UTC 18:26:52 SAST 20:26 SFI 124 Kp 1 20 M GOOD 10 M FAIR 145.650 · 438.650 · 438.200 · 438.300 · 439.175
RAE practice questions

Chapter 20: Frequency Translation

Free practice questions on Frequency Translation, chapter 20 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:

  • 20.1 The frequency multiplier
  • 20.2 The frequency divider
  • 20.3 The Phase Locked Loop frequency synthesiser
  • 20.4 The mixer

77 questions

  1. 1. Which of these is NOT one of the frequency translation circuits that the manual describes?
    Show answer

    The Colpitts oscillator

    Frequency translation circuits make a new frequency from signals at other frequencies: multipliers, dividers, synthesisers and mixers. An oscillator such as the Colpitts only generates a signal; the divider is a translation circuit too.

  2. 2. Which of these waveforms contains no harmonics at all?
    Show answer

    A perfect sine wave

    Only a perfect sine wave is free of harmonics; every other shape contains some. A triangle wave looks smooth, but it still contains odd harmonics of its fundamental.

  3. 3. What is the frequency of the third harmonic of a 7,05 MHz signal?
    Show answer

    21,15 MHz

    The nth harmonic is n times the fundamental: 3 × 7,05 = 21,15 MHz. Counting three steps above the fundamental gives 4 × 7,05 = 28,2 MHz, which is the fourth harmonic.

    Try it: QRM Hunting
  4. 4. Which option is nearest to the fifth harmonic of a 3,65 MHz signal?
    Show answer

    18 MHz

    The fifth harmonic is 5 × 3,65 = 18,25 MHz, so 18 MHz is nearest. Taking it as the fifth step above the fundamental gives 6 × 3,65 = 21,9 MHz, nearest the 22 MHz option.

    Try it: QRM Hunting
  5. 5. A 3,6 MHz signal has a harmonic at 14,4 MHz. Which harmonic is it?
    Show answer

    The fourth harmonic

    Since 14,4 ÷ 3,6 = 4, it is the fourth harmonic; the fundamental itself counts as the first. Calling it the third counts only the steps above the fundamental.

    Try it: QRM Hunting
  6. 6. How does a frequency multiplier produce its output?
    Show answer

    It distorts the input, then a bandpass filter selects one harmonic

    The input is distorted on purpose so that it is rich in harmonics, and a bandpass filter tuned to the wanted harmonic passes it alone. A lowpass filter would also pass the fundamental and the lower harmonics.

  7. 7. How is a frequency multiplier stage generally operated?
    Show answer

    Non-linearly, so that it creates harmonics

    A multiplier must distort the signal to create harmonics, so its stage is run non-linearly: in the manual’s doubler the diodes do this, and a tripler may use class C. A linear stage adds no harmonics to select.

  8. 8. A 12 MHz oscillator drives a tripler, which drives a quadrupler. What frequency comes out of the quadrupler?
    Show answer

    144 MHz

    Each stage multiplies the frequency it is given: 12 × 3 = 36 MHz, then 36 × 4 = 144 MHz, an overall factor of 12. Adding the factors, 3 + 4 = 7, gives the wrong 84 MHz.

  9. 9. A transmitter’s crystal oscillator is followed by two triplers in a row, giving 145,8 MHz. Which option is nearest to the crystal frequency?
    Show answer

    16 MHz

    The two triplers multiply by 3 × 3 = 9, so the crystal runs at 145,8 ÷ 9 = 16,2 MHz, nearest 16 MHz. Adding the factors to 6 gives 24,3 MHz, nearest the 24 MHz option.

  10. 10. Which kind of frequency multiplier does a full-wave rectifier suit, and why?
    Show answer

    A doubler, as its output repeats twice per input cycle

    A full-wave rectified sine repeats twice in every input cycle, so it is rich in the second harmonic and suits a doubler. It has almost no third harmonic; the manual uses class C for its tripler.

  11. 11. What does the manual say makes a class C amplifier useful in a frequency multiplier?
    Show answer

    Its output has a strong third harmonic

    A class C stage conducts in short pulses, for less than half of each cycle, so its output has a high third-harmonic content that suits a tripler. Conducting for more than half of each cycle is class AB, or A.

  12. 12. In the manual’s frequency doubler, what does the coil L1 of the parallel tuned circuit L1-C1 do to the DC part of the signal?
    Show answer

    It passes it straight through to ground

    For DC a coil is almost a short circuit, so L1 conducts the DC part to ground while L1-C1 resonates at the output frequency. A coil passes DC; blocking DC is what a capacitor does.

