333 lines
15 KiB
TeX
333 lines
15 KiB
TeX
\hypertarget{chapter-2}{%
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\section{Chapter 2}\label{chapter-2}}
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\hypertarget{maxwells-equations-in-differential-form}{%
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\subsection{Maxwell's Equations in Differential
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Form}\label{maxwells-equations-in-differential-form}}
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\[\int \vec{E} \cdot d\vec{s} = \frac{Q_{enc}}{\varepsilon_0} \xleftrightarrow{\text{divergence}} \nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}\]
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\[\int \vec{B} \cdot d\vec{s} = 0 \xleftrightarrow{} \nabla \cdot \vec{B} = 0\]
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\[\int \vec{E} \cdot d\vec{l} = -\frac{d}{dt}\vec{B} \xleftrightarrow{\text{stokes}} \nabla \times \vec{E} = -\frac{d}{dt}\vec{B}\]
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\[\int \vec{B} \cdot d\vec{l} = \mu_0\left( \vec{J} + \varepsilon_0 \frac{d\vec{E}}{dt}\right) \xleftrightarrow{} \nabla \times \vec{B} = \mu_0 J + \mu_0 \varepsilon_0 \frac{\partial \vec{E}}{\partial t}\]
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Before Maxwell (only true for magnetostatics):
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\[\nabla \times \vec{B} = \mu_0\vec{J}\]
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\[\nabla \cdot (\nabla \times \vec{B}) = \mu_0(\nabla \cdot \vec{J})\]
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\[\nabla \cdot \vec{J} = 0\]
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For changing magnetic fields:
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\[\nabla \cdot \vec{J} = -\frac{\partial}{\partial t} \rho_v\]
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\[\nabla \cdot \vec{J} + \frac{\partial}{\partial t} \rho_v = 0\]
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\hypertarget{example-2.1}{%
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\subsubsection{Example 2.1}\label{example-2.1}}
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\[\vec{J} = e^{-x^2}\hat{x}\] Find the time rate of change of charge
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densisty at \(x=1\):
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\[\nabla \cdot \vec{J} = -\frac{\partial}{\partial t} \rho_v\]
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\[\frac{\partial \rho}{\partial t} = -\nabla \cdot \vec{J} = -\frac{d}{dx}e^{-x^2} = 2xe^{-x^2}\]
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\[\therefore 2 e^{-1}\]
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\hypertarget{example-2.2}{%
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\subsubsection{Example 2.2}\label{example-2.2}}
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For some spherical current density:
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\[\vec{J} = \frac{J_0 e^{-t/\tau}}{\rho}\hat{\rho}\] Find the total
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current that leaves the surface of radius, \(t = \tau\):
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\[I = \int \vec{J} \cdot d\vec{s} = 4\pi a^2 \left(\frac{J_0 e^{-t/\tau}}{a}\right)\]
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\[I = 4\pi a J_0 e^{-1}\]
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Find \(\rho_v(\rho, t)\):
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\[\nabla \cdot \vec{J} = -\frac{\partial \rho}{\partial t} = \frac{1}{\rho^2} \frac{\partial}{\partial \rho}(\rho^2 J)\]
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Plug in for \(J\):
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\[\nabla \cdot \vec{J} = \frac{1}{\rho^2} \frac{\partial}{\partial \rho}\left(\rho^2 \frac{J_0 e^{-t/\tau}}{\rho}\right) = \frac{J_0}{\rho^2} e^{-t/\tau} = -\frac{\partial}{\partial t} \rho_v\]
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Solve for \(\rho_v\):
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\[\rho_v = \int -\frac{J_0}{\rho^2}e^{-t/\tau} dt\]
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\[\rho_v = \frac{J_0}{\rho^2} e^{-t/\tau} \tau\]
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\hypertarget{wave-propagation}{%
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\subsection{Wave Propagation}\label{wave-propagation}}
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\[\nabla \times \vec{E} = -\frac{\partial}{\partial t} \vec{B}\]
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\[\nabla \times \vec{B} = \mu_0 (\vec{J} + \varepsilon_0 \frac{\partial}{\partial t} \vec{E})\]
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\[\nabla \cdot \vec{E} = \frac{\rho_v}{\varepsilon_0}\]
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\[\nabla \cdot \vec{B} = 0\]
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Take the curl of the first equation:
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\[\nabla \times (\nabla \times \vec{E}) = -\frac{\partial}{\partial t}(\nabla \times \vec{B})\]
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\[\nabla (\nabla \cdot \vec{E}) - \nabla^2\vec{E} = -\frac{\partial}{\partial t}(\nabla \times \vec{B})\]
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Substitute the second equation into the one directly above:
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\[\nabla(\nabla \cdot \vec{E}) - \nabla^2\vec{E} = -\frac{\partial}{\partial t} \left( \mu_0 \vec{J} + \mu_0\varepsilon_0 \frac{\partial}{\partial t}\vec{E} \right)\]
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\hypertarget{homogenious-vector-wave-for-vece-fields}{%
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\paragraph{\texorpdfstring{Homogenious vector wave for
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\(\vec{E}\)-fields}{Homogenious vector wave for \textbackslash vec\{E\}-fields}}\label{homogenious-vector-wave-for-vece-fields}}
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Lets say we are considering wave propagation in a source free region
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(\(\rho_v =0\)).
