According to Noise in Mixers, Oscillators, Samplers, and Logic (Joel Phillips and Kent Kundert, The Designer's Guide Community, May 2000, p. 7-8), the phase noise from the receiver's local oscillator can mix with a strong interfering signal from a neighboring channel and swamp out the signal from the desired, weaker channel in an effect known as reciprocal mixing. Because the excessive phase noise in the local oscillator section can cause strong signals on nearby frequencies to interfere with weaker incoming signals, the answer is: It can cause strong signals on nearby frequencies to interfere with reception of weak signals.

Furthermore, because the receiver's ability to receive strong signals is not limited, but enhanced (mixed), answer (A) is eliminated. It may be viewed that the receiver's sensitivity to the weaker incoming signal might be reduced, but overall the phase noise has little effect on its sensitivity, eliminating answer (B). Finally, intermodulation distortion is filtered in stages prior to mixing with the strong interfering signal, so the possible reduction in dynamic range does not affect the result of the phase noise mixing with the strong signal, eliminating answer (C). (Note that answer order may be scrambled on the question where you read this.)

-nojiratz

Mnemonic: Noise Interferes. The correct choice is the only answer including the word 'interferes.'

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When combined with the signal of the local oscillator, two different input signals may generate the intermediate frequency (by sum or difference). The input signal we are NOT interested in is called the image.

We can design the IF so that the image falls outside of the amateur band. A front end filter is a band-pass filter for a whole amateur band. With this, we ensure that only the signal we are interested in is translated to the IF.

Product detector or a narrow IF filters would not work, as by then, the image already interfered with the desired signal. A notch filter could work, but it has limited applicability, as it can only reject around one frequency.

Hint: The question and the correct answer each have two hyphens in them.

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Capture effect is an FM phenomenon in which given two signals at or near the same frequency, a receiver will demodulate only the stronger of the two. The weaker signal is effectively suppressed.

Desensitization occurs when a transmitter in close proximity and frequency to a receiver, without adequate isolation, causes interference and makes reception of weaker signals difficult.

Memory tip: strong warriors capture weaker ones.

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Hint: Since were talking about ratios, the correct answer is the only one with dB mentioned in it.

See Wikipedia for more information: https://en.wikipedia.org/wiki/Noise_figure

Hint: The question and only the correct answer have the word receiver in them.

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Memorize: Every room (in answer) has a floor (in question)

floor/room = DONE!

You're welcome

The inherent (average) energy in relation to absolute temperature is given by

\[E = kT\]

in which:

• \(k\) is Boltzmann's Constant (\(1.38 \times 10^{-23} \frac{\text{J}}{\text{K}}\)), and
• \(T\) is the absolute temperature in Kelvins (\(\text{K}\))

The inherent (average) power (measured in Watts, or \(\frac{\text{J}}{\text{s}}\)) for a given bandwidth \(B\) (measured in Hz, which is actually \(\text{s}^{-1}\)) is therefore:

\[P = kTB\]

The theoretical noise floor is defined as the power Pm (the power in mW) from a bandwidth of \(1\) Hz at room temperature (\(290\text{ K}\)).

From "Planar Microwave Engineering" by Thomas H. Lee, 2004 (Cambridge University Press, Cambridge, UK; ISBN 0-521-83526-7) p. 441, section 13.2.1, footnote 1 states

"...290 K as the reference temperature had particular appeal in an era of slide-rule computation, and it was adopted rapidly by engineers and ultimately by standards committees."

