Noise in Analog Systems Questions and Answers

Q: Which measurement determines a receiver's noise figure?

The Y-factor method compares the output with two known noise sources at different temperatures, which yields the noise figure without needing an absolute power calibration. A swept generator measures frequency response. Audio output with no antenna gives residual noise but no reference. And the detector's direct output measures signal strength. Any measurement of a degradation needs two known conditions, which is what the two sources provide.

Q: What does a noise diode provide in such a measurement?

It supplies a known excess noise, quoted as an excess noise ratio, which serves as the hot source in the Y-factor method. Reference frequencies come from crystals. A modulated carrier is a signal generator's output. And a matched termination is the cold source, at ambient temperature. Each instrument in a noise measurement supplies one known quantity, and the arithmetic then yields the unknown.

Q: Which quantity is unchanged by inserting a matched attenuator ahead of a receiver?

The ratio at the antenna is a property of what arrives there and is unaffected by anything downstream, whereas the attenuator worsens the system noise figure and reduces both signal and noise power at the receiver. That distinction matters because it explains why attenuation before the gain is damaging while attenuation after it is harmless: the ratio delivered to the detector is what counts, not the ratio available at the antenna.

Q: Why is a signal-to-noise ratio quoted with its bandwidth?

Because noise power is density times bandwidth, a ratio quoted without a bandwidth is ambiguous: the same signal gives very different ratios in a kilohertz and a megahertz. Signal power is fixed by the transmitter and the path. The ratio is indeed dimensionless but that is not why the bandwidth matters. And the noise figure is a device property independent of the bandwidth chosen. Any noise figure quoted without its conditions should be treated with suspicion.

Q: Three schemes are compared at the same channel signal-to-noise ratio. SchemeFigure of meritBandwidth AM at unity index0.332 f mSSB1.0f mFM at index 545012 f m Which conclusion follows?

Reading across, FM buys a figure of merit of 450 by spending twelve times the message bandwidth, which is the trade in one line. SSB has the best bandwidth efficiency rather than the best output ratio. AM is worst on both counts and offers no compromise at all beyond receiver simplicity. And the figures plainly differ. Tabulating the schemes against both resources is the compact way to hold the comparison.

Q: Which noise source would a directional antenna help against?

A directional antenna can null an external interferer arriving from a particular bearing, which is why it is the first remedy for local electrical noise. The other three are generated inside the receiver and no antenna can influence them. Separating external from internal noise decides whether the remedy lies in siting and antennas or in devices and circuit design, and it is the first question to ask about any noise complaint. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Q: What is the effect of doubling a receiver's noise bandwidth?

Noise power is proportional to bandwidth, so doubling it raises the floor by 3 dB and halves the signal-to-noise ratio for the same signal. Six decibels would correspond to a fourfold increase. The floor certainly changes. And the noise figure is a device property independent of bandwidth, so it does not double. This is why a receiver's bandwidth is kept no wider than the signal requires, and why the matched filter is optimal.

Q: Which is the correct order of increasing figure of merit?

The order is AM at a third, SSB and DSB-SC at unity, and wideband FM at hundreds, which is the exact reverse of the bandwidth ordering. That inversion is the central trade of analogue communication: every improvement in noise performance is bought with spectrum, and every saving in spectrum is paid for in noise. The reversed and jumbled orders each misplace at least one scheme.

Q: What does the bandwidth expansion factor measure?

It is the ratio of transmitted to message bandwidth, so it is two for AM and DSB, one for SSB, and about twelve for FM broadcasting. The output-over-channel ratio is the figure of merit, its natural partner in any comparison. The noise-to-3 dB bandwidth ratio is a filter detail. And deviation over message frequency is the modulation index. Plotting figure of merit against expansion factor is how the schemes are properly compared. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

41.

Which measurement determines a receiver's noise figure?

A.Measuring the audio output with the antenna disconnected
B.Reading the direct voltage at the detector output
C.Comparing outputs with a hot and cold source
D.Sweeping a signal generator across the passband

Answer: C

The Y-factor method compares the output with two known noise sources at different temperatures, which yields the noise figure without needing an absolute power calibration. A swept generator measures frequency response. Audio output with no antenna gives residual noise but no reference. And the detector's direct output measures signal strength. Any measurement of a degradation needs two known conditions, which is what the two sources provide.

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42.

What does a noise diode provide in such a measurement?

A.A reference frequency for the local oscillator
B.A modulated carrier of known depth
C.A matched termination of known resistance
D.A calibrated excess noise source

Answer: D

It supplies a known excess noise, quoted as an excess noise ratio, which serves as the hot source in the Y-factor method. Reference frequencies come from crystals. A modulated carrier is a signal generator's output. And a matched termination is the cold source, at ambient temperature. Each instrument in a noise measurement supplies one known quantity, and the arithmetic then yields the unknown.

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43.

Which quantity is unchanged by inserting a matched attenuator ahead of a receiver?

