Question

Using trigonometric relation $\sin A \sin B=1 / 2[\cos (A-B)]-\cos (A+B)$, we can write $c_{m}(t)=A_{c} \sin \omega_{c} t+\mu A_{c} \sin \omega_{m} t \sin \omega_{c} t$ as follows

• A. $c_{m}(t)=A_{c} \sin \omega_{m} t+\frac{\mu A_{c}}{2} \cos \left(\omega_{c}-\omega_{m}\right) t -\frac{\mu A_{c}}{2} \cos \left(\omega_{c}+\omega_{m}\right) t$
• B. $c_{m}(t)=A_{c} \sin \omega_{c} t+\frac{\mu A_{m}}{2} \cos \left(\omega_{c}-\omega_{m}\right) t-\frac{\mu A_{m}}{2} \cos \left(\omega_{c}+\omega_{m}\right) t$
• C. $c_{m}(t)=A_{c} \sin \omega_{c} t+\frac{\mu A_{c}}{2} \cos \left(\omega_{c}-\omega_{m}\right) t -\frac{\mu A_{c}}{2} \cos \left(\omega_{c}+\omega_{m}\right) t$
• D. $c_{m}(t)=A_{c} \sin \omega_{c} t+\frac{\mu A_{c}}{2} \cos \left(\omega_{c}+\omega_{m}\right) t-\frac{\mu A_{c}}{2}\left(\omega_{c}-\omega_{m}\right) t$

Answer 1

Answer: (C)

Using the trigonometric relation $\sin A \sin B=1 / 2$ $[\cos (A-B)-\cos (A+B)]$, we can write $c_{m}(t)$ of equation $c_{m}(t)=A_{c} \sin \omega_{c} t+\mu A_{c} \sin \omega_{m} t \sin \omega_{c} t$ as $c_{m}(t)=A_{c} \sin \omega_{c} t+\frac{\mu A_{c}}{2} \cos \left(\omega_{c}-\omega_{m}\right) t$
$-\frac{\mu A_{c}}{2} \cos \left(\omega_{c}+\omega_{m}\right) t$
Here, $\omega_{c}-\omega_{m}$ and $\omega_{e}+\omega_{m}$ are respectively called the lower side and upper sideband frequencies.