析

信噪比 (SNR) is defined as the ratio of useful-signal power to noise power (A). It is a power ratio, not a peak-voltage ratio (B, D) nor the signal-over-total (signal+noise) fraction (C).

Candidate Tips 考生提示

US–China difference: N/A — SNR is defined the same way everywhere.

Common pitfall: Using voltage ratios or confusing SNR with the signal/(signal+noise) fraction.

Real on-air practice: You read an RST report’s “599” partly by judging the signal’s strength above the noise.

ABCQ55 3.6.3 U0555
中文原题 Original (Chinese)

对于需要接收微弱信号的业余通信,例如EME通联,接收机的噪声系数Fn(以比值的形式表示)是一项重要技术指标。它是指:

  1. A接收机输入端信噪比Si/Ni对输出端信噪比So/No的比值
  2. B接收机输入端无信号时,输出端的噪声功率电平
  3. C接收机输出端噪声功率电平与输入端噪声功率电平的比值
  4. D接收机输入端噪声功率电平与输出端噪声功率电平的比值
English Translation

For amateur communication that must receive very weak signals, such as EME contacts, the receiver noise figure Fn (expressed as a ratio) is an important technical indicator. It refers to:

  1. Athe ratio of the input signal-to-noise ratio Si/Ni to the output signal-to-noise ratio So/No
  2. Bthe output noise-power level when there is no signal at the receiver input
  3. Cthe ratio of the output noise-power level to the input noise-power level
  4. Dthe ratio of the input noise-power level to the output noise-power level
Correct answer: A
Knowledge Point Analysis 知识点解析

噪声系数 (noise figure, Fn) is the ratio of input SNR to output SNR (Fn = (Si/Ni)/(So/No)) (A). It measures how much the receiver degrades the signal-to-noise ratio. B is just an output noise reading; C/D are noise-power ratios, not the SNR-degradation definition.

Candidate Tips 考生提示

US–China difference: N/A — Fn is an ITU/IEEE standard definition used worldwide.

Common pitfall: Confusing noise figure (SNR ratio) with a simple output/input noise-power ratio.

Real on-air practice: A low-Fn preamp ahead of a higher-Fn receiver preserves the weak EME signal.

ABCQ56 3.6.3 U0556
中文原题 Original (Chinese)

接收机的静噪灵敏度是指:

  1. A能够使静噪电路退出静噪状态的射频信号最小输入电平
  2. B关闭静噪功能之后所能接收到的射频信号最小输入电平
  3. C带有静噪功能的接收机开启静噪功能后,按照灵敏度定义测得的灵敏度
  4. D带有静噪功能的接收机关闭静噪功能后,按照灵敏度定义测得的灵敏度
English Translation

The squelch sensitivity of a receiver is:

  1. Athe minimum input level of an RF signal that can cause the squelch circuit to come out of (release) the squelched state
  2. Bthe minimum input level of an RF signal that can be received after the squelch function is turned off
  3. Cthe sensitivity measured, per the sensitivity definition, for a receiver with a squelch function after the squelch is enabled
  4. Dthe sensitivity measured, per the sensitivity definition, for a receiver with a squelch function after the squelch is disabled
Correct answer: A
Knowledge Point Analysis 知识点解析

静噪灵敏度 (squelch sensitivity) is the smallest RF input that will open (release) the squelch (A). It is a squelch-opening threshold, distinct from ordinary sensitivity measured with squelch off (B, D) or a generic enabled-squelch measurement (C).

Candidate Tips 考生提示

US–China difference: N/A — squelch sensitivity is defined the same way.

Common pitfall: Equating squelch sensitivity with ordinary receiver sensitivity.

Real on-air practice: You set the squelch threshold so weak noise doesn’t open the speaker but real signals do.

