CRAC Class C (2025) — Questions 1171–1180

U1119. What should you pay attention to when choosing an SWR meter for your FM station?

为自己的FM电台选择驻波比表应注意什么问题？

Exam tips · 考试提示

The practical criteria for an SWR meter (驻波比表) are that its frequency range and power range (频率范围和功率量程) cover your FM band and transmitter power. Extreme temperature ratings (B, C) are irrelevant for normal station use, and "constant-amplitude FM" (D) is not a meaningful selection criterion.

U1120. If you use an SWR meter to measure the output power of a transmitter, where should you place the meter?

如果用驻波比表测量发射机的输出功率，你应将仪表装在哪里？

Exam tips · 考试提示

To measure the transmitter's output power (not the antenna system), the SWR/power meter must be in the RF path right at the transmitter output — between the transmitter and the feed line (发射机与馈线之间). Placing it at the antenna feed point (D) measures power delivered there (after feed-line loss); B/C are on the DC supply, which is wrong.

U1121. An amplifier has a signal gain of 20 dB. This means:

一个放大器具有20dB的信号增益，其意义是：

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20 dB power gain = 10^(20/10) = 100×. Crucially, an amplifier does not create energy from the input — it draws energy from the power supply and delivers 100× the input's power to the load (A). B/C/D wrongly imply the input energy itself is "amplified/multiplied," which violates energy conservation (the extra energy comes from the supply).

U1122. When an RF signal passes through a certain circuit, a loss of 20 dB is produced. This lost energy:

射频信号通过某电路时产生了20dB的损耗。这部分被损耗的能量：

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By energy conservation, the 20 dB of lost power is not destroyed — it is transformed into heat (in resistances) or radiated away (leakage). B and D claim energy "disappears" (impossible); C says it returns to the source, which describes reflection, not dissipation loss.

U1123. The output signal power of a circuit is 100 times the input signal power. The gain of this circuit is:

某电路输出信号功率是输入信号功率的100倍。该电路的增益为：

Exam tips · 考试提示

Power gain in dB = 10·log₁₀(Pout/Pin) = 10·log₁₀(100) = 10×2 = 20 dB. Options B/C/D use wrong factors.

U1124. The output signal power of a circuit is 1,000,000 times the input signal power. The gain of this circuit is:

某电路输出信号功率是输入信号功率的100万倍。该电路的增益为：

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10·log₁₀(10⁶) = 10×6 = 60 dB. Multiplying the ratio by 10⁶ adds 60 dB, not 100 dB or absurd million-dB values.

U1125. The output signal power of a circuit is 5 times the input signal power. The gain of this circuit is approximately:

某电路输出信号功率是输入信号功率的5倍。该电路的增益约为：

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10·log₁₀(5) ≈ 10×0.699 = 6.99 ≈ 7 dB. B is half that (would be a voltage ratio misapplied); C confuses the ratio with dB; D uses 20·log.

U1126. The input signal power of a circuit is half the output signal power. The gain of this circuit is approximately:

某电路输入信号功率是输出信号功率的一半。该电路的增益约为：

Exam tips · 考试提示

Input = ½ output means output = 2× input. Gain = 10·log₁₀(2) ≈ 3.01 ≈ 3 dB (positive — the circuit provides gain). B is the inverse (a 2× loss); C/D are far off.

U1127. The output signal voltage of a circuit is 100 times the input signal voltage. The gain of this circuit is:

某电路输出信号电压是输入信号电压的100倍。该电路的增益为：

Exam tips · 考试提示

For a voltage ratio, gain in dB = 20·log₁₀(Vout/Vin) = 20·log₁₀(100) = 20×2 = 40 dB. B/C/D apply wrong factors (power formula or raw ratio).

U1128. The output signal voltage of a circuit is 10,000 times the input signal voltage. The gain of this circuit is: ("x^m" means "x to the power m")

某电路输出信号电压是输入信号电压的1万倍。该电路的增益为：（“x＾m”表示“x的m次方”）

Exam tips · 考试提示

Voltage gain = 20·log₁₀(10⁴) = 20×4 = 80 dB. Each factor of 10 in voltage is +20 dB, so 10⁴ → 80 dB. B/C/D wrongly treat the numeric value as dB.