TransmissionLineEq

Electromagnetic Wave Transmission Equation (Transmission Line Analogy)#

Maxwell's Equations#

$\nabla \cdot E = \frac{\rho}{\varepsilon_0} = 0$

$\nabla \cdot B = 0$

$\nabla \times E = -\frac{\partial B}{\partial t}$

$\nabla \times B = \mu_0 \varepsilon_0 \frac{\partial E}{\partial t}$

Derivation:

$\nabla \times (\nabla \times E) = -\frac{\partial}{\partial t}(\nabla \times B) = -\frac{\partial}{\partial t}(\mu_0 \varepsilon_0 \frac{\partial E}{\partial t}) = -\mu_0 \varepsilon_0 \frac{\partial^2 E}{\partial t^2}$

Tensor form:

$\nabla \times (\nabla \times E) = \varepsilon_{imn} \partial_m (\varepsilon_{njk} \partial_j E_k) = (\varepsilon_{imn} \varepsilon_{njk}) \partial_m \partial_j E_k = (\delta_{ij} \delta_{mk} - \delta_{ik} \delta_{mj}) \partial_m \partial_j E_k = \partial_j \partial_m E_m - (\nabla \cdot \nabla) E$

Final wave equation:

$-(\nabla \cdot \nabla) E = -\mu_0 \varepsilon_0 \frac{\partial^2 E}{\partial t^2}$

$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right) E = \mu_0 \varepsilon_0 \frac{\partial^2 E}{\partial t^2}$

Wave speed:

$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$

Plane wave solution:

$E = E_0 e^{i\omega t - i\frac{\omega}{c} z}$

$B = B_0 e^{i\omega t - i\frac{\omega}{c} z}$

$E, B$ are instantaneous quantities, $E_0, B_0$ are perpendicular to each other.

$-\frac{i\omega}{c} E_0 = -B_0 \cdot i\omega$

$E_0 = cB_0$

Transmission Line#

Telegraph Equations (Transmission Line Equations)#

Telegraph equation illustration

Equation Derivation#

$u(t,x) = u(t,x+\Delta x) + L\Delta x\frac{\partial i(t,x)}{\partial t} + R\Delta x \cdot i(t,x)$

$i(t,x) = i(t,x+\Delta x) + C\Delta x \frac{\partial u(t,x+\Delta x)}{\partial t} + G\Delta x \cdot u(t,x+\Delta x)$

$\begin{cases} -\frac{\partial u}{\partial x} = Ri + L\frac{\partial i}{\partial t} \\ -\frac{\partial i}{\partial x} = Gu + C\frac{\partial u}{\partial t} \end{cases}$

Lossless Line ( $R=0, G=0$ )#

$\begin{cases} -\frac{\partial u}{\partial x} = L\frac{\partial i}{\partial t} \\ -\frac{\partial i}{\partial x} = C\frac{\partial u}{\partial t} \end{cases}$

$\frac{\partial^2 u}{\partial x^2} = LC\frac{\partial^2 u}{\partial t^2}$

Wave speed:

$v = \frac{1}{\sqrt{LC}}$

Transmission line diagram

Boundary Conditions#

$\begin{cases} u(t,x) - i(t,x)Z_s = 0 \\ u(t,L) = i(t,L)Z_L \end{cases}$

Initial values:

$u(t=0,x) = 0, \quad i(t=0,x) = 0$

$\frac{\partial u}{\partial t}(t=0,x) = 0, \quad \frac{\partial i}{\partial t}(t=0,x) = 0$

Laplace Transform#

$\mathcal{L}[i(t,x)] = \tilde{I}(s,x)$

$\mathcal{L}[u(t,x)] = \tilde{U}(s,x)$

$\begin{cases} -\frac{\partial \tilde{U}}{\partial x} = sL\tilde{I} \\ -\frac{\partial \tilde{I}}{\partial x} = sC\tilde{U} \end{cases}$

$\frac{\partial^2 \tilde{U}}{\partial x^2} = s^2LC\tilde{U}$

Solution:

$\tilde{U}(s,x) = Ae^{-s\sqrt{LC}x} + Be^{+s\sqrt{LC}x}$

$\tilde{I}(s,x) = \frac{1}{Z_c}(Ae^{-s\sqrt{LC}x} - Be^{+s\sqrt{LC}x})$

Where:

$