An unmodulated carrier is 300 $VP-P$. Calculate the percentage of modulation when its maximum p-p value reac

Solved Examples · Example 8

V_{P-P}

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With $V_C = 150$ V and $V_{max} = 250$ V (peak), $m_a = (V_{max}-V_C)/V_C = 0.666$ , i.e. 66.6% modulation.

The amplitudes are quoted peak-to-peak, so each is first halved to get the peak value. The unmodulated carrier fixes $V_C$ ; when modulated, the envelope peak reaches $V_{max} = V_C + m_a V_C = V_C(1 + m_a)$ , which rearranges to $m_a = (V_{max} - V_C)/V_C$ .

Given $V_{C(P-P)} = 300$ V and $V_{max(P-P)} = 500$ V: …

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Q.An unmodulated carrier is 300 $V_{P-P}$ . Calculate the percentage of modulation when its maximum p-p value reaches 500V.

Concept understanding — AM Modulation Index

Q.An unmodulated carrier is 300 VP−PV_{P-P}VP−P​. Calculate the percentage of modulation when its maximum p-p value reaches 500V.

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Q.An unmodulated carrier is 300 $V_{P-P}$ . Calculate the percentage of modulation when its maximum p-p value reaches 500V.

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