The maximum peak to peak voltage of an AM wave is 26 mV and minimum peak to peak voltage is 4 mV. Find modulation index.

Solution

Maximum voltage of AM wave = 26 mV
Minimum peak to peak voltage = 4 mV

Therefore,

Maximum voltage of AM wave,

V_{\max} = \frac{26}{2} mV = 13 mV

Minimum voltage of Am wave,

V_{\min} = \frac{4}{2} mV = 2 mV

∴ Modulation index, m = \frac{V_{\max} - V_{\min}}{V_{\max} + V_{\min}} = \frac{13 - 2}{13 + 2} = \frac{11}{15} = 0.73 .

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