Noise in Analog Systems Questions and Answers

Answer: A

Because an amplifier raises signal and noise by the same factor, the ratio between them is unchanged, and this is the single most useful thing to know about receiver noise: the ratio is decided at the point where the noise enters, and nothing downstream can recover it. Later gain does add a little noise but that is a second-order effect divided down by the gain ahead of it. Friis quantifies the contribution rather than forbidding gain. And gain does not change bandwidth. Once a ratio is lost it stays lost, which is why front-end design receives so much attention.

Answer: B

The figure of merit is defined at equal transmitted power and equal message bandwidth, with the channel bandwidth left free to differ - which is what makes the comparison meaningful, since the extra channel bandwidth is precisely what FM spends to buy its advantage. Holding channel bandwidth equal would forbid FM the resource it trades in. Equalising the output ratio asks a different question, namely what power each scheme needs. And receiver noise figure is common to both and irrelevant to the comparison. Stating the conditions of a comparison is as important as the numbers it yields.

Answer: C

FM spends channel bandwidth, roughly twice the deviation plus twice the highest message frequency, and receives an output ratio improvement growing with the cube of the index. Transmitter power is held constant in the comparison, which is what makes the trade worth having. Dynamic range is relaxed rather than spent, since a limiter tolerates amplitude variation. And FM does not need a linear modulator, another of its practical advantages. Naming the resource spent and the benefit obtained in the same sentence is the compact way to hold any scheme comparison in mind.

Answer: D

Because an FM signal's amplitude is constant by design, clipping it discards noise without discarding information, whereas an AM signal's amplitude is the message and a limiter would destroy it. That single asymmetry is the origin of most of FM's noise advantage. Peak-to-average ratio, bandwidth and detector linearity are all real differences between the schemes but none of them decides whether limiting is permissible. Every noise-reduction technique exploits an asymmetry between signal and noise, and here the asymmetry is that one is constant in amplitude and the other is not.

Answer: A

The allocated channel bandwidth caps the index, since bandwidth grows with deviation and a station may not spill into its neighbours. That is why broadcast FM settled on an index of about five rather than the far larger value the noise arithmetic would favour. Audio linearity affects distortion rather than the index. Noise figure sets sensitivity. And oscillator stability concerns frequency accuracy. Regulatory allocation, rather than physics, is what stops the bandwidth-for-noise trade being pushed further, and that is worth remembering when the arithmetic suggests otherwise.

Answer: B

The index is five, so applying the wideband result:

\[ \begin{aligned} \text{FoM} &= 3\beta^{2}(\beta+1) \\ &= 3(25)(6) = 450 \end{aligned} \]

giving about 450, some thirty decibels better than single sideband at the same transmitted power. 150 drops the factor of \( \beta+1 \), which is the commonest slip. 75 comes from \( 3\beta^{2} \) with \( \beta \) mistaken. And 1500 overcounts by the index. The cubic growth in the index is what makes the trade so attractive, and it is why the bandwidth limit rather than the arithmetic sets the design.

Answer: C

Putting the noisy stage first makes its full 10 dB the dominant term, while the quiet stage's advantage is then divided away:

\[ \begin{aligned} \text{as built} &: 2 + \tfrac{10-1}{10} = 2.9 \\ \text{reversed} &: 10 + \tfrac{2-1}{100} \approx 10 \end{aligned} \]

so reversing is much worse, about 10 dB against 4.6 dB. Order matters enormously in a cascade, which is why the quietest available device always goes first even when it has modest gain. And the effect is independent of signal level.

Answer: D

A passive loss ahead of the first amplifier adds directly:

\[ 1 + 3 = 4 \ \mathrm{dB} \]

so 4 dB, and a decibel of feeder has cost a decibel of sensitivity. 3 dB ignores the feeder, which is the mistake the Friis formula exists to prevent. 2 dB has the sign wrong. And 6 dB double-counts. The simplicity of this addition is exactly why low-loss feeders and mast-head amplifiers are worth their cost on any weak-signal installation.

Answer: A

Two changes add in decibels:

\[ \begin{aligned} \Delta &= 3 \ (\text{bandwidth}) + 6 \ (\text{noise figure}) \\ &= 9 \ \mathrm{dB} \end{aligned} \]

so the floor moves to -101 dBm. -104 dBm counts only one change. -107 dBm counts only the bandwidth. And -113 dBm has the sign reversed, which would mean the receiver improved. Doubling a decibel figure doubles it in decibels, whereas doubling a bandwidth adds three, and keeping those two operations distinct is what makes the arithmetic come out.

Answer: B

The first is true and the second is false: flicker noise rises as frequency falls, which is why it is also called one-over-f noise and why it matters at audio and direct-coupled frequencies rather than at radio frequencies. Thermal noise really is flat to well beyond any frequency an analogue radio uses. The contrast is what decides where each mechanism must be designed against, and it explains why chopper and modulation techniques move a low-frequency measurement up in frequency to escape flicker noise entirely.