Question

Consider a frequency-modulated (FM) signal \[ f(t) = A_c \cos(2\pi f_c t + 3 \sin(2\pi f_1 t) + 4 \sin(6\pi f_1 t)), \] where \( A_c \) and \( f_c \) are, respectively, the amplitude and frequency (in Hz) of the carrier waveform. The frequency \( f_1 \) is in Hz, and assume that \( f_c>100 f_1 \). The peak frequency deviation of the FM signal in Hz is _________.

• A. \( 15f_1 \)
• B. \( 12f_1 \)
• C. \( 4f_1 \)
• D. \( 2f_1 \)

Hint:

For frequency modulation, the peak frequency deviation is the sum of the deviations due to all modulating signals. Each modulating signal contributes to the frequency deviation based on its amplitude and frequency.

Answer 1

The Correct Option is A

The general form of the FM signal is: \[ f(t) = A_c \cos(2\pi f_c t + \Delta \omega \cdot m(t)), \] where \( \Delta \omega \) is the frequency deviation, and \( m(t) \) is the modulating signal.
The given FM signal has two modulating signals: \( 3 \sin(2\pi f_1 t) \) and \( 4 \sin(6\pi f_1 t) \).
The frequency deviation caused by the first modulating signal \( 3 \sin(2\pi f_1 t) \) is given by the amplitude of the signal multiplied by the frequency of the modulating signal:
\[ \Delta \omega_1 = 3 \cdot f_1. \] The frequency deviation caused by the second modulating signal \( 4 \sin(6\pi f_1 t) \) is: \[ \Delta \omega_2 = 4 \cdot 3f_1 = 12f_1. \] The total peak frequency deviation is the sum of the deviations from both modulating signals: \[ \Delta \omega_{{total}} = 3f_1 + 12f_1 = 15f_1. \] Thus, the peak frequency deviation is \( 15f_1 \).