Suppose that the modulating signal is `m(t)=2cos(2pif_(m)t)` and the carrier signal is `x_(c)(t)=A_(c)cos(2pif_(c)t)` which one of the following is a conventional `AM` signal without over modulation

Options

A. `x(t)=A_(c)m(t)cos(2pif_(c)t)`
B. `x(t)=A_(c)(1+m(t))cos(2pif_(c)t)`
C. `x(t)=A_(c)cos(2pif_(c)t)+(A_(c))/(4)m(t)cos(2pif_(c)t)`
D. `x_(c)(t)=A_(c)cos(2pif_(m)t)+cos(2pif_(c)t)+A_(c)sin(2pif_(m)t)sin(2pif_(c)t)`

Answer

C

Explanation

Convention `AM` signal withour over modulation means `mlt1`
`x(t)=A_(c)(1+m(t))cos(2pif_(c)t)`
For option `C`
`x(t)=A_(c)(1+(1)/(4)m(t))cos(2pif_(c)t)`, `:. M=(1)/(4)`