Peak voltage of a modulating signal is 2 V. The carrier wave is represented by \( C(t) = 4\sin(8\pi t)\,\mathrm{V} \). The modulation index of modulated signal is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.5
In amplitude modulation the modulation index is the ratio of the peak amplitude of the modulating (message) signal to the amplitude of the carrier wave.
\( \displaystyle \mu = \frac{A_m}{A_c} \)
From \( C(t) = 4\sin(8\pi t) \), the carrier amplitude is \( A_c = 4\,\mathrm{V} \), and the peak of the modulating signal is \( A_m = 2\,\mathrm{V} \).
\( \displaystyle \mu = \frac{2\,\mathrm{V}}{4\,\mathrm{V}} = 0.5 \)
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Amplitude Modulation – “… Let \( c(t) = A_c \sin\omega_c t \) represent carrier wave and \( m(t) = A_m \sin\omega_m t \) represent the message or the modulating signal where \( \omega_m = 2\pi f_m \) is the angular frequency of the message signal.”
Amplitude Modulation – “… Here \( \mu = A_m/A_c \) is the modulation index; in practice, \( \mu \) is kept \( \le 1 \) to avoid distortion.”