ABCQ57 3.6.3 U0557
中文原题 Original (Chinese)

对于需要接收微弱信号的业余通信,例如EME通联,接收机的噪声系数Fn(以比值的形式表示)是一项重要技术指标。关于Fn的一些基本常识是:

  1. AFn一定大于1;在同样的灵敏度下,Fn越接近1越好
  2. BFn一定小于1;在同样的灵敏度下,Fn越接近0越好
  3. CFn一定大于1;在同样的灵敏度下,Fn越大越好
  4. DFn一定小于1;在同样的灵敏度下,Fn越接近1越好
English Translation

For amateur communication that must receive very weak signals, such as EME contacts, the receiver noise figure Fn (expressed as a ratio) is an important technical indicator. Some basic knowledge about Fn is:

  1. AFn is always greater than 1; at the same sensitivity, the closer Fn is to 1 the better
  2. BFn is always less than 1; at the same sensitivity, the closer Fn is to 0 the better
  3. CFn is always greater than 1; at the same sensitivity, the larger Fn the better
  4. DFn is always less than 1; at the same sensitivity, the closer Fn is to 1 the better
Correct answer: A
Knowledge Point Analysis 知识点解析

Because any real receiver degrades the SNR, Fn is always > 1, and a value closer to 1 means less added noise, hence better (A). Options claiming Fn < 1 (B, D) are impossible for a passive-or-active real receiver; "larger is better" (C) is wrong.

Candidate Tips 考生提示

US–China difference: N/A — Fn > 1 is a universal rule.

Common pitfall: Forgetting that Fn can never be below 1 for a real device.

Real on-air practice: A good VHF preamp might have Fn ≈ 1.2–1.5; an ideal one would be exactly 1.

ABCQ58 3.6.3 U0558
中文原题 Original (Chinese)

对于需要接收微弱信号的业余通信,例如EME通联,接收机的噪声指数NF(以对数表示)是一项重要技术指标。它是指:

  1. A接收机输入端信噪比Si/Ni对输出端信噪比So/No的比值的对数形式
  2. B接收机输入端无信号时,输出端的噪声功率电平的对数形式
  3. C接收机输出端噪声功率电平与输入端噪声功率电平的比值的对数形式
  4. D接收机输入端噪声功率电平与输出端噪声功率电平的比值的对数形式
English Translation

For amateur communication that must receive very weak signals, such as EME contacts, the receiver noise figure NF (expressed logarithmically) is an important technical indicator. It refers to:

  1. Athe logarithmic form of the ratio of the input signal-to-noise ratio Si/Ni to the output signal-to-noise ratio So/No
  2. Bthe logarithmic form of the output noise-power level when there is no signal at the receiver input
  3. Cthe logarithmic form of the ratio of the output noise-power level to the input noise-power level
  4. Dthe logarithmic form of the ratio of the input noise-power level to the output noise-power level
Correct answer: A
Knowledge Point Analysis 知识点解析

噪声指数 (noise figure, NF) is simply the noise figure expressed in decibels: NF = 10·log10(Fn) = 10·log10[(Si/Ni)/(So/No)] (A). It is the dB form of the SNR-degradation ratio; B/C/D describe unrelated noise readings.

Candidate Tips 考生提示

US–China difference: N/A — NF in dB is the standard worldwide.

Common pitfall: Treating NF as a plain output-noise ratio rather than 10·log10 of Fn.

Real on-air practice: A spec “NF 0.5 dB” means an excellent, very low-noise front end.

ABCQ59 3.6.3 U0559
中文原题 Original (Chinese)

对于需要接收微弱信号的业余通信,例如EME通联,接收机的噪声指数NF(以对数表示)是一项重要技术指标。关于NF的一些基本常识是:

  1. ANF一定大于0;在同样的灵敏度下,NF越接近0越好
  2. BNF一定小于0;在同样的灵敏度下,NF越接近0越好
  3. CNF一定大于1;在同样的灵敏度下,NF越大越好
  4. DNF一定处于0和1之间;在同样的灵敏度下,NF越接近1越好
English Translation

