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Exercise Problems · Q14

Q.An antenna has an impedance of 50Ω. An un modulated AM signal produces a current of 4.8 A. The percentage of modulation is 90. Calculate

(a) the carrier power
(b) the total power and
(c) sideband power.
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[!TLDR]

The unmodulated current gives a carrier power of 1152 W; at 90 % modulation the total power is about 1613 W and the sidebands together carry about 461 W.

  1. The carrier power is the unmodulated power dissipated in the antenna:

    Pc=Ic2R=(4.8)2×50=23.04×50=1152 WP_c = I_c^{2}R = (4.8)^{2}\times50 = 23.04\times50 = 1152\ \text{W}

  2. The total AM power exceeds the carrier by the factor (1+ma2/2)(1+m_a^{2}/2). With ma=0.9m_a = 0.9, ma2/2=0.405≈0.4m_a^{2}/2 = 0.405 \approx 0.4: Pt=Pc(1+ma22)≈1152×1.4=1613 WP_t = P_c\left(1+\dfrac{m_a^{2}}{2}\right) \approx 1152\times1.4 = 1613\ \text{W} …
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