Correct Answer - Option 2 : 95.5 MHz and 995 Hz
Concept:
The standard equation of FM is given as:
Sfm(t) = Ac cos [2πfct + 2π kf ∫ m(t)dt]
Assume m(t) = Am cos 2π fmt. The FM signal will be:
\({S_{fm}}\left( t \right) = {A_c}\cos \left[ {2\pi \;{f_c}t + 2\pi {k_f}\;\frac{{{A_m}\cos 2\pi {f_m}t}}{{2\pi {f_m}}}} \right]\;\)
\({S_{fm}}\left( t \right) = {A_c}\cos \left[ {2\pi {f_c}t + \frac{{{k_f}{A_m}}}{{{f_m}}}\sin 2\pi {f_m}t} \right]\)
Sfm (t) = Ac cos [2πfc t + β sin 2π fm t] ---(1)
β = modulation Index given by:
\(\beta = \frac{{{k_f}{A_m}}}{{{f_m}}}\)
kfAm = Frequency deviation
\(\beta = \frac{{frequency\;deviation}}{{Message\;freq.}} = \frac{{{\rm{\Delta }}f}}{{{f_m}}}\) ---(2)
Application:
Given:
V = 12 sin (6 × 108t + 5 sin 1250t)
Comparing this equation with Equation (1), we get:
\(2\pi {f_c} = 6 \times {10^8}\)
\({f_c} = \frac{{6\; \times \;{{10}^8}}}{{2\pi }} = 95.54\;MHz\)
β = 5 and
2πfm = 1250
fm = 199 Hz
Using Equation (2), we can write:
Δf = β × fm
Δf = 5 × 199 Hz
Δf = 995 Hz