Modern Physics 7 Question 20
21. An amplitude modulates signal is given by $v(t)=10\left[1+0.3 \cos \left(2.2 \times 10^{4} t\right)\right] \sin \left(5.5 \times 10^{5} t\right)$.
Here, $t$ is in seconds. The sideband frequencies (in kHz) are Take, $\pi=\frac{22}{7}$
(Main 2019, 11 Jan I)
(a) 892.5 and 857.5
(c) 178.5 and 171.5
(b) 89.25 and 85.75
(d) 1785 and 1715
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Solution:
- $v(t)=10\left[1+0.3 \cos \left(2.2 \times 10^{4} t\right)\right] \quad\left[\sin \left(5.5 \times 10^{5} t\right)\right]$
Upper band angular frequency
$$ \begin{aligned} \omega _{\nu} & =\left(2.2 \times 10^{4}+5.5 \times 10^{5}\right) rad / s \\ & =572 \times 10^{3} rad / s \end{aligned} $$
Similarly, lower band angular frequency.
$$ \begin{aligned} \omega _L & =\left(5.5 \times 10^{5}-2.2 \times 10^{4}\right) rad / s \\ & =528 \times 10^{3} rad / s \end{aligned} $$
$\therefore$ Side band frequency are,
$$ \begin{aligned} f _u & =\frac{\omega _u}{2 \pi}=\frac{572}{2 \pi} kHz \simeq 91 kHz \\ \text { and } \quad f _L & =\frac{\omega _L}{2 \pi}=\frac{528}{2 \pi} kHz \simeq 84 kHz \end{aligned} $$
