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RAE practice questions

Chapter 22: The Transmitter

Free practice questions on The Transmitter, chapter 22 of the SARL's Class A manual, for South Africa's Radio Amateur Examination (RAE). Every question has the right answer and an explanation. Sections covered:

  • 22.1 A single-band CW transmitter
  • 22.2 An amplitude-modulated (AM) transmitter
  • 22.3 A simple SSB transmitter
  • 22.4 A frequency-synthesised VHF FM transmitter

74 questions

  1. 1. What does the manual give as the purpose of a transmitter?
    Show answer

    To generate a modulated RF signal for an antenna

    A transmitter must produce a radio-frequency signal that carries information, through modulation, ready for the antenna. A carrier with no modulation carries no information, and recovering audio is the receiver’s job.

  2. 2. In what order are the stages of the manual’s single-band CW transmitter connected, from the oscillator to the antenna?
    Show answer

    VFO, buffer, PA, lowpass filter

    The VFO sets the frequency, the buffer shields it, the keyed PA builds the power and the lowpass filter comes last. A buffer after the PA could no longer shield the VFO from the PA’s changing load.

  3. 3. What is the job of the buffer amplifier in the manual’s CW transmitter?
    Show answer

    To give the VFO a light, steady load

    With a light, steady load the VFO is unaffected as the PA is keyed, so its frequency holds and there is no chirp. Changes in supply voltage need a regulated supply; a buffer cannot cure them.

  4. 4. Which stage of the manual’s CW transmitter does the keying circuit switch on and off?
    Show answer

    The power amplifier (PA)

    The keying block drives the PA, while the VFO and buffer run all the time, so the frequency stays steady. Starting and stopping the VFO for every character would make its frequency wander as it settles.

  5. 5. Why can the PA of the manual’s CW transmitter run in class C?
    Show answer

    A keyed carrier needs no linear amplification

    Keying only switches the carrier on and off, so the PA need not copy the RF waveform faithfully, and efficient class C suits it. Class C makes plenty of harmonics, which the output lowpass filter removes.

  6. 6. What does the lowpass filter at the output of the manual’s CW transmitter do?
    Show answer

    It weakens the harmonics that the PA produces

    The PA’s distortion creates harmonics at whole multiples of the signal frequency, all above it, so a lowpass filter passes the signal and stops them. Blocking the fundamental and passing harmonics is a highpass filter.

    Try it: Low-pass Filter Designer
  7. 7. The PA of a CW transmitter works on 1,83 MHz. What is the lowest harmonic frequency that its output lowpass filter must stop?
    Show answer

    3,66 MHz

    Harmonics lie at whole multiples of the signal frequency, so the lowest is the second: 2 × 1,83 = 3,66 MHz. Class C makes even harmonics too, so the third, 5,49 MHz, is not the lowest.

  8. 8. A CW transmitter covers 1,810 to 1,850 MHz. Which cutoff frequency best suits its output lowpass filter?
    Show answer

    About 2,2 MHz

    The cutoff must lie above 1,85 MHz, so the signal passes, yet well below the second harmonic at 3,62 to 3,70 MHz. A 3,5 MHz cutoff sits just under that harmonic, so it would barely weaken it.

    Try it: Low-pass Filter Designer
  9. 9. In the manual’s CW transmitter, what does the waveform shaper prevent?
    Show answer

    Key clicks

    Switching the carrier on or off too sharply spreads keying sidebands, heard as clicks on nearby frequencies, so the shaper ramps it. Chirp, a shift of frequency as the key closes, is prevented by the buffer.

  10. 10. By chapter 21, about how long should the CW transmitter’s keying waveform shaper take to ramp the carrier up or down?
    Show answer

    About 5 ms

    Chapter 21 gives about 5 ms as the best rise and fall time: slow enough to keep the keying sidebands narrow, fast enough for Morse. A 50 µs edge is almost a square switch and would click.

