💥 [EMERGENCY / 极限警报] 🚨 本站服务器正随“星舰”前往火星,距离升空: [ T-MINUS ---D --:--:--.-- ]

CRAC Bilingual Manual: Digital Logic Gates | 数字逻辑门 (22 questions)

CRAC Bilingual Manual › Part: Electrical Basics for Amateur Radio

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

22bilingual questions·Classes C

This section covers Digital Logic Gates with 22 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.

CQ1 4.5.1 U1299
中文原题 Original (Chinese)

具有两个输入端的与门(AND)是最简单的数字逻辑电路之一。如果这两个输入的组合分别为00、01、10、11,则对应的输出为:

  1. A0、0、0、1
  2. B0、1、1、1
  3. C0、1、0、1
  4. D1、0、1、0
English Translation

The two-input AND gate is one of the simplest digital logic circuits. If the two inputs are respectively 00, 01, 10, 11, the corresponding outputs are:

  1. A0, 0, 0, 1
  2. B0, 1, 1, 1
  3. C0, 1, 0, 1
  4. D1, 0, 1, 0
Correct answer: A
Knowledge Point Analysis 知识点解析

An AND gate outputs 1 only when both inputs are 1. So 00→0, 01→0, 10→0, 11→1, i.e. 0,0,0,1 (A). B is OR, C is XOR, D is a none-standard pattern.

Candidate Tips 考生提示

US–China difference: AND truth table is identical in US digital logic.

Common pitfall: Mixing up AND with OR/XOR output patterns.

Real on-air practice: An AND of PTT and “mic-present” enables transmit in firmware.

CQ2 4.5.1 U1300
中文原题 Original (Chinese)

具有两个输入端的或门(OR)是最简单的数字逻辑电路之一。如果这两个输入的组合分别为00、01、10、11,则对应的输出为:

  1. A0、1、1、1
  2. B0、1、1、0
  3. C0、1、0、1
  4. D1、0、1、0
English Translation

The two-input OR gate is one of the simplest digital logic circuits. If the two inputs are respectively 00, 01, 10, 11, the corresponding outputs are:

  1. A0, 1, 1, 1
  2. B0, 1, 1, 0
  3. C0, 1, 0, 1
  4. D1, 0, 1, 0
Correct answer: A
Knowledge Point Analysis 知识点解析

An OR gate outputs 1 if either input is 1: 00→0, 01→1, 10→1, 11→1, i.e. 0,1,1,1 (A). B is NOR-like, C is XOR, D is a non-standard pattern.

Candidate Tips 考生提示

US–China difference: OR truth table is the same worldwide.

Common pitfall: Forgetting that 11 gives 1 for OR (unlike XOR).

Real on-air practice: OR of two band-select lines can gate a shared filter bank.

CQ3 4.5.1 U1301
中文原题 Original (Chinese)

具有两个输入端的异或门(XOR)是最简单的数字逻辑电路之一。如果这两个输入的组合分别为00、01、10、11,则对应的输出为:

  1. A0、1、1、0
  2. B0、1、0、1
  3. C0、1、1、1
  4. D1、0、1、0
English Translation

The two-input XOR (exclusive-OR) gate is one of the simplest digital logic circuits. If the two inputs are respectively 00, 01, 10, 11, the corresponding outputs are:

  1. A0, 1, 1, 0
  2. B0, 1, 0, 1
  3. C0, 1, 1, 1
  4. D1, 0, 1, 0
Correct answer: A
Knowledge Point Analysis 知识点解析

An XOR gate outputs 1 when the inputs differ: 00→0, 01→1, 10→1, 11→0, i.e. 0,1,1,0 (A). B is XNOR, C is OR, D is non-standard.

Candidate Tips 考生提示

US–China difference: XOR truth table is universal.

Common pitfall: Confusing XOR (0,1,1,0) with XNOR (1,0,0,1).

Real on-air practice: XOR is used in balanced modulators to generate DSB.

