Modern Physics Ques 266
- In an amplitude modulator circuit, the carrier wave is given by $\quad C(t)=4 \sin (20000 $ $\pi t)$ while modulating signal is given by, $m(t)=2 \sin (2000$ $ \pi t)$. The values of modulation index and lower side band frequency are
(Main 2019, 12 April II)
(a) $0.5$ and $10$ $ kHz$
(b) $0.4$ and $10$ $ kHz$
(c) $0.3$ and $9 $ $kHz$
(d) $0.5$ and $9 $ $kHz$
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Answer:
Correct Answer: 266.(d)
Solution:
- Given, carrier wave,
$ C(t)=4 \sin (20000 $ $\pi t) $
Modulating signal,
$ m(t)=2 \sin (2000 $ $\pi t) $
So, carrier wave’s amplitude and frequency are
$ \begin{aligned} & A _c=4 V, \omega _c=20000 \pi=2 \pi \times 10^{4} rad / s \\ \Rightarrow & f _c=\frac{\omega _c}{2 \pi}=10^{4} Hz=10 kHz \end{aligned} $
and modulating signal’s amplitude and frequency are
$ A _m=2 V, \omega _m=2000 \pi=2 \pi \times 10^{3} $ $rad / s $
$\Rightarrow f _m=\frac{\omega _m}{2 \pi}=10^{3} Hz=1 $ $kHz$
So, modulating index is $m=\frac{A _m}{A _c}=\frac{2}{4}=0.5$
and lower side band frequency is,
$ f _{LSB}=f _c-f _m=10-1=9 $ $kHz $
BETA
