The ratio of the maximum amplitude to the min | vedclass

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(B) The modulation index $m_a$ is given by the formula: $m_a = \frac{E_{\max} - E_{\min}}{E_{\max} + E_{\min}}$ Given that the ratio of maximum amplit

Vedclass · Accepted Answer

$0.6$

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(B) The modulation index $m_a$ is given by the formula: $m_a = \frac{E_{\max} - E_{\min}}{E_{\max} + E_{\min}}$ Given that the ratio of maximum amplitude to minimum amplitude is: $\frac{E_{\max}}{E_{\min}} = \frac{4}{1}$ Let $E_{\max} = 4k$ and $E_{\min} = 1k$. Substituting these values into the formula: $m_a = \frac{4k - 1k}{4k + 1k} = \frac{3k}{5k} = \frac{3}{5} = 0.6$ Therefore,the modulation index is $0.6$.

The ratio of the maximum amplitude to the minimum amplitude of an amplitude-modulated wave is $4$. What is the modulation index of this wave?

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