Question

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 $\pi = 22/7$]

• A. 1785 and 1715
• B. 892.5 and 857.5
• C. 89.25 and 85.75
• D. 178.5 smf 171.5

Answer 1

The Correct Option is C

$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 $