Modern Physics 7 Question 54
2. 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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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,
$$ f _{LSB}=f _c-f _m=10-1=9 kHz $$