  13. 13. What do C2 and C4 do in the manual’s frequency doubler?
    Show answer

    They block DC but pass the RF signal

    C2 stops L1 shorting Q1’s base bias, set by R1 and R2, to ground, and C4 keeps Q1’s collector DC out of the next stage; both pass the RF. Neither is wired as a supply decoupler.

  14. 14. The manual’s frequency doubler has a second parallel tuned circuit, L2-C3, at the output of Q1. What is it there for?
    Show answer

    To weaken the unwanted harmonics further

    Tuned to the output frequency, L2-C3 further weakens the harmonics left after L1-C1, such as three and four times the input. It is a filter; a stage that oscillated would be useless as a multiplier.

  15. 15. What signal comes out of the manual’s frequency doubler?
    Show answer

    A sine wave at twice the input frequency

    The tuned circuits remove the DC and the other harmonics, leaving a sine wave at twice the input frequency. A square wave would be full of harmonics, which those tuned circuits are there to remove.

  16. 16. In the manual’s frequency doubler, how is transistor Q1 connected?
    Show answer

    As a common-emitter amplifier

    Q1 takes its input at the base and gives its output at the collector, through L2-C3 and C4, with R1, R2 and R3 setting its bias: a common-emitter amplifier. An emitter follower takes its output from the emitter.

  17. 17. In UHF transmitters, RF power is often fed to a diode with tuned circuits to raise the signal’s frequency. What is such a stage called?
    Show answer

    A varactor multiplier

    At VHF and UHF a varicap (varactor) diode is often the non-linear part of a multiplier, and tuned circuits select the wanted harmonic. A varicap-tuned oscillator, such as a VCO, makes its own signal; it takes no RF input.

  18. 18. Which of these signals can safely be passed through a frequency multiplier?
    Show answer

    A phase-modulated signal

    The manual allows multipliers for unmodulated carriers, CW, FM and phase modulation. An AM signal, even with its full carrier, holds a range of frequencies that the distortion would intermodulate.

  19. 19. Which of these signals should NOT be passed through a frequency multiplier?
    Show answer

    A double-sideband AM signal

    An AM signal holds several frequencies at once, which the multiplier’s distortion would intermodulate. FM, CW and a plain carrier are safe, though multiplying an FM signal multiplies its deviation too.

  20. 20. Why is an SSB signal unsuitable for a frequency multiplier?
    Show answer

    It would create IMD too close to filter out

    The distortion makes the many frequencies in an SSB signal intermodulate, and the IMD products fall too near the wanted ones to be filtered out. A multiplier works on whatever frequencies it is fed, carrier or not.

  21. 21. A 7 MHz oscillator is 60 Hz above its marked frequency, and its output is tripled to 21 MHz. How far off is the 21 MHz signal?
    Show answer

    180 Hz

    A multiplier multiplies every frequency it is fed, so a 60 Hz error at 7 MHz becomes 3 × 60 = 180 Hz at 21 MHz. The error is not left at 60 Hz; it grows with the multiplication factor.

  22. 22. An oscillator’s output swings ±350 Hz either side of 8,8 MHz, and it is then multiplied 6 times to 52,8 MHz. How far either side of 52,8 MHz does the output swing?
    Show answer

    ±2,1 kHz

    Multiplying the frequency multiplies every swing in it too: 6 × 350 Hz = 2,1 kHz either side, which is how an FM signal’s deviation grows in a multiplier chain. The swing does not stay at ±350 Hz.

  23. 23. Why is a frequency multiplier unsuited to multiplying by a large number such as 100?
    Show answer

    No filter could pick out the wanted harmonic from its neighbours

    The 99th, 100th and 101st harmonics lie only 1% apart, so no practical filter could separate them, and a multiplier makes most harmonics. Transistor speed is not the limit the manual gives.

  24. 24. What waveform does a digital frequency divider IC typically give at its output?
    Show answer

    A square wave

    A digital divider switches between two levels, so its output is typically a square wave, rich in harmonics. A sawtooth ramps steadily up and drops back, which a switching circuit does not do.

  25. 25. A digital divider’s output is at 25 kHz. At which frequencies are its strongest harmonics?
    Show answer

    75 kHz, 125 kHz and 175 kHz

    The divider’s square wave is strongest in the odd harmonics, at three, five and seven times its frequency: 75, 125 and 175 kHz. A symmetrical square wave has almost no even harmonics.

