Electronics · Ch 8 — Modulation and Demodulation

Power relations in AM wave

8.3b

Power relations in AM wave

Note

In the printed textbook this section is also numbered 8.3, the same number given earlier to Analysis of Amplitude Modulated Wave. To keep every entry uniquely identifiable here it is shown as 8.3b, but on the page it appears simply as section 8.3.

The signal-to-noise ratio at a receiver depends on the received signal power, which must be much larger than the noise power. The power carried by an AM wave is therefore an important design parameter.

An AM wave is made of a carrier and two sidebands, so its total radiated power is the sum of the three:

P_T = P_C + P_{LSB} + P_{USB} \qquad (8.16) = \frac{V_C^2}{R} + \frac{V_{USB}^2}{R} + \frac{V_{LSB}^2}{R} \qquad (8.17)

where the voltages are rms values and $R$ is the antenna resistance in which the power is dissipated. The rms amplitude of each sideband is

V_{LSB} = V_{USB} = \frac{m_a V_C}{2} \qquad (8.18)

Using $V_C$ as the peak carrier voltage, the carrier power is

P_C = \frac{V_{rms}^2}{R} = \frac{(V_C/\sqrt2)^2}{R} = \frac{V_C^2}{2R} \qquad (8.19)

and each sideband contributes

P_{LSB} = P_{USB} = \frac{(m_a V_C/2\sqrt2)^2}{R} = \frac{m_a^2 V_C^2}{8R} \qquad (8.20)

Substituting (8.19) and (8.20) into the total gives the central result

P_T = \frac{V_C^2}{2R} + \frac{m_a^2 V_C^2}{4R} = \frac{V_C^2}{2R}\left(1 + \frac{m_a^2}{2}\right) = P_C\left(1 + \frac{m_a^2}{2}\right) \qquad (8.21)

So the carrier power in the modulated wave is exactly the same as in the unmodulated carrier; the extra power is contributed entirely by the sidebands, which carry all the useful information. The large carrier power carries no information at all — the carrier is only a vehicle. At 100% modulation ( $m_a = 1$ ), $P_T = 1.5\,P_C$ .

Power in the sidebands

The total power in both sidebands is

P_{SB} = P_T - P_C = P_C\left(1 + \frac{m_a^2}{2}\right) - P_C = \frac{m_a^2}{2}P_C = \frac{m_a^2}{2 + m_a^2}\,P_T \qquad (8.22)

For $m_a = 1$ this is $P_{SB} = \tfrac{1}{3}P_T = \tfrac{1}{2}P_C$ . The power in each individual sideband is half of this:

P_{USB} = P_{LSB} = \frac{1}{2}P_{SB} = \frac{m_a^2}{4}P_C = \frac{1}{2}\cdot\frac{m_a^2}{2 + m_a^2}\,P_T \qquad (8.23)

which equals $\tfrac{1}{2}P_C = P_T/6$ at 100% modulation.

Transmission efficiency

Transmission efficiency is the ratio of the power carried by the sidebands (the useful power) to the total transmitted power:

\eta = \frac{P_{SB}}{P_T} = \frac{(m_a^2/2)P_C}{P_C(1 + m_a^2/2)} = \frac{m_a^2}{2 + m_a^2} \qquad (8.24)

Even at 100% modulation ( $m_a = 1$ ) this gives only $\eta = 33.33\%$ . In other words, the maximum transmission efficiency of AM is about one-third: only $1/3$ of the total power is carried by the information-bearing sidebands, and the remaining $2/3$ is wasted on the carrier, which carries no information.

Modulation index in terms of currents …

Formula 1Total power in an AM wave

$P_T = P_C\left(1 + \dfrac{m_a^2}{2}\right)$ (8.21): carrier power plus sideband power. At 100% modulation $P_T = 1.5\,P_C$ . The carrier power is unchanged by modulation; the ex …

Formula 2Power in the sidebands

Total sideband power $P_{SB} = \dfrac{m_a^2}{2}P_C = \dfrac{m_a^2}{2+m_a^2}P_T$ (8.22); power in each sideband $P_{USB} = P_{LSB} = \dfrac{m_a^2}{4}P_C$ (8.23). At $m_a = 1$ , $P_{SB} = P_T …

Definition 3Transmission efficiency of AM

The ratio of useful sideband power to total transmitted power, $\eta = \dfrac{m_a^2}{2+m_a^2}$ (8.24). Its maximum value is only 33.33% (at 100% modulation), because two-thirds of the power sta …

Formula 4Modulation index in terms of antenna currents

$I_t = I_c\sqrt{1 + \dfrac{m_a^2}{2}}$ (8.25), where $I_c$ is the rms unmodulated carrier current and $I_t$ the rms total current — lets $m_a$ be found from current measurements. Since power is proportional to current squared, $P_T/P_C = (I_t/I_c)^2 = 1 + m_a^2/2$ , so measuring only the carrier-only and modula …

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