A.The signal-to-noise ratio at the antenna
B.The system noise figure
C.The signal power reaching the receiver input
D.The noise power reaching the receiver input

Answer: A

The ratio at the antenna is a property of what arrives there and is unaffected by anything downstream, whereas the attenuator worsens the system noise figure and reduces both signal and noise power at the receiver. That distinction matters because it explains why attenuation before the gain is damaging while attenuation after it is harmless: the ratio delivered to the detector is what counts, not the ratio available at the antenna.

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44.

Why is a signal-to-noise ratio quoted with its bandwidth?

A.Bandwidth determines the noise figure of the receiver
B.Noise power depends on the bandwidth measured
C.Signal power depends on the bandwidth measured
D.The ratio is dimensionless and needs a reference

Answer: B

Because noise power is density times bandwidth, a ratio quoted without a bandwidth is ambiguous: the same signal gives very different ratios in a kilohertz and a megahertz. Signal power is fixed by the transmitter and the path. The ratio is indeed dimensionless but that is not why the bandwidth matters. And the noise figure is a device property independent of the bandwidth chosen. Any noise figure quoted without its conditions should be treated with suspicion.

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45.

Three schemes are compared at the same channel signal-to-noise ratio.

Scheme	Figure of merit	Bandwidth
AM at unity index	0.33	2 f m
SSB	1.0	f m
FM at index 5	450	12 f m

Which conclusion follows?

A.AM offers the best compromise of the three
B.All three deliver the same output ratio
C.FM trades bandwidth for output ratio
D.SSB gives the best output ratio of the three

Answer: C

Reading across, FM buys a figure of merit of 450 by spending twelve times the message bandwidth, which is the trade in one line. SSB has the best bandwidth efficiency rather than the best output ratio. AM is worst on both counts and offers no compromise at all beyond receiver simplicity. And the figures plainly differ. Tabulating the schemes against both resources is the compact way to hold the comparison.

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46.

Which noise source would a directional antenna help against?

A.Thermal noise in the receiver's first stage
B.Shot noise in the mixer diode
C.Flicker noise in the audio amplifier
D.Man-made interference from one direction

Answer: D

A directional antenna can null an external interferer arriving from a particular bearing, which is why it is the first remedy for local electrical noise. The other three are generated inside the receiver and no antenna can influence them. Separating external from internal noise decides whether the remedy lies in siting and antennas or in devices and circuit design, and it is the first question to ask about any noise complaint. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Discuss in Forum

47.

What is the effect of doubling a receiver's noise bandwidth?

A.The noise floor rises by 3 dB
B.The noise floor rises by 6 dB
C.The noise floor is unchanged
D.The noise figure doubles in decibels

Answer: A

Noise power is proportional to bandwidth, so doubling it raises the floor by 3 dB and halves the signal-to-noise ratio for the same signal. Six decibels would correspond to a fourfold increase. The floor certainly changes. And the noise figure is a device property independent of bandwidth, so it does not double. This is why a receiver's bandwidth is kept no wider than the signal requires, and why the matched filter is optimal.

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48.

Which is the correct order of increasing figure of merit?

A.AM, wideband FM, SSB
B.AM, SSB, wideband FM
C.Wideband FM, SSB, AM
D.SSB, AM, wideband FM

Answer: B

The order is AM at a third, SSB and DSB-SC at unity, and wideband FM at hundreds, which is the exact reverse of the bandwidth ordering. That inversion is the central trade of analogue communication: every improvement in noise performance is bought with spectrum, and every saving in spectrum is paid for in noise. The reversed and jumbled orders each misplace at least one scheme.

Discuss in Forum

49.

What does the bandwidth expansion factor measure?

A.Noise bandwidth over 3 dB bandwidth
B.Deviation over message frequency
C.Transmitted bandwidth over message bandwidth
D.Output ratio over channel ratio

Answer: C

It is the ratio of transmitted to message bandwidth, so it is two for AM and DSB, one for SSB, and about twelve for FM broadcasting. The output-over-channel ratio is the figure of merit, its natural partner in any comparison. The noise-to-3 dB bandwidth ratio is a filter detail. And deviation over message frequency is the modulation index. Plotting figure of merit against expansion factor is how the schemes are properly compared. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

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50.

Consider the following statements about noise in analogue systems.

1. Thermal noise power depends on temperature and bandwidth only

2. The first stage dominates a cascade's noise figure

3. SSB has a better figure of merit than DSB-SC

Which of the statements given above is or are correct?

A.1 and 3 only
B.2 and 3 only
C.1, 2 and 3
D.1 and 2 only

Answer: D

The first two are standard results. The third is the one to watch: SSB and DSB-SC both achieve unity, because SSB's halved sideband power is exactly offset by its halved receiver bandwidth. SSB's real advantage is spectral rather than in noise, and remembering that keeps the comparison honest. Expecting a noise advantage from every efficiency is the intuition this result corrects.

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