Q.A frequency modulated signal is given by 10sin[6×10⁸t + 5sin1250t] Determine

Comparing the given signal with the standard FM form $v = V_c\sin(\omega_c t + m_f\sin\omega_m t)$ reads off the carrier and modulating angular frequencies, the peak voltage and the modulation index directly, and the rest follow from them. …

[!TLDR]

Matching the signal to the standard FM equation gives $V_c = 10\ \text{V}$, $\omega_c = 6\times10^{8}$, $m_f = 5$, $\omega_m = 1250$; from these $f_c = 95.48\ \text{MHz}$, $f_m = 199\ \text{Hz}$, $\Delta f = 995\ \text{Hz}$, deviation ratio $= 377$ and the power in a 5 $\Omega$ load is 10 W.

The standard FM voltage equation is $v = V_c\sin(\omega_c t + m_f\sin\omega_m t)$. Comparing it with the given $10\sin[6\times10^{8}t + 5\sin(1250t)]$:

1. Carrier frequency: $f_c = \dfrac{\omega_c}{2\pi} = \dfrac{6\times10^{8}}{2\pi} = 95.48\ \text{MHz}$.
2. Modulating frequency: $f_m = \dfrac{\omega_m}{2\pi} = \dfrac{1250}{2\pi} = 199\ \text{Hz}$.
3. Modulation index: read directly from the coefficient of the inner sine, $m_f = 5$.
4. Maximum deviation: $\Delta f = m_f\,f_m = 5\times199 = 995\ \text{Hz}$. …