Q. An amplitude modulated signal is given by $V(t) = 10[1 + 0.3 \cos(2.2 \times 10^4)] \sin(5.5 \times 10^5t)$.
Here t is in seconds. The sideband frequencies (in kHz) are, [Given $\pi$ = 22/7]

Solution:

$V\left(t\right) = 10 + \frac{3}{2} \left[2\cos A \sin B\right] $
$ =10+ \frac{3}{2} \left[\sin\left(A+B\right)-\sin\left(A-B\right)\right] $
$=10 + \frac{3}{2} \left[\sin\left(57.2 \times10^{4} t\right) -\sin\left(52.8 \times10^{4} t\right)\right] $
$\omega_{1} = 57.2 \times10^{4} = 2\pi f_{1} $
$ f_{1} = \frac{57.2 \times10^{4}}{2 \times\left(\frac{22}{7}\right) } = 9.1 \times10^{4} $
$ \simeq 91KHz $
$ f_{2} = \frac{52.8 \times10^{4}}{2 \times\left(\frac{22}{7}\right)} $
$\simeq 84 KHz $
Side band frequency are
$ f_{1} = f_{c} -f_{w} = \frac{52.8\times10^{4}}{2\pi} \approx 85.00 kHz $
$ f_{2} =f_{c} +f_{w} = \frac{57.2 \times10^{4}}{2\pi} \approx 90.00 kHz $

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