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

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).

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) \(\{1 + k_{m}(t)\}\, A \sin (w_{c}t)\)

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.

Signal Representation in Communication Questions and Answers

Q: A signal described in the frequency domain is expressed as

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.

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

Column I	Column 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

A.P-Z, Q-W, R-Y, S-X
B.P-W, Q-Z, R-Y, S-X
C.P-X, Q-W, R-Z, S-Y
D.P-Y, Q-Z, R-W, S-X

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

List I	List 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) \(\{1 + k_{m}(t)\}\, A \sin (w_{c}t)\)

A.a-iv, b-i, c-iii, d-ii
B.a-iii, b-iv, c-i, d-ii
C.a-iii, b-i, c-ii, d-iv
D.a-iv, b-iii, c-i, d-ii

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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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

A.4, 2, 1, 3
B.2, 1, 3, 4
C.1, 2, 3, 4
D.3, 4, 1, 2

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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List - I	List - 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)

A.(i) (ii) (iii) (iv)
B.(iv) (iii) (ii) (i)
C.(iii) (i) (ii) (iv)
D.(ii) (iv) (i) (iii)

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

A.integration
B.band pass filter
C.differentiation
D.AND gates

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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Demodulation is a process

A.done at the transmitter
B.of filtering the carrier
C.of removing the sidebands
D.similar to modulation

Exam Year(s)

UGC NET - 2013,

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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Modulation is a process of

A.Reducing distortions
B.Improving thermal stability
C.Combining audio and radio frequency waves at the transmitter
D.Generating constant frequency waves

Exam Year(s)

UGC NET - 2013,

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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Audio signals are not transmitted by EM waves because

(i) Antennas used will be unimaginably lengthy
(ii) Audio signals do not radiate
(iii) Interference is due to simultaneous transmissions.
(iv) Usable frequency range is too low and narrow. Which of the following is correct ?

A.(i) & (ii)
B.(ii) & (iii)
C.(i) & (iii)
D.(i), (iii) & (iv)

Exam Year(s)

UGC NET - 2013,

Answer: C

The keyed answer is statements (i) and (iii), option (C). Audio frequencies are too low to be radiated efficiently, since an antenna must be a sizeable fraction of the wavelength and would have to be unimaginably long, and if several stations sent audio directly they would interfere with each other, as they would all use the same band.

Statement (ii), that audio signals do not radiate, is not accepted as a reason. Option D adds (iv), the low and narrow usable frequency range, which the key does not accept. A reviewer may wish to check whether (iv) is also a fair reason.

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Arrange the following communication systems in increasing order of their available commercial bandwidth :

1. Satellite Communication System
2. Amplitude Modulated System
3. Frequency Modulated System
4. Two-line transmission system Correct sequence is

A.4, 2, 3, 1
B.1, 2, 3, 4
C.4, 3, 2, 1
D.1, 3, 2, 4

Exam Year(s)

UGC NET - 2014,

Answer: A

The increasing order of commercial bandwidth is 4, 2, 3, 1. A two-line transmission system carries only a few kilohertz to some hundreds of kilohertz, an AM channel is about 10 kHz wide, an FM channel about 200 kHz, and a satellite link supplies hundreds of megahertz, so satellite has the largest bandwidth.

Options that put AM after FM, or place the two-line system after the others, break this order. Option (B) lists the order the wrong way round, starting with the satellite system.

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10.

A signal described in the frequency domain is expressed as

A.Amplitude and phase against frequency
B.Amplitude against time, sampled at the Nyquist rate
C.Power against time, averaged over one period of the carrier
D.Energy against distance travelled along the transmission path

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.

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