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.