Ulaby problem 1.14

  • Ulaby
  • Electromagnetics
  • FE Exam
  • PE Exam
  • Wave math

posted on 31 Aug 2016

Ulaby Problem 1.14

Problem Statement

A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10m, and an amplitude of 81.87 (V/m) at a depth of 100 m. What is the attenuaton constant of seawater? 1

Analysis

When we have a sinusoidal wave in a lossy medium, we need to use and attenuation factor:

\[y(x,t) = Ae^{-\alpha x} \cos{(\omega t - \beta x + \phi_0)}\]

Where \(-\alpha\) is the attenuation constant with units of \(1/m\). The units of nepers per meter, \(Np/m\), refer to the attenuation constant.

Solution

Let’s set the two observations equal to each other, and then solve for the attenuation constant.

\[\begin{aligned} 98.02e^{-\alpha 10} &= 81.87e^{-\alpha 100} \\ \frac{98.02e^{-\alpha 10}}{81.87} &= e^{-\alpha 100} \\ \frac{98.02}{81.87} &= \frac{e^{-\alpha 100}}{e^{-\alpha 10}} \\ \frac{98.02}{81.87} &= e^{(-\alpha 100 - -\alpha 10)} \\ \frac{98.02}{81.87} &= e^{-\alpha (100 - 10)} \\ \ln{\left( \frac{98.02}{81.87} \right)} &= -\alpha 90 \\ \frac{\ln{\left( \frac{98.02}{81.87} \right)}}{-90} &= \alpha \\ \alpha &= \boxed{0.02 \; (Np/m)} \\\end{aligned}\]

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