Question

An amplitude modulated signal is given by $V\left(t\right)=10[1+0.3\cos (2.2\times {10}^4\left)\right]\sin (5.5\times {10}^{5}t)$. here $t$ is in seconds. The side band 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 $89.25$ and $85.75$

$V\left(t\right)=10+\frac{3}{2}\left[2\cos A\sin B\right]$

$=10+\frac{3}{2}\left[\sin \right(A+B)-\sin (A-B\left)\right]$

$10+\frac{3}{2}\left[\sin \right(57.2\times {10}^{4}t)-\sin (52.8\times {10}^{4}t\left)\right]$

${\omega }_1=57.2\times {10}^4=2\pi f_1$

$f_1=\frac{57.2\times {10}^4}{2\times \left(\frac{22}{7}\right)}=9.1\times {10}^4$

$\approx 91KHz$

$f_2=\frac{52.8\times {10}^4}{2\times \left(\frac{22}{7}\right)}$

$\approx 84KHz$

Side band frequency are

$f_1=f_c-f_w=\frac{52.8\times {10}^4}{2\pi }\approx 85.00kHz$

$f_2=f_c+f_w=\frac{57.2\times {10}^4}{2\pi }\approx 90.00kHz$