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The modulation index of an AM wave is :

  • A.\(m = (V_{max} - V_{min})/(V_{max} + V_{min})\)
  • B.\(m = (V_{max} + V_{min})/(V_{max} - V_{min})\)
  • C.\(m = (V_{max} \times V_{min})/(V_{max} + V_{min})\)
  • D.\(m = (V_{max} \times V_{min})/(V_{max} - V_{min})\)

Answer: A

The modulation index is \(m = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}\).

An AM wave has envelope \(V_{c}(1 + m\cos\omega_{m}t)\), so its maximum and minimum amplitudes are \(V_{max} = V_{c}(1 + m)\) and \(V_{min} = V_{c}(1 - m)\). Subtracting and adding:

\[ \begin{aligned} V_{max} - V_{min} &= 2mV_{c} \\ V_{max} + V_{min} &= 2V_{c} \\ m &= \frac{V_{max} - V_{min}}{V_{max} + V_{min}} \end{aligned} \]

The inverted ratio would always exceed 1, and the product forms have the units of voltage, not a pure number.

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