Question

The maximum peak to peak voltage of an AM wave is 24 mV and the minimum peak to peak voltage is 8 mV, then the modulation factor (${m_a}$) is:
(A) 50 %
(B) 75 %
(C) 25 %
(D) 10 %

Answer 1

Hint We can say that in general terms the ratio of the peak variation in the modulation actually is applicable for the transmitter to the maximum variation for which the transmitter is specifically designed.

Complete step by step answer
We know that,
${V_{\max }} = \dfrac{{24}}{2} = 12mV$
And the value of ${V_{\min }} = \dfrac{8}{4} = 4mV$
So now the value of modulation factor is:
$\dfrac{{{V_{\max }} - {V_{\min }}}}{{{V_{\max }} + {V_{\min }}}} = \dfrac{{12 - 4}}{{12 + 4}} = \dfrac{8}{{16}} = \dfrac{1}{2}$ $\dfrac{{{V_{\max }} - {V_{\min }}}}{{{V_{\max }} + {V_{\min }}}} = \dfrac{{12 - 4}}{{12 + 4}} = \dfrac{8}{{16}} = \dfrac{1}{2}$
So now the percentage is given as:
0.5 or 50 %

Note It should be known to us that the modulation index is defined as the ratio between the frequency derivation to the modulating frequency. From this formula and also the explanation of the modulation index, we will see that there is no term present which will be including the carrier frequency. This signifies that it will be totally independent of the carrier frequency.