  26. 26. How can a digital divider’s output be turned into a clean sine wave at the divided frequency?
    Show answer

    A lowpass filter, or a bandpass one tuned to it

    The harmonics all lie above the wanted frequency, so a lowpass filter, or a bandpass filter tuned to the wanted frequency, leaves a clean sine. A bandstop filter there would remove the very signal that is wanted.

    Try it: Low-pass Filter Designer
  27. 27. A 6 MHz crystal oscillator feeds a ÷10 divider, which feeds a ÷50 divider. What frequency comes out of the second divider?
    Show answer

    12 kHz

    Each divider divides the frequency it is fed: 6 MHz ÷ 10 = 600 kHz, then 600 kHz ÷ 50 = 12 kHz, an overall ÷500. Leaving out the first divider gives 6 MHz ÷ 50 = 120 kHz.

  28. 28. A 7 MHz oscillator feeds a ÷10 divider and then a ÷12 divider. Which option is nearest to the final output frequency?
    Show answer

    60 kHz

    The overall division is 10 × 12 = 120, and 7 MHz ÷ 120 = 58,3 kHz, so 60 kHz is nearest. Leaving out the ÷10 gives 7 MHz ÷ 12 = 583 kHz, nearest the 580 kHz option.

  29. 29. A 10 MHz crystal oscillator is 50 Hz high, and its output is divided by 1 000. How far from 10 kHz is the divided output?
    Show answer

    0,05 Hz

    A divider divides every frequency it is fed, errors included: 50 Hz ÷ 1 000 = 0,05 Hz. The output is wrong by the same fraction, five parts in a million, not by the same 50 Hz.

  30. 30. What is the main drawback of a variable-frequency oscillator (VFO) as a transmitter’s main oscillator?
    Show answer

    It drifts with heat and jumps if knocked

    A VFO is hard to make really stable: its frequency drifts as the temperature changes and can jump if it is knocked. A tiny tuning range is the crystal oscillator’s drawback, not the VFO’s.

  31. 31. Why is a crystal oscillator on its own unsuited to be the main oscillator of a transmitter that must cover a whole amateur band?
    Show answer

    Its frequency can be pulled only slightly

    A crystal oscillator is very stable against heat and knocks, but its frequency can be moved only a little, far less than a band. Drifting with temperature is the VFO’s weakness, which the crystal avoids.

  32. 32. Which circuit forms the heart of most frequency synthesisers in modern amateur equipment?
    Show answer

    A phase locked loop

    Modern transceivers are mostly built around phase locked loop synthesisers, which hold a VCO at N times a crystal reference. A phase-shift oscillator simply oscillates; nothing ties it to a reference.

  33. 33. In the manual’s synthesiser, what is fed to the second input of the phase comparator?
    Show answer

    The VCO’s output, divided by N

    The VCO’s output passes through the ÷N divider, and the comparator matches that against the reference. With the reference divided by N instead, the output would settle at a fraction of the reference.

  34. 34. What does the phase comparator do in the manual’s synthesiser?
    Show answer

    It gives a voltage set by its inputs’ phase difference

    The comparator compares the phases of the reference and the divided VCO signal and gives a voltage that depends on their difference. Dividing by N is the divider’s job, and it divides the VCO’s output, not the reference.

  35. 35. Why does the manual’s synthesiser have a lowpass filter between the phase comparator and the VCO?
    Show answer

    To keep the comparator’s fast ripple away from the VCO

    Most phase comparators also give out ripple at their input frequencies; the lowpass filter removes it, leaving the slow phase-error voltage to steer the VCO. It filters the control voltage, not the output signal.

  36. 36. What kind of feedback holds the manual’s synthesiser on its set frequency?
    Show answer

    Negative feedback

    Any drift of the VCO produces a comparator voltage that pushes it back the other way, which is negative feedback. Positive feedback is what makes an oscillator start and keep oscillating.

  37. 37. In the manual’s synthesiser, the VCO drifts a little high. What happens next?
    Show answer

    It is brought back to its set frequency

    The drift changes the phase difference at the comparator, and the new voltage brings the VCO back down to N times the reference. Pushing it further away would be positive feedback, and it would never settle.

  38. 38. How stable is a synthesiser’s output compared with its crystal reference?
    Show answer

    Nearly as stable as the reference itself

    The VCO is held at N times the reference, so its stability nearly matches the crystal’s, though it adds some phase noise. The circuit cannot make it more stable than the crystal it follows.

