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## Q. 1.1.15

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

(a) $x(t)=A_{c} m(t) \cos \left(2 \pi f_{c} t\right)$

(b) $x(t)=A_{c}[1+m(t)] \cos \left(2 \pi f_{c} t\right)$

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

(d) \begin{aligned}x(t) &=A_{C} \cos \left(2 \pi f_{m} t\right) \cos \left(2 \pi f_{c} t\right) \\&+A_{C} \sin \left(2 \pi f_{m} t\right) \sin \left(2 \pi f_{c} t\right)\end{aligned}

## Verified Solution

Conventional AM signal is

\begin{aligned}x(t) &=A_{ C }[1+m(t)] \cos \left(w_{ c } t\right) \\&=A_{ C } \cos \left(w_{ c } t\right)+A_{ C } m(t) \cos w_{ c } t\end{aligned}

In option b,

Modulation index  $=\frac{2}{1}=2$

So, it is conventional A.M signal but with over modulation.
In option c ,

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

Here, modulation index

$=\frac{2}{4}=\frac{1}{2}$

Therefore, it is conventional AM signal without over modulation.

Hence, the correct option is (c).