CQ4 4.5.1 U1302
中文原题 Original (Chinese)

具有两个输入端的与非门(NAND)是最简单的数字逻辑电路之一。如果这两个输入的组合分别为00、01、10、11,则对应的输出为:

  1. A1、1、1、0
  2. B0、1、1、1
  3. C0、1、0、1
  4. D1、0、1、0
English Translation

The two-input NAND (NOT-AND) gate is one of the simplest digital logic circuits. If the two inputs are respectively 00, 01, 10, 11, the corresponding outputs are:

  1. A1, 1, 1, 0
  2. B0, 1, 1, 1
  3. C0, 1, 0, 1
  4. D1, 0, 1, 0
Correct answer: A
Knowledge Point Analysis 知识点解析

A NAND gate is the inversion of AND: 00→1, 01→1, 10→1, 11→0, i.e. 1,1,1,0 (A). B is OR, C is XOR, D is non-standard.

Candidate Tips 考生提示

US–China difference: NAND truth table is identical worldwide; NAND is a universal gate.

Common pitfall: Thinking NAND of 00 is 0 — it is 1 (only 11 gives 0).

Real on-air practice: NAND gates build the control logic in many CW keyers.

CQ5 4.5.1 U1303
中文原题 Original (Chinese)

具有两个输入端的或非门(NOR)是最简单的数字逻辑电路之一。如果这两个输入的组合分别为00、01、10、11,则对应的输出为:

  1. A1、0、0、0
  2. B0、1、1、1
  3. C0、1、0、1
  4. D1、0、1、0
English Translation

The two-input NOR (NOT-OR) gate is one of the simplest digital logic circuits. If the two inputs are respectively 00, 01, 10, 11, the corresponding outputs are:

  1. A1, 0, 0, 0
  2. B0, 1, 1, 1
  3. C0, 1, 0, 1
  4. D1, 0, 1, 0
Correct answer: A
Knowledge Point Analysis 知识点解析

A NOR gate is the inversion of OR: 00→1, 01→0, 10→0, 11→0, i.e. 1,0,0,0 (A). B is OR, C is XOR, D is non-standard.

Candidate Tips 考生提示

US–China difference: NOR truth table is the same everywhere; NOR is also a universal gate.

Common pitfall: Mixing NOR with OR (only 00 gives 1 for NOR).

Real on-air practice: NOR of two fault signals can force a hardware shutdown.

CQ6 4.5.1 U1304
中文原题 Original (Chinese)

具有两个输入端的异或非门(NXOR)是最简单的数字逻辑电路之一。如果这两个输入的组合分别为00、01、10、11,则对应的输出为:

  1. A1、0、0、1
  2. B0、1、1、1
  3. C1、1、0、0
  4. D0、1、1、0
English Translation

The two-input XNOR (exclusive-NOR / equivalence) gate is one of the simplest digital logic circuits. If the two inputs are respectively 00, 01, 10, 11, the corresponding outputs are:

  1. A1, 0, 0, 1
  2. B0, 1, 1, 1
  3. C1, 1, 0, 0
  4. D0, 1, 1, 0
Correct answer: A
Knowledge Point Analysis 知识点解析

An XNOR (equivalence) gate outputs 1 when inputs are equal: 00→1, 01→0, 10→0, 11→1, i.e. 1,0,0,1 (A). B is OR, C/D are non-standard.

Candidate Tips 考生提示

US–China difference: XNOR truth table is universal.

Common pitfall: Confusing XNOR (1,0,0,1) with XOR (0,1,1,0).

Real on-air practice: XNOR can detect when two digital mode indicators agree.

