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 as compared to an "ideal" receiver, which exists only in theory.

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

Hint: The only the correct answer has the words receiver and theoretical in it.

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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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Receiver noise floor increases directly with increased receiver bandwidth since a wider bandwidth allows proportionally more background noise into the receiver. This becomes a log calculation of the dB change from 50Hz to 1000Hz as follows;

\[ 10\log{\frac{1000\text{Hz}}{50\text{Hz}}} = 10\log{20} = 13.01 dB\]

If you'd rather avoid the log calculation and estimate the answer, you can consider that:

\begin{align} 50\text{Hz} \times 10 &= 500\text{Hz (10dB)}\\ &+\\ 500\text{Hz} \times 2 &= 1000\text{Hz (3dB)}\\ &= 13\text{dB} \end{align}

So 13dB total increase in noise floor.

Funny Hint: it's unlucky (13) to be near a noisy floor

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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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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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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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Received signals are mixed with a local oscillator (LO) and down-converted to an intermediate frequency (IF). In the IF stage amplification and filtering are applied. The IF shift control allows the operator to move the IF filtering window up and down without changing the tuning of the radio. In this way the operator can slide the filter edge over a nearby interfering signal while allowing the desired signal through.

The other answers refer to changing to or listening to different frequencies, which IF shifting does not do.

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