FE Electrical and Computer Communications Practice Problems

The current FE Electrical and Computer specification assigns 5-8 of the 110 exam questions to Communications. Form A contains 6 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A conventional double-sideband, full-carrier amplitude-modulated transmitter delivers a total average power of 15 kW to its antenna while modulated by a single sinusoidal tone at a modulation index of 0.80. The average power contained in one sideband is most nearly:

(A)1.82 kW
(B)2.40 kW
(C)3.64 kW
(D)11.36 kW

Show worked solution

Answer: (A)

The carrier absorbs 11.4 kW of the 15 kW transmitted, and each of the two sidebands takes 1.82 kW of the remainder.

Total power in a tone-modulated full-carrier signal divides between an unmodulated carrier and two sidebands, so the carrier level is recovered from the total once the index is known.

P_{t} = P_{c} (1 + \frac{m ^{2}}{2})

P_{c} = \frac{15}{1 + \frac{0.8 0 ^{2}}{2}} = \frac{15}{1.32} = 11.36 kW

Each sideband carries one quarter of the carrier power scaled by the square of the index, the pair together making up the modulation-dependent remainder.

P_{S B} = \frac{m ^{2}}{4} P_{c} = 0.16 \times 11.36 = 1.82 kW

Sideband power is thus only 12 percent of the transmitted total even at 80 percent modulation, the well-known power penalty of retaining the carrier.

\frac{P _{S B}}{P _{t}} = \frac{m ^{2} /4}{1 + m ^{2} /2} = \frac{0.16}{1.32} = 0.121

FE Reference Handbook — Communication Theory and Concepts: Amplitude Modulation, Power in a Double-Sideband Full-Carrier Signal

Why the other choices appear

(B)Using the total power in place of the carrier power in the sideband expression: 0.16 x 15 = 2.40 kW.
(C)Reporting the combined power of both sidebands: 2 x 1.82 = 3.64 kW.
(D)Reporting the carrier power, 15/1.32 = 11.36 kW, rather than a sideband.