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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.

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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.

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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.

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104.

Match the Column I with Column II and select the correct answer using the codes given below :

Column IColumn II
(P) \(Km(t) A \sin(\omega_{c}t)\)(W) PM
(Q) \(A \sin[\omega_{c}t + Km(t)]\)(X) FM
(R) \([1 + Km(t)] A \sin(\omega_{c}t)\)(Y) AM
(S) \(A \sin\left[\omega_{c}t + K\int_{-\infty}^{t} m(t)\, d\tau\right]\)(Z) DSBSC

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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).

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105.

Match List I with List II and select the correct answer using codes given below :

List IList II
(a) DSB-SC modulation(i) \(A \sin \left(w_{c}t + k\int_{-\infty}^{t} m(z)\, dz\right)\)
(b) Amplitude modulation(ii) \(A \sin (w_{c}t + k_{m}(t))\)
(c) Frequency modulation(iii) \(k.m(t)\, A \sin (w_{c}t)\)
(d) Phase modulation(iv) \(\{1 + k_{m}(t)\}\, A \sin (w_{c}t)\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

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106.

Consider the following communication systems :

1. FM Broadcast
2. AM Broadcast
3. Microwave Communication
4. Optical Fiber Communication The sequence of the communication systems in the order of their increasing carrier frequency is

Exam Year(s)

UGC NET - 2012,

Answer: B

The carrier frequencies increase in the order AM broadcast (about 0.5 to 1.6 MHz), FM broadcast (88 to 108 MHz), microwave communication (gigahertz) and optical fibre communication (hundreds of terahertz), that is 2, 1, 3, 4, option B. Option A starts with the optical fibre, which is the highest of all, option C places FM below AM, and option D puts the microwave link below the broadcast bands, so all three break the order.

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107.

List - IList - II
(a) Frequency Modulation(i) Balanced Modulator
(b) Double sideband suppressed carrier(ii) Companding
(c) PCM(iii) Pre-emphasis and De-emphasis
(d) AM(iv) Envelope Detection

Codes :
(a) (b) (c) (d)

Exam Year(s)

UGC NET - 2013,

Answer: C

The matching is (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv), which is option (C). Frequency modulation uses pre-emphasis and de-emphasis to improve the high-frequency signal-to-noise ratio. A double sideband suppressed carrier signal is generated with a balanced modulator. PCM uses companding to give small signals finer quantization. An AM signal is recovered by envelope detection. Options (A), (B) and (D) each mismatch at least two of these pairs, for example pairing companding with FM or envelope detection with DSBSC.

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108.

To separate channels in an FDM receiver, it is necessary to use

Exam Year(s)

UGC NET - 2013,

Answer: B

In a frequency-division multiplex receiver every channel occupies its own band of frequencies, so the receiver separates them with band pass filters, each one tuned to pass its own channel and reject the others. Integration and differentiation are mathematical operations and do not select frequencies, and AND gates are digital circuits, which are used to separate channels in time-division, not frequency-division, systems. The answer is a band pass filter.

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Answer: B

Demodulation recovers the modulating (message) signal from the modulated carrier, and the key describes it as filtering the carrier, that is, getting rid of the carrier so that the baseband signal remains. It is done at the receiver, not at the transmitter. It is the reverse of modulation and not similar to it, and removing the sidebands would remove the message itself. So the second option is the keyed answer.

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Answer: C

Modulation is the process of combining the audio (message) signal with the radio-frequency carrier at the transmitter so that the information can be sent over long distances, which is option (C).

It is not a method of reducing distortion (option A) or of improving thermal stability (option B), which are matters of circuit design. It is also not the generation of constant-frequency waves (option D), which is the job of an oscillator and only gives the carrier that is later modulated.

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