Let the sinusoidal electric field polarised in the $\hat{x}$ direction be $\overline{\mathcal{E}}(x, y, z, t) = \hat{x}A(x, y, z)\cos(\omega t + \phi)$, where $A$ is the amplitude, $\omega$ is the radian frequency, and $\phi$ is the phase reference of the wave at $t = 0$. Now let $\overline{E}(x, y, z) = \hat{x} A(x, y, z) e^{j \phi}$. Therefore, $\overline{\mathcal{E}}(x, y, z, t) = \text{Re}\left\{ \overline{E}(x, y, z) e^{j \omega t} \right\}$.
I am then told that the average of the square of the magnitude of an electric field, given as
$$\overline{\mathcal{E}} = \hat{x} E_1 \cos(\omega t + \phi_1) + \hat{y} E_2 \cos(\omega t + \phi_2) + \hat{z} E_2 \cos(\omega t + \phi_3),$$
has the phasor form
$$\overline{E} = \hat{x} E_1 e^{j \phi_1} + \hat{y} E_2 e^{j \phi_2} + \hat{z} E_3 e^{j \phi_3},$$
can be calculated as
$$\begin{align} \left\vert \overline{\mathcal{E}} \right\vert^2_{\text{avg}} &= \dfrac{1}{T} \int_0^T \overline{\mathcal{E}} \cdot \overline{\mathcal{E}} \ dt \\ &= \dfrac{1}{T} \int_0^T \left[ E_1^2 \cos^2(\omega t + \phi_1) + E_2^2 \cos^2(\omega t + \phi_2) + E_3^2 \cos^2(\omega t + \phi_3) \right] \ dt \\ &= \dfrac{1}{2} (E_1^2 + E_2^2 + E_3^2) = \dfrac{1}{2} \left\vert \overline{E} \right\vert^2 = \dfrac{1}{2} \overline{E} \cdot \overline{E}^* \end{align}$$
It's unclear to me how we get from $$\dfrac{1}{T} \int_0^T \left[ E_1^2 \cos^2(\omega t + \phi_1) + E_2^2 \cos^2(\omega t + \phi_2) + E_3^2 \cos^2(\omega t + \phi_3) \right] \ dt$$ to $$\dfrac{1}{2} (E_1^2 + E_2^2 + E_3^2)$$
I have that $$\dfrac{1}{T} \int_0^T E^2 \cos^2(\omega t + \phi) = \dfrac{1}{T} E^2 \left\{ \dfrac{1}{2} (\omega t + \phi) + \dfrac{1}{4} \sin[2(\omega t + \phi)] \right\}^T_0 \\ = \dfrac{1}{T} E^2 \left\{ \dfrac{1}{2} (\omega T + \phi) + \dfrac{1}{4} \sin(2\omega T + 2 \phi) – \left[ \dfrac{1}{2} \phi + \dfrac{1}{4} \sin(2\phi) \right] \right\},$$
but it seems to me that this doesn't get us the expected answer. So how do we get from $$\dfrac{1}{T} \int_0^T \left[ E_1^2 \cos^2(\omega t + \phi_1) + E_2^2 \cos^2(\omega t + \phi_2) + E_3^2 \cos^2(\omega t + \phi_3) \right] \ dt$$ to $$\dfrac{1}{2} (E_1^2 + E_2^2 + E_3^2)$$?
Best Answer
Have another look at the integration:
$$\frac{1}{T}\int_0^TE^2\cos^2(\omega t+\phi)dt=\frac{E^2}{2T}\int_0^T(1+\cos2(\omega t+\phi))dt$$ $$=\frac{E^2}{2T}\left[t+\frac{1}{2\omega}\sin2(\omega t+\phi)\right]^T_0$$ $$=\frac{E^2}{2T}\left[T+\frac{1}{2\omega}\sin2(\omega T+\phi)-\frac{1}{2\omega}\sin 2\phi\right]$$ $$=\frac{E^2}{2T}\left[T+\frac{1}{2\omega}(2\cos(\omega T+2\phi)\sin\omega T)\right]$$
but $T=\frac{2\pi}{\omega}$ So for this (and similarly with the other integrals) you get the required result $\frac12E^2$.