Noise in Analog Systems Questions and Answers

Q: Why does an FM discriminator's output noise rise with frequency?

Frequency is the derivative of phase, so a discriminator differentiates and therefore multiplies the noise spectrum by frequency, giving the triangular output noise that pre-emphasis exists to counter. Integration would produce the opposite slope, which is what a phase detector effectively delivers. The tuned circuits pass the whole baseband. And the limiter acts on amplitude rather than on baseband frequency. One mathematical operation explains the noise shape and hence a whole audio processing convention.

Q: What bounds the FM noise advantage at low signal levels?

Below threshold the output collapses and the improvement vanishes abruptly, so the figure of merit applies only above it. That is why FM coverage boundaries are sharp where AM's fade gradually. Capture concerns two signals rather than one weak one. The pre-emphasis constant shapes the audio. And channel spacing is an allocation. Every gain has a condition attached, and for FM the condition is adequate carrier-to-noise ratio at the detector.

Q: What does the capture effect suppress?

It suppresses the weaker co-channel signal, so the listener hears one programme cleanly rather than two mixed. Baseband noise is dealt with by de-emphasis, amplitude modulation by the limiter, and the image by preselection. Four mechanisms for four different impairments, and attributing an effect to the wrong one is the commonest error in questions on FM noise performance.

Q: A deep-space link has ample spectrum but very little transmitter power. Which scheme suits it?

Wideband FM converts spare bandwidth into output signal-to-noise ratio, which is exactly the trade a power-limited, bandwidth-rich link wants, and its constant envelope allows an efficient saturated amplifier as well. Full-carrier AM wastes power. DSB-SC merely breaks even on noise. And vestigial sideband is chosen for spectral reasons in television. Matching the scheme to which resource is scarce is the whole of modulation selection.

Q: Which scheme would you choose where bandwidth is scarce and power is plentiful?

SSB occupies the message bandwidth alone, the narrowest of any analogue scheme, and its unity figure of merit is acceptable when power can be increased instead. Wideband FM is the opposite choice. Narrowband FM occupies rather more than SSB for no noise advantage. And full-carrier AM wastes both resources. Stating which resource binds before choosing a scheme is what prevents circular arguments about which modulation is best. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Q: A receiver has a \( 200 \ \mathrm{kHz} \) noise bandwidth and a \( 6 \ \mathrm{dB} \) noise figure. Its noise floor is about

Building it up from the reference density:\[ \begin{aligned} P &= -174 + 10\log_{10}(2\times10^{5}) + 6 \\ &= -174 + 53 + 6 = -115 \ \mathrm{dBm} \end{aligned} \]so about -115 dBm. -121 dBm omits the noise figure. -108 dBm double-counts it. And -174 dBm is the density per hertz before any bandwidth or noise figure is applied. Assembling a noise floor from density, bandwidth and noise figure is the standard three-term calculation of every link budget.

Q: A signal of \( -95 \ \mathrm{dBm} \) arrives at that receiver. The signal-to-noise ratio is about

Subtracting the noise floor from the signal:\[ -95 - (-115) = 20 \ \mathrm{dB} \]giving 20 dB, comfortably above the threshold any analogue scheme needs. 10 dB and 30 dB would follow from a noise floor ten decibels adrift either way, and 6 dB is the noise figure quoted back. Working in decibels turns the whole calculation into subtraction, which is precisely why link budgets are drawn up logarithmically.

Q: Two uncorrelated noise sources each contribute \( 4 \ \mathrm{\mu V} \). Together they give

Uncorrelated voltages add in quadrature:\[ v = \sqrt{4^{2}+4^{2}} = 5.7 \ \mathrm{\mu V} \]so 5.7 microvolts, a rise of 3 dB rather than 6. 8 microvolts adds them arithmetically, which applies only to fully correlated sources. 4 microvolts ignores one. And 16 is their product. Whether quantities add in power, in voltage or in quadrature is decided by their correlation, and getting it wrong is worth a factor of two. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

31.

Why does an FM discriminator's output noise rise with frequency?

A.It differentiates the phase
B.It integrates the phase
C.Its tuned circuits attenuate low frequencies
D.The limiter removes low-frequency components

Answer: A

Frequency is the derivative of phase, so a discriminator differentiates and therefore multiplies the noise spectrum by frequency, giving the triangular output noise that pre-emphasis exists to counter. Integration would produce the opposite slope, which is what a phase detector effectively delivers. The tuned circuits pass the whole baseband. And the limiter acts on amplitude rather than on baseband frequency. One mathematical operation explains the noise shape and hence a whole audio processing convention.

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32.

What does pre-emphasis and de-emphasis together achieve?

A.Elimination of the capture effect
B.A net improvement in high-frequency signal-to-noise ratio
C.A reduction in the transmitted bandwidth
D.Immunity to the threshold effect

Answer: B

Boosting treble before modulation and cutting it after detection leaves the signal unchanged while attenuating the noise that rose with frequency, a net gain of several decibels at the top of the band. Bandwidth increases slightly rather than reducing, since the boost deepens deviation on bright material. Threshold and capture are unaffected, being properties of the limiter and discriminator. Any matched pre- and post-filter pair works this way when the noise spectrum is not flat.