  11. 11. Below what frequency does the manual say a VFO is usually best kept, for good stability?
    Show answer

    10 MHz

    The manual advises keeping VFOs below about 10 MHz, since the higher an LC oscillator runs, the harder it is to hold steady. 30 MHz, the top of HF, would be far too high.

  12. 12. For which bands does the manual say its single-band CW design suits best, and why?
    Show answer

    80 m or 40 m, keeping the VFO low enough to be stable

    Here the VFO runs on the output frequency, and VFOs are best kept below 10 MHz, so 3,5 MHz or 7 MHz suits. On 20 m the VFO itself would have to run at 14 MHz.

  13. 13. The PA of a CW transmitter is 65% efficient and delivers 50 W of RF. Which option is nearest to the DC power it draws from its supply?
    Show answer

    75 W

    Efficiency = RF output ÷ DC input, so the input is 50 ÷ 0,65 = 76,9 W, and 75 W is nearest. Multiplying instead, 50 × 0,65 = 32,5 W, would give less input than output, which is impossible.

  14. 14. An AM transmitter generates its AM signal at low level and then amplifies it. What does this demand of the amplifiers that follow the modulator?
    Show answer

    They must all be linear amplifiers

    An AM signal holds the carrier and its sidebands at once, so non-linear stages would cause intermodulation distortion; every amplifier after the modulator must be linear. Class C suits only signals of constant amplitude: a plain carrier, CW or FM.

  15. 15. Why do modern multimode transceivers usually generate AM at low level?
    Show answer

    Low-level AM uses the same circuits that make SSB

    Making SSB is simplest at low level, and the same circuits give AM too, given a filter wide enough for both sidebands. AM cannot pass the narrow SSB filter, which keeps only one sideband.

  16. 16. What is the main advantage of high-level modulation in an AM transmitter?
    Show answer

    Its RF stages can be efficient class C amplifiers

    Until it is modulated the carrier is a single frequency, so class C stages can amplify it efficiently, and the audio is added at full power. Either way the AM signal’s bandwidth is the same, set by the audio.

  17. 17. Why can class C stages amplify the RF of a high-level AM transmitter without causing intermodulation distortion?
    Show answer

    The RF is a single frequency until it is modulated

    Intermodulation needs two or more frequencies to mix; the unmodulated carrier is just one, so class C only adds harmonics, which the lowpass filter removes. Intermodulation products near the signal could never be filtered.

  18. 18. In the manual’s high-level AM transmitter, through which stages does the microphone’s signal pass, in order?
    Show answer

    Preamplifier, filter, audio PA, modulator

    The weak microphone signal is pre-amplified, filtered to the speech band, built up by the audio power amplifier and then drives the modulator. The manual filters at low level, ahead of the audio PA, not after it.

  19. 19. Why is the microphone signal filtered before it reaches the modulator of the manual’s AM transmitter?
    Show answer

    To remove audio above about 3 kHz and keep the signal narrow

    Communications speech needs only about 300 Hz to 3 kHz, and each audio frequency reappears that far either side of the carrier, so higher audio would widen the signal. Speech without its lows would be unintelligible.

  20. 20. What audio range does the manual give as needed for communications-grade speech in its AM transmitter?
    Show answer

    300 Hz to 3 kHz

    Speech stays intelligible with only 300 Hz to 3 kHz, so the filter removes the rest and keeps the signal narrow. The voice itself spans about 50 Hz to 20 kHz (chapter 21), far more than communication needs.

  21. 21. An AM transmitter like the manual’s has its audio limited to 2,5 kHz. What bandwidth does its signal occupy?
    Show answer

    5 kHz

    Each audio frequency appears both above and below the carrier, so the signal spans 2,5 kHz either side: 2 × 2,5 = 5 kHz. Only a single-sideband signal would fit into 2,5 kHz.

