SIGNALS AND EMISSIONS
SIGNALS AND EMISSIONS
Frequency changing; bandwidths of various modes; deviation; intermodulation
Which mixer input is varied or tuned to convert signals of different frequencies to an intermediate frequency (IF)?
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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What is the term for interference from a signal at twice the IF frequency from the desired signal?
In Software Defined Radios (SDRs), "image response" refers to interference that occurs when there is an unwanted signal present at a frequency that is double the Intermediate Frequency (IF) of the desired signal. The IF is a lower frequency to which the received radio signal is converted before further processing in an SDR.
The mixing process in an SDR can sometimes unintentionally create these interference signals at the double IF frequency. Image response interference can degrade the quality and clarity of the desired signal, making it harder to receive and interpret the intended communication.
Silly hint: I for IF, I for image. Also, since this is an SDR thing, you can remember images being transmitted by software.
For further study: https://en.m.wikipedia.org/wiki/Software-defined_radio
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What is another term for the mixing of two RF signals?
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?
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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Which intermodulation products are closest to the original signal frequencies?
If the original frequencies are near each other, then the difference frequency is relatively close to DC (0). So the odd frequencies like 2f1 - f2 can be rewritten as f1 + (f1-f2), which is relatively close to f1.
Silly hint: Original -> Odd-Order or O O O it's Magic, you know!
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What is the total bandwidth of an FM phone transmission having 5 kHz deviation and 3 kHz modulating frequency?
\[(f_{\text{dev}} + f_{\text{mod}}) \times 2 = \text{FM Bandwidth}\]
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:
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?
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: 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?
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?
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?
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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What combination of a mixer’s Local Oscillator (LO) and RF input frequencies is found in the output?
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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What process combines two signals in a non-linear circuit to produce unwanted spurious outputs?
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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Which of the following is an odd-order intermodulation product of frequencies F1 and F2?
For even order intermodulation products, the absolute values of the coefficients to the frequencies will add up to an even number--so in the case of 5F1-3F2 the order is 5+3=8, which is even, and for 3F1-1F2 the order is 3+1=4, which is also even. The order for 2F1-1F2 is 2+1=3, so it is an odd-order IM product.
Silly Hint: It's odd that its the even one.
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