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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Communication Systems
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An AM signal is represented by the equation $25+5\sin (2\pi\times 8\times 10^3\;t) \times \sin (2\pi\times 0.25\times10^6\;t)$. Calculate the bandwidth, the modulating factor and the amplitude of each side frequency?

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Given the signal: $25+5\sin (2\pi\times 8\times 10^3\;t) \times \sin (2\pi\times 0.25\times10^6\;t)$
This is of the form: $v = V_c \big [ $$\sin (\omega_c t) + \large\frac{1}{2}$$m (\cos ((\omega_c - \omega_m)t) + \cos ((\omega_c+\omega_m)t)$$\big ]$
Comparing this to the give signal, we get the following:
$(1) \; \omega_c = 2\pi f_c = 2\pi \times 0.25 \times 10^6$
$(2) \; \omega_m = 2\pi f_m = 2\pi \times 8 \times 10^3$
$(3)\; V_c = 25\;V$
$(4)\; V_m = 5\;V$
$\Rightarrow$ Carrier Frequence $f_c = 0.25 \times 10^6 = 0.25\;MHz$
$\Rightarrow$ Modulating Frequency $f_m = 8 \times 10^3 = 8\;kHz$
$\Rightarrow $ Bandwidth $ = 2f_m = 16\;kHz$
$\Rightarrow$ Modulation index $m = \large\frac{V_m}{V_c} $$ = \large\frac{5}{25} $$= 0.2$
Therefore, the amplitude of each side frequency can be calculated as $\large\frac{m \times V_c}{2}$$ = 0.2 \times \large\frac{25}{2}$$ = 2.5\;V$
answered Jul 13, 2014 by balaji.thirumalai
 

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