For amateur communication that must receive very weak signals, such as EME contacts, the receiver noise figure NF (expressed logarithmically) is an important technical indicator. Some basic knowledge about NF is:

  1. ANF is always greater than 0; at the same sensitivity, the closer NF is to 0 the better
  2. BNF is always less than 0; at the same sensitivity, the closer NF is to 0 the better
  3. CNF is always greater than 1; at the same sensitivity, the larger NF the better
  4. DNF is always between 0 and 1; at the same sensitivity, the closer NF is to 1 the better
Correct answer: A
Knowledge Point Analysis 知识点解析

Since Fn > 1, NF = 10·log10(Fn) is always > 0, and a value closer to 0 dB means less added noise, hence better (A). Options with NF < 0 (B) or NF > 1 in dB terms (C) or “between 0 and 1” (D) misstate the dB relationship.

Candidate Tips 考生提示

US–China difference: N/A — NF > 0 dB universally.

Common pitfall: Mixing up Fn (a ratio >1) with NF (its dB value >0).

Real on-air practice: An ideal receiver has NF = 0 dB; real ones are a fraction of a dB to several dB.

ABCQ60 3.6.3 U0560
中文原题 Original (Chinese)

在无线电通信领域中,描述信号源、放大器或接收机等设备或系统组件的内部噪声大小时常用”噪声温度”指标。以接收机为例,等效噪声温度Te的意义是:

  1. A接收机的内部噪声功率等于一个接在天线输入端的优质匹配电阻在产生相同的热噪声功率时该电阻所具有的绝对温度
  2. B接收机内部噪声在输出端的功率可以使一个接在输出端上的匹配电阻发热的相对温度
  3. C接收机内部噪声在输出端的功率可以使一个接在输出端上的匹配电阻升温的绝对温度
  4. D接收机信噪比符合技术指标时所要求的设备工作环境的温度
English Translation

In radio communication, the “noise temperature” indicator is often used to describe the internal-noise magnitude of signal sources, amplifiers, receivers, or other equipment/system components. Taking a receiver as an example, the meaning of equivalent noise temperature Te is:

  1. Athe absolute temperature at which a high-quality matched resistor connected at the antenna input would have to be, in order to produce the same thermal-noise power as the receiver’s internal noise power
  2. Bthe relative temperature to which a matched resistor connected at the output could be heated by the power of the receiver’s internal noise at the output
  3. Cthe absolute temperature to which a matched resistor connected at the output could be raised by the power of the receiver’s internal noise at the output
  4. Dthe equipment operating-environment temperature required for the receiver’s signal-to-noise ratio to meet the technical specification
Correct answer: A
Knowledge Point Analysis 知识点解析

等效噪声温度 (equivalent noise temperature, Te) represents the receiver’s internal noise as the physical temperature a matched input resistor would need to have to generate the same thermal noise power (A). It is an input-referred equivalent, not an output heating temperature (B, C) nor an ambient operating temperature (D).

Candidate Tips 考生提示

US–China difference: N/A — noise temperature (Kelvin) is an international concept.

Common pitfall: Thinking Te is a real temperature of the output or ambient environment.

Real on-air practice: Low-earth-orbit satellite systems quote antenna+system noise temperature in Kelvin to judge EME-grade receivers.

ABCQ61 3.6.3 U0561
中文原题 Original (Chinese)

不产生任何内部噪声的理想放大器或接收机的噪声系数Fn、噪声指数NF和噪声温度Te分别为:

  1. A1,0dB,0°K
  2. B0,0dB,-273°K
  3. C0,1dB,17°K
  4. D0,0dB,-275°K
English Translation

For an ideal amplifier or receiver that produces no internal noise, the noise figure Fn, noise figure NF, and noise temperature Te are respectively:

  1. A1, 0 dB, 0 °K
  2. B0, 0 dB, −273 °K
  3. C0, 1 dB, 17 °K
  4. D0, 0 dB, −275 °K
Correct answer: A
Knowledge Point Analysis 知识点解析

An ideal noiseless device adds no noise, so its input and output SNRs are equal: Fn = 1; NF = 10·log10(1) = 0 dB; and Te = 0 K (absolute zero) (A). Values of Fn = 0 (B, C, D) are impossible.

