Login or Register for FREE!
Subelement G8

SIGNALS AND EMISSIONS

Section G8B

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)?

  • Image frequency
  • Correct Answer
    Local oscillator
  • RF input
  • Beat frequency oscillator

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.

Last edited by edponce@yahoo.com. Register to edit

Tags: none

What is the term for interference from a signal at twice the IF frequency from the desired signal?

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

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.

For further study: https://en.m.wikipedia.org/wiki/Software-defined_radio

Last edited by mprindle. Register to edit

Tags: none

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

  • Correct Answer
    Heterodyning
  • Synthesizing
  • Frequency inversion
  • Phase inversion

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

Last edited by katzelover. Register to edit

Tags: none

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
  • Balanced converter
  • 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

Last edited by gconklin. Register to edit

Tags: none

Which intermodulation products are closest to the original signal frequencies?

  • Second harmonics
  • Even-order
  • Correct Answer
    Odd-order
  • Intercept point

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!

Last edited by turnncough. Register to edit

Tags: none

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:

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


Wikipedia ► Carson bandwidth rule

Last edited by myzhang1029. Register to edit

Tags: none

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

Last edited by dogshed. Register to edit

Tags: none

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 that could exceed the transmitter’s average power rating
  • To allow time for the other station to break in during a transmission
  • To prevent overmodulation

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

Last edited by kd7bbc. Register to edit

Tags: none

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

Last edited by kd7bbc. Register to edit

Tags: none

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 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

Last edited by pjmotor12613. Register to edit

Tags: none

What combination of a mixer’s Local Oscillator (LO) and RF input frequencies is found in the output?

  • The ratio
  • The average
  • Correct Answer
    The sum and difference
  • The arithmetic product

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

Last edited by edponce@yahoo.com. Register to edit

Tags: none

What process combines two signals in a non-linear circuit to produce unwanted spurious outputs?

  • Correct Answer
    Intermodulation
  • Heterodyning
  • Detection
  • Rolloff

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.

Last edited by edponce@yahoo.com. Register to edit

Tags: none

Which of the following is an odd-order intermodulation product of frequencies F1 and F2?

  • 5F1-3F2
  • 3F1-F2
  • Correct Answer
    2F1-F2
  • All these choices are correct

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.

Last edited by nschroe. Register to edit

Tags: none

Go to G8A Go to G8C