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33.

What bounds the FM noise advantage at low signal levels?

A.The pre-emphasis time constant
B.The channel spacing in use
C.The threshold effect
D.The capture effect

Answer: C

Below threshold the output collapses and the improvement vanishes abruptly, so the figure of merit applies only above it. That is why FM coverage boundaries are sharp where AM's fade gradually. Capture concerns two signals rather than one weak one. The pre-emphasis constant shapes the audio. And channel spacing is an allocation. Every gain has a condition attached, and for FM the condition is adequate carrier-to-noise ratio at the detector.

Discuss in Forum

34.

What does the capture effect suppress?

A.The noise in the upper part of the baseband
B.Amplitude modulation on the wanted carrier
C.The image frequency ahead of the mixer
D.The weaker of two co-channel signals

Answer: D

It suppresses the weaker co-channel signal, so the listener hears one programme cleanly rather than two mixed. Baseband noise is dealt with by de-emphasis, amplitude modulation by the limiter, and the image by preselection. Four mechanisms for four different impairments, and attributing an effect to the wrong one is the commonest error in questions on FM noise performance.

Discuss in Forum

35.

A deep-space link has ample spectrum but very little transmitter power. Which scheme suits it?

A.Wideband FM
B.Full-carrier AM
C.DSB-SC
D.Vestigial sideband

Answer: A

Wideband FM converts spare bandwidth into output signal-to-noise ratio, which is exactly the trade a power-limited, bandwidth-rich link wants, and its constant envelope allows an efficient saturated amplifier as well. Full-carrier AM wastes power. DSB-SC merely breaks even on noise. And vestigial sideband is chosen for spectral reasons in television. Matching the scheme to which resource is scarce is the whole of modulation selection.

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36.

Which scheme would you choose where bandwidth is scarce and power is plentiful?

A.Full-carrier AM
B.SSB
C.Wideband FM
D.Narrowband FM

Answer: B

SSB occupies the message bandwidth alone, the narrowest of any analogue scheme, and its unity figure of merit is acceptable when power can be increased instead. Wideband FM is the opposite choice. Narrowband FM occupies rather more than SSB for no noise advantage. And full-carrier AM wastes both resources. Stating which resource binds before choosing a scheme is what prevents circular arguments about which modulation is best. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

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37.

A receiver has a \( 200 \ \mathrm{kHz} \) noise bandwidth and a \( 6 \ \mathrm{dB} \) noise figure. Its noise floor is about

A.-108 dBm
B.-174 dBm
C.-115 dBm
D.-121 dBm

Answer: C

Building it up from the reference density:

\[ \begin{aligned} P &= -174 + 10\log_{10}(2\times10^{5}) + 6 \\ &= -174 + 53 + 6 = -115 \ \mathrm{dBm} \end{aligned} \]

so about -115 dBm. -121 dBm omits the noise figure. -108 dBm double-counts it. And -174 dBm is the density per hertz before any bandwidth or noise figure is applied. Assembling a noise floor from density, bandwidth and noise figure is the standard three-term calculation of every link budget.

Discuss in Forum

38.

A signal of \( -95 \ \mathrm{dBm} \) arrives at that receiver. The signal-to-noise ratio is about

A.10 dB
B.30 dB
C.6 dB
D.20 dB

Answer: D

Subtracting the noise floor from the signal:

\[ -95 - (-115) = 20 \ \mathrm{dB} \]

giving 20 dB, comfortably above the threshold any analogue scheme needs. 10 dB and 30 dB would follow from a noise floor ten decibels adrift either way, and 6 dB is the noise figure quoted back. Working in decibels turns the whole calculation into subtraction, which is precisely why link budgets are drawn up logarithmically.

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39.

Two uncorrelated noise sources each contribute \( 4 \ \mathrm{\mu V} \). Together they give

A.5.7 microvolts
B.8 microvolts
C.4 microvolts
D.16 microvolts

Answer: A

Uncorrelated voltages add in quadrature:

\[ v = \sqrt{4^{2}+4^{2}} = 5.7 \ \mathrm{\mu V} \]

so 5.7 microvolts, a rise of 3 dB rather than 6. 8 microvolts adds them arithmetically, which applies only to fully correlated sources. 4 microvolts ignores one. And 16 is their product. Whether quantities add in power, in voltage or in quadrature is decided by their correlation, and getting it wrong is worth a factor of two. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

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40.

A stage has \( 20 \ \mathrm{dB} \) gain and \( 2 \ \mathrm{dB} \) noise figure, followed by one of \( 10 \ \mathrm{dB} \) noise figure. The overall figure is close to

A.2.9 dB
B.2.2 dB
C.12 dB
D.6 dB

Answer: B

The second stage's contribution is divided by a gain of a hundred:

\[ F = 1.58 + \frac{10-1}{100} = 1.67 \]

which is about 2.2 dB, close to the first stage's own figure. That is the Friis formula's practical message: ample gain in a quiet first stage renders later stages almost irrelevant. Adding the decibel figures gives 12 dB and ignores the gain entirely. 6 dB and 2.9 dB follow from partial substitutions.

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