CQ7 4.5.1 U1305
中文原题 Original (Chinese)

D触发器是一种时序逻辑电路,具有数据输入端D、时钟输入端CK和数据输出端Q。假设D触发器由CK的上升沿触发,则其工作情况为:

  1. A每次CK上升沿到来时,Q端都变得与D端一致
  2. B每次CK上升沿到来时,Q端都与D端反相
  3. C每次CK下降沿到来时,Q端都与D端反相
  4. D每次CK下降沿到来时,Q端都变得与D端一致
English Translation

A D flip-flop is a sequential logic circuit with data input D, clock input CK, and data output Q. Suppose the D flip-flop is triggered by the rising edge of CK; its operation is:

  1. Aeach time a CK rising edge arrives, Q becomes equal to D
  2. Beach time a CK rising edge arrives, Q is inverted from D
  3. Ceach time a CK falling edge arrives, Q is inverted from D
  4. Deach time a CK falling edge arrives, Q becomes equal to D
Correct answer: A
Knowledge Point Analysis 知识点解析

A rising-edge-triggered D flip-flop copies D to Q on each rising clock edge (A). B would be a D with inversion (not a plain D); C/D describe falling-edge behavior, which contradicts the stated rising-edge trigger.

Candidate Tips 考生提示

US–China difference: D-flip-flop edge behavior is identical in US digital design.

Common pitfall: Ignoring the trigger edge (rising vs falling) stated in the question.

Real on-air practice: A D flip-flop in your rig samples a digital mode bit on the clock edge.

CQ8 4.5.1 U1306
中文原题 Original (Chinese)

D触发器是一种时序逻辑电路,具有数据输入端D、时钟输入端CK和数据输出端Q。若将D触发器的Q端反相之后连到D端,则在每次CK上升沿到来时会发生什么?

  1. A2分频;若CK的频率为f,则Q端输出信号的频率为f/2
  2. B2倍频;若CK的频率为f,则Q端输出信号的频率为2f
  3. C移相:Q端输出信号与CK相差90°
  4. D反相:Q端输出信号与CK相差180°
English Translation

A D flip-flop has data input D, clock input CK, and data output Q. If the inverted Q is fed back to D, what happens on each CK rising edge?

  1. Afrequency divide by 2; if CK frequency is f, the Q output frequency is f/2
  2. Bfrequency multiply by 2; if CK frequency is f, the Q output frequency is 2f
  3. Cphase shift: the Q output is 90° from CK
  4. Dinversion: the Q output is 180° from CK
Correct answer: A
Knowledge Point Analysis 知识点解析

Feeding /Q back to D makes the flip-flop toggle every clock edge, so Q changes once per full CK period — a divide-by-2 (A). It cannot multiply frequency (B); the output is a 50% square wave at f/2, not a fixed 90°/180° phase of CK (C/D).

Candidate Tips 考生提示

US–China difference: The toggle-flip-flop divider is a standard US building block.

Common pitfall: Thinking feedback of /Q multiplies rather than divides the clock.

Real on-air practice: Cascaded /Q-feedback D flip-flops make the clock dividers in your frequency counter.

CQ9 4.5.1 U1307
中文原题 Original (Chinese)

制作SDR作品,特别是在制作直接射频采样的数字收发信机时,我们会使用一种叫做FPGA的现场可编程数字逻辑器件。FPGA在SDR系统中可以用来:

  1. A实现宽带数字混频、数字滤波、信道化样本抽取或采样率变换
  2. B在器件内创建CPU,或利用器件的内置CPU完整实现SDR系统
  3. C提供并行运算功能,辅助提升外部计算机系统的信号处理能力
  4. D充当固态硬盘,帮助外部计算机系统保存电台日志和比赛录音
English Translation

When building SDR (software-defined radio) projects, especially direct-RF-sampling digital transceivers, we use a field-programmable logic device called an FPGA. In an SDR system, an FPGA can be used to: (Choose all that apply.)

  1. Aimplement wideband digital mixing, digital filtering, channelized sample decimation, or sample-rate conversion
  2. Bcreate a CPU inside the device, or use the device’s built-in CPU to fully implement the SDR system
  3. Cprovide parallel computing functions, assisting in improving the signal-processing capability of an external computer system
  4. Dact as a solid-state drive, helping the external computer system save station logs and contest recordings
Correct answer: A, B, C
Knowledge Point Analysis 知识点解析

An FPGA provides reconfigurable parallel hardware: it does the DSP (mixing, filtering, decimation/rate conversion) (A), can embed a soft-CPU to run the SDR (B), and offloads parallel work from a host PC (C). D (acting as storage/SSD) is not an FPGA function, so it is not selected.

