Question

Suppose that the modulating signal is $x\left(t\right)=2cos\left(2\pi f_{m}t\right)$ and the carrier signal is $x_c\left(t\right)=A_{c}cos\left(2\pi f_{c}t\right)$. Which one of the following is a conventional AM signal without over-modulation?

• A. $x\left(t\right)=A_{c}m\left(t\right)cos\left(2\pi f_{c}t\right)$
• B. $x\left(t\right)=A_c[1+m(t\left)\right]cos\left(2\pi f_{c}t\right)$
• C. $x\left(t\right)=A_{c}cos\left(2\pi f_{c}t\right)+\frac{A_c}{4}m\left(t\right)cos\left(2\pi f_{c}t\right)$
• D. $x\left(t\right)=A_{c}m\left(t\right)cos\left(2\pi f_{m}t\right)$

Answer 1

The correct option is C $x\left(t\right)=A_{c}cos\left(2\pi f_{c}t\right)+\frac{A_c}{4}m\left(t\right)cos\left(2\pi f_{c}t\right)$

Conventional am signal is Double side band full carrier and the standard expression for the modulated wave sis
$e\left(t\right)=A_c[1+m(t\left)\right]cos\left(w_{c}t\right)$