Discussion
Suppose the modulating signal is \(m(t) = 2\cos(2\pi f_m t)\) and the carrier signal is \(y_c(t) = A_c \cos(2\pi f_c t)\), which one of the following is a conventional AM signal without over modulation ?
Answer: A
The conventional AM signal without overmodulation is \(x(t) = A_c\cos(2\pi f_c t) + \dfrac{A_c}{4} m(t) \cos(2\pi f_c t)\).
Substituting \(m(t) = 2\cos(2\pi f_m t)\):
\[ x(t) = A_c[1 + 0.5\cos(2\pi f_m t)]\cos(2\pi f_c t) \]
This is a carrier plus sidebands with modulation index 0.5, less than 1, so there is no overmodulation.
\(A_c m(t)\cos(2\pi f_c t)\) has no carrier term, so it is DSB-SC, not conventional AM. \(A_c[1 + m(t)]\cos(2\pi f_c t)\) has modulation index 2, which is overmodulation. The last option contains no separate carrier term either, so it is not conventional AM.
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