Candidate Tips 考生提示

US–China difference: US SDR rigs (e.g., Hermes/Angelia) likewise use FPGAs for front-end DSP.

Common pitfall: Treating an FPGA as storage — it is logic fabric, not memory.

Real on-air practice: The FPGA in your SDR does the DDC so the PC only handles the demodulated audio.

CQ10 4.5.2 U1308
中文原题 Original (Chinese)

实现数字信号处理(DSP)的第一步是将模拟信号转换为数字信号。为确保模拟信号的各频率成分不会在采集的样本中发生混叠,模数转换器(ADC)的采样率必须高于:

  1. A输入信号中的最高频率分量的2倍
  2. B输入信号的波形复杂程度
  3. C输入信号中的最低频率分量的1/2
  4. D输入信号的噪声相对带宽
English Translation

The first step of digital signal processing (DSP) is converting the analog signal to a digital signal. To ensure the frequency components of the analog signal do not alias in the captured samples, the sampling rate of the analog-to-digital converter (ADC) must be higher than:

  1. Atwice the highest frequency component in the input signal
  2. Bthe complexity of the input signal’s waveform
  3. Cone half of the lowest frequency component in the input signal
  4. Dthe noise relative bandwidth of the input signal
Correct answer: A
Knowledge Point Analysis 知识点解析

This is the Nyquist–Shannon sampling theorem: to avoid aliasing, the ADC sample rate must exceed 2× the highest input frequency (A). B/C/D are unrelated quantities.

Candidate Tips 考生提示

US–China difference: The Nyquist rule is taught identically in US DSP courses.

Common pitfall: Applying the 2× rule to the lowest frequency instead of the highest.

Real on-air practice: Sampling a 30 MHz IF at >60 MHz (with an anti-alias filter) keeps the SDR clean.

CQ11 4.5.2 U1309
中文原题 Original (Chinese)

实现数字信号处理 (DSP) 的第一步是将模拟信号转换为数字信号。所用模数转换器(ADC)的量化精度取决于:

  1. A输入信号的波形复杂程度
  2. B输入信号中的最高频率分量的2倍
  3. C输入信号中的最低频率分量的1/2
  4. D输入信号的噪声相对带宽
English Translation

The first step of digital signal processing (DSP) is converting the analog signal to a digital signal. The quantization precision of the analog-to-digital converter (ADC) used depends on:

  1. Athe complexity of the input signal’s waveform
  2. Btwice the highest frequency component in the input signal
  3. Cone half of the lowest frequency component in the input signal
  4. Dthe noise relative bandwidth of the input signal
Correct answer: A
Knowledge Point Analysis 知识点解析

Quantization precision (the ADC’s bit depth) sets the amplitude resolution; the required precision is governed by the signal’s dynamic range / waveform complexity (how finely you must resolve amplitude detail), so A is marked. Note that the ADC’s inherent resolution is a fixed hardware property, while B (2× highest frequency) is the separate Nyquist sampling-rate requirement, not quantization precision.

Candidate Tips 考生提示

US–China difference: ADC bits vs sample rate are distinct concepts in US SDR design too.

Common pitfall: Conflating quantization precision (bits) with the sampling-rate (Nyquist) rule from the previous question.

Real on-air practice: A 16-bit ADC captures weak-signal detail that an 8-bit one would quantize away.

CQ12 4.5.2 U1310
中文原题 Original (Chinese)

进行数模转换时,为了将模拟信号正确还原出来,数模转换器(DAC)的采样率应当至少等于输入样本率。但是如果在数模转换时通过插值获得了一个更高的采样率,则:

  1. A数模转换的频率响应将变得更为平坦
  2. B数模转换之后残存的杂散更容易滤除
  3. C采样时丢失的波形细节会得以再现
  4. D因采样不完美而残存的混叠会复原为频率正确的有用信号
English Translation

During digital-to-analog conversion, to correctly reconstruct the analog signal the DAC’s sample rate should be at least the input sample rate. But if interpolation is used in the DAC to obtain a higher sample rate, then: (Choose all that apply.)

