Question

The maximum and minimum amplitude of an amplitude-modulated wave is $16V,8V$respectively. The modulation index for this amplitude modulated wave is$x\times {10}^{–2}.$ The value of$x$ is

Answer 1

Step 1. Given data:

The maximum amplitude of an amplitude-modulated wave $\begin{array}{rcl}& & A\end{array}\begin{array}{rcl}& & \end{array}$.

The minimum amplitude of an amplitude-modulated wave $\begin{array}{rcl}& & A\end{array}\begin{array}{rcl}& & \end{array}$.

The modulation index$(M.I.)=x\times {10}^{–2}$

Step 2: Calculating the value of$x$,

The modulating index is defined as the ratio of the amplitude of modulating signal to the amplitude of the carrier signal.

$\begin{array}{rcl}Modulationindex\left(mi\right)& =& \frac{A_m}{A_c}\\ & =& \frac{\left[{\displaystyle \frac{\left(A_{max}–A_{min}\right)}{2}}\right]}{\left[\frac{\left(A_{max}+A_{min}\right)}{2}\right]}\\ & =& \frac{(A_{max}–A_{min})}{(A_{max}+A_{min})}\end{array}$

$\begin{array}{rcl}M.I& =& \frac{(16-8)}{(16+8)}\\ & =& \frac{8}{24}\\ & =& \frac{1}{3}\\ & =& 0.33\\ & =& 33\times {10}^{-2}\\ M.I& =& x\times {10}^{-2}\\ x& =& 33\end{array}$

Thus, the value of$x$ is $33$