Q.An AM wave represented as $V = 5(1 + 0.5\sin 3960\,t)\sin 35\times10^5\,t$ volts. Determine the maximum and minimum amplitudes of AM wave.

The modulation index (or modulation factor, or depth of modulation) tells us how strongly the message controls the carrier amplitude. For AM it is the ratio of peak modulating voltage to peak carrier voltage:

and it is often quoted as a percentage, $M = (V_m/V_C)\times 100$. For distortion-free modulation $V_m$ must be less than $V_C$, so $m_a$ lies between 0 and 1, with a maximum permitted value of 1.0 (100% modulation). If $V_m > V_C$ then $m_a > 1$, giving over-modulation: the negative peaks are clipped, the envelope no longer matches the message, and distortion known as splattering appears.

When an AM wave is displayed on an oscilloscope, the index is read directly from the largest and smallest envelope amplitudes:

which rearrange to the standard working formula

$m_a = \frac{V_{max}-V_{min}}{V_{max}+V_{min}}$ …