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In as Amplitude Modulated wave ,The ratio between the bandwidth of the modulated wave and the frequency of message signal is
Option: 1
2

Option: 2
1

Option: 3
3

Option: 4
4

Answers (1)

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As we have learned

Voltage equation for AM wave -

$e_{c}= E_{c}\cos \omega _{c}t$

$e_{m}= E_{m}\sin \omega _{m}t$

- wherein

Resultant Modulated wave

$e= \left ( E_{c} +e_{m}\sin \omega _{m}t\right )\cdot \sin \omega _{e}t$

If message signal is $e_m = E_m \sin w_m t$ and carrier waves $e_c = E_c \sin w_c t$

Amplitude modulatedwave e = $(E_c + E_m \sin w_c t ) \sin w_ct$

$e = E_c \sin w_c t + E_m \sin w_m t \cdot \sin w_ct$

$= E_c \sin w_c t + \frac{V_m}{2}[ \cos (w_c - w_m)t - \cos (w_c+w_m)t]$

Here we find three frequency $w_c , w_c -w_m , w_c +w_m$

$w_c =$ angular frequency of carrier

$w_c -w_m =$ angular frequency of lower side band

$w_c +w_m =$ angular frequency of upper side band

NOw , $\frac{Bandwidth }{ferquency\: \: of \: \: message \: \: signal}= \frac{(w_c-w_m)-(w_c-w_m)}{w_m}$

$\frac{2 w_m }{w_m}$ = 2

Posted by

Suraj Bhandari

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