Candidate Tips 考生提示

US–China difference: N/A — the ideal values are universal.

Common pitfall: Writing Fn = 0; noise figure is a ratio that bottoms out at 1, not 0.

Real on-air practice: Cryogenically cooled preamps approach Te ≈ 0 K and NF ≈ 0 dB for moonbounce.

ABCQ62 3.6.3 U0562
中文原题 Original (Chinese)

假设一个用于卫星业余业务的天线放大器工作在标准温度(17℃)下,其输入端已连接良好匹配的天线。如果放大器所产生的内部噪声与输入的热噪声等效,则该放大器的噪声系数Fn、噪声指数NF和噪声温度Te分别为:

  1. A2,3dB,290°K
  2. B1,1dB,0°K
  3. C2,0dB,17°K
  4. D1,0dB,-273°K
English Translation

Suppose an antenna amplifier used for the amateur-satellite service operates at standard temperature (17 °C), with its input already connected to a well-matched antenna. If the internal noise produced by the amplifier is equivalent to the input thermal noise, then the amplifier’s noise figure Fn, noise figure NF, and noise temperature Te are respectively:

  1. A2, 3 dB, 290 °K
  2. B1, 1 dB, 0 °K
  3. C2, 0 dB, 17 °K
  4. D1, 0 dB, −273 °K
Correct answer: A
Knowledge Point Analysis 知识点解析

If the amplifier’s added noise equals the input thermal noise, the total output noise is doubled relative to the ideal, so Fn = 2; NF = 10·log10(2) ≈ 3 dB; and Te = (Fn−1)·T0 ≈ 1×290 K = 290 K (A). The “standard temperature” reference is T0 ≈ 290 K (≈17 °C), which is why Te lands at 290 K.

Candidate Tips 考生提示

US–China difference: N/A — the 290 K reference temperature is international.

Common pitfall: Confusing the 17 °C operating temperature with the 290 K noise-reference temperature.

Real on-air practice: A 3 dB NF preamp for satellite work is a common, modest-performance choice.

BCQ63 3.6.2 U0963
中文原题 Original (Chinese)

必要带宽(necessary bandwidth)是指:对给定的发射类别而言,其恰好足以保证在相应速率及在指定条件下具有所要求质量的信息传输的所需带宽。业余电台单边带话音通信SSB、低速莫尔斯电码通信CW、调频话音通信FM和残余边带业余电视VSB ATV的必要带宽分别是:

  1. A3000Hz、400Hz、12.5kHz、5MHz以上
  2. B3000Hz、400Hz、5MHz以上、12.5kHz
  3. C5MHz、3000Hz、400Hz、12.5kHz
  4. D12.5kHz、5MHz以上、400Hz、2700Hz
English Translation

The necessary bandwidth is: for a given emission class, the bandwidth just sufficient to ensure the required quality of information transmission at the corresponding rate and under the specified conditions. The necessary bandwidths of amateur single sideband (SSB) voice, low-speed Morse code (CW), frequency-modulation (FM) voice, and vestigial-sideband amateur television (VSB ATV) are respectively:

  1. A3000 Hz, 400 Hz, 12.5 kHz, above 5 MHz
  2. B3000 Hz, 400 Hz, above 5 MHz, 12.5 kHz
  3. C5 MHz, 3000 Hz, 400 Hz, 12.5 kHz
  4. D12.5 kHz, above 5 MHz, 400 Hz, 2700 Hz
Correct answer: A
Knowledge Point Analysis 知识点解析

The necessary bandwidth (必要带宽) for typical amateur emissions: SSB voice ≈ 3000 Hz (often 2.7 kHz channel), CW ≈ 400 Hz (narrow keying), FM voice ≈ 12.5 kHz (channel spacing), and vestigial-sideband ATV (残余边带业余电视) ≈ >5 MHz (wide video). The order in A matches SSB, CW, FM, ATV exactly.

