Frequency Modulation Questions and Answers

Answer: C

The disadvantage of FM is that higher bandwidth is required.

By Carson's rule, an FM signal needs a bandwidth of about \(2(\Delta f + f_{m})\). Broadcast FM, with a deviation of 75 kHz and audio up to 15 kHz, occupies about 180-200 kHz per channel, while AM needs only twice the highest audio frequency, about 10 kHz. FM therefore has to use VHF bands and fewer channels fit in a given spectrum.

Needing less modulating power and giving better noise immunity are advantages of FM, not disadvantages. 'Carrier is of any shape' is not a property of either system.

Answer: B

The statement that is not an advantage of FM over AM is that a lower bandwidth is required, option (B). Frequency modulation spreads the signal over a much wider band than amplitude modulation, which is why commercial FM uses a channel width of 200 kHz, against 10 kHz for AM. Better noise immunity, option (A), is the chief advantage of FM, since noise adds mainly to amplitude, which an FM receiver removes with a limiter. In FM the transmitted power is more useful, option (C), because all the power goes into the information-carrying sidebands and carrier at constant amplitude. Less modulating power, option (D), is required, since the modulator works at a low level.

Answer: C

The key takes statements I and III as correct, so option (C) is right, which makes statements II and IV the false ones. The Armstrong system generates frequency modulation indirectly: a crystal-controlled carrier is phase modulated by a signal that has been integrated, which is the phase and frequency modulation of statement I, and the small frequency deviation is then raised by frequency multiplication, statement III. Statement II is false, since the system is not amplitude modulation, although an amplitude component is removed with a limiter. Statement IV is false, because the integrator in the modulating path, a form of equalisation, is necessary. Options (A), (B) and (D) include a false statement as a correct one.

Answer: C

The approximate bandwidth is 3.2 kHz. The message is \(5\sin(2\pi\,100\,t)\), so \(f_{m} = 100\) Hz, and the modulated signal carries a phase deviation of 15 rad, so the modulation index is \(\beta = 15\). Carson's rule gives

\[ BW \approx 2(\beta + 1) f_{m} = 2(15 + 1)(100) \]

\[ BW = 3200\ \text{Hz} = 3.2\ \text{kHz} \]

Options (A) and (B) are of the order of the message frequency alone and ignore the many significant sidebands that a modulation index of 15 produces. Option (D), 4 kHz, does not follow from Carson's rule for these values. The 100 kHz carrier frequency does not enter the bandwidth.

Answer: B

When the frequency deviation in FM is doubled, the modulation index is doubled.

The modulation index is the frequency deviation divided by the modulating frequency. If the modulating frequency is unchanged, doubling the deviation doubles the index.

Decreasing it, halving it or leaving it unchanged would each require the deviation to fall, to be halved or to stay the same, which is not what is stated.

Answer: D

Frequency modulation uses pre-emphasis and de-emphasis to improve the noise performance (iv); a double sideband suppressed carrier signal is produced by a balanced modulator (iii); PCM uses companding to improve the quantization of weak signals (ii); and amplitude modulation is demodulated by an envelope detector (i). So the code is a-iv, b-iii, c-ii, d-i. The answer is a-iv, b-iii, c-ii, d-i.

Answer: A

The correct match is envelope detector with AM demodulation (iv), Foster-Seeley discriminator with FM detection (iii), varactor diode with frequency modulation (ii) and vestigial side band with TV transmission (i), which is the code iv iii ii i, option (A).

The envelope detector recovers an AM wave, the Foster-Seeley circuit demodulates FM, a varactor gives a voltage-controlled capacitance used in an FM modulator, and VSB is used in television. Options B, C and D mismatch at least one pair.

Answer: B

The bandwidth needed is 16 KHz. Carson's rule gives the bandwidth of an FM signal as twice the sum of the peak deviation and the highest modulating frequency. The highest audio frequency is 5 KHz and the peak deviation is 3 KHz, so

\[ B = 2(\Delta f + f_{m}) = 2(3 + 5) = 16\ \text{KHz} \]

6 KHz is just twice the deviation and ignores the audio bandwidth. 10 KHz would be 2 x 5 KHz, the bandwidth of an AM signal, not an FM one. 9.6 KHz does not follow from Carson's rule with any of the given data.

Answer: D

Techniques 1 and 4 generate FM. The Armstrong method produces FM indirectly: a narrow-band phase modulated signal is made and then multiplied in frequency. The reactance modulator produces FM directly by varying the capacitance or inductance in an oscillator's tank circuit with the message.

The Foster-Seeley discriminator is a demodulator, converting FM to AM or voltage at the receiver, so it does not generate FM. A balanced modulator produces a double sideband suppressed carrier AM signal. Options with 2 or 3 are therefore wrong, and only 1 and 4 remain.

Answer: C

The modulation index is 2.5 and the bandwidth is 14 KHz. The printed stem is missing the modulating frequency, but the options settle it: the modulation index is \(m_{f} = \Delta f / f_{m}\) and Carson's rule gives the bandwidth \(BW = 2(\Delta f + f_{m})\).

\[ m_{f} = \frac{5}{f_{m}} = 2.5 \]

\[ f_{m} = 2\ \text{KHz} \]

\[ BW = 2(5 + 2) = 14\ \text{KHz} \]

Only option C pairs a modulation index with a bandwidth that agree under these two relations for a deviation of 5 KHz. For option A, \(m_{f}\) = 1 would give \(BW = 2(5+5) = 20\) KHz, not 10, and the same check fails for B and D.