  39. 39. A synthesiser has a 100 Hz reference, and its divider is set to N = 70 350. What frequency does it produce?
    Show answer

    7,035 MHz

    The VCO settles at N times the reference: 70 350 × 100 Hz = 7 035 000 Hz = 7,035 MHz. Dividing N by the reference instead gives the wrong 703,5 Hz.

    Try it: Unit Converter
  40. 40. A synthesiser with a 12,5 kHz reference must produce 145,525 MHz. What divider ratio N does it need?
    Show answer

    11 642

    N = output ÷ reference = 145 525 000 Hz ÷ 12 500 Hz = 11 642. Taking the reference as 1,25 kHz, one decimal place out, gives the wrong 116 420.

    Try it: Unit Converter
  41. 41. A synthesiser with a 50 Hz reference must tune from 7,000 MHz to 7,200 MHz. Over what range must its divider ratio N run?
    Show answer

    140 000 to 144 000

    N = output ÷ reference: 7 000 000 Hz ÷ 50 Hz = 140 000, and 7 200 000 ÷ 50 = 144 000. Leaving the output in kilohertz gives the wrong 140 to 144.

    Try it: Unit Converter
  42. 42. A synthesiser uses a 25 Hz reference. By how much must its divider ratio N change to move the output by 1 kHz?
    Show answer

    40

    Each step of N moves the output by the reference, 25 Hz, so 1 kHz needs 1 000 ÷ 25 = 40 steps. Taking each step of N as 1 Hz gives 1 000, the step-size slip.

  43. 43. A synthesiser with a 10 Hz reference has its divider changed from N = 1 420 000 to N = 1 420 150. How far does its output move?
    Show answer

    1,5 kHz

    Each unit of N is worth one reference step of 10 Hz, so 150 × 10 Hz = 1 500 Hz = 1,5 kHz. Taking each unit of N as 1 Hz gives only 150 Hz.

  44. 44. In a modern synthesised radio, what sets the divider ratio N as the operator turns the tuning dial?
    Show answer

    A microprocessor, from the user’s choice

    The dial is read by a microprocessor, which sets N and with it the output frequency. The phase comparator only gives a phase-error voltage to steer the VCO; it never chooses N.

  45. 45. Why does the manual’s synthesiser make its 10 Hz reference by dividing down a crystal oscillator, rather than using a 10 Hz crystal?
    Show answer

    A quartz crystal for 10 Hz would be impractically large

    The lower a crystal’s frequency, the bigger it must be, and 10 Hz is far too low for a crystal of sensible size. Thin, fragile crystals are the high-frequency ones, so the crystal runs higher and is divided down.

  46. 46. A synthesiser divides a 12,8 MHz crystal oscillator by 4 096 to make its reference. Which option is nearest to the reference frequency?
    Show answer

    3 kHz

    12 800 000 Hz ÷ 4 096 = 3 125 Hz, so 3 kHz is nearest. Slipping one decimal place gives about 31 kHz, nearest the 30 kHz option.

  47. 47. Why does a synthesiser made for very small frequency steps tend to change frequency slowly?
    Show answer

    Small steps need a low filter cutoff, which responds slowly

    Small steps need a low reference frequency, which needs a low cutoff in the lowpass filter, and that limits how fast the synthesiser can follow a change. Nothing makes the VCO visit every step on its way.

  48. 48. A synthesiser’s phase comparator works with a 2 kHz reference. Which cutoff frequency best suits its lowpass filter?
    Show answer

    200 Hz

    The filter must remove the comparator’s outputs at the 2 kHz reference and above, so its cutoff lies below the reference: 200 Hz suits. A 4 kHz cutoff would let the 2 kHz components through to the VCO.

  49. 49. A synthesiser combines one section that makes large frequency jumps quickly with another that fills in the fine steps. What does the manual call it?
    Show answer

    A multiple-loop synthesiser

    The manual calls this a multiple-loop synthesiser: a coarse loop with a high reference makes the big jumps, and a fine loop with a small step fills in. No multiplier chain is involved.

  50. 50. What harm did the phase noise of early synthesisers, such as those in amateur gear of the early eighties, do?
    Show answer

    It spoilt both received and transmitted signals

    Phase noise spreads the synthesiser’s signal, and since a transceiver uses it on receive and on transmit, it degraded both. Slow drift with temperature is a VFO fault, not phase noise.