Candidate Tips 考生提示

US–China difference: N/A — these necessary-bandwidth values follow the same ITU emission-designator logic used by the FCC; US FM repeaters also use 12.5 kHz / 25 kHz channels.

Common pitfall: Swapping FM (12.5 kHz) with ATV (>5 MHz) order (option B); ATV video is by far the widest.

Real on-air practice: On SSB you tune in 100 Hz steps; an FM repeater needs a 12.5 kHz-wide channel; ATV needs MHz of bandwidth, hence its own allocations.

BCQ64 3.6.2 U0964
中文原题 Original (Chinese)

以CW方式进行速度为25WPM的摩尔斯电码通信,如果考虑传播衰落等因素,必要带宽通常不大于:

  1. A200Hz
  2. B2700Hz
  3. C6.25kHz
  4. D12.5kHz
English Translation

For Morse-code (CW) communication at a speed of 25 WPM, accounting for propagation fading and similar factors, the necessary bandwidth is normally not greater than:

  1. A200 Hz
  2. B2700 Hz
  3. C6.25 kHz
  4. D12.5 kHz
Correct answer: A
Knowledge Point Analysis 知识点解析

CW (等幅电报 / continuous wave) at 25 WPM has a very narrow necessary bandwidth — roughly keying rate dependent, typically about 100–200 Hz. Allowing for fading and practical filter skirts, it is normally within 200 Hz. 2700 Hz is an SSB voice figure; 6.25/12.5 kHz are digital/FM figures.

Candidate Tips 考生提示

US–China difference: N/A — CW bandwidth ≈ keying bandwidth is universal; US CW contests likewise use <500 Hz filters.

Common pitfall: Reaching for the SSB 2700 Hz figure (B); CW is far narrower.

Real on-air practice: Many HF operators use 250–500 Hz CW filters; a 25 WPM QSO fits comfortably in 200 Hz.

BCQ65 3.6.2 U0965
中文原题 Original (Chinese)

用单边带方式进行RTTY通信,速度为50波特,频偏为170Hz,必要带宽通常不大于:

  1. A450Hz
  2. B200Hz
  3. C2700Hz
  4. D12.5kHz
English Translation

For RTTY communication using single sideband, at a speed of 50 baud with a frequency shift of 170 Hz, the necessary bandwidth is normally not greater than:

  1. A450 Hz
  2. B200 Hz
  3. C2700 Hz
  4. D12.5 kHz
Correct answer: A
Knowledge Point Analysis 知识点解析

RTTY (无线电传) at 50 baud / 170 Hz shift: the necessary bandwidth is approximately the shift plus twice the baud rate (170 + 2×50 ≈ 270 Hz) plus filter margins, commonly taken as about 450 Hz. So 450 Hz (A) is correct; 200 Hz is CW-like and too narrow, while 2700/12.5 kHz are voice/FM figures.

Candidate Tips 考生提示

US–China difference: N/A — the 170 Hz-shift / 45.45–50 baud RTTY standard and its ~500 Hz bandwidth are used by US HF digital operators too.

Common pitfall: Using CW’s 200 Hz (B) or SSB’s 2700 Hz (C); RTTY needs the mark/space shift plus baud-rate allowance.

Real on-air practice: On 14.080 MHz you tune RTTY in a ~500 Hz window; modern software like MMTTY decodes it cleanly.

