Difficulty: Easy
Correct Answer: x(t) = A_c [1 + m(t)] cos(2π f_c t)
Explanation:
Introduction / Context:Conventional AM (double sideband with carrier) multiplies a carrier by a DC-plus-message term. Recognizing the correct mathematical form is foundational for understanding spectra, envelope detection, and constraints such as overmodulation (|μ| ≤ 1).
Given Data / Assumptions:
Concept / Approach:The presence of a nonzero DC term ensures the envelope never crosses zero when |μ| ≤ 1, enabling simple diode envelope detection. Expressions without the DC term correspond to DSB-SC. Quadrature combinations correspond to phase or frequency translation, not standard AM with carrier.
Step-by-Step Solution:
Identify conventional AM: a carrier scaled by [1 + message term].Match with options: only option (b) has a carrier multiplied by [1 + m(t)].Note about modulation index: with m(t) = 2 cos(⋯), the implicit μ would be 2, which exceeds 1. In practice, one would scale m(t) to keep |μ| ≤ 1. The form, however, is uniquely correct for conventional AM with carrier.Verification / Alternative check:
Expanding x(t) = A_c cos(2π f_c t) + A_c m(t) cos(2π f_c t) produces a carrier term at f_c and symmetric sidebands at f_c ± f_m—exactly the spectrum of standard DSB-AM with carrier.Why Other Options Are Wrong:
(a) A_c m(t) cos(2π f_c t) is DSB suppressed-carrier; no DC term (no carrier).(c) Uses the identity cosα cosβ + sinα sinβ = cos(α − β), which yields frequency translation, not AM with carrier.(d) Pure baseband; no modulation.(e) Linear addition of baseband and carrier is not amplitude modulation.Common Pitfalls:
Confusing the correct AM form with DSB-SC; overlooking the modulation index limit and the need to scale m(t) in practice.Final Answer:
x(t) = A_c [1 + m(t)] cos(2π f_c t)
Discussion & Comments