Noise in Analog Systems Questions and Answers

Answer: C

A passive loss ahead of the receiver adds its own decibels to the noise figure, so

\[ NF = 3 + 3 = 6 \ \mathrm{dB} \]

giving about 6 dB, which is why feeder loss at the front of a system is so damaging and why the amplifier is moved to the antenna. Keeping the receiver's own 3 dB ignores the feeder. 4.5 dB averages them. And 9 dB double-counts. That a passive component can dominate a noise budget surprises people until they apply the formula.

Answer: D

It is the width of an ideal rectangular filter passing the same noise power as the real one, and it differs a little from the 3 dB bandwidth because real filters have sloping skirts. Using the 3 dB figure in a noise calculation introduces a small error, usually a few per cent. The transmitted bandwidth is a property of the signal. And noise is measurable everywhere. Defining an equivalent rectangle is the standard way of reducing a shaped response to one number.

Answer: A

The figure of merit is the ratio of the output signal-to-noise ratio to the signal-to-noise ratio at the channel, so it measures what the demodulation itself contributes. Power ratios describe the link budget. The bandwidth ratio is the bandwidth expansion factor, a related but separate quantity. And noise figure and noise temperature are two forms of the same receiver property. Isolating the demodulator's own contribution is what makes the schemes comparable. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

Answer: B

With \( \gamma = \mu^{2}/(2+\mu^{2}) \) and \( \mu = 1 \):

\[ \gamma = \frac{1}{3} \]

so a third, and less at any shallower modulation. That is the quantitative statement of AM's inefficiency: two thirds of the power sits in a carrier that contributes nothing to the output ratio. Unity is the suppressed-carrier result. Two thirds and three do not follow from the expression. The figure of merit is the cleanest way to rank the schemes on noise. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Answer: C

It is unity: coherent detection of a suppressed-carrier double-sideband signal delivers exactly the channel signal-to-noise ratio, which is three times better than full-carrier AM at best. A third is the AM figure. Three would imply a gain the scheme does not have. And two is not a standard result here. The improvement over AM comes entirely from not wasting power on a carrier, which is why the two suppressed-carrier schemes share the same figure.

Answer: D

SSB also achieves unity, matching DSB-SC, because although it transmits half the sidebands it needs only half the receiver bandwidth, and the two effects cancel exactly. That equality is a satisfying result and often misremembered as SSB being better or worse. Its real advantage over DSB-SC is bandwidth rather than noise. A half or a third would imply a penalty the algebra does not produce.

Answer: A

The standard result is \( \gamma = 3\beta^{2}(\beta+1) \), which for a large index grows as the cube, so a modest increase in deviation buys a large improvement. That is the bandwidth-for-noise trade in algebra. Proportionality to the index alone would understate it badly. The deviation appears through the index rather than on its own. And the message bandwidth enters through the index too. The cube law is why FM broadcasting sounds so much quieter than AM.

Answer: B

Substituting into the expression:

\[ \gamma = 3 \times 25 \times 6 = 450 \]

so about 450, some 26 decibels, against unity for the suppressed-carrier schemes and a third for AM. That single comparison is the whole case for FM broadcasting. 75 uses \( 3\beta^{2} \) without the extra factor. 150 and 25 follow from partial substitutions. The price is a bandwidth of 180 kHz against AM's 10, which is the other half of the trade. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

Answer: C

SSB achieves unity, neither better nor worse than the channel, and DSB-SC matches it. Both angle schemes exceed unity substantially at large index, which is their whole point. And the last option is self-contradictory, since an index of five is wideband rather than narrowband. Recognising that the amplitude family's best case is merely to break even is the clearest way to see why angle modulation displaced it wherever bandwidth allowed. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: D

The carrier takes at least two thirds of the transmitted power and carries no information, so the output ratio is worse than the channel ratio by that factor. Envelope detection does introduce nonlinearity below threshold, but the factor of three applies well above it. The double bandwidth is shared with DSB-SC, which nevertheless achieves unity. And the quadrature argument describes the detail rather than the dominant cause. Following where the power goes explains the figure of merit directly.