BCQ66 3.6.2 U0966
中文原题 Original (Chinese)

用单边带方式进行话音通信并间或传输速度为50波特,频偏为70Hz的RTTY消息,必要带宽约为:

  1. A2700Hz
  2. B200Hz
  3. C6.25kHz
  4. D12.5kHz
English Translation

For single-sideband voice communication that intermittently also transmits RTTY messages at 50 baud with a 70 Hz shift, the necessary bandwidth is approximately:

  1. A2700 Hz
  2. B200 Hz
  3. C6.25 kHz
  4. D12.5 kHz
Correct answer: A
Knowledge Point Analysis 知识点解析

Since the channel is primarily SSB voice, the necessary bandwidth is governed by the voice component (~2700 Hz). The occasional narrow 50-baud/70 Hz RTTY (≈70 + 2×50 ≈ 170 Hz) easily fits inside the SSB passband, so the overall necessary bandwidth is about 2700 Hz, not the narrow RTTY figure. Hence A.

Candidate Tips 考生提示

US–China difference: N/A — an SSB voice channel’s 2.7–3 kHz bandwidth dominates; US phone patches with digital bursts work the same way.

Common pitfall: Letting the RTTY sub-text pull the answer down to 200 Hz (B); the channel must accommodate the voice, so it stays ~2700 Hz.

Real on-air practice: An SSB net that occasionally sends a short digital text still occupies the full SSB channel width.

BCQ67 3.6.2 U0967
中文原题 Original (Chinese)

用通常的调频方式进行话音通信,必要带宽约为:

  1. A12.5kHz
  2. B2700Hz
  3. C200Hz
  4. D6.25kHz
English Translation

For ordinary frequency-modulation (FM) voice communication, the necessary bandwidth is approximately:

  1. A12.5 kHz
  2. B2700 Hz
  3. C200 Hz
  4. D6.25 kHz
Correct answer: A
Knowledge Point Analysis 知识点解析

Conventional FM (调频) voice, with ±5 kHz deviation and 3 kHz audio, occupies about 12.5 kHz of necessary bandwidth (the standard narrowband FM channel). SSB voice is 2700 Hz (B), CW ~200 Hz (C); 6.25 kHz (D) is a narrower digital channel. So 12.5 kHz is correct for typical FM.

Candidate Tips 考生提示

US–China difference: Both countries use 12.5 kHz (or 25 kHz) narrowband FM channels on VHF/UHF; the US also mandates 12.5 kHz narrowband for many services.

Common pitfall: Using the SSB 2700 Hz figure (B); FM is much wider due to deviation.

Real on-air practice: A 144 MHz FM repeater uses 12.5 kHz (or 25 kHz) channel spacing — that is why repeaters are spaced 12.5/25 kHz apart.

BCQ68 3.6.2 U0968
中文原题 Original (Chinese)

商用业余单边带电台的语音信号通道为300-3000Hz,高音频部分衰减很大。如果希望对设备进行改装,以将信号带宽扩展到30-16000Hz,大大提升信号质量。可行的方案是:

  1. A放弃该目标,不应超越业务性质所允许的必要带宽最低值
  2. B配用广播级超高音质话筒
  3. C将原单边带信道的晶体滤波器更换为带宽更宽的滤波器
  4. D在话筒电路中增加分段可控音频均衡电路以提升话筒信号的高音分量
English Translation

A commercial amateur single-sideband transceiver has a voice channel of 300–3000 Hz, with the high audio frequencies strongly attenuated. If one wishes to modify the equipment to extend the signal bandwidth to 30–16000 Hz and greatly improve signal quality, the feasible approach is:

  1. Aabandon this goal — one should not exceed the minimum necessary bandwidth permitted by the nature of the service
  2. Bfit a broadcast-grade, ultra-high-fidelity microphone
  3. Creplace the original SSB channel’s crystal filter with a wider-bandwidth filter
  4. Dadd a segment-controlled audio equalizer in the microphone circuit to boost the high-frequency content of the mic signal
Correct answer: A
Knowledge Point Analysis 知识点解析

The necessary bandwidth (必要带宽) for the amateur service is limited by regulation; widening SSB to 30–16000 Hz would exceed the allowed bandwidth and cause harmful interference (有害干扰) to adjacent channels. Therefore the correct approach is to abandon the modification (A). Options B–D are technical tweaks that would still violate the bandwidth limit.

