Q.The output of a transmitter is given by $400[1 + 0.4\sin(6280)t]\sin(3.14\times10^7\,t)$. This voltage is fed to an antenna of resistance 500 $\Omega$. Determine

Comparing the output with the standard AM form gives $V_C = 400$ V, $m_a = 0.4$, $\omega_c = 3.14\times10^7$ and $\omega_m = 6280$; the frequencies follow from $f = \omega/2\pi$ and the powers from $P_C = V_C^2/2R$, $P_T = P_C(1+m_a^2/2)$ and $P_{peak} = V_{peak}^2/2R$ …

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By comparison $V_C = 400$ V, $m_a = 0.4$; then $f_c = 5$ MHz, $f_m = 1$ kHz, $P_C = 160$ W, $P_T = 173$ W and $P_{peak} = 313.6$ W.

Matching the given output to the standard AM equation $v_{AM} = V_C(1 + m_a\sin\omega_m t)\sin\omega_c t$ identifies $V_C = 400$ V, $m_a = 0.4$, $\omega_m = 6280$ rad/s and $\omega_c = 3.14\times10^7$ rad/s. The antenna resistance is $R = 500\ \Omega$.

1. Carrier frequency:

   $f_c = \frac{\omega_c}{2\pi} = \frac{3.14\times10^7}{2\times3.142} = 5\ \text{MHz}$

2. Modulating frequency:

   $f_m = \frac{\omega_m}{2\pi} = \frac{6280}{2\times3.142} = 1\ \text{kHz}$

3. Carrier power (the carrier is a sinusoid of peak $V_C$, so $P_C = V_C^2/2R$):

   $P_C = \frac{V_C^2}{2R} = \frac{(400)^2}{2\times500} = 160\ \text{W}$

4. Mean (total) power output: …