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\[\nabla \cdot \vec{E} = \frac{\rho_v}{\varepsilon_0} =0\]
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\[\vec{J} = -\frac{\partial}{\partial t} \rho_v = 0\]
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Now we have:
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\[-\nabla^2\vec{E} = -\mu_0\varepsilon_0 \frac{\partial}{\partial t} \vec{E}\]
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\[\therefore \nabla^2\vec{E}-\mu_0\varepsilon_0 \frac{\partial}{\partial t}\vec{E} = 0\]
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\hypertarget{homogeneous-vector-wave-for-vecb-fields}{%
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\paragraph{\texorpdfstring{Homogeneous vector wave for
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\(\vec{B}\)-fields}{Homogeneous vector wave for \textbackslash vec\{B\}-fields}}\label{homogeneous-vector-wave-for-vecb-fields}}
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\[\nabla^2\vec{B} - \mu_0\varepsilon_0 \frac{\partial^2}{\partial t^2}\vec{B} = 0\]
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\hypertarget{phasor-representations}{%
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\subsubsection{Phasor representations}\label{phasor-representations}}
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In general, all fields and their sources vary as a function of position
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and time. If the time variations are sinusoidal, with angular frequency,
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\(\omega\), then each of their quantities can be represented by a time
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independent phasor.
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\[\vec{E}(x,y,z,t) = \Re\{\vec{E}(x,y,z)e^{j \omega t}\}\]
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\[\vec{E}(r,t) = \Re\{\hat{E}(r) e^{j \omega t}\}\]
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\[\vec{B}(r,t) = \Re\{\hat{B}(r) e^{j \omega t}\}\]
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\(\hat{E}\) and \(\hat{B}\) are the complex time variations in vector
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form. (\(\hat{E} = \hat{\vec{E}}\)) .
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If \(\hat{E}(r) = E_0 e^{j \theta}\),
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\[\vec{E}(r, t) = \Re\{E_0 e^{j \theta} e^{j \omega t}\} = E_0\cos(\omega t + \theta)\]
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Consider a plane wave in the x-direction only (\(E_y = E_z = 0\)), and
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does ont vary the x or y direction
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(\(\frac{\partial}{\partial x}\hat{E} = \frac{\partial}{\partial y}\hat{E} = 0\)).
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Assume free space for propagation (\(\hat{J} = \hat{\rho}_v = 0\)).
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\[\nabla \times \vec{E} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \hat{E}_x & \hat{E}_y & \hat{E}_z \end{vmatrix} = -j\omega (\hat{B}_x\hat{x} + \hat{B}_y \hat{y} + \hat{B}_z \hat{z}\]
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\[\therefore \begin{Bmatrix} -\frac{\partial}{\partial z}\hat{E} = -j\omega\hat{B}_x \\ \frac{\partial}{\partial z}\hat{E} = -j\omega\hat{B}_y \\ 0 = -j\omega\hat{B}_z \end{Bmatrix}\]
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Similarly, Ampere's Law:
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\[\begin{Bmatrix} -\frac{\partial \hat{B}_y}{\partial z} = j\omega\varepsilon_0\mu_0\hat{E}_x \\ \frac{\partial \hat{B}_x}{\partial z} = j\omega\varepsilon_0\mu_0\hat{E}_y \\ 0 = j\omega\varepsilon_0\mu_0\hat{E}_z\end{Bmatrix}\]
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Therefore, \(B_x\) and \(E_y\) are related to each other. \(B_x\) acts
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as the source for generating \(E_y\), and \(E_y\) acts as the source for
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generating \(B_x\). Solving for \(\hat{E}_x\) and \(\hat{B}_y\):
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\[\frac{\partial^2 \hat{E}_x}{\partial z^2} = -j\omega\frac{\partial \hat{B}_y}{\partial z} = -\omega^2\mu_0\varepsilon_0\hat{E}_x\]
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\[\therefore \frac{\partial^2}{\partial z^2}\hat{E}_x + \mu_0\varepsilon_0\omega^2\hat{E}_x = 0\]
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The general solution:
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\[\hat{E}_x = \hat{C}_1 e^{-j \beta_0 z} + \hat{C}_2 e^{j \beta_0 z}\]
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Where \(\hat{C}_1\) and \(\hat{C}_2\) are complex.
