The maximum peak-to-peak voltage of an AM wav | vedclass

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(D) The maximum peak-to-peak voltage is $V_{pp,max} = 24 \, mV$,so the maximum amplitude is $V_{max} = \frac{24}{2} = 12 \, mV$.The minimum peak-to-pe

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$50$

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(D) The maximum peak-to-peak voltage is $V_{pp,max} = 24 \, mV$,so the maximum amplitude is $V_{max} = \frac{24}{2} = 12 \, mV$.The minimum peak-to-peak voltage is $V_{pp,min} = 8 \, mV$,so the minimum amplitude is $V_{min} = \frac{8}{2} = 4 \, mV$.The modulation factor $m$ is given by the formula:$m = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}$Substituting the values:$m = \frac{12 - 4}{12 + 4} = \frac{8}{16} = 0.5$To express this as a percentage:$m = 0.5 	imes 100\% = 50\%$

The maximum peak-to-peak voltage of an $AM$ wave is $24 \, mV$ and the minimum peak-to-peak voltage is $8 \, mV$. The modulation factor is.....$\%$

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