Question

An amplitude modulated signal is given by
$V\left(t\right)=10[1+0.6cos(2.2\times {10}^{4}t\left)sin\right(5.5\times {10}^5\left)t\right].$ Here $t$ is in seconds. The sideband frequencies (in $kHz$) are $[Given\pi =\frac{22}{7}]$

• A. $1785$ and $1715$
• B. $178.5$ and $171.5$
• C. $91$ and $84$
• D. $892.5$ and $857.5$

Answer 1

The correct option is C $91$ and $84$

when you superimpose a signal on the carrier by AM or FM, you produce sidebands at the sum and difference of the carrier frequency $f_c$ and modulation frequency $f_m$ .
$f_{sb}=f_c\pm f_m$
$=\frac{{\omega }_c\pm {\omega }_m}{2\pi }$ $(\because {\omega }_c=5.5\times {10}^5,{\omega }_m=0.22\times {10}^5)$
$=\frac{(5.5\pm 0.22)\times {10}^5}{2\times \frac{22}{7}}$
$=91,84kHz$