Hint: the local oscillator is the TUNER (VFO) on your receiver

In the front end of a receiver, the incoming signal is at a fixed frequency. The only thing we have control over is the LO frequency. As this local oscillator is varied, a different incoming RF signal is mixed with it and presented to the IF. This facilitates "tuning" the radio to different RF frequencies.

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

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

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Wikipedia ► Carson bandwidth rule

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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}

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

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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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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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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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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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Hint: SUM of both gives you DIFFERENCE ( -10 + 5)

A mixer combines two input frequencies to produce an output signal containing both frequencies. For example, a Local Oscillator signal of frequency A is combined with an RF signal of frequency B. When mixed, the output signal will contain a signal of frequencies (A + B) and abs(A - B).

Mixers have many applications, such as in superheterodyne receivers, which convert very high frequencies down to lower frequencies that are more easily handled by the radio. In that case, where you want a lower frequency, the A+B signal will be filtered out.

Sometimes you may want to produce a signal with a higher frequency; in that case you can use a mixer to generate A+B and A-B signals, and simply filter out the A-B signal.

https://en.m.wikipedia.org/wiki/Frequency_mixer

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Many situations can result in intermodulation: two signals mixing in the preamp as they are received, two transmitters combining their outputs into a single antenna, two sources of RF near a non-linear junction (the rusty bolt syndrome), or other situations in which two (or more) sources of RF are present in the same area. The result is that the non-linear junction produces the same mixing effect as a mixer, and creates a third (or fourth) signal which is then heard by a receiver.

Hint: Unwanted / Inter(upted)modulation

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