  22. 22. In the manual’s high-level AM transmitter, on what frequency does the VFO run?
    Show answer

    On the output frequency itself

    The VFO feeds the buffer and the class C PA directly, with no mixer or multiplier, so it runs on the output frequency. A doubler would need the VFO at half the output, which this design does not use.

  23. 23. What share of the power in a 100% modulated AM signal is in the carrier?
    Show answer

    Two-thirds

    At 100% modulation the two sidebands together carry half as much power as the carrier, so the carrier holds 1 part in 1,5: two-thirds. One-third is the sidebands’ combined share.

  24. 24. A high-level modulated AM transmitter has a 60 W carrier. Ignoring the PA’s losses, what audio power must the modulator supply for 100% modulation?
    Show answer

    30 W

    The audio supplies the sideband power, half the carrier at 100% modulation: 0,5 × 60 = 30 W, more in practice to cover the PA’s losses. Taking a third of the carrier gives the wrong 20 W.

  25. 25. The audio amplifier of a high-level AM transmitter can supply 23 W. Ignoring losses, which option is nearest to the largest carrier power it can modulate fully?
    Show answer

    45 W

    Full modulation needs audio of half the carrier power, so the carrier can be 2 × 23 = 46 W (less after real PA losses): 45 W is nearest. Halving instead gives 11,5 W, nearest 12 W.

  26. 26. An AM transmitter’s 70 W carrier is modulated 100%. What is the total power of the AM signal?
    Show answer

    105 W

    At 100% modulation the sidebands add half the carrier power: 70 + 35 = 105 W, two-thirds of it in the carrier. Adding a third of the carrier instead gives 70 + 23,3 = 93 W.

  27. 27. In the manual’s high-level AM transmitter, which stage provides the energy that goes into the sidebands?
    Show answer

    The audio power amplifier

    In high-level modulation the sideband power comes from the audio power amplifier, through the modulator, so it must be large. The RF power amplifier only turns it into RF; its own DC supply provides the carrier.

  28. 28. In a high-level AM transmitter like the manual’s, the audio drives the modulator beyond 100% modulation. What results?
    Show answer

    Distortion and splatter onto nearby frequencies

    Beyond 100% the carrier is cut off on the troughs and the modulation is clipped, which distorts it and spreads splatter onto nearby channels (chapter 21). The signal grows wider, not usefully louder.

  29. 29. In a high-level AM modulator like the manual’s, what does the modulation transformer do?
    Show answer

    It varies the final stage’s supply with the audio

    Its secondary carries the supply to the final stage, so the audio adds to and subtracts from the supply voltage and the RF output’s amplitude follows. It carries audio, not RF; varying the frequency would give FM.

  30. 30. In the manual’s high-level modulation circuit, where does the RF from the earlier stages enter?
    Show answer

    At the transistor’s base

    The RF drive goes to the base, while the audio drives the transformer’s primary, whose secondary feeds the collector supply through the RF choke. The emitter is earthed, and +VCC is the supply end.

  31. 31. What does the RF choke in the collector supply of the manual’s high-level modulated stage do?
    Show answer

    It passes DC and audio but blocks the RF

    A choke has little reactance at DC and audio but a high reactance at RF, so the modulated supply reaches the collector while the RF stays out of the supply. Passing RF and blocking DC is the output capacitor’s job.

    Try it: Reactance & Resonance
  32. 32. What does the capacitor between the collector and the output of the manual’s high-level modulated stage do?
    Show answer

    It passes the modulated RF and blocks the DC supply

    A capacitor’s reactance is low at RF and infinite at DC, so the modulated RF reaches the output while the collector’s DC supply is kept off it. Passing DC and blocking RF is the choke’s job.

    Try it: Reactance & Resonance
  33. 33. How does the manual say its simple SSB transmitter could be made into a low-level AM transmitter?
    Show answer

    Use an unbalanced modulator and a wider crystal filter

    An unbalanced modulator lets the carrier through, and a crystal filter wide enough for both sidebands passes the whole AM signal. The narrow filter would keep one sideband and reject the carrier just outside its passband.