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\[\therefore \hat{E}_x = \hat{E}_m^+ e^{-j \beta_0 z} + \hat{E}_m^- e^{j \beta_0 z}\]
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Where \(E_m^+\) and \(E_m^-\) are wave amplitudes (volts per meter).
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\hypertarget{the-phase-constant}{%
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\subparagraph{The phase constant:}\label{the-phase-constant}}
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\[\beta_0 = \omega \sqrt{\mu_0 \varepsilon_0} = \frac{2\pi}{\lambda} = \frac{2\pi f}{C}\]
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\hypertarget{real-time-form-of-the-solution}{%
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\paragraph{Real time form of the
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solution:}\label{real-time-form-of-the-solution}}
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\[E_x(z,t) = \Re\{\hat{E}_x e^{j \omega t}\} = \Re\{E_m^+ e^{j(\omega t - \beta_0 z)} + E_m^- e^{j(\omega t + \beta_0 z)}\}\]
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\[\therefore E_x(z,t) = E_m^+ \cos(\omega t - \beta_0 z) + E_m^- \cos(\omega t + \beta_0 z)\]
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If \(E_m^+ = E_m^-\), the magnitude and direction is \(E_0\).
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\[\therefore E_x(z,t) = E_0\cos(\omega t \pm \beta_0 z)\] Where:
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\(\omega = 2\pi f\)
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\(v_0\) is the phase velocity
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\hypertarget{the-phase-velocity}{%
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\subparagraph{The phase velocity}\label{the-phase-velocity}}
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\[v_0 = \frac{1}{\sqrt{\mu \varepsilon}}\] In a vacuum: \[v_0 = c\]
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\hypertarget{also}{%
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\paragraph{Also}\label{also}}
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\[\hat{B}_y = \frac{\hat{E}_x}{c}\]
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Common \(\vec{E}\) and \(\vec{B}\) field ratio is:
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\[\mu_0 \hat{H}_y = \frac{\hat{E}_x}{c}\]
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\[\therefore \frac{\hat{E}_x}{\hat{H}_y} = \mu_0 c = \frac{\mu_0}{\sqrt{\mu_0 \varepsilon_0}} = \sqrt{\frac{\mu_0}{\varepsilon_0}} = \eta_0 \approx 120\pi = 377\Omega\]
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Where \(\eta_0\) is the intrinsic wave impedance.
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\[\therefore \hat{E}_x = \hat{H}_y \eta_0\]
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\[\therefore \hat{H}_y = \frac{\hat{E}_x}{\eta_0}\]
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Therefore, both \(\vec{H}\) and \(\vec{E}\) fields are in phase.
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In general: \[\hat{H} = \frac{\hat{k}}{\eta_0} \times \hat{E}\]
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\[\hat{E} = -\eta_0 \hat{k} \times \hat{H}\]
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\hypertarget{example-2.3}{%
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\subsubsection{Example 2.3}\label{example-2.3}}
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\[\vec{E} = 50\cos(10^8t + \beta_0 x)\hat{y} \text{[v/m]}\]
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\[E_y = E_0\cos(\omega t + \beta x)\] \(E_y\) propagates in the
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\(-\hat{x}\) direction.
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\[\beta_0 = \omega \sqrt{\mu_0 \varepsilon_0} = \frac{\omega}{c} = \frac{10^8}{3 \times 10^8} = \frac{1}{3}\]
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What is the time it takes to travel a distance of \(\frac{\lambda}{2}\)?
