Question

An amplitude modulated signal is given by $V(t)=10[1+0.3\cos (2.2\times 10^4)]\sin (5.5\times 10^{5}t)$. 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

Answer: (C)

$V\left(t\right)=10+\frac{3}{2}\left[2\cos A\sin B\right]$
$=10+\frac{3}{2}\left[\sin \left(A+B\right)-\sin \left(A-B\right)\right]$
$=10+\frac{3}{2}\left[\sin \left(57.2\times 10^{4}t\right)-\sin \left(52.8\times 10^{4}t\right)\right]$
${\omega }_1=57.2\times 10^4=2\pi f_1$
$f_1=\frac{57.2\times 10^4}{2\times \left(\frac{22}{7}\right)}=9.1\times 10^4$
$\simeq 91KHz$
$f_2=\frac{52.8\times 10^4}{2\times \left(\frac{22}{7}\right)}$
$\simeq 84KHz$
Side band frequency are
$f_1=f_c-f_w=\frac{52.8\times 10^4}{2\pi }\approx 85.00kHz$
$f_2=f_c+f_w=\frac{57.2\times 10^4}{2\pi }\approx 90.00kHz$