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An FM signal is represented by v = 12 sin (6 × 108 t + 5 sin 1250 t). The carrier frequency fc and frequency deviation δ, respectively, are
1. 191 MHz and 665 Hz
2. 95.5 MHz and 995 Hz
3. 191 MHz and 995 Hz
4. 95.5 MHz and 665 Hz

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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     

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