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\[\omega = 2\pi f = \frac{2\pi}{T}\] \[T = \frac{2\pi}{\omega}\]
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\[\therefore t_{\lambda/2} = \frac{T}{2} = \frac{\pi}{\omega} = \frac{\pi}{10^8} \approx 31.42\text{[ns]}\]
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The refractive index of a medium is given by the ratio of \(c\) and
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\(v_p\): \[c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}\]
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\[v_p = \frac{1}{\sqrt{\mu \varepsilon}}\]
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\[n = \frac{c}{v_p} = \sqrt{\frac{\mu \varepsilon}{\mu_0 \varepsilon_0}} = \sqrt{\mu_r \varepsilon_r}\]
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For a non-magnetic material: \[\mu_r \approx 1\]
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\[\therefore n = \sqrt{\varepsilon_r}\]
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\hypertarget{polarization}{%
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\subsection{Polarization}\label{polarization}}
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\[\begin{rcases}
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\nabla \cdot \vec{E} = {\rho \over \varepsilon_0} \\
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\nabla \cdot \vec{B} = 0
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\end{rcases} \text{Gauss}\]
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\[\begin{rcases}
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\nabla \times \vec{E} = -{\partial \over \partial t} \vec{B}
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\end{rcases} \text{Faraday}\]
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\[\begin{rcases}
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\nabla \times B = \mu_0 \vec{J} + \mu_0 \varepsilon_0{\partial \over \partial t}\vec{E}
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\end{rcases} \text{Ampere}\]
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Polarization of a uniform plane wave describes the shape and olcus of
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the tip of the \(\vec{E}\)-field vector in the plane orthoganal to the
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direction of propagation. There are two pairs of \(\vec{E}\) and
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\(\vec{B}\) fields (\(\hat{E}_x\), \(\hat{H}_y\)) and (\(\hat{E}_y\),
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\(\hat{H}_x\)). When both pairs are present, we can evaluate
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polarization of plane waves. The \(\vec{E}\)-field has components in the
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\(\hat{x}\) and \(\hat{y}\) directions and travels in \(\hat{z}\).
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\[\hat{E} = (\hat{E}_x \hat{x} + \hat{E}_y \hat{y})e^{-j \beta z}\]
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\[\hat{E}_x = \lvert \hat{E}_{x_0} \rvert e^{-j \beta a}\]
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\[\hat{E}_y = \lvert \hat{E}_{y_0} \rvert e^{-j \beta b}\]
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\hypertarget{locus}{%
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\paragraph{Locus}\label{locus}}
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The shape the tip of the \(\vec{E}\)-field vector traces out while in
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motion.
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\hypertarget{phase}{%
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\paragraph{Phase}\label{phase}}
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Typically defined relative to a reference point such as \(z=0\) or
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\(t=0\) or some combination.
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\hypertarget{polarization-characteristics}{%
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\subsubsection{Polarization
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Characteristics}\label{polarization-characteristics}}
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\hypertarget{linear-polarization}{%
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\paragraph{Linear polarization}\label{linear-polarization}}
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\(\hat{E}_x\) and \(\hat{E}_y\) have the same phase angle: \(a = b\), so
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the x and y components of the \(\vec{E}\)-field will be in phase.
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\[\hat{E} = (\lvert \hat{E}_x \rvert \hat{x} + \lvert \hat{E}_y \rvert \hat{y} ) e^{-j( \beta z - a)}\]
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In real time:
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\[\hat{E} = (\lvert \hat{E}_x \rvert \hat{x} + \lvert \hat{E}_y \rvert \hat{y} ) \cos(\omega t - \beta z + a)\]
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As the wave continues to propagate in the \(\hat{z}\) direction, the
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\(\vec{E}\)-field vector maintains its direction with angle, \(\theta\),
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with respect to the y-axis.
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\[\tan(\theta) = {\lvert \hat{E}_x \rvert \over \vert \hat{E}_y \rvert}\]
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When \(z=0\), the \(\vec{E}\) field is given by:
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\[\vec{E}(0, t) = (\lvert \hat{E}_x \rvert \hat{x} + \lvert \hat{E}_y \rvert \hat{y} ) \cos(\omega t + a)\]
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\[\vec{E}(0, 0) = (\lvert \hat{E}_x \rvert \hat{x} + \lvert \hat{E}_y \rvert \hat{y} ) \cos(a)\]
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\hypertarget{elliptical-polarization}{%
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\paragraph{Elliptical Polarization}\label{elliptical-polarization}}
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\(\hat{E}_x\) and \(\hat{E}_y\) have different phase angles. \(\vec{E}\)
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is no longer in one plate.
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\[\hat{E} = \hat{x} \lvert \hat{E}_x \rvert e^{j (a - \beta z)} + \hat{y} \lvert \hat{E}_y \rvert e^{j(b - \beta z)}\]
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Where:
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\(E_x = \lvert \hat{E}_x \rvert \cos(\omega t - a + \beta z)\)
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\(E_y = \lvert \hat{E}_y \rvert \cos(\omega t - b + \beta z)\)
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If \(a=0\) and \(b = {\pi \over 2}\):
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\[\vec{E}_x(z,t) = \lvert \hat{E}_x \rvert \cos(\omega t - a + \beta z)\]
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\[\vec{E}_y(z,t) = \lvert \hat{E}_y \rvert \cos(\omega t - b + \beta z)\]
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\hypertarget{circular-polarization}{%
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\paragraph{Circular Polarization}\label{circular-polarization}}
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\(\hat{E}_x\) and \(\hat{E}_y\) have the same magnitude with a phase
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angle difference of \({\pi \over 2}\).
