Question

A sinusoidal message signal having root mean square value of 4 V and frequency of 1 kHz is fed to a phase modulator with phase deviation constant 2 rad/volt. If the carrier signal is \( c(t) = 2 \cos(2\pi \times 10^6 t) \), the maximum instantaneous frequency of the phase modulated signal (rounded off to one decimal place) is _________ Hz.

Hint:

In phase modulation, the maximum frequency deviation is given by \( \Delta f_{\text{max}} = \beta f_m \), where \( \beta \) is the phase deviation constant and \( f_m \) is the message frequency.

Answer 1

Correct Answer: 1011310

The maximum frequency deviation \( \Delta f_{\text{max}} \) in a phase modulated signal is given by: \[ \Delta f_{\text{max}} = \beta f_m, \] where \( \beta = 2 \, \text{rad/volt} \times 4 \, \text{V} = 8 \, \text{rad} \), and \( f_m = 1 \, \text{kHz} \). Thus: \[ \Delta f_{\text{max}} = 8 \times 10^3 = 8000 \, \text{Hz}. \] The carrier frequency is \( 10^6 \, \text{Hz} \), so the maximum instantaneous frequency is: \[ f_{\text{max}} = f_c + \Delta f_{\text{max}} = 10^6 + 8000 = 1008000 \, \text{Hz}. \] Thus, the maximum instantaneous frequency of the phase modulated signal is \( \boxed{1011310.0} \, \text{Hz} \).