Candidate Tips 考生提示

US–China difference: Both the FCC (Part 97) and China’s rules limit emission bandwidth; US amateurs also must stay within the allocated bandwidth and may not broaden SSB beyond ~2.8 kHz.

Common pitfall: Being tempted by the “improve audio quality” engineering fixes (B/C/D) and missing the regulatory constraint on necessary bandwidth.

Real on-air practice: Wideband “Hi-Fi” SSB is frowned upon because it splatters into neighboring 3 kHz channels and irritates other operators.

BCQ69 3.6.4 U0969
中文原题 Original (Chinese)

现代收发信机大多基于DSP技术设计并制造。因主要功能由软件实现,便获得软件无线电(SDR)设备的称谓。相比模拟收发信机,SDR设备的优势是:

  1. A噪声更低,几乎仅包含数字系统的量化噪声
  2. B精度更高,具有更低的信号失真和更好的滤波性能
  3. C很少因元器件老化而出现参数漂移等软故障
  4. D处理信号的时间延迟较模拟设备更短
English Translation

Modern transceivers are mostly designed and built on DSP technology. Because their main functions are implemented in software, they are called software-defined radio (SDR) equipment. Compared with analog transceivers, the advantages of SDR equipment are: (Choose all that apply.)

  1. Alower noise, consisting almost only of the quantization noise of the digital system
  2. Bhigher precision, with lower signal distortion and better filtering performance
  3. Crarely suffer soft faults such as parameter drift due to component aging
  4. Dthe signal processing time delay is shorter than that of analog equipment
Correct answer: A, B, C
Knowledge Point Analysis 知识点解析

SDR (软件无线电) moves functions into software/DSP, giving lower noise (mainly quantization noise), higher precision with less distortion and better (arbitrary) filtering, and freedom from analog-component aging drift (A, B, C). Digital processing typically adds some latency rather than reducing it, so D is not a correct advantage per the key.

Candidate Tips 考生提示

US–China difference: N/A — SDR advantages (noise, precision, stability) are the same for US and Chinese operators using FlexRadio, SDRplay, etc.

Common pitfall: Including D — digital processing usually introduces latency, not less; keep to A, B, C.

Real on-air practice: An SDR like an SDRplay or Hermes gives you panadapter spectrum and clean filters that analog rigs can’t match.

BCQ70 3.6.4 U0970
中文原题 Original (Chinese)

SDR收发信机大多提供波段频谱显示器,其基本用途是:

  1. A观察波段内的信号活动,以快速切换频率与之通联
  2. B观察信号的强度、发射质量和频率组成,以向对方提供更精确的信号描述
  3. C观察信号的衰落情况,以快速了解当前工作波段的传播情况
  4. D观察信号的音色并向对方提供多频段话音频谱均衡器的调整建议
English Translation

Most SDR transceivers provide a band spectrum display; its basic uses are: (Choose all that apply.)

  1. Aobserve signal activity across the band, to quickly switch frequency and contact stations
  2. Bobserve the signal’s strength, transmission quality, and frequency composition, to give the other station a more precise signal description
  3. Cobserve the fading of signals, to quickly understand the current band’s propagation conditions
  4. Dobserve the timbre of signals and provide the other station with advice on adjusting a multi-band voice equalizer
Correct answer: A, B, C
Knowledge Point Analysis 知识点解析

The panadapter/spectrum display of an SDR serves to spot band activity and tune quickly (A), to assess a signal’s strength/quality/spectrum for reporting (B), and to watch fading as a propagation indicator (C). Option D describes subjective “timbre” and equalizer coaching, which is not a basic function of the display, so it is excluded.

Candidate Tips 考生提示

US–China difference: N/A — SDR panadapters are used identically by US and Chinese hams for spotting and propagation.

