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Subelement G8

Signals and Emissions

Section G8B

Frequency mixing; multiplication; bandwidths of various modes; deviation; duty cycle

What receiver stage combines a 14.250 MHz input signal with a 13.795 MHz oscillator signal to produce a 455 kHz intermediate frequency (IF) signal?

  • Correct Answer
    Mixer
  • BFO
  • VFO
  • Discriminator

(A). The mixer is the receiver stage which would combine a 14.250 MHz input signal with a 13.795 MHz oscillator signal to produce a 455 kHz intermediate frequency (IF) signal.

For single sideband transmissions a mixer is used to combine the two input signals (14.250 MHZ, and 13.795) to produce a new carrier signal which is an intermediate median frequency. Along with this new IF signal, two side signals are produced, the first being the sum of the two input frequencies, which becomes the upper sideband frequency (28.045 MHz), and the second being the difference between the input frequencies which becomes the lower sideband frequency (0.455 MHz = 455 kHz). Filters, etc are then used to suppress the undesired portions of the signal and allow the desired sideband (455 kHz) to pass for SSB transmittion.

For more info see Wikipedia: Single-sideband modulation, Frequency Mixer

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Tags: arrl chapter 5 arrl module 18

If a receiver mixes a 13.800 MHz VFO with a 14.255 MHz received signal to produce a 455 kHz intermediate frequency (IF) signal, what type of interference will a 13.345 MHz signal produce in the receiver?

  • Quadrature noise
  • Correct Answer
    Image response
  • Mixer interference
  • Intermediate interference

Image response would be the type of interference produced. The mixer produces intermediate frequencies based on the sum (USB) and the difference (LSB) between the two input frequencies.

The difference between the 14.255 MHz and 13.800 MHz signals would produce an LSB intermediate frequency of 0.455 MHz = 455 kHz. However, when the 13.345 MHz signal is encountered, the difference between 13.800 MHz and 13.345 MHz is ALSO 0.455 MHz = 455 kHz.

When the interference is produced at the same frequency as the desired frequency, this is called image response interference. The remedy is to use input filtration prior to the mixing stage.

For more info see Wikipedia: image response interference

Silly Hint: With all the math, this question is an ugly image of a question, so it makes you think of an ugly "Image response"

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Tags: arrl chapter 5 arrl module 20

What is another term for the mixing of two RF signals?

  • Correct Answer
    Heterodyning
  • Synthesizing
  • Cancellation
  • Phase inverting

Heterodyning is the term listed which refers to the mixing of two RF signals. The heterodyning system uses a local HF oscillator, a mixer, and detector to modulate the carrier signal producing upper and lower sidebands.

For more info see Wikipedia: heterodyning

Hint: Hetero means different or other like 2 different signals.

Memory aid: When you are dining (dyning) somewhere there may be a mix of foods.

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What is the stage in a VHF FM transmitter that generates a harmonic of a lower frequency signal to reach the desired operating frequency?

  • Mixer
  • Reactance modulator
  • Pre-emphasis network
  • Correct Answer
    Multiplier

Hint: A harmonic has MULTIPLE sounds

The frequency multiplier is the stage in a VHF FM transmitter that generates a harmonic of a lower frequency signal to reach the desired operating frequency. The frequency multiplier produces signals at harmonic multiples (double, triple, etc) of the modulated signal to bring the frequency to the desired output level.

For more info see Wikipedia: Frequency multiplier

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What is the approximate bandwidth of a PACTOR3 signal at maximum data rate?

  • 31.5 Hz
  • 500 Hz
  • 1800 Hz
  • Correct Answer
    2300 Hz

PACTOR-III's maximum uncompressed speed is 2722 bps. Using online compression, up to 5.2 kbps is achievable. This requires an audio passband from 400 Hz to 2600 Hz (for PACTOR-III speedlevel 6). The occupied bandwidth at the --40 dB points is 2.4 kHz (from 300 Hz to 2700 Hz).

Also if helpful think PACTOR 3 --> 2(3)00 Hz

or

PACTOR has 2 syllables, PACTOR 3 --> 2300 Hz

for more information on PACTOR III: PACTOR III

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Tags: arrl chapter 6 arrl module 26

What is the total bandwidth of an FM phone transmission having 5 kHz deviation and 3 kHz modulating frequency?