  1. Athe DAC’s frequency response becomes flatter
  2. Bthe residual spurs after DAC are easier to filter out
  3. Cwaveform details lost during sampling will be recovered
  4. Dresidual aliasing from imperfect sampling will be restored to correctly-frequency useful signals
Correct answer: A, B
Knowledge Point Analysis 知识点解析

Oversampling/interpolation in a DAC pushes the image spurs to higher frequencies, giving a flatter in-band response (A) and making the residual images easier to remove with a simpler post-filter (B). C and D are impossible — interpolation cannot recover information already lost at sampling or undo aliasing.

Candidate Tips 考生提示

US–China difference: Oversampling DACs are standard in US SDR transmit paths for the same reasons.

Common pitfall: Believing oversampling recovers lost detail or fixes aliasing — it only eases filtering.

Real on-air practice: Your SDR’s transmit DAC oversamples so a gentle LPF cleans up the analog output.

CQ13 4.5.2 U1311
中文原题 Original (Chinese)

将频率为f且最大电压为U的模拟信号转换为数字信号。如果数字信号需要反映相当于0.1%U的信号幅度细节和相当于f的8次谐波的时间细节,则模数转换器应满足要求:

  1. A采样率大于16f,量化精度不低于10位
  2. B采样率大于8f,量化精度不低于10位
  3. C采样率大于4f,量化精度不低于8位
  4. D采样率大于10f,量化精度不低于9位
English Translation

Convert an analog signal of frequency f and maximum voltage U into a digital signal. If the digital signal must reflect amplitude detail equivalent to 0.1% of U and time detail equivalent to the 8th harmonic of f, the ADC must satisfy:

  1. Asample rate greater than 16f, quantization precision no lower than 10 bits
  2. Bsample rate greater than 8f, quantization precision no lower than 10 bits
  3. Csample rate greater than 4f, quantization precision no lower than 8 bits
  4. Dsample rate greater than 10f, quantization precision no lower than 9 bits
Correct answer: A
Knowledge Point Analysis 知识点解析

To capture the 8th harmonic (8f) without aliasing, Nyquist needs a sample rate > 2×8f = 16f. To resolve 0.1% = 1/1000 of full scale you need at least 10 bits (2^10 = 1024 levels ≈ 0.1%). So A is correct; B samples too slowly for the 8th harmonic, C/D have insufficient rate or resolution.

Candidate Tips 考生提示

US–China difference: Same Nyquist + bit-resolution arithmetic used in US SDR spec’ing.

Common pitfall: Setting the sample rate at just 2× the fundamental f instead of 2× the highest harmonic (8f).

Real on-air practice: Sampling a 1 kHz tone plus its 8th harmonic needs >16 kHz and ~10 bits for clean capture.

CQ14 4.5.3 U1312
中文原题 Original (Chinese)

编写SDR软件时,爱好者们经常使用一种中心频率为0Hz的I/Q序列,利用正、负两个频率区域来完整表达信号的基带信息。这种形式的数字信号也叫”复信号”,其优点是:

  1. A将序列的样本率降至最低,节省信号处理的资源开销
  2. B简化信号的幅度和相位运算,降低软件的复杂性
  3. C便于实现复杂数字调制和常见模拟调制
  4. D获取射频输入或还原射频输出仅需一个ADC或DAC
English Translation

When writing SDR software, hams often use an I/Q sequence centered at 0 Hz, using both positive and negative frequency regions to fully express the signal’s baseband information. This form of digital signal is also called a “complex signal”; its advantages are: (Choose all that apply.)

