Signal Representation in Communication Questions and Answers

Answer: C

The suffix m fixes the reference at one milliwatt, so 0 dBm is 1 mW, 30 dBm is 1 W and -30 dBm is 1 microwatt. One watt is 30 dBm rather than zero. A voltage reference gives dBV or dBmV, which are different units entirely and cannot be compared with dBm without knowing the impedance. One microwatt is -30 dBm. Absolute decibel units always carry their reference in the suffix, and mixing two of them is a reliable source of thirty-decibel errors.

Answer: D

Inverting the power decibel relation:

\[ \mathrm{SNR} = 10^{20/10} = 100 \]

so the signal power is one hundred times the noise power. 20 is the decibel figure repeated as a ratio. 10 would correspond to 10 dB. 400 would follow from treating the 20 dB as a voltage ratio and then squaring, which double-counts the conversion. The 10 dB per decade of power is worth memorising in both directions, since link budgets are worked in decibels and thresholds are quoted as ratios.

Answer: A

In practice bandwidth is the band containing the bulk of the signal's power, with the exact fraction fixed by convention: 3 dB points, 99 per cent power, first null, or an occupied-bandwidth rule from a regulator. The mathematical highest frequency is often infinite, as it is for any pulse with sharp edges, so it is useless as a specification. The reciprocal of duration is an order-of-magnitude estimate rather than a definition. Harmonic spacing is the fundamental frequency itself.

Answer: B

Three decibels is a factor of two in power, so the 3 dB points are where the output power has halved and the voltage has fallen to \( 1/\sqrt{2} \) of its peak. A factor of three would be about 4.8 dB. Nothing rises above the peak by definition. A tenth of the power is 10 dB down. The half-power convention is used because it coincides with the point where the reactive and resistive parts of a simple resonant circuit are equal, which makes it a natural measure rather than an arbitrary one.

Answer: C

An impulse contains every frequency in equal measure, so its transform is flat - the frequency-domain statement of the fact that a perfectly short event has infinite bandwidth. An impulse in frequency corresponds to a constant in time, which is the dual pair. A sinc arises from a rectangular pulse of finite width. And a transform that vanished everywhere would describe no signal at all. This pair underpins impulse-response testing, since exciting a system with an impulse probes all frequencies at once.

Answer: D

Multiplication in time is convolution in frequency, and the dual holds too: convolution in time is multiplication in frequency. That single property explains modulation, since multiplying a message by a carrier convolves the message spectrum with a pair of impulses and so shifts it up to the carrier frequency. Multiplying spectra corresponds to filtering in time. Addition of spectra corresponds to adding the signals. Differentiation is unrelated to the product rule here.

Answer: A

A delay multiplies the spectrum by a complex exponential, which is a linear phase term; the amplitude spectrum is unchanged. That is why a delay is inaudible in isolation but destroys the waveform of a wideband signal if it varies with frequency, which is precisely the phase distortion a real channel imposes. Amplitude scaling comes from a gain change. A frequency shift comes from multiplication by a sinusoid, not from a delay. And the spectrum is certainly affected, in phase if not in magnitude.

Answer: B

Parseval equates the energy computed from the time waveform with the energy computed from the spectrum, which is the formal statement that a transform preserves energy. Conjugate symmetry of a real signal's spectrum is a separate and equally useful property. The time-bandwidth relation is a consequence of the transform but is not Parseval's statement. And the convolution theorem is a third distinct property, often confused with this one because both relate the two domains.

Answer: C

Real signals have conjugate-symmetric spectra: the amplitude is even and the phase odd about zero frequency. That is why only the positive-frequency half is ever plotted, the negative half carrying no additional information. Odd amplitude would allow negative amplitudes, which is not what the magnitude spectrum means. Periodicity affects whether the spectrum is discrete, not its symmetry. And symmetric phase would imply the signal is also even in time, a much stronger condition.

Answer: D

At zero lag the autocorrelation reduces to the mean square, which is the average power of the signal. That gives it a physical anchor and makes it the natural starting point for describing random signals, whose power spectral density is the Fourier transform of this function. Peak amplitude is not recoverable from the autocorrelation. Bandwidth follows from the shape of the transform rather than from a single value. And the last expression has the wrong dimensions.