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

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

floor/room = DONE!

You're welcome

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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.

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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

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

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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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This means that the receiver could detect weaker signals

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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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The desirable amount of selectivity for an amateur RTTY HF receiver is 300 Hz.

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The bandwidth of a RTTY transmission on the HF bands is about 300Hz. A smaller receiver bandwidth would result in loss of some of the sidebands while a larger bandwidth would allow more noise and interference to be received.

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The bandwidth of a SSB signal is about 2.4 kHz. A smaller receiver bandwidth would result in loss of some of the sideband with consequent loss of intelligibility while a larger bandwidth would allow more noise and interference to be received.

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Receiver bandwidth is often a compromise between signal intelligibility and interference reduction especially on crowded HF bands.

Remember "Undesirable leads to Undesired."

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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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According to CRC Handbook of Atmospherics, "High Frequency Radio Noise" (E.A. Lewis, CRC Press, 1982), p. 251-288, the main sources of HF noise is that generated from impulse-type electromagnetic radiation (sometimes called atmospherics), or atmospheric noise.

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