Introduction
AC Analysis
For a total impedance \(Z(s)\) consists of \(R, L\) and \(C\), assume all the initial conditions of \(L\) and \(C\) are \(0\), then
\[V(s) = I(s)Z(s)\]assume when \(t\ge 0\), we apply current \(i(t)=I \cos(\omega t + \varphi)\), then the voltage response can be calculated by
\[v(t) = \int_0^t z(\tau) I \cos\left(\omega(t-\tau) + \varphi\right)d\tau\]we are interested in the response when \(t\to\infty\). Instead of directly calculating the above integration, we calculate
\[y(t) = I \int_0^t z(\tau)e^{j\omega(t-\tau)+\varphi}d\tau = e^{j\omega t} I e^{j\varphi} \int_0^t z(\tau)e^{-j\omega \tau} d\tau\]when \(t \to \infty\)
\[y(t) \to e^{j\omega t} I e^{j\varphi} \int_0^\infty z(\tau)e^{-j\omega \tau}d\tau = e^{j\omega t} I e^{j\varphi} Z(j\omega)\]then
\[\begin{cases} v(t) = \mathrm{Re}\left(y(t)\right) \to V \cos(\omega t + \theta)\\ V e^{j\theta} = I e^{j\varphi} \cdot Z(j\omega) \end{cases}\]Average Power Consumption
Assume
\[\begin{cases} v(t) = V \cos(\omega t)\\ i(t) = I \cos(\omega t + \theta) \end{cases}\] \[\overline{P} = \dfrac{1}{T} \int_0^{T} VI\cos(\omega t) \cos(\omega t + \theta) dt =\]Plane Wave
In Lossless Media
Example 1
assume \(E_y = E_z = \dfrac{\partial E_x}{\partial x} = \dfrac{\partial E_x}{\partial y} = 0\)
\[\dfrac{d^2 E_x(z)}{dz^2} + k^2 E_{x}(z) = 0\] \[E_x(z) = E_0^+ e^{-jkz} + E_0^{-} e^{jkz}\]Then the magnatic filed can be derived from Maxwell’s equation, we don’t have to solve the differential equation again
\[\nabla \times \vec{E} = -j\omega\mu \vec{H}\]For example, if \(E_x(z) = E_0^+ e^{-jkz} = E_x^+\)
\[\begin{cases} H_x^+ = 0\\ H_y^+ = \dfrac{k}{\omega \mu}E_x^+ = \dfrac{1}{\eta} E_x^+\\ H_z^+ = 0 \end{cases}\]where
\[\eta = \sqrt{\dfrac{\mu}{\eta}}\]