  51. 51. What is the phase noise of a synthesiser, as the manual describes it?
    Show answer

    Tiny rapid wobbles in the output’s phase and frequency

    As the circuit keeps adjusting the VCO to hold it to the reference, its phase and frequency wobble very slightly: that is phase noise. Harmonics are separate signals at multiples of the frequency, not noise.

  52. 52. How does a synthesiser’s phase noise appear on a spectrum display?
    Show answer

    As fuzz on both sides of the signal line

    Phase noise widens the spectrum: instead of a single clean line there is fuzz above and below it. Clean, separate extra lines would be spurious signals, a different problem.

  53. 53. Which identity does the manual use to explain what a mixer does?
    Show answer

    2 sin A cos B = sin(A + B) + sin(A − B)

    With A = 2πf1t and B = 2πf2t, multiplying two waves gives waves at f1 + f2 and f1 − f2 added together. The minus form is the identity for 2 cos A sin B instead.

  54. 54. A mixer’s two main output products are at 2 MHz and 20 MHz. What frequencies were fed to it?
    Show answer

    9 MHz and 11 MHz

    The inputs satisfy f1 + f2 = 20 MHz and f1 − f2 = 2 MHz, so f1 = 11 MHz and f2 = 9 MHz. The 2 MHz and 20 MHz signals are the products, not the inputs.

  55. 55. A mixer is to turn a 9 MHz signal into 14,25 MHz, as the sum of its two inputs. What must the LO frequency be?
    Show answer

    5,25 MHz

    The sum product is 9 MHz + fLO = 14,25 MHz, so fLO = 14,25 − 9 = 5,25 MHz. An LO of 23,25 MHz would also give 14,25 MHz, but as the difference, not the sum.

  56. 56. A 28,5 MHz signal is mixed with a 116 MHz LO, and the sum is kept. What frequency is that?
    Show answer

    144,5 MHz

    The sum product is 116 + 28,5 = 144,5 MHz, which lands in the 2 m band. The difference, 116 − 28,5 = 87,5 MHz, is also produced, but it is not the sum that was asked for.

  57. 57. A 7,15 MHz signal is mixed with a 17,85 MHz LO. What is the difference frequency at the mixer’s output?
    Show answer

    10,7 MHz

    The difference product is 17,85 − 7,15 = 10,7 MHz. Adding them gives 25 MHz, the sum product, which is also present but is not the difference asked for.

  58. 58. A 21,1 MHz signal is mixed with a 7,45 MHz LO, and the difference is used. Which option is nearest to that difference frequency?
    Show answer

    14 MHz

    The difference is 21,1 − 7,45 = 13,65 MHz, so 14 MHz is nearest. Adding the two instead gives the sum, 28,55 MHz, nearest the 29 MHz option.

  59. 59. A mixer is fed with 145,000 MHz and a 144,955 MHz LO. What is the difference frequency?
    Show answer

    45 kHz

    The difference is 145,000 − 144,955 = 0,045 MHz, and 0,045 MHz × 1 000 = 45 kHz. Moving the decimal comma one place too far gives 450 kHz, ten times too much.

    Try it: Unit Converter
  60. 60. A mixer’s LO runs at 10 MHz. Which two RF input frequencies would each give a 3 MHz mixing product?
    Show answer

    7 MHz and 13 MHz

    13 − 10 = 3 and 10 − 7 = 3: both give a 3 MHz product. The second sits on the LO’s other side (receivers call it the image), not at 13 + 2 × 3 = 19 MHz.

  61. 61. Why do many practical mixers switch one signal with the other, rather than truly multiplying the two?
    Show answer

    It is hard to multiply accurately without distortion

    Multiplying two sine waves accurately without adding distortion is hard, so switching mixers chop one signal with the other instead. It is the switching mixer, not true multiplication, that needs a strong LO.

  62. 62. How does a switching mixer such as the manual’s diode ring do its work?
    Show answer

    It uses one input to flip the other’s polarity

    The LO switches the diodes so that, on alternate half cycles, the RF reaches the IF port with its polarity reversed. Accurate multiplication of sine waves is exactly what switching mixers avoid.

  63. 63. In the manual’s diode ring mixer, the LO reverses polarity while the RF voltage stays the same. What happens at the IF port?
    Show answer

    It changes to the opposite polarity

    The other pair of diodes now conducts and earths the opposite end of T1’s winding, so the same RF voltage drives the IF port the other way. The LO thus flips the RF’s polarity every half cycle.

  64. 64. In the manual’s diode ring mixer, where is the IF output taken from?
    Show answer

    The centre tap of the RF transformer’s winding

    The IF port is the centre tap of T1’s secondary, the RF transformer. The centre tap of T2, the LO transformer, is earthed, which is what the conducting diode pair is balanced against.