  34. 34. In the manual’s simple SSB transmitter, which stage comes straight after the balanced modulator?
    Show answer

    The crystal filter

    The crystal filter follows the balanced modulator, then come the mixer, bandpass filter, PA and lowpass filter. The crystal filter works only at the fixed 9 MHz, so it cannot follow the mixer, whose output moves with the VFO.

  35. 35. What does the balanced modulator of the manual’s SSB transmitter put out?
    Show answer

    Both sidebands, with the carrier suppressed

    Balancing cancels the carrier and leaves the upper and lower sidebands: a double-sideband suppressed-carrier signal. An unbalanced modulator would keep the carrier as well, giving AM.

  36. 36. Which stage of the manual’s SSB transmitter mainly removes the carrier?
    Show answer

    The balanced modulator

    The balanced modulator cancels the carrier and leaves only the two sidebands. The crystal filter’s job is to remove one sideband, though in practice it also trims any carrier left over.

  37. 37. In the manual’s SSB block diagram, what does a cross drawn inside a circle stand for?
    Show answer

    A mixer, which multiplies two signals

    The circle with a cross is the mixer symbol; the manual uses it for the balanced modulator and for the mixer, since both multiply two signals. Simply adding two signals would create no new frequencies.

  38. 38. What does the crystal filter in the manual’s SSB transmitter do?
    Show answer

    It passes the upper sideband and blocks the lower

    The carrier has already gone at the balanced modulator; the very narrow crystal filter passes only the upper sideband, 300 Hz to 3 kHz above 9 MHz, and rejects the lower. This is how the filter method makes SSB.

  39. 39. What name does the manual give to the way its simple SSB transmitter generates its SSB signal?
    Show answer

    The filter method

    A balanced modulator removes the carrier and a narrow crystal filter then removes one sideband: the filter method. The phasing method, which cancels the unwanted sideband instead, is a different technique, not used here.

  40. 40. A 9,000 MHz carrier is modulated by audio tones from 1 kHz to 2 kHz. Which frequencies does the lower sideband occupy?
    Show answer

    8,998 to 8,999 MHz

    The lower sideband lies below the carrier by each audio frequency: 9,000 − 0,002 = 8,998 MHz up to 9,000 − 0,001 = 8,999 MHz. Adding instead gives the upper sideband, 9,001 to 9,002 MHz.

  41. 41. An SSB signal carries speech from 250 Hz to 2,9 kHz. Which option is nearest to its bandwidth?
    Show answer

    2,7 kHz

    One sideband spans the speech band itself, shifted in frequency: 2,9 kHz − 250 Hz = 2,65 kHz, so 2,7 kHz is nearest. Doubling the top audio frequency gives 5,8 kHz, nearest 6 kHz, the width of AM.

  42. 42. Why does the manual’s SSB transmitter work at a fixed 9 MHz and then mix up to 14 MHz, rather than modulate at 14 MHz directly?
    Show answer

    Its crystal filter cannot be retuned

    Filters selective enough to pass one sideband and reject the other are fixed-tuned, so the signal is formed at their frequency and then shifted by a mixer. Balanced modulators work over a wide range of frequencies.

  43. 43. According to the manual, what is gradually ending the need to generate SSB at a fixed frequency?
    Show answer

    Digital signal processing (DSP)

    A footnote says growing use of DSP will soon make the fixed-tuned rule untrue: a digital filter can be retuned in software. Better quartz crystals would still give fixed-tuned filters.

  44. 44. In an SSB transmitter like the manual’s, the 9,000 MHz signal is mixed with the VFO and the sum is used. The VFO is set to 5,215 MHz. What is the output frequency?
    Show answer

    14,215 MHz

    The sum product is 9,000 + 5,215 = 14,215 MHz, in the 20 m phone segment. Subtracting gives 9,000 − 5,215 = 3,785 MHz, the difference product, which the bandpass filter rejects.