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\hypertarget{example}{%
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\subsubsection{Example}\label{example}}
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Determine the real-valued \(\vec{E}\)-field.
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\[\hat{E}(z) = -3j\hat{x} e^{-j\beta z}\]
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\[\vec{E}(z, t) = -3j\hat{x} e^{-j \beta z} e^{j\omega t}\]
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\[3 \hat{x} e^{-j \beta z} e^{j \omega t} e^{-\pi \over 2}\]
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\[3 \cos\left(\omega t- \beta z - {\pi \over 2}\right)\]
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\hypertarget{example-1}{%
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\subsubsection{Example}\label{example-1}}
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What are \(E_x\) and \(E_y\)
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\[\hat{E}(z) = (3\hat{x} + 4\hat{y})e^{j\beta z}\]
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\[\vec{E}(z, t) = (3\hat{x} + 4\hat{y})e^{j\beta z}e^{j\omega t}\]
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\[E_x = 3\cos(\omega t + \beta z)\] \[E_y = 4\cos(\omega t + \beta z)\]
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\hypertarget{example-2}{%
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\subsubsection{Example}\label{example-2}}
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\[\hat{E}(z) = (-4\hat{x} + 3\hat{y})e^{-j\beta z}\]
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\[\hat{E}(z, t) = (-4\hat{x} + 3\hat{y})e^{-j\beta z}e^{j\omega t}\]
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\[E_x = 4\cos(\omega t - \beta z + \pi)\]
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\[E_y = 3\cos(\omega t - \beta z)\]
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\hypertarget{non-sinusoidal-waves}{%
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\subsection{Non-Sinusoidal Waves}\label{non-sinusoidal-waves}}
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Analytical solution of a 1-D traveling wave.
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\[{\partial^2 \over \partial z^2}\vec{E} - {1\over c^2}{\partial^2 \over \partial t^2}\vec{E} = 0\]
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\hypertarget{dalemberts-solution}{%
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\subsubsection{D'Alemberts Solution}\label{dalemberts-solution}}
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\[\vec{E}(z, t) = E(z - ct) + E'(z + ct)\]
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Show that the function, \(F(z - ct) = F_0 e^{-(z+ct)^2}\) is a solution
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of the wave equation.
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Let \(\gamma = z - ct\), and \({\partial \gamma \over \partial z} = 1\):
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\[F(z-ct) = F_0 e^{-\gamma^2}\] \[F'(z + ct) = 0\]
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\[{\partial F \over \partial z} = {\partial F \over \partial \gamma} {\partial \gamma \over \partial z} = F_0 e^{-\gamma^2}(-2\gamma)\]
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\[{\partial^2 F \over \partial z^2} = {\partial \over \partial z}\left({\partial F \over \partial \gamma}{\partial \gamma \over \partial z}\right)\]
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\[G = \left({\partial F \over \partial \gamma}{\partial \gamma \over \partial z}\right)\]
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Don't forget chain rule:
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\[{\partial G \over \partial \gamma}{\partial \gamma \over \partial z} = {\partial \over \partial z}\left({\partial F \over \partial \gamma}{\partial \gamma \over \partial z}\right){\partial \gamma \over \partial z}\]
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\[{\partial \over \partial \gamma} - 2\gamma F_0 e^{-\gamma^2} = -2F_0 {\partial \over \partial \gamma} \gamma e^{-\gamma^2}\]
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By product rule:
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\[-2F_0 (\gamma (-2\gamma e^{-\gamma^2}) + e^{-\gamma^2})\]
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\[F_0 (4\gamma^2 e^{-\gamma^2} - 2e^{-\gamma^2}) = F_0(-2 + 4\gamma^2)e^{-\gamma^2}\]
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\[={\partial^2 \over \partial \gamma^2}F \left({\partial \gamma \over \partial z}\right)^2\]
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\[\therefore {\partial^2 \over \partial t^2}F = {\partial^2 F \over \partial \gamma^2} \left({\partial \gamma \over \partial t}\right)^2 = F_0(-2 + 4\gamma^2)e^{-\gamma^2}(C^2)\]
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This solution satisfies:
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\[{\partial^2 \over \partial z^2}\vec{E} - {1\over c^2}{\partial^2 \over \partial t^2}\vec{E} = 0\]
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