Amplitude Modulation Questions and Answers

Answer: C

The relation is \( P_t = P_c(1 + \mu^{2}/2) \), the two sidebands contributing \( \mu^{2}/4 \) each. It is the most used formula in the topic and every efficiency argument follows from it. A linear dependence on the index would overstate the sideband power badly at small index. \( P_c\mu^{2} \) omits the carrier entirely. And dividing has the relation upside down, implying that modulating a transmitter reduces its output.

Answer: D

Applying the power relation:

\[ \begin{aligned} P_t &= 1000\left(1 + \frac{0.64}{2}\right) \\ &= 1000 \times 1.32 = 1320 \ \mathrm{W} \end{aligned} \]

so 1320 watts. 1800 W would follow from using \( 1 + \mu \). 1080 W halves the sideband contribution twice over. 1640 W uses \( \mu^{2} \) without the factor of two. The sideband power here is 320 W in total, or 160 W in each sideband, which is the figure a sideband filter would have to handle.

Answer: A

With \( \mu = 1 \) the total is \( 1.5P_c \) and the sidebands carry \( 0.5P_c \), so their share is

\[ \frac{0.5}{1.5} = \frac{1}{3} \]

which is a third, the best full-carrier AM can achieve. Two thirds of the transmitter's output goes into a carrier that carries no information at all, which is the central criticism of the scheme and the motive for suppressed-carrier systems. A half or two thirds would require an index above unity, and a quarter corresponds to an index of about 0.82.

Answer: B

Substituting the index:

\[ \eta = \frac{0.36}{2.36} = 0.153 \]

so about 15 per cent of the radiated power carries information. 36 per cent is \( \mu^{2} \) quoted as a percentage without the denominator. 18 per cent halves \( \mu^{2} \) and forgets the denominator entirely. 30 per cent is close to the figure for full modulation rather than for this index. Efficiency falls fast as modulation is reduced, which is why depth of modulation is maintained as high as distortion allows.

Answer: C

Current follows the square root of power:

\[ \begin{aligned} I_t &= 10\sqrt{1 + \frac{0.36}{2}} \\ &= 10\sqrt{1.18} = 10.9 \ \mathrm{A} \end{aligned} \]

giving about 10.9 amperes. 11.8 A is the power ratio applied directly to the current, forgetting the root. 10.3 A halves the increase. 13.0 A corresponds to a much deeper modulation. Measuring the antenna current with and without modulation is the classic field method of determining the index.

Answer: D

Indices from independent tones combine in quadrature:

\[ \mu_{eff} = \sqrt{0.4^{2}+0.3^{2}} = 0.5 \]

so 0.5, and the power relation then uses that figure. 0.7 adds the indices arithmetically, which would overstate the power and could suggest over-modulation where none exists. 0.35 is their mean. 0.12 is their product. The quadrature rule is what allows a multi-tone or speech signal to be handled with the same simple power formula.

Answer: A

Because at best two thirds of the radiated power sits in a carrier that conveys nothing, the information residing entirely in the sidebands. That is the argument for DSB-SC and SSB, which remove the carrier and, in the latter case, one sideband too. Propagation does not discriminate between carrier and sidebands. Class C operation is actually efficient in the amplifier sense, which is a separate meaning of the word. And an envelope detector uses both sidebands rather than discarding one.

Answer: B

The classic detector is a diode and an RC network: the diode rectifies and the capacitor holds the peaks while the resistor lets the envelope fall, so the output follows the envelope. Its cheapness is the reason AM broadcasting reached the mass market. A multiplier with a local oscillator is a synchronous or coherent detector, needed for suppressed-carrier signals. A phase-locked loop with a limiter is an FM discriminator. And a square-law device needs a low-pass output filter, not a band-pass one at the carrier.

Answer: C

The constant must satisfy both inequalities at once: long enough to smooth the carrier ripple and short enough to follow the envelope. Too short and carrier ripple appears at the output; too long and the detector cannot follow the falling envelope, which is diagonal clipping. The reversed condition would smooth away the message and pass the carrier. Equality with the carrier period leaves heavy ripple. And a constant longer than the message period is precisely the failure case.

Answer: D

It arises when the capacitor cannot discharge fast enough on the falling side of the envelope, so the output leaves the envelope on a straight diagonal line and the recovered waveform is distorted. Shortening the time constant cures it, at the cost of more carrier ripple. A low forward drop is desirable rather than harmful. Over-modulation distorts too, but by clipping the troughs symmetrically rather than diagonally. And gain affects level rather than the detector's ability to track.