Signal Representation in Communication Questions and Answers

Answer: A

It forces the amplifier to be operated well below its maximum output. The peaks must pass without clipping, so the mean power is set far below the amplifier's capability and the efficiency suffers accordingly, which matters enormously for battery life and for cooling. Multicarrier signals are the difficult case and a good deal of engineering goes into reducing their peaks. Bandwidth is a spectral property that the crest factor does not directly set, though clipping the peaks certainly does spread the spectrum. Synchronisation and antenna impedance are unrelated to the peak-to-mean ratio.

Answer: B

Because aliasing is irreversible. Once a component above half the sampling rate has folded down, it occupies exactly the same frequency as a legitimate component and no subsequent processing can tell them apart or separate them. The only remedy is to remove the offending energy before the fold happens, which is what the anti-alias filter does. Filters operate on discrete-time signals perfectly well, which is the whole of digital filtering, so the third option is simply false. Nothing damages the filter. And a delay is a benign and correctable effect, quite unlike an irreversible loss of information.

Answer: C

Time-reversal is exactly what makes the contributions add coherently. Convolving a pulse with its own time-reverse brings every frequency component into phase at one instant, so the signal builds to a peak there while the noise, being uncorrelated, adds in power only - which is why the signal-to-noise ratio at that instant is the best any linear filter can achieve. It cannot remove noise; it optimises the ratio, which is a different and achievable goal. Bandwidth is set by the transmitted pulse. And the output shape is the pulse's autocorrelation, which is generally rounded rather than rectangular.

Answer: A

The correct matching is P-Z, Q-W, R-Y, S-X.

• \(Km(t)A\sin(\omega_{c}t)\) is the message times the carrier with no carrier term left: DSBSC (Z).
• \(A\sin[\omega_{c}t + Km(t)]\) has the phase proportional to the message: PM (W).
• \([1 + Km(t)]A\sin(\omega_{c}t)\) adds the carrier back, so the envelope follows the message: AM (Y).
• Phase proportional to the integral of the message means instantaneous frequency proportional to the message: FM (X).

The main trap is FM against PM: the integral inside the phase is the signature of FM, because differentiating the phase gives a frequency deviation proportional to m(t).

Answer: B

The correct matching is a-iii, b-iv, c-i, d-ii.

• (a) DSB-SC is the product of the message and the carrier, \(k\,m(t) A \sin(w_{c}t)\), with no carrier term (iii).
• (b) AM adds a carrier: the amplitude is \(\{1 + k\,m(t)\}A\) (iv).
• (c) FM makes the instantaneous frequency follow m(t), so the phase contains the integral of m(t) (i).
• (d) PM makes the phase itself follow m(t): \(A \sin(w_{c}t + k\,m(t))\) (ii).

The other codes confuse AM with DSB-SC or FM with PM.