  1. Ait lowers the sequence’s sample rate to the minimum, saving signal-processing resource overhead
  2. Bit simplifies the signal’s amplitude and phase computation, reducing software complexity
  3. Cit facilitates implementation of complex digital modulation and common analog modulation
  4. Dacquiring the RF input or restoring the RF output needs only one ADC or DAC
Correct answer: A, B, C
Knowledge Point Analysis 知识点解析

I/Q complex (baseband) representation centers the spectrum at 0 Hz, so the sample rate need only cover the signal bandwidth (not the RF carrier) (A); magnitude/phase become simple complex math (B); and any modulation is easy to synthesize/analyze (C). D is false — you still need quadrature mixing (two paths or an RF front end) to get/emit the actual RF, not a single ADC/DAC at RF.

Candidate Tips 考生提示

US–China difference: I/Q complex baseband is the standard SDR representation in US software (GNU Radio, etc.).

Common pitfall: Thinking one ADC/DAC at the antenna can directly produce I/Q RF — quadrature down/up-conversion is still needed.

Real on-air practice: Your SDR shows a 0 Hz-centered waterfall; tuning just shifts the complex LO.

CQ15 4.5.3 U1313
中文原题 Original (Chinese)

SDR收发信机的主要功能均由数字信号处理算法实现,包括:

  1. A混频
  2. B滤波
  3. C调制
  4. D解调
English Translation

In an SDR transceiver, the main functions are all implemented by digital signal-processing algorithms, including: (Choose all that apply.)

  1. Amixing
  2. Bfiltering
  3. Cmodulation
  4. Ddemodulation
Correct answer: A, B, C, D
Knowledge Point Analysis 知识点解析

In an SDR, mixing, filtering, modulation, and demodulation are all performed in software/DSP rather than with analog hardware, so A, B, C, and D are all correct.

Candidate Tips 考生提示

US–China difference: US SDRs (FlexRadio, SDRplay, etc.) implement exactly these functions in DSP.

Common pitfall: Assuming some of these (e.g., mixing) still require analog stages in an SDR.

Real on-air practice: Switching your SDR from SSB to FM just loads a different demod algorithm.

CQ16 4.5.3 U1314
中文原题 Original (Chinese)

既然模拟滤波器早已支持业余无线电通信成功运作,我们为什么还要研发使用数字滤波器?

  1. A数字滤波器最灵活,可以定制出各种通带形状并保持最优性能
  2. B数字滤波器可用作高级接收机的高性能波段预选器
  3. C数字滤波器不存在电容击穿或电感过热问题,功率耐受能力强
  4. D数字滤波器无需数学运算单元的支持,运作成本低
English Translation

Since analog filters already support amateur radio communication successfully, why do we still develop and use digital filters?

  1. Adigital filters are the most flexible; they can be customized into various passband shapes while maintaining optimal performance
  2. Bdigital filters can serve as high-performance band preselectors in advanced receivers
  3. Cdigital filters have no capacitor breakdown or inductor overheating, so they tolerate high power
  4. Ddigital filters need no math-computation-unit support, so they have low operating cost
Correct answer: A
Knowledge Point Analysis 知识点解析

The key advantage of digital filters is flexibility — arbitrary passband shapes with repeatable, optimal performance, easily changed in software (A). B is not their role (HF preselection needs analog front ends); C is false (digital filters handle no RF power); D is false (they require DSP/math hardware).

Candidate Tips 考生提示

US–China difference: US SDRs emphasize digital-filter flexibility for band-pass shape and roofing filters.

Common pitfall: Thinking a digital filter can do high-power RF preselection — that stays analog.

Real on-air practice: Your SDR lets you dial in a 500 Hz CW filter or a 2.4 kHz SSB filter instantly.

CQ17 4.5.4 U1315
中文原题 Original (Chinese)

在SDR系统中,对I/Q信号进行混频可使用算法:

  1. A复数乘法
  2. B复数加法
  3. C复数减法
  4. D复数开方
English Translation

In an SDR system, mixing (frequency translation) of I/Q signals can be done using the algorithm:

  1. Acomplex multiplication
  2. Bcomplex addition
  3. Ccomplex subtraction
  4. Dcomplex square root
Correct answer: A
Knowledge Point Analysis 知识点解析

Frequency translation of complex (I/Q) baseband signals is performed by complex multiplication with a complex exponential (A). Addition/subtraction just shift levels; square root is unrelated to mixing.