  45. 45. An SSB transmitter like the manual’s uses the sum of its 9,000 MHz signal and the VFO frequency. Over what range must the VFO tune to cover 14,000 to 14,350 MHz?
    Show answer

    5,000 to 5,350 MHz

    The VFO supplies the difference between output and 9 MHz: 14,000 − 9,000 = 5,000 MHz up to 14,350 − 9,000 = 5,350 MHz. Adding 9 MHz to the output gives 23,000 to 23,350 MHz, the wrong way round.

  46. 46. In an SSB transmitter like the manual’s, the 9,000 MHz signal meets a VFO set to 5,170 MHz in the mixer. At what frequency is the unwanted difference product?
    Show answer

    3,830 MHz

    The difference product is 9,000 − 5,170 = 3,830 MHz. 5,170 MHz is the VFO’s own signal, which the bandpass filter also rejects; the manual (p199) calls that VFO range the unwanted product.

  47. 47. In an SSB transmitter like the manual’s, the VFO is tuned up from 5,200 to 5,300 MHz. What happens to the difference product, 9 MHz minus the VFO?
    Show answer

    It moves down, from 3,800 to 3,700 MHz

    The difference falls from 9,000 − 5,200 = 3,800 MHz to 9,000 − 5,300 = 3,700 MHz as the VFO rises: the inversion the manual mentions. Only the tuning runs backwards; the signal is still USB.

  48. 48. What does the bandpass filter between the mixer and the PA of the manual’s SSB transmitter do?
    Show answer

    It passes the transmitter’s 20 m output, blocking the rest

    Passing only 14,100 to 14,350 MHz, this LC filter keeps the sum and rejects both the 3,65 to 3,9 MHz difference and the VFO’s own 5,1 to 5,35 MHz signal, which p199 calls the unwanted product.

  49. 49. What best explains why the bandpass filter after the mixer of the manual’s SSB transmitter can be an ordinary LC filter?
    Show answer

    The unwanted products lie MHz away from the wanted one

    The nearest unwanted mixer output lies several MHz from 14,1 to 14,35 MHz, so a gentle LC filter can reject it. Sidebands only kHz apart need a crystal filter; the signal is still weak here, so power is no concern.

  50. 50. Which kind of power amplifier does the manual’s SSB transmitter use?
    Show answer

    A linear amplifier, probably in class AB

    An SSB signal’s amplitude follows the speech, so the PA must be linear; the manual suggests class AB, linear enough yet reasonably efficient. A class C stage would distort it into splatter and intermodulation.

  51. 51. Why does the manual’s SSB transmitter reach 14 MHz with a mixer rather than with frequency multipliers?
    Show answer

    A multiplier’s distortion would ruin the SSB signal

    A multiplier works by distortion, which would intermodulate the many frequencies in the SSB signal into products too close to filter out; a properly designed mixer shifts it cleanly. Multiplying swaps no sideband.

  52. 52. The manual’s SSB transmitter already has a bandpass filter after its mixer. Why does it still need a lowpass filter after the PA?
    Show answer

    The PA creates new harmonics of its own

    Even a linear class AB PA adds harmonics, from 28,2 MHz up, after the bandpass filter, so a lowpass filter must follow it. A VFO leak at 5 MHz lies below the signal, which a lowpass filter passes.

    Try it: Low-pass Filter Designer
  53. 53. An SSB transmitter like the manual’s covers 14,170 to 14,330 MHz. Which option is nearest to the lowest harmonic frequency that its final lowpass filter must reject?
    Show answer

    28 MHz

    The lowest harmonic is twice the lowest frequency: 2 × 14,170 = 28,34 MHz, so 28 MHz is nearest. The third harmonic, 3 × 14,170 = 42,51 MHz, is not the lowest; even harmonics occur too.

