An amplitude modulated signal is given by V(t)=10[1+0.3 . (2.2 ×...

Equation given
$$<br>$$\begin{array}{l}
\mathrm{V}(\mathrm{t})=10\left[1+0.3 \cos \left(2.2 \times 10^{4}\right)\right] \\
\sin \left(5.5 \times 10^{5} \mathrm{t}\right) \\
=10+1.5\left[\sin \left(57.2 \times 10^{4} \mathrm{t}\right)+\sin \left(52.8 \times 10^{4} \mathrm{t}\right)\right] \\
\omega_{\mathrm{c}}+\omega_{\mathrm{w}}=57.2 \times 10^{4}=2 \pi \mathrm{f}_{1}
\end{array}$$<br>$$
$$<br>\mathrm{f}_{1}=\frac{57.2 \times 10^{4}}{2 \times\left(\frac{22}{7}\right)}=9.1 \times 10^{4} \simeq 91 \mathrm{KHz}<br>$$
$$<br>\omega_{\mathrm{c}}-\omega_{\mathrm{w}}=52.8 \times 10^{4}<br>$$
$$<br>\mathrm{f}_{2}=\frac{52.8 \times 10^{4}}{2 \times\left(\frac{22}{7}\right)} \simeq 84 \mathrm{KHz}<br>$$

Upper side band frequency $\left(f_{1}\right)$ is
$$<br>\mathrm{f}_{1}=\mathrm{f}_{\mathrm{c}}-\mathrm{f}_{\mathrm{w}}=\frac{52.8 \times 10^{4}}{2 \pi} \approx 85.00 \mathrm{kHz}<br>$$
Lower side band frequency $\left(\mathrm{f}_{2}\right)$ is
$$<br>\mathrm{f}_{2}=\mathrm{f}_{\mathrm{c}}+\mathrm{f}_{\mathrm{w}}=\frac{57.2 \times 10^{4}}{2 \pi} \approx 90.00 \mathrm{kHz}<br>$$