  65. 65. In the manual’s diode mixer, the output port is marked IF. What do the letters stand for?
    Show answer

    Intermediate frequency

    IF stands for intermediate frequency, a name that makes sense in receivers, where the mixer’s output lies between the antenna and audio stages. The IF port is the output, not an input.

  66. 66. Why must the four diodes in the manual’s ring mixer be closely matched?
    Show answer

    So the conducting pair’s midpoint sits at earth

    With identical forward voltages, the junction of the conducting pair sits at the potential of the LO transformer’s centre tap, which is earth, and the mixer stays balanced. The diodes always conduct in pairs.

  67. 67. Reversing a signal’s polarity on alternate half cycles of the LO is mathematically the same as doing what to it?
    Show answer

    Multiplying it by a ±1 square wave

    Multiplying by +1 leaves the signal as it is and multiplying by −1 reverses it, so the switching is multiplication by a ±1 square wave. Adding a square wave would not mix the two signals at all.

  68. 68. A switching mixer has an 8 MHz LO and a 19 MHz RF input. The LO also has a harmonic at 24 MHz, which mixes with the RF too. Which of these products comes from that harmonic?
    Show answer

    43 MHz

    The harmonic and the RF give 24 + 19 = 43 MHz (and 24 − 19 = 5 MHz). 57 MHz is three times the RF, while 27 MHz and 11 MHz come from the LO’s fundamental, 8 MHz.

  69. 69. How are a mixer’s unwanted output products normally removed?
    Show answer

    By filters placed after the mixer

    Unwanted mixing products, including those from the LO’s harmonics in a switching mixer, are removed by suitable filters after the mixer. A stronger LO does not remove them.

  70. 70. A diode mixer like the manual’s needs about +7 dBm at its LO port. Roughly what power is that?
    Show answer

    5 mW

    A level of 0 dBm is 1 mW, and +7 dBm is 7 dB above it, a factor of about 5, so about 5 mW. Reading dBm as milliwatts gives the wrong 7 mW.

    Try it: dBm, Watts & dB
  71. 71. A diode mixer’s conversion loss is quoted as 6 to 7 dB. What does that figure mean?
    Show answer

    Each product comes out 6 to 7 dB below the RF input

    Conversion loss compares each output product with the RF input: here each comes out 6 to 7 dB weaker. A mixer with this loss gives no gain, and the LO level is a separate matter.

  72. 72. A diode mixer with a conversion loss of 6 dB is fed a 30 µW RF signal. Which option is nearest to the power of each output product?
    Show answer

    8 µW

    A 6 dB loss divides the power by about 4, so 30 µW ÷ 3,98 = 7,5 µW and 8 µW is nearest. Treating the 6 dB as a gain gives 119 µW, nearest the 120 µW option.

    Try it: dBm, Watts & dB
  73. 73. Why is the manual’s diode ring mixer called double balanced?
    Show answer

    Its output holds neither the RF nor the LO input

    Double balanced means that neither input signal, RF or LO, appears at the IF output; single balanced keeps out only one. Every mixer gives the sum and difference, so that is not what the name means.

  74. 74. An ideal double-balanced mixer has an 11 MHz LO and a 21 MHz signal at its RF port. Which are the main frequencies at its output?
    Show answer

    10 MHz and 32 MHz

    The products are 21 − 11 = 10 MHz and 21 + 11 = 32 MHz, and a double-balanced mixer keeps both inputs out of its output. With the 11 and 21 MHz inputs present it would be unbalanced.

  75. 75. Apart from the mixing products, what else does a single-balanced mixer let through to its output?
    Show answer

    One input, usually the RF signal

    A single-balanced mixer keeps one input out of its output but lets the other through, typically the weaker RF signal. Both inputs get through only in an unbalanced mixer.

  76. 76. Which signals reach the output of an unbalanced mixer, besides the mixing products?
    Show answer

    Both the RF and the LO inputs

    An unbalanced mixer does nothing to keep its inputs out, so the RF and LO inputs both appear at its output. Letting only one of them through is the single-balanced case.

  77. 77. Which circuit can move an AM signal to a new frequency without spoiling its modulation?
    Show answer

    A frequency mixer

    A properly designed mixer adds little distortion, so it can shift AM, SSB and even audio signals. A synthesiser generates its own signal from a reference; it cannot carry an existing AM signal to a new frequency.