  54. 54. Why does the manual base its VHF FM transmitter on a frequency synthesiser rather than a VFO?
    Show answer

    A VFO is hard to keep stable enough at VHF

    At VHF an LC oscillator drifts too much for channelised FM, while a synthesiser holds its output almost as steady as its crystal reference. A VFO can be frequency-modulated quite easily, with a varicap.

  55. 55. Why does a simple single-loop synthesiser suffice for the manual’s FM transmitter?
    Show answer

    Its channels lie far apart, so coarse steps will do

    With channels 12,5 or 25 kHz apart the reference can be that large, and one loop then steps quickly. Multiple loops are needed only where small steps are wanted, since a low reference slows a single loop (chapter 20).

  56. 56. In the manual’s FM transmitter, why is the microphone’s signal passed through a lowpass filter?
    Show answer

    To keep the audio below about 3 kHz, so the FM stays narrow

    Speech needs audio up to about 3 kHz, and higher audio would widen the FM signal, so the lowpass filter removes it before the modulation. A 300 Hz cutoff would remove most of the speech itself.

  57. 57. Where is the speech applied as frequency modulation in the manual’s synthesised FM transmitter?
    Show answer

    To the 12,8 MHz crystal oscillator (VCXO)

    The filtered audio drives the VCXO, whose 12,8 MHz output is divided down to become the reference, so the reference carries the FM. Many synthesised rigs modulate the VCO instead, but not the manual’s design.

  58. 58. How does the VCXO in the manual’s FM transmitter follow the audio?
    Show answer

    A varicap diode pulls its frequency slightly

    The audio voltage changes the varicap’s capacitance, and with it the crystal’s frequency, by a small amount: pulling the crystal. Switching between two frequencies would be frequency-shift keying, used for data.

  59. 59. The speech in the manual’s FM transmitter is never fed to the VCO directly. Why is the VCO’s output still frequency-modulated?
    Show answer

    It is locked to a reference that carries the FM

    The loop holds the VCO at N times the reference, so when the reference swings with the speech, the VCO swings with it, N times as far in hertz. Stray coupling would be a fault, not the design.

  60. 60. In a synthesiser like the manual’s, N = 4 200 and the speech swings the 12,8 MHz VCXO by ±350 Hz. Which option is nearest to the VCO’s deviation?
    Show answer

    ±1,4 kHz

    The ÷1 024 divides the swing and the loop multiplies it by N: 350 × 4 200 ÷ 1 024 = 1 436 Hz, nearest ±1,4 kHz. It is not passed on unchanged; leaving out the ÷1 024 gives ±1,5 MHz.

    Try it: Unit Converter
  61. 61. The 12,8 MHz VCXO of a synthesiser like the manual’s runs 64 Hz high, and N = 4 160. How far off frequency is the VCO?
    Show answer

    260 Hz

    The ÷1 024 passes on 64 ÷ 1 024 = 0,062 5 Hz of error, which the loop multiplies by N: 0,062 5 × 4 160 = 260 Hz. The error does not stay at 64 Hz.

  62. 62. What does the ÷1 024 stage do in the manual’s FM transmitter?
    Show answer

    It divides the VCXO’s output down to the reference

    It divides the frequency-modulated 12,8 MHz VCXO signal by 1 024 to give the 12,5 kHz reference. Dividing the VCO’s output for the comparator is the job of the ÷N divider, which the microprocessor sets.

  63. 63. A synthesiser like the manual’s divides a 6,4 MHz VCXO by 1 024 to make its reference. What is the reference frequency?
    Show answer

    6,25 kHz

    6 400 000 Hz ÷ 1 024 = 6 250 Hz, or 6,25 kHz, a channel step some radios use. Dividing by 102,4 instead, one decimal place out, gives 62,5 kHz.

    Try it: Unit Converter
  64. 64. A transmitter like the manual’s keeps its 12,8 MHz VCXO but needs a 25 kHz reference. By what ratio must it divide the VCXO?
    Show answer

    512

    The ratio is VCXO ÷ reference = 12 800 kHz ÷ 25 kHz = 512, half the manual’s 1 024. Keeping ÷1 024 would give the manual’s 12,5 kHz reference, half of what is wanted.

