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An amplitude modulate wave is represented by $10\left[1+0.6 \sin \left(40 \times 10^3 t\right)\right] \sin \left(4 \times 10^6 t\right)$ volt where $t$ is in seconds. Then the ratio of the upper to the lower side band frequencies is
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$101: 99$
An Amplitude modulate wave is represented by $\mathrm{C}_{\mathrm{m}}$ $=10\left[1+0.6 \sin \left(40 \times 10^3 \mathrm{t}\right)\right] \sin \left(4 \times 10^6 \mathrm{t}\right) \mathrm{V}$
Compair with
$\begin{aligned} & C_m=A_C\left[1+\frac{A_m}{A_c} \sin \omega_m\right] \sin \omega_c t v \\ & \frac{A_m}{A_c}=0.6 \omega_m=40 \times 10^3 \omega_c=4 \times 10^6\end{aligned}$
$f_m=\frac{40 \times 10^3}{2 \pi} H z ; f_c=\frac{4 \times 10^6}{2 \pi} H z$
$\begin{aligned} & \text { Ratio }=\frac{\text { Upper side bond }}{\text { Lower side bond }}=\frac{f_c+f_m}{f_c-f_m}=\frac{4 \times 10^6+40 \times 10^3}{4 \times 10^6-40 \times 10^3} \\ & \text { Ratio }=\frac{4040000}{3960000}=\frac{101}{99}\end{aligned}$
Compair with
$\begin{aligned} & C_m=A_C\left[1+\frac{A_m}{A_c} \sin \omega_m\right] \sin \omega_c t v \\ & \frac{A_m}{A_c}=0.6 \omega_m=40 \times 10^3 \omega_c=4 \times 10^6\end{aligned}$
$f_m=\frac{40 \times 10^3}{2 \pi} H z ; f_c=\frac{4 \times 10^6}{2 \pi} H z$
$\begin{aligned} & \text { Ratio }=\frac{\text { Upper side bond }}{\text { Lower side bond }}=\frac{f_c+f_m}{f_c-f_m}=\frac{4 \times 10^6+40 \times 10^3}{4 \times 10^6-40 \times 10^3} \\ & \text { Ratio }=\frac{4040000}{3960000}=\frac{101}{99}\end{aligned}$
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