CRAC Class C (2025) — Questions 1081–1090

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U1250. The hobbyist Xiao-qiang was experimenting with a circuit. When powered on, a nearby receiver picked up a bird-like sound; when the power was cut, the sound disappeared. What function does Xiao-qiang's circuit perform?

爱好者小强实验某个电路。接通电源时,近旁的收信机收到了 类似鸟鸣的声音。断开电源,鸟鸣消失。小强的电路具备哪种 功能?

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The circuit generates a signal on its own only when powered — that is 振荡 (oscillation). A self-excited oscillator radiates a tone (here warbling like a birdie) that the receiver picks up; removing power stops it. Beat/detection/mixing all require an incoming RF to act on, but here the circuit itself is the source.

U1251. Two signals of frequencies f1 and f2 passing through the same nonlinear element intermodulate with each other; the third-order intermodulation products may become false signals in a receiver or a kind of spurious output from a transmitter. The frequencies of the third-order intermodulation products are calculated as:

频率为f1、f2的两个信号流经同一个非线性元件会彼此互调, 其中的三阶互调产物可能成为收信机中的假信号或发信机输出 的一种杂散。三阶互调产物的频率可以计算为:

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Third-order intermodulation (三阶互调) products from two tones are at 2f1−f2, 2f1+f2, 2f2−f1, 2f2+f1 (the combinations whose order sums to 3). These are the ones that fall close to the wanted band and cause interference, so A is correct. B lists harmonics; C lists sum/difference (mixing, second-order); D lists fifth-order products.

U1252. The term "beat" (差拍) refers to a phenomenon: (Choose all that apply.)

术语“差拍”是指一种现象:

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差拍 (beat) is the difference-frequency product of two signals through a nonlinear/multiplicative stage (A correct). Mixing (变 频) and second-order intermodulation (二阶互调) are the same mathematical product (f1−f2 and f1+f2); only the context differs (B correct). C is too broad (not every whistle is a beat), and D describes harmonics, not a beat.

U1253. Amateur radio operators often use a frequency multiplier to obtain an output at an integer multiple of the input frequency. The operating principle of a frequency multiplier is:

业余无线电爱好者常用倍频器获取输入信号整倍频率的输出信 号。倍频器的工作原理为:

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A 倍频器 (frequency multiplier) works by driving an active device into nonlinearity so the output contains harmonics of the input, then a tuned circuit selects the desired harmonic (e.g., 2×, 3×) (A correct). B is wrong because a linear amplifier does not generate harmonics; C is wrong because a passive tuned circuit cannot create a harmonic that is not present; D is wrong because a crystal oscillator sets its own frequency, not a multiplication of the input.

U1254. Through which of the following circuits does a sinusoidal AC signal generate higher-order harmonics? (Choose all that apply.)

正弦交流信号通过下列哪些电路时会产生高次谐波?

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A diode rectifier (A), diode detector (B), and diode ring modulator (C) are all nonlinear and therefore create harmonics from a sine input. A passive 选频网络 (tuned network, D) only selects existing components; it cannot generate new harmonics, so it is incorrect.

U1255. Among the following modulation modes, those in which the modulated signal's amplitude remains constant are: (Choose all that apply.)

在下列调制方式中,已调信号的振幅恒定不变的是:

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All four are angle/digital constant-envelope modulations: FSK, FM, PM, and PSK all keep a constant amplitude (the information is in frequency or phase, not envelope), so all four are correct. This is why nonlinear (Class C/D) amplifiers can be used for them.

U1256. Among the following modulation modes, those in which applying a modulating signal of a specific frequency and specific amplitude can make the carrier component disappear are: (Choose all that apply.)

在下列调制方式中,施加特定频率和特定幅度的调制信号可使 载波分量消失的是:

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For angle modulation (FM and PM), the carrier amplitude coefficient is the Bessel function J0(β); at certain modulation indices β (specific frequency/deviation) the carrier term J0 goes to zero — a Bessel null — so the carrier component disappears (A, B correct). SSB/DSB suppress the carrier by design (not by a specific modulation frequency/amplitude), and OOK keeps a full carrier while keyed on, so C and D are not the intended answers here.

U1257. Among the following modulation modes, those in which applying a modulating signal of a specific frequency and specific amplitude can make one pair of sidebands disappear are: (Choose all that apply.)

在下列调制方式中,施加特定频率和特定幅度的调制信号可使 某一对边带消失的是:

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In angle modulation the sideband amplitudes are Bessel functions Jn(β); at certain modulation indices a specific sideband pair Jn vanishes (A, B correct). For AM the two sidebands are always symmetric and present whenever modulating (C wrong); for DSB both sidebands are always present (D wrong). Thus only FM and PM fit the statement.

U1258. Among the following modulation modes, the one in which the modulated signal's carrier phase remains constant is:

在下列调制方式中,已调信号的载波相位恒定不变的是:

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In 幅度调制 (AM), only the envelope amplitude varies with the message while the carrier phase stays constant (A correct). FM changes frequency (hence phase), PSK and FSK change phase/frequency, so their carrier phase is not constant (B/C/D wrong).

U1259. Among the following modulation modes, the one in which the modulated signal's carrier amplitude remains constant is:

在下列调制方式中,已调信号的载波幅度恒定不变的是:

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Per the examination key, 幅度调制 (AM) is the answer: in AM the carrier component itself retains a fixed amplitude and the information rides on the envelope, whereas FM/PM/PSK/FSK are treated here as angle/digital modulations whose "carrier" is not a fixed-amplitude reference component in the same sense. (Note: FM/PM/FSK/PSK are also constant-envelope in practice; this item keys on AM as the modulation whose carrier amplitude is the defining constant reference.)