Common pitfall: Adding D; the spectrum display shows frequency/amplitude, not “timbre” advice.

Real on-air practice: You scan the 20 m panadapter, click a clear PSK31 trace, and read its bandwidth to give an accurate RST-style report.

BCQ71 3.6.4 U0971
中文原题 Original (Chinese)

有些SDR收发信机可以提供CW、RTTY或FT8等信号的编解码功能。这是因为:

  1. A利用设备内部的计算机或与设备相连的外部PC机,用软件实现了相应功能
  2. B制造商向第三方购买编解码模块,并将之插入了收发信机内部的相应插槽
  3. C通过云数据中心,经高速联网和AI识别获得了相应的编解码结果
  4. D一种屏保,通过3D动画或虚拟现实实现了正在解码的某种视觉效果
English Translation

Some SDR transceivers can provide encoding/decoding functions for signals such as CW, RTTY, or FT8. This is because:

  1. Ausing the computer inside the equipment or an external PC connected to it, the functions are implemented in software
  2. Bthe manufacturer bought codec modules from a third party and plugged them into the corresponding slots inside the transceiver
  3. Cthrough a cloud data center, via high-speed networking and AI recognition, the corresponding codec results are obtained
  4. Dit is a screensaver that realizes, through 3D animation or virtual reality, some visual effect of decoding in progress
Correct answer: A
Knowledge Point Analysis 知识点解析

By definition, an SDR (软件无线电) performs modulation/demodulation in software on an internal or attached computer, so it can add CW/RTTY/FT8 codec modes by software update rather than hardware. Options B (plug-in modules), C (cloud/AI), and D (screensaver) misdescribe how SDR codecs work.

Candidate Tips 考生提示

US–China difference: N/A — SDR codecs running on internal/external CPUs are universal (e.g., IC-7300, Elecraft, Flex).

Common pitfall: Imagining hardware modules (B) or cloud processing (C); the essence of SDR is software on a local processor.

Real on-air practice: An IC-7300 decodes CW/RTTY internally; for FT8 most hams run WSJT-X on a PC connected to the rig.

BCQ72 3.6.4 U0972
中文原题 Original (Chinese)

相比模拟收发信机,为什么多数SDR收发信机可以提供更加灵活的滤波功能?

  1. A数字滤波无需物理元件支持,只要设备的计算能力足够便可现场调节滤波特性
  2. BSDR收发信机大量内置机械滤波器组,以通过插值运算实现带宽的连续调节
  3. CSDR收发信机用移相法实现调制与解调。只需改变电阻便可改变信号的带宽
  4. DSDR收发信机运用高速联网和云数据中心,可对任意信号进行随心所欲的滤波
English Translation

Compared with analog transceivers, why can most SDR transceivers provide more flexible filtering?

  1. Adigital filtering needs no physical-component support; as long as the device’s computing power is sufficient, the filter characteristics can be adjusted on the fly
  2. BSDR transceivers contain large banks of mechanical filters and achieve continuous bandwidth adjustment through interpolation
  3. CSDR transceivers use the phase-shift method for modulation and demodulation; one only needs to change a resistor to alter the signal bandwidth
  4. DSDR transceivers use high-speed networking and cloud data centers to filter arbitrary signals at will
Correct answer: A
Knowledge Point Analysis 知识点解析

SDR filtering is implemented digitally, so filter shape and bandwidth are set in software and can be changed freely with adequate DSP computing power — no physical crystal/mechanical filters needed (A). B describes analog filter banks, C confuses phase-shift methods, and D invokes nonexistent cloud filtering.

Candidate Tips 考生提示

US–China difference: N/A — digital filtering flexibility is the same selling point of US SDR rigs.

Common pitfall: Thinking SDR still relies on mechanical filter banks (B) like old analog rigs.

Real on-air practice: You can switch an SDR from a 500 Hz CW filter to a 2.4 kHz SSB filter instantly in software, with adjustable shape factor.

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