  • 3 kHz
  • 5 kHz
  • 8 kHz
  • Correct Answer
    16 kHz

\[(f_{\text{dev}} + f_{\text{mod}}) \times 2 = \text{FM Bandwidth}\]

To calculate the total bandwidth of an FM phone transmission, use Carson's bandwidth rule by adding together the frequency deviation and the modulating frequency then multiplying the sum by \(2\):

\[BW = (f_{\text{dev}} + f_{\text{mod}}) \times 2\]

Where:

  • \(BW\) is the total FM phone bandwidth
  • \(f_{\text{dev}}\) is the frequency deviation
  • \(f_{\text{mod}}\) is the modulating frequency

So for this question:

\[f_{\text{dev}} = 5 \text{kHz}\] \[f_{\text{mod}} = 3 \text{kHz}\]

Therefore:

\[BW = (5 \text{kHz} + 3 \text{kHz}) \times 2\] \[BW = 8 \text{kHz} \times 2\] \[BW = 16 \text{kHz}\]

Memory aid: \(5 \times 3 = 15\) and 16 is the closest


Wikipedia ► Carson bandwidth rule

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What is the frequency deviation for a 12.21 MHz reactance modulated oscillator in a 5 kHz deviation, 146.52 MHz FM phone transmitter?

  • 101.75 Hz
  • Correct Answer
    416.7 Hz
  • 5 kHz
  • 60 kHz

416.7 Hz is the frequency deviation for a 12.21 MHz reactance modulated oscillator in a 5-kHz deviation, 146.52 MHz FM phone transmitter.

To calculate the frequency deviation, first calculate the multiplication factor of the FM transmitter:

\begin{align} \small \text{Multiplication Factor} &= \small { \frac{\text{Transmitter Frequency}}{\text{HF Oscillator Frequency}}}\\ &= \frac{146.52\ \text{MHz}}{12.21\ \text{MHz}}\\ &= 12 \end{align}

Next, divide the transmitter deviation by the multiplication factor:

\begin{align} \small \text{Frequency deviation} &= \small{ \frac{\text{Transmitter Deviation}}{\text{Multiplication Factor}}}\\ &= \frac{ 5\ \text{kHz}}{12}\\ &= \frac{ 5000\ \text{Hz}}{12}\\ &= 416.7\ \text{Hz} \end{align}


Silly Hint: 146 think 416, and 146.52, so 5 + 2 = 7, and 416.7.


Alt: The standard FM deviation allowed is 5 KHz so, 5 ÷ 12 = 0.416 KHz, or 416 Hz

Alt: 5/150 simplifies to 1/30. 1/30th of 12 is 0.4

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Why is it important to know the duty cycle of the mode you are using when transmitting?

  • To aid in tuning your transmitter
  • Correct Answer
    Some modes have high duty cycles which could exceed the transmitter's average power rating
  • To allow time for the other station to break in during a transmission
  • All of these choices are correct

It is important to know the duty cycle of the data mode you are using when transmitting because some modes have high duty cycles which could exceed the transmitter's average power rating.

Data modes vary in the percentage of time that they are actually transmitting at full power versus the amount of "off time" between signals. This is referred to as the duty cycle. As an example, the intermittant dots and dashes of CW mean that the transmitter is actually only operating at full power for somewhere around 40 to 50% of the time. Some of the RTTY data modes on the other hand, can actually run at full power for close to 100% of the transmission time. Because these modes have such high duty cycles, it may be necessary to reduce the power used so that the transmitter's average power rating is not exceeded.

SILLY HINT: Do your "duty" "duty"

For more info see Wikipedia: Duty cycle

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Why is it good to match receiver bandwidth to the bandwidth of the operating mode?

  • It is required by FCC rules
  • It minimizes power consumption in the receiver
  • It improves impedance matching of the antenna
  • Correct Answer
    It results in the best signal to noise ratio

Matching bandwidths reduces the amount of noise outside the desired frequency range.

Doing so means less energy is lost in filtering, resulting in a better signal-to-noise ratio.

For more info see Wikipedia: Signal-to-noise ratio

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What is the relationship between transmitted symbol rate and bandwidth?

  • Symbol rate and bandwidth are not related
  • Correct Answer
    Higher symbol rates require wider bandwidth
  • Lower symbol rates require wider bandwidth
  • Bandwidth is always half the symbol rate

The relationship between transmitted symbol rate and bandwidth is that higher symbol rates require higher amounts of bandwidth.

As the symbol rate for a data transmisson increases (baud rate), the amount of bandwidth required to send that signal must also increase in order to maintain a low signal-to-noise ratio.

Hint: More information requires more space.

For more info see Wikipedia: Symbol rate

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