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I am new to communication and studying amplitude modulation. Let us assume an amplitude modulated wave given by $$\big(m(t)+ A\big) \cos(2\pi f_c t)$$ Now we have formula for efficiency as $$ \eta=\text{useful power / Total power}=\frac{\overline{m(t)^2}/2}{A^2/2+\overline{m(t)^2}/2}$$ I am not able to understand why total power is not equal to $$\overline{(m(t)+ A)^2}$$ where $\bar{.}$ denotes average power. I mean where does the term for $A^2\overline{m(t)^2}$ go in the denominator of the efficiency expression. Do we assume modulating signals and carrier signal to be uncorrelated?

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The total power of an amplitude modulated signal is

$$\begin{align}\overline{s^2_{AM}(t)}&=\overline{\big(A+m(t)\big)^2\cos^2(2\pi f_ct)}\\&=\frac12\overline{\big(A+m(t)\big)^2}+\frac12\overline{\big(A+m(t)\big)^2\cos(4\pi f_ct)}\tag{1}\end{align}$$

The second term on the right-hand side of $(1)$ is zero if $m(t)$ is a lowpass signal, and if $f_c$ is sufficiently large (i.e., greater than the highest frequency in $m(t)$).

Consequently, we have

$$\begin{align}\overline{s^2_{AM}(t)}&=\frac12\overline{\big(A+m(t)\big)^2}\\&=\frac12\big(A^2+2A\cdot\overline{m(t)}+\overline{m^2(t)}\big)\tag{2}\end{align}$$

The term $\overline{m(t)}$ is the DC value of the message signal, which is assumed to be zero. With this assumption we finally arrive at

$$\overline{s^2_{AM}(t)}=\frac12\big(A^2+\overline{m^2(t)}\big)\tag{3}$$

In sum, the assumptions for which $(3)$ is valid are that $m(t)$ is a lowpass signal, that the carrier frequency is sufficiently high, and that $m(t)$ has a DC value equal to zero. It is usually straightforward to satisfy all of these assumptions to a very high degree of accuracy.

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  • $\begingroup$ @ Matt L. Regarding the assumption that .....The second term on the right-hand side of (1) is zero if m(t) is a lowpass signal, and if fc is sufficiently large (i.e., greater than the highest frequency in m(t))......Intuitively, we mean that m(t) is taken to be slowly varying with respect to cos(2πfct) and therefore can be assumed as constant, and not affecting the integral of (1). Thank you for your response. $\endgroup$ – Userhanu Feb 23 at 11:31
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    $\begingroup$ @Userhanu: Yes, that's a way to see it intuitively. $\endgroup$ – Matt L. Feb 23 at 11:50

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