Signal Representation in Communication Questions and Answers

Answer: C

Varying group delay disperses a signal, so pulse energy spreads into neighbouring symbol slots and intersymbol interference results, which is what limits the data rate on many real channels and what an equaliser corrects. Power is preserved by a phase-only distortion. Carrier drift is an oscillator matter. And the receiver's noise figure is a property of its own front end, unaffected by the channel's phase behaviour.

Answer: D

For a flat noise density, power is density times bandwidth, so it rises in direct proportion to the bandwidth admitted. That is why a receiver's noise bandwidth is kept no wider than the signal requires, and why the matched filter is optimal. Independence of bandwidth would make filtering pointless. A falling relationship inverts the physics. And the carrier frequency affects the noise temperature of some sources but is not what sets the power in a given band.

Answer: A

A single tone is affected only by the gain and a delay, so it emerges the same from both; a wideband pulse depends on the relative phases and so is distorted differently by each. That contrast is the cleanest demonstration of why the phase response matters. Identical behaviour for every input would require identical transfer functions. Power delivery follows the amplitude response. And noise figure is unrelated to phase.

Answer: B

Because the pair is the complete description. The amplitude spectrum says how much of each frequency is present and the phase spectrum says how those components line up in time; together they invert exactly back to the waveform, and either one alone does not. Two signals can share an amplitude spectrum and look nothing alike, which is why a magnitude-only plot from a spectrum analyser cannot tell you what the waveform is doing. Phase does not set the frequency of anything; the frequency axis does that. It does not set the bandwidth either, which is read from the amplitude spectrum's extent. And the channel's noise has no bearing on which parts of the description are needed.

Answer: C

Modulation exists to translate the message spectrum to a band where radiation, propagation and channel allocation all work: antennas of practical size, a channel the regulator has assigned, and a propagation mechanism that reaches the receiver. A three-kilohertz speech signal cannot be radiated efficiently by any antenna anyone would build, which is the whole problem. Bandwidth is generally preserved or increased rather than reduced, and in wideband schemes it is deliberately increased in exchange for noise performance. Nothing about the translation improves the source's own signal-to-noise ratio. And conversion between analogue and digital is a separate operation that may or may not precede it.

Answer: D

Because it splits the signal into the two parts that behave quite differently. The carrier is fast and carries no information; the envelope and phase vary slowly and carry all of it, so representing the signal as a slowly varying complex function times a known exponential is both compact and physically meaningful. Every simulation and every software radio works this way, sampling at a rate suited to the bandwidth rather than the carrier. It removes no noise, since noise within the band is carried through unchanged. It makes the representation complex rather than real. And frequency resolution depends on observation length, not on the representation chosen.

Answer: A

Two things together. An infinitely sharp cut-off would require a non-causal impulse response, one that responds before the input arrives, so it cannot be realised at all; and every degree of approach to it costs filter order, which means components, loss, group-delay ripple and cost. The practical design therefore trades transition width against complexity and against phase behaviour in the passband. The sampling theorem constrains the sampling rate rather than the filter. The following stage's noise figure matters to sensitivity but not to achievable selectivity. And impedance affects matching rather than the shape of the response.

Answer: B

Because of the power it would imply. Flat density multiplied by infinite bandwidth gives infinite power, so the model must always be qualified as flat over the band of interest - which is a perfectly good approximation, since thermal noise is essentially flat up to frequencies far above any communication band. Real noise does have a flat region, so the second option is too strong. The quantum correction eventually rolls the density off, but not in a way that matters at radio frequencies. And random signals are described by spectral density precisely through the Wiener-Khinchin relation, which is what makes the whole framework work.

Answer: C

It states a floor, not an equality. The product of a signal's effective duration and its effective bandwidth cannot fall below a bound set by the Fourier transform itself, so any attempt to localise a signal sharply in one domain spreads it in the other. That is why radar trades pulse length against bandwidth, why a short observation cannot resolve two close tones, and why the Gaussian, which attains the bound, is of special interest. The relation is an inequality rather than the exact reciprocal, and the product varies with waveform shape while never going below the limit. Longer signals in fact tend to occupy less bandwidth, which is the opposite of the last option.

Answer: D

Purely for arithmetic convenience, and it is a large convenience. Because the decibel is logarithmic, every amplifier gain, cable loss, antenna gain and path loss in a chain is added rather than multiplied, so a link budget is a column of figures that can be checked by eye and modified by inspection. Accuracy is identical, since the conversion is exact. Absolute levels are expressible either way, with dBm and dBW being simply logarithmic forms of milliwatts and watts. And a linear ratio below unity represents a loss perfectly well; it is just harder to accumulate across twenty stages without error.