Analog Communication Questions and Answers

Answer: A

A frequency-domain description gives amplitude and phase as functions of frequency, which together carry exactly the same information as the time waveform and can be inverted back to it. Amplitude alone is not enough, since discarding phase loses the waveform's shape. The second option describes a sampled time-domain record. Power against time is an instantaneous power waveform, still a time-domain quantity. Energy against distance describes propagation loss along a path, which is a link-budget matter rather than a signal representation.

Answer: B

A Fourier series applies to periodic signals, decomposing them into a discrete set of harmonics of the fundamental. Aperiodic signals need the Fourier transform instead, which replaces the discrete harmonics with a continuous spectrum. Finite-energy decaying signals are exactly the aperiodic case the transform handles. Random signals are described by their power spectral density, obtained from the autocorrelation function, since no single deterministic spectrum exists for them.

Answer: C

Periodicity forces the spectrum to be discrete: lines at the fundamental and its harmonics, and nothing in between. That is the defining structural difference from an aperiodic signal, whose spectrum is continuous. A single component describes a pure sinusoid, which is one special case of a periodic signal rather than the general one. And a deterministic periodic signal has a fixed spectrum, so it does not vary between observations.

Answer: D

The transform of a rectangular pulse is a sinc function whose first zero is at \( 1/\tau \), so a one-microsecond pulse has its first null at one megahertz. The main lobe width is the usual practical estimate of the bandwidth needed to carry the pulse. \( 2/\tau \) is the width between the first nulls either side of the centre, which is a different measure. \( \tau \) itself has units of time. And \( 1/2\tau \) is half the correct figure, sometimes reached by confusing the null spacing with the half-width.

Answer: A

Both statements are correct, and the second is the general principle behind the first. The time-bandwidth product has a lower bound, so compressing a signal in time necessarily spreads it in frequency and vice versa. It is why radar uses short pulses and pays for them in bandwidth, and why a narrowband channel cannot carry fast transitions. The relationship is a property of the Fourier transform itself rather than of any particular waveform, which is what makes it inescapable.

Answer: B

An energy signal has finite total energy, which forces its average power over all time to be zero; a single pulse is the standard example. A power signal is the complement, with infinite energy and finite non-zero average power, which is what a sinusoid or any periodic signal is. Boundedness of amplitude is a separate property that neither implies. And band limitation concerns the spectrum's extent rather than the signal's energy.

Answer: C

A continuous sinusoid runs for all time, so its energy is infinite while its average power is finite and non-zero: the definition of a power signal. The other three are all finite-duration or decaying, hence finite energy and zero average power. The distinction matters because energy and power signals are analysed with different tools, the Fourier transform for the first and the power spectral density for the second.

Answer: D

The mean square of a sinusoid is half the square of its amplitude, so the power is \( A^{2}/2 \) and the root mean square value is \( A/\sqrt{2} \). \( A^{2} \) would be the peak instantaneous power. \( A/2 \) is a voltage rather than a power. \( A^{2}/4 \) results from squaring the mean rather than taking the mean of the square, an easy slip. The half factor propagates through every modulation power calculation in the subject, so it is worth holding firmly.

Answer: A

For power ratios the decibel is ten times the logarithm:

\[ 10\log_{10}(100) = 20 \ \mathrm{dB} \]

so 20 dB. 10 dB would be a ratio of ten. 40 dB uses twenty times the logarithm, which is the form for voltage or current ratios and would be right only if 100 were a voltage ratio. 100 dB is the ratio quoted as though it were already logarithmic. Keeping the factor of ten for power and twenty for amplitude straight is the single most useful habit in this subject.

Answer: B

Amplitude ratios use twenty times the logarithm, because power goes with the square of voltage:

\[ 20\log_{10}(10) = 20 \ \mathrm{dB} \]

giving 20 dB, the same figure a power ratio of 100 produces - which is exactly the point, since ten times the voltage is a hundred times the power. 10 dB would follow from using the power formula on a voltage ratio. 1 dB and 40 dB do not follow from either convention.