# Background

### Electromagnetic Waves

As I understand it, an electromagnetic wave, like any other wave, is capable of delivering power from a source to a destination. To quantify how much power is delivered from the source to the destination, a quantity called the Poynting vector is used. The Poynting vector points in the direction of propagation. Mathematically, it is given by

$$\vec{S} = \vec{E} \times \vec{H}$$

This quantity, $\vec{S}$, has units of (instantaneous) power per unit area. If it is assumed that both the electric field $\vec{E}$ and the magnetic field $\vec{H}$ are sinusoidally varying functions of time, then it is possible to determine an expression for the average power per unit area delivered $\left<\vec{S}\right>$. Consider this Wikipedia link for the short derivation. The final result is

$$\left<\vec{S}\right> = \frac{1}{2}\text{Re}\left(\vec{\widetilde{E}} \times \vec{\widetilde{H}}^*\right)$$

This is the average power per unit area. To get the average power delivered to a region, simply integrate over the region's surface area.

$$P_{\text{avg}} = \int_A \left<\vec{S}\right> \cdot d\vec{A}$$

# Question

### Power Delivered By A Transmission Line

In Elements of Electromagnetics (Sadiku, 3rd edition, Section 11.4), the author defines the average input power at a distance $\ell$ from the load to be given by

$$P_{\text{avg}} = \frac{1}{2} \text{Re}\left(V_s\left(\ell\right)I_s^*\left(\ell\right)\right)$$

Where again $P_{\text{avg}}$ is the average power, $V_s\left(\ell\right)$ is the time-harmonic solution to the Telegrapher's Equations solved for voltage, $I_s^*\left(\ell\right)$ is the complex conjugate of the time-harmonic solution to the Telegrapher's Equations solved for current. Further it is said that the $1/2$ term comes from the fact that peak values are used rather than RMS values.

So, is there a way to get from the average power equation given in terms of the Poynting vector, which is in terms of the electric field and the magnetic field, to this formulation involving voltage and current?

# Attempt At A Solution

Note that I have used the same notation as is found on the Poynting vector link here. I assume a coaxial cable as the area over which to integrate. Somehow I see two separate line integrals will need to crop up in my solution because units of electric field times units of distance will give me units of electrical potential (voltage) and units of magnetic field times units of distance will give me units of electrical current (current).

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