Candidate Tips 考生提示

US–China difference: Complex multiply for frequency shift is standard in US SDR math (GNU Radio “multiply by CW”).

Common pitfall: Thinking simple addition shifts frequency — only multiplication rotates the phase.

Real on-air practice: Tuning your SDR is just multiplying the I/Q stream by e^{j2πft}.

CQ18 4.5.4 U1316
中文原题 Original (Chinese)

在SDR系统中,解调AM复信号可使用算法:

  1. A求复数的模值
  2. B求复数的辐角
  3. C求复数的余弦
  4. D求复数的指数
English Translation

In an SDR system, demodulating an AM complex signal can be done using the algorithm:

  1. Acompute the magnitude (modulus) of the complex number
  2. Bcompute the argument (phase angle) of the complex number
  3. Ccompute the cosine of the complex number
  4. Dcompute the exponential of the complex number
Correct answer: A
Knowledge Point Analysis 知识点解析

AM’s envelope equals the magnitude of the complex baseband sample, so demodulation is the complex modulus |I + jQ| (A). The argument/phase (B) is used for FM/PM; cosine/exponential are not the AM detector.

Candidate Tips 考生提示

US–China difference: Magnitude-detection of I/Q is the standard SDR AM demod in US software.

Common pitfall: Using the phase (argument) for AM — that recovers angle-modulated, not amplitude-modulated, info.

Real on-air practice: Your SDR’s AM mode just takes the envelope magnitude of each I/Q sample.

CQ19 4.5.4 U1317
中文原题 Original (Chinese)

SDR系统中的数字滤波器具有多种类型。常见的有:

  1. A有限冲激响应(FIR)滤波器
  2. B无限冲激响应(IIR)滤波器
  3. C萨伦-基滤波器
  4. D状态变量滤波器
English Translation

Digital filters in an SDR system come in several types. The common ones are: (Choose all that apply.)

  1. Afinite impulse response (FIR) filter
  2. Binfinite impulse response (IIR) filter
  3. CSallen–Key filter
  4. Dstate-variable filter
Correct answer: A, B
Knowledge Point Analysis 知识点解析

The two mainstream digital-filter structures are FIR (finite impulse response) (A) and IIR (infinite impulse response) (B). C (Sallen–Key) and D (state-variable) are analog active-filter topologies, not digital-filter types, so they are not selected.

Candidate Tips 考生提示

US–China difference: FIR/IIR are the digital-filter classes in every US SDR toolkit.

Common pitfall: Listing analog filter types (Sallen–Key, state-variable) as digital filters.

Real on-air practice: Your SDR’s roofing filter is typically an FIR implemented in the FPGA.

CQ20 4.5.4 U1318
中文原题 Original (Chinese)

制作SDR发信机时,生成SSB信号的算法可以是:

  1. A滤波法
  2. B移相法
  3. C维弗法
  4. D限幅法
English Translation

When building an SDR transmitter, the algorithms that can generate an SSB (single sideband) signal are: (Choose all that apply.)

  1. Athe filter method
  2. Bthe phasing method
  3. Cthe Weaver method
  4. Dthe limiting method
Correct answer: A, B, C
Knowledge Point Analysis 知识点解析

SSB can be generated by the filter method (A), the phasing method (B), and the Weaver method (C) — all common in SDR. The limiting method (D) is used for FM/improving carrier purity, not for producing SSB, so it is not selected.

Candidate Tips 考生提示

US–China difference: Filter/phasing/Weaver SSB generation is standard in US SDR transmit chains.

Common pitfall: Including “limiting” as an SSB method — limiting is for FM/keying, not sideband selection.

Real on-air practice: Most SDRs use the filter (or Weaver) method to put your voice on 14.070 MHz USB.

