Signal Representation in Communication Questions and Answers

Answer: A

The matched filter is the one whose impulse response is the time-reverse of the expected pulse, and it maximises the signal-to-noise ratio at the instant the output is sampled, which is exactly what minimises the probability of a detection error. It does not widen the available bandwidth; if anything it restricts the noise bandwidth to what the signal needs. Phase linearity is a channel property. And passing more harmonics would admit more noise for no gain in signal.

Answer: B

Flat spectral density transforms into an impulse at zero lag, so samples taken at any two different instants are uncorrelated: the formal meaning of whiteness. A constant at every lag would describe a direct-current signal. A sinusoidal autocorrelation belongs to a narrowband process, and indeed band-limited noise does show one. Zero at zero lag would mean zero power, so the noise would not exist.

Answer: C

The standard regulatory definition is the band containing ninety-nine per cent of the transmitted power, with half a per cent excluded at each edge, and it exists so that channel allocations can be enforced consistently. Fifty per cent would be far too permissive and bears no relation to interference. Everything above the noise floor is not a workable definition, since it depends on the victim receiver. And the first lobe is an engineering estimate rather than a regulatory one.

Answer: D

Narrowing the resolution bandwidth reveals finer spectral detail and lowers the displayed noise floor, at the cost of a much longer sweep, which is the central trade in every measurement. Maximum input power is set by the attenuator and the mixer. Frequency accuracy comes from the reference oscillator. Input impedance is fixed by the hardware, usually at fifty ohms. Reporting a noise measurement without stating the resolution bandwidth makes it meaningless, which is why it always appears alongside.

Answer: A

The thermal noise floor is \( -174 \ \mathrm{dBm/Hz} \) at this temperature, so in a megahertz:

\[ \begin{aligned} P &= -174 + 10\log_{10}(10^{6}) \\ &= -174 + 60 = -114 \ \mathrm{dBm} \end{aligned} \]

giving about -114 dBm, a figure worth memorising as the noise floor in one megahertz. -174 dBm is the density per hertz rather than the power in the band. -84 dBm would correspond to a gigahertz of bandwidth. -144 dBm corresponds to a kilohertz.

Answer: B

Noise figure is the ratio by which the stage worsens the signal-to-noise ratio, expressed in decibels, and it is defined against a standard reference temperature so that stages can be compared. Absolute noise power depends on the bandwidth and is a different quantity. Gain is separate, although it strongly affects how much the following stages contribute through the cascade formula. And usable bandwidth is a further independent specification.

Answer: C

The decibel expresses a ratio, so it carries no dimensions; the volt, watt and hertz are all dimensional units. That is why an absolute level must be written with a reference attached, as dBm or dBW or dBV, and why adding two plain decibel figures is meaningful while adding two dBm figures is not. Keeping ratio units and absolute units distinct removes most of the confusion beginners have with link budgets.

Answer: D

Shannon's result makes capacity proportional to bandwidth and logarithmic in the signal-to-noise ratio, so both matter and bandwidth is the more powerful lever. Depending on either alone would miss half the result. Carrier frequency does not enter the expression at all, though in practice higher carriers make wider bandwidths easier to obtain, which is why the two are often correlated in real systems.

Answer: A

It is a baseband signal. Passband describes the same information after modulation has translated it up to a carrier. An analytic signal is the one-sided complex representation constructed with the Hilbert transform. Quadrature refers to a component in phase opposition of ninety degrees, one half of a complex envelope. The baseband and passband pair is the fundamental division in the subject, since every modulation scheme is a rule for moving between them.

Answer: B

Energy is preserved, which is the content of Parseval's relation and the reason a spectrum can be used to compute power. Bandwidth and duration are not preserved in any sense; they trade against one another. Peak amplitude in one domain bears no simple relation to the peak in the other, as the impulse and its flat spectrum show most starkly. Energy conservation is what makes the two representations genuinely equivalent rather than merely related.