In an anisotropic medium, permittivity varies with electric field direction, and is a property seen in most crystals. Consider a uniform plane wave propagating in the \(z\) direction in such a medium, and which enters the material with equal field components along the \(x\) and \(y\) axes. The field phasor will take the form: $$\mathbf{E}_{s}(z)=E_{0}\left(\mathbf{a}_{x}+\mathbf{a}_{y} e^{j \Delta \beta z}\right) e^{-j \beta z}$$ where \(\Delta \beta=\beta_{x}-\beta_{y}\) is the difference in phase constants for waves that are linearly polarized in the \(x\) and \(y\) directions. Find distances into the material (in terms of \(\Delta \beta\) ) at which the field is ( \(a\) ) linearly polarized and \((b)\) circularly polarized. (c) Assume intrinsic impedance \(\eta\) that is approximately constant with field orientation and find \(\mathbf{H}_{s}\) and \(<\mathbf{S}>\).

Step by step solution

01

Linearly Polarized Distance

To find the distance \(z\) at which the field is linearly polarized, we must find the value of \(z\) when the electric field takes the form of \(\alpha \mathbf{a}_{x} + \beta \mathbf{a}_{y}\), with constant \(\alpha\) and \(\beta\). Since the electric field components have opposite exponential factors, this will occur when the two components have equal magnitudes and opposite phases, meaning that the imaginary exponential term must be equal to \(-1\) or \(1\). Therefore, we need to find the values of \(z\) that satisfy the equation: $$ e^{j \Delta \beta z} = 1 \ \ \text{or} \ \ e^{j \Delta \beta z} = -1$$

02

Solving for Linearly Polarized Distance

To solve for the values of \(z\) that satisfy the linearly polarized equation, we can use the Euler's formula: $$ e^{j \Delta \beta z} = \cos(\Delta \beta z) + j \sin(\Delta \beta z)$$ We can now compare this to the real and imaginary components of \(1\) and \(-1\): $$\cos(\Delta \beta z) = 1, \ -1$$, and $$\sin(\Delta \beta z) = 0$$. For the cosine term to be \(1\) or \(-1\), \(\Delta \beta z\) must be a multiple of \(\pi\). Thus, we have that the linearly polarized distances are given by: $$z = \frac{n \pi}{\Delta \beta} \ \ (n = 1, 2, 3, ... )$$

03

Circularly Polarized Distance

To find the distance \(z\) at which the field is circularly polarized, we need the field to take the form \(\alpha \mathbf{a}_{x} + \beta j \mathbf{a}_{y}\), with \(\alpha\) and \(\beta\) being equal magnitudes. In this case, the imaginary exponential term must be equal to \(j\), so we should have: $$ e^{j \Delta \beta z} = j$$

04

Solving for Circularly Polarized Distance

Using Euler's formula, and comparing the real and imaginary components, we obtain: $$\cos(\Delta \beta z) = 0$$, and $$\sin(\Delta \beta z) = 1$$. For the sin term to be \(1\), \(\Delta \beta z\) must be an odd multiple of \(\frac{\pi}{2}\). Hence, we have: $$z = \frac{(2n - 1)\pi}{2 \Delta \beta} \ \ (n = 1, 2, 3, ... )$$

05

Calculate \(\mathbf{H}_{s}\) and \(\)

With intrinsic impedance \(\eta\), the magnetic field phasor can be found by: $$\mathbf{H}_{s}(z) = \frac{1}{\eta}\left(\mathbf{a}_{x}+\mathbf{a}_{y} e^{j \Delta \beta z}\right)e^{-j \beta z}$$ To compute the average Poynting vector, we first calculate the complex Poynting vector as: $$\mathbf{S} = \frac{1}{2}\mathbf{E}_s \times \mathbf{H}_s^*$$ Taking the real part of the complex Poynting vector gives us the average Poynting vector: $$<\mathbf{S}> = \text{Re}(\mathbf{S})$$

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!