Modern Physics Ques 267
- 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 $857.5$
(b) $89.25 $ and $85.75$s
(c) $178.5 $ and $171.5$
(d) $1785 $ and $ 1715$
Show Answer
Answer:
Correct Answer: 267.(b)
Solution:
Formula:
- $v(t)=10\left[1+0.3 \cos \left(2.2 \times 10^{4} t\right)\right] \quad\left[\sin \left(5.5 \times 10^{5} t\right)\right]$
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} $
BETA
