Noise in Analog Systems Questions and Answers

Answer: C

Converting through \( T_e = (F-1)T_0 \) with \( F = 2 \):

\[ T_e = (2-1) \times 290 = 290 \ \mathrm{K} \]

so 290 kelvin, which is the reference temperature itself and a useful anchor: 3 dB noise figure equals 290 K. 580 K would be 4.8 dB, and 145 K about 1.8 dB. Satellite and radio astronomy work uses noise temperature because the numbers are more informative at low noise figures, where decibels compress the differences.

Answer: D

Applying the cascade formula:

\[ \begin{aligned} F &= F_1 + \frac{F_2-1}{G_1} \\ &= 2 + \frac{3}{10} = 2.3 \end{aligned} \]

so 2.3, about 3.6 dB. Adding the factors directly gives 6.0 and ignores the first stage's gain entirely. 2.0 ignores the second stage. And 4.2 follows from no correct substitution. The formula's structure is the important part: the second stage's contribution is divided by the gain ahead of it. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: A

Because later contributions are divided by the gain preceding them, the first stage dominates, which is why a low-noise amplifier is placed as close to the antenna as possible, sometimes at the antenna itself. Equal contribution would make the ordering irrelevant, which it emphatically is not. Optimising the last stage buys almost nothing. And adequate early gain is precisely what suppresses later noise. One formula dictates the whole architecture of a sensitive receiver.

Answer: B

Feeder loss adds directly to the system noise figure if it comes before the gain, so placing the amplifier at the antenna puts the gain ahead of the loss and the Friis formula then divides the loss's contribution. Radiation pattern is fixed by the antenna's geometry. Standing wave ratio depends on matching. And power flows the other way, up the feeder to the amplifier. Where a stage sits in the chain can matter as much as how good it is.

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.