CQ21 4.5.4 U1319
中文原题 Original (Chinese)

在SDR系统中,快速傅里叶变换(FFT)可以用来:

  1. A分析信号的频率成分
  2. B生成FIR滤波器系数
  3. C实现调制和解调功能
  4. D实现快速卷积运算
English Translation

In an SDR system, the fast Fourier transform (FFT) can be used to: (Choose all that apply.)

  1. Aanalyze the frequency components of a signal
  2. Bgenerate FIR filter coefficients
  3. Cimplement modulation and demodulation functions
  4. Dimplement fast convolution operations
Correct answer: A, B, C, D
Knowledge Point Analysis 知识点解析

The FFT is used for spectrum analysis (A), designing FIR coefficients via frequency sampling (B), OFDM/FFT-based modulation and demodulation (C), and fast convolution (overlap-save/add) (D). All four are valid FFT applications in SDR.

Candidate Tips 考生提示

US–China difference: FFT use for panadapter, filtering, and OFDM is identical in US SDR software.

Common pitfall: Thinking the FFT is only for “looking at” the spectrum — it underpins filtering and modulation too.

Real on-air practice: The waterfall on your SDR screen is an FFT of the incoming I/Q samples.

CQ22 4.5.4 U1320
中文原题 Original (Chinese)

使用SDR收发信机时,我们常在系统设置中看到与频谱显示有关的窗函数选项,例如汉明、升余弦或布莱克曼等。这些主要用来:

  1. A降低FFT分析的谱泄漏,呈现信号功率谱的原本特征
  2. B提高FFT分析的幅度或频率精度,从而提升观测精度
  3. C防止FFT运算的结果超出屏幕上频谱显示窗口的宽度
  4. D提升CW或RTTY等幅度或频率键控信号的观测精度
English Translation

When using an SDR transceiver, we often see window-function options related to the spectrum display in the system settings, such as Hamming, raised cosine, or Blackman. These are mainly used to: (Choose all that apply.)

  1. Areduce spectral leakage in the FFT analysis and reveal the true characteristics of the signal’s power spectrum
  2. Bimprove the amplitude or frequency precision of the FFT analysis, thereby enhancing observation accuracy
  3. Cprevent the result of the FFT computation from exceeding the width of the spectrum display window on the screen
  4. Dimprove the observation precision of amplitude- or frequency-keyed signals such as CW or RTTY
Correct answer: A, B
Knowledge Point Analysis 知识点解析

A window function (窗函数), such as Hamming (汉明), raised cosine (升余弦), or Blackman (布莱克曼), is multiplied with the sampled data before performing the FFT (a fast algorithm for the discrete Fourier transform). Its primary purpose is to taper the data block and reduce spectral leakage (谱泄漏) caused by the discontinuity at the block edges, so the true shape of the signal’s power spectrum is revealed (A). A good window also sharpens the measured peaks, improving the precision of amplitude and frequency estimates and thus the observation accuracy (B). C is wrong because windowing does not change the display-window width — that is a screen/span setting; D is wrong because windowing is a general spectral-analysis tool, not a technique specific to keyed modes like CW or RTTY.

Candidate Tips 考生提示

US–China difference: Same DSP principle worldwide; US hams using SDR software (e.g., SDR#, HDSDR) see the identical Hamming/Blackman/raised-cosine window menus.

Common pitfall: Thinking the window changes the on-screen window/span size (option C) or that it is a mode-specific filter for CW/RTTY (option D); it is about leakage, not display or modulation type.

Real on-air practice: When a nearby strong carrier smears across your 2 m waterfall, choosing a Blackman window tightens its skirt and reduces leakage onto weak signals.

💬 Have questions about this topic, or FCC / CRAC exam preparation?
对本篇内容或 FCC / CRAC 备考有疑问?

👉 Join the Discussion / 前往统一留言板参与讨论

☕ 请作者喝杯咖啡 / Buy me a coffee

如果本站对你有帮助,欢迎赞赏支持,谢谢!
If this site helped you, tips are warmly appreciated.

微信赞赏码

微信赞赏 / WeChat

Ko-fi赞赏码

Ko-fi 赞赏 / Ko-fi

滚动至顶部