Question

An amplitude modulated signal is given by $V\left(t\right)=0.4\cos (2.2\times {10}^{4}t)\left]\sin \right(5.5\times {10}^{5}t).$ Here $t$ is in seconds. The sideband frequencies $\text{(in kHz)}$ are, $[\text{Given}\pi =\frac{22}{7}]$

• A. $1785\text{and}1715$
• B. $178.5\text{and}171.5$
• C. $90\text{and}85$
• D. $892.5\text{and}857.5$

Answer 1

The correct option is C $90\text{and}85$

The given equation is,

$V\left(t\right)=0.4\cos (2.2\times {10}^{4}t)\sin (5.5\times {10}^{5}t).$

$\text{Using,}2\cos A\sin B=\sin (A+B)-\sin (A-B)$

$\Rightarrow V\left(t\right)=0.2\left[\sin \right(2.2+55)\times {10}^{4}t-\sin (2.2-55)\times {10}^{4}t]$

$\Rightarrow V\left(t\right)=0.2\left[\sin \right(57.2)\times {10}^{4}t+\sin (52.8)\times {10}^{4}t]$

From the above expression,

${\omega }_c+{\omega }_w=57.2\times {10}^4$

${\omega }_c-{\omega }_w=52.8\times {10}^4$

Upper sideband frequency $\left(f_1\right)$ is
$f_1=f_c-f_w=\frac{52.8\times {10}^4}{2\pi }\approx 85.00\text{kHz}$

Lower side band frequency $\left(f_2\right)$ is
$f_2=f_c+f_w=\frac{57.2\times {10}^4}{2\pi }\approx 90.00\text{kHz}$


Hence, option $\left(C\right)$ is correct.