CRAC Class C (2025) — Questions 741–750
Unofficial study reference only. Not affiliated with the FCC, NCVEC or CRAC — official sources prevail. 非官方学习参考，与 FCC、NCVEC、CRAC 无隶属关系；以官方发布为准。
U0523. In a superheterodyne amateur transceiver, the component responsible for suppressing image-frequency interference is:
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：
Exam tips · 考试提示
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.
U0524. 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:
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：
Exam tips · 考试提示
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.
U0525. 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:
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：
Exam tips · 考试提示
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.
U0526. 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:
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：
Exam tips · 考试提示
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).
U0527. 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:
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：
Exam tips · 考试提示
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.
U0528. The efficiency of an amateur radio transmitter means:
业余无线电发射机的效率是指：
Exam tips · 考试提示
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.
U0529. The efficiency of an amateur radio transmitter is always significantly below 1. The lost portion of energy:
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：
Exam tips · 考试提示
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.
U0530. 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:
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：
Exam tips · 考试提示
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.
U0531. 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:
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：
Exam tips · 考试提示
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.
U0532. 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:
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：
Exam tips · 考试提示
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.