The theoretical noise floor power in mW is therefore

\begin{align} Pm &=\left(1.38\times 10^{-23} \frac{\text{J}}{\text{K}}\right)(290 \text{ K})(1 \text{ Hz})\left(1000 \frac{\text{mW}}{\text{ W}}\right) \\ &= 4.002 \times 10^{-18} \text{ mW}\\ \end{align}

This converts to:

\[\text{Theoretical noise floor}_{\text{(in dB relative to mW)}}\]

\[ \begin{align} &= 10\log{(\frac{4.002 \times 10^{-18} \text{ mW}}{1 \text{ mW}})} \\ &= 10\log{(4.002 \times 10^{-18})} \\ &= -173.977 \text{ dBm} \\ &≈ -174 \text{ dBm} \\ \end{align} \]

This results in \(-174 \text{ dBm}\) for each \(\text{Hz}\) of bandwidth, or \(-174 \frac{\text{dBm}}{\text{Hz}}\)

-wileyj2956

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Calculate the ratio of 400 Hz to 1 Hz in dB:

\[10\log{\left(\frac{400 \text{ Hz}}{1 \text{ Hz}}\right)} = 26 \text{ dB}\]

Now just add:

\[-174 \text{ dBm} + 26 \text{ dB} = -148 \text{ dBm}\]

This could also be done by converting -174 dBm to milliwatts and dividing by 400. But dB is easier.

The thermal noise value is given per Hz. A receiver with \(400\text{ Hz}\) bandwidth will therefore have a noise floor \(400\) times higher than this. Converting this to decibels gives \(10 \log{(400)} = 26\text{ dB}\). So the receiver noise floor is \(26\) dB higher than the \(-174 \frac{\text{dBm}}{\text{Hz}}\) value, giving \(-148\text{ dBm}\).

Since the noise floor is also the level of the minimum detectable signal, \(-148\text{ dBm}\) is correct.

See also: https://en.wikipedia.org/wiki/Minimum_detectable_signal

CW - continuous wave
AGC - automatic gain control
SNR - signal-to-noise ratio

Silly trick.

174... 4 is last.

400... 4 is first.

only one answer with "4" in the middle

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According to Modern Communication Circuits (Jack Smith, McGraw Hill, 1998), p. 82, the MDS is the minimum detectable signal or minimum discernible signal, by definition.

You can also check out Minimum detectible signal on the Wikipedia

Hint: You receive a signal.

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An SDR receiver is overloaded when input signals exceed the reference voltage of the analog-to-digital-converter.

Overload of a software defined radio (SDR) is reached when the combination of all signals at the receiver's analog-to-digital converter exceed the maximum level, also called "clipping," for which the converter generates a unique digital value. Maximum signal value is controlled by the converter's reference voltage.

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According to The Art of Electronics (Cambridge University Press, 2006) p. 886, the basic idea behind the superheterodyne receiver is that, because it's possible for more than one signal to enter the IF (Intermediate Frequency) amplifier, the unwanted (mirror image) signals must be eliminated. But according to The Technician's Radio Receiver Handbook (Joseph J. Carr, Newnes, 2001) p. 8-9, the superheterodyne design suffers from the difficulty of image rejection with increasing RF frequency due to the sharp filtering necessary to reject the unwanted signals while maintaining appropriate gain. According to RF Components and Circuits (Joseph J. Carr, Newnes, 2002) ch. 3, this is largely overcome by selecting a high frequency for the IF, thereby reducing the need for a sharply tuned circuit prior to the mixer, making it easier for the front-end circuitry to reject the mirror images.

Hint: 2 questions have 'Front-End' answers, if it has '-' in middle 'Front-End' is the answer, no hyphen not the answer ; )

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RF signals occupy more than just a single frequency, but a range of frequencies. The width of that range is called "bandwidth". The width depends on the type of modulation and the amount of information contained within the signal. The ability to adapt the receiver's bandwidth to the width of the signal to be received reduces extraneous noise that is received thus reducing interference from other nearby signals and improving the signal-to-noise ratio.

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One useful thing to remember about HF radio is that as the frequency decreases, the noise from geomagnetic activity (caused by space weather) increases.

So as you get to the lower HF bands like 80m and 160m, Atmospheric noise is generally greater than internally generated noise even after attenuation. (Just remember that lower HF bands have more atmospheric noise and you will get this one right.)

As you crank up the attenuation, band noise and signals move down together. As long as you don't attenuate the band noise to below the noise floor of the receiver, the SNR will not be affected.

(Diagram by VE2HEW.)

