or
Amateur Extra Class (2016-2020)
Subelement E4
AMATEUR PRACTICES
Section E4C
Receiver performance characteristics, phase noise, noise floor, image rejection, MDS, signal-to-noise-ratio; selectivity; effects of SDR receiver non-linearity
What is an effect of excessive phase noise in the local oscillator section of a receiver?
• It decreases receiver third-order intermodulation distortion dynamic range
It can cause strong signals on nearby frequencies to interfere with reception of weak signals

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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Which of the following portions of a receiver can be effective in eliminating image signal interference?
A front-end filter or pre-selector
• A narrow IF filter
• A notch filter
• A properly adjusted product detector

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.

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What is the term for the blocking of one FM phone signal by another, stronger FM phone signal?
• Desensitization
• Cross-modulation interference
Capture effect
• Frequency discrimination

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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How is the noise figure of a receiver defined?
• The ratio of atmospheric noise to phase noise
• The ratio of the noise bandwidth in Hertz to the theoretical bandwidth of a resistive network
• The ratio of thermal noise to atmospheric noise
The ratio in dB of the noise generated by the receiver to the theoretical minimum noise

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

-KE0IPR

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What does a value of -174 dBm/Hz represent with regard to the noise floor of a receiver?
• The minimum detectable signal as a function of receive frequency
The theoretical noise at the input of a perfect receiver at room temperature
• The noise figure of a 1 Hz bandwidth receiver
• The galactic noise contribution to minimum detectable signal

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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A CW receiver with the AGC off has an equivalent input noise power density of -174 dBm/Hz. What would be the level of an unmodulated carrier input to this receiver that would yield an audio output SNR of 0 dB in a 400 Hz noise bandwidth?
• -174 dBm
• -164 dBm
• -155 dBm
-148 dBm

A signal-to-noise ratio of $0$ dB means $1$:$1$ [because $10\log{\left(\frac{1}{1}\right)} = 0$], so 400 Hz of bandwidth translates to a 400 Hz of output level.

This output level is:

$10\log{\left(400 \frac{\text{Hz}}{1}\right)} = 26 \text{ dBm}$

relative to the noise floor, which results in a carrier input level of

$-174 \text{ dBm} + 26 \text{ dBm} = -148 \text{ dBm}$

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.

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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What does the MDS of a receiver represent?
• The meter display sensitivity
The minimum discernible signal
• The multiplex distortion stability
• The maximum detectable spectrum

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

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• One-half the maximum sample rate
• One-half the maximum sampling buffer size
The maximum count value of the analog-to-digital converter
• The reference voltage of the analog-to-digital converter

An SDR receiver is overloaded when input signals exceed the maximum count value of the analog-to-digital converter, which is determined by The reference voltage of the analog-to-digital-converter.

In an SDR receiver, after the analog signal is converted to a digital representation, the amplitude of the signal is stored as a "count value." There are just so many bits that can be used to represent the amplitude, and if you need more bits than you have to represent the amplitude, you can no longer see any difference -- stronger signals can only have that maximum count value. When the SDR has no more bits to represent the signal's amplitude, the receiver is considered overloaded.

The maximum count value is reached at the reference voltage, which is considered to be represented by the maximum count value.

Hint: It's overloaded when it's "more than you can count." N5Lab

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Which of the following choices is a good reason for selecting a high frequency for the design of the IF in a conventional HF or VHF communications receiver?
• Fewer components in the receiver
• Reduced drift
Easier for front-end circuitry to eliminate image responses

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.

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Which of the following is a desirable amount of selectivity for an amateur RTTY HF receiver?
• 100 Hz
300 Hz
• 6000 Hz
• 2400 Hz

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

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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Which of the following is a desirable amount of selectivity for an amateur SSB phone receiver?
• 1 kHz
2.4 kHz
• 4.2 kHz
• 4.8 kHz

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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What is an undesirable effect of using too wide a filter bandwidth in the IF section of a receiver?
• Output-offset overshoot
• Filter ringing
• Thermal-noise distortion
Undesired signals may be heard

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

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How does a narrow-band roofing filter affect receiver performance?
• It improves sensitivity by reducing front end noise
• It improves intelligibility by using low Q circuitry to reduce ringing
It improves dynamic range by attenuating strong signals near the receive frequency
• All of these choices are correct

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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What transmit frequency might generate an image response signal in a receiver tuned to 14.300 MHz and which uses a 455 kHz IF frequency?
• 13.845 MHz
• 14.755 MHz
• 14.445 MHz
15.210 MHz

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.

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What is usually the primary source of noise that is heard from an HF receiver with an antenna connected?
• Detector noise
• Induction motor noise
Atmospheric noise

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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Which of the following is caused by missing codes in an SDR receiver's analog-to-digital converter?
Distortion
• Loss of sensitivity
• Excess output level

Distortion is the condition that overloads an SDR receiver to the point that its internal analog to digital buffer becomes too full to process additional signal. It can't keep up with the signal's input - thus missing codes. Thus attenuating your input signal will bring the throughput of the buffer to a more manageable level.

-KE0IPR

When you overload a SDR, you'll get DISTORTION because the A/D converter is running out of code values (bits) to convert the signal with such high amplitude.

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Which of the following has the largest effect on an SDR receiver's linearity?
• CPU register width in bits
• Anti-aliasing input filter bandwidth
• RAM speed used for data storage
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.