    Try it: Unit Converter
  65. 65. Which blocks lie inside the dashed box marked “Frequency Synthesiser” in the manual’s FM transmitter diagram?
    Show answer

    Phase comparator, lowpass filter, VCO and ÷N

    The loop itself is boxed: phase comparator, lowpass filter, VCO and ÷N divider. The microprocessor that sets N, and the VCXO and ÷1 024 that make the reference, lie outside the box.

  66. 66. What job does the microprocessor do in the manual’s FM transmitter?
    Show answer

    It sets the ÷N ratio for the chosen frequency

    The microprocessor reads the user’s choice and loads the matching ratio into the ÷N divider, from 11 521 to 11 679 across the 2 m band. The ÷1 024 ratio is fixed.

  67. 67. A PLL synthesiser with a 25 kHz reference must tune its VCO from 430,000 to 440,000 MHz. Over what range must its ÷N ratio run?
    Show answer

    17 200 to 17 600

    N = VCO frequency ÷ reference: 430 000 kHz ÷ 25 kHz = 17 200, and 440 000 ÷ 25 = 17 600. Dividing by 12,5 kHz instead gives the wrong 34 400 to 35 200.

    Try it: Unit Converter
  68. 68. A PLL synthesiser like the one in the manual’s FM transmitter has a 12,5 kHz reference, and its ÷N is set to 34 800. What frequency does its VCO produce?
    Show answer

    435,000 MHz

    The VCO settles at N × reference = 34 800 × 12 500 Hz = 435 000 000 Hz, or 435,000 MHz. Taking the reference as 1,25 kHz, one decimal place out, gives the wrong 43,500 MHz.

    Try it: Unit Converter
  69. 69. A PLL synthesiser with a 12,5 kHz reference must put its VCO on 438,650 MHz. What ÷N ratio does it need?
    Show answer

    35 092

    N = VCO frequency ÷ reference = 438 650 kHz ÷ 12,5 kHz = 35 092. Taking the reference as 1,25 kHz, one decimal place out, gives the wrong 350 920.

    Try it: Unit Converter
  70. 70. In a synthesiser like the manual’s, with its 12,5 kHz reference, N goes up by 3. How far does the VCO’s frequency move?
    Show answer

    37,5 kHz

    Each step of N moves the VCO by one reference step, so 3 × 12,5 kHz = 37,5 kHz. Taking the whole change as a single 12,5 kHz step ignores how far N moved.

  71. 71. In the manual’s FM synthesiser, what does the lowpass filter inside the loop do?
    Show answer

    It smooths the comparator’s output to steer the VCO

    The comparator’s output carries ripple at the 12,5 kHz reference besides the phase-error voltage; the filter removes the ripple, leaving a smooth voltage, speech swing included, to steer the VCO. It does not filter the RF.

  72. 72. At about what frequency does the VCO of the manual’s FM transmitter run?
    Show answer

    On the output frequency, in the 2 m band

    The VCO is locked at N times 12,5 kHz, from 144,012 5 to 145,987 5 MHz, so it runs on the output frequency itself. No tripler follows it; the PA amplifies its output directly.

  73. 73. What follows the synthesiser in the manual’s FM transmitter, on the way to the antenna?
    Show answer

    A power amplifier, then a lowpass filter

    The VCO already runs on the output frequency, so the PA simply amplifies it and the lowpass filter removes its harmonics. Older crystal-controlled FM rigs multiplied up, but this design needs no multiplier.

  74. 74. By chapter 18, which PA class best suits the manual’s synthesised FM transmitter?
    Show answer

    Class C

    An FM signal’s amplitude stays constant, so the PA need not be linear and efficient class C suits it, as chapter 18 says; the lowpass filter removes its harmonics. Linear class AB is needed for SSB or low-level AM.