Another way to think of it is, say SNR is 9:1. Attenuating that by 50% is \(0.5 \frac{9}{1} = \frac{4.5}{0.5} = 9:1\), which is still the same radio as 9:1 and holds true as long as the internal noise is not greater than 0.5, otherwise if the internal noise floor was say 0.7 then you'd end up with \(\frac{4.5}{0.7} = 6.428:1\)

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The dynamic range is the ratio between the full scale value and smallest value that the ADC is designed to measure. It is directly related to the number of bits that are used to digitize the analog signal. For a given N-bit ADC, say the minimum value that can be detected is \(v_{0}\). The full scale value then is \(\left (2^N-1\right)\) times the \(v_{0}\). Hence, in terms of decibels, the dynamic range of the ADC will be:

\[DR = 20\times \log_{10}\left ( \frac{\left(2^N-1\right )\times v_{0}}{v_{0}} \right )\\ \approx 6.021\times N \]

For example, a 10-bit ADC has the dynamic range of 60dB, whereas for a 12-bit ADC dynamic range is 72dB. Every additional bit increases the dynamic range of an ADC by \(\approx\) 6dB.

Memory trick: The largest effect is the largest answer.

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A roofing filter is placed early in the IF amplifier chain and shields the later IF filter stages from strong signals.

Silly mnemonic device ... the shape of a roof ^ is an image of a strong signal in the middle and attenuated or lower signals on either side.

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Quick tip:

An image response signal may be located at the tuned frequency \(\pm 2 \times\) the intermediate frequency:

\[14.300\text{ MHz} + (455\text{ kHz} \cdot 2) = 15.210\text{ MHz}\] \[14.300\text{ MHz} - (455\text{ kHz} \cdot 2) = 13.390\text{ MHz}\]

Of the two results, only one is in the list of possible answers: \(15.210\text{ MHz}\).

Explanation:

An input signal can be mixed with a local oscillator (LO) to produce the difference of the two signals. By changing the LO's frequency, different frequencies in the input signal are "shifted" (aka tuned) to another constant output frequency, called the intermediate frequency (IF), which simplifies processing.

The problem is that there are two frequencies which produce the IF in this situation: \(f_1\) minus the LO, and the LO minus \(f_2\). One is the desired frequency, and the other is called an image.

The trick here is that we don't know the frequency of the LO, nor whether it's higher or lower than the desired frequency. All we know is that the difference is \(455\text{ kHz}\). So first we find the possible LOs:

\[14.500\text{ MHz} \pm 455\text{ kHz} =\]

\[14.755\text{ MHz or } 13.845\text{ MHz}\]

If the LO is above the desired frequency, the image will be above the LO by the same distance, and vice versa.

If LO is \(14.755\) (higher): \[14.755\text{ MHz} + 455\text{ kHz} = 15.210\text{ MHz}\] If LO is \(13.845\text{ MHz}\) (lower):

\[13.845\text{ MHz} - 455\text{ kHz} = 13.390\text{ MHz}\]

Only \(15.210\text{ MHz}\) is in the list of possible answers.

For more information, Alan Wolke, W2AEW, has an excellent video explaining as well as demonstrating this concept. View it here: https://www.youtube.com/watch?v=Mm7WfVzr1ao

Pay close attention to his initial notes when he talks about "3rd Order", as well as the image signal demonstration at the 12 minute mark.

This assumes no IF filter

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This is a rather obscure question and a full explanation is a bit much for this flashcard, but here is some info and some ways to remember the answer.

The key term to remember here is phase noise. Reciprocal mixing is a term for Local oscillator phase noise mixing with adjacent strong signals to create interference to desired signals.

A large but narrow interfering signal can have its energy spread over a broad range of frequencies by the phase noise of the receiver's LO, and this is what happens in reciprocal mixing.

Phase noise could be thought of as a sort of phase jitter, or as noise on the Q component of an IQ signal.

Hint: Only look at answers that have the word "mixing" in it, and you will narrow this down to two possible answers.

More information is available in this article on phase noise.

Hint: You mix to create something.

Hint - the question has reciprocal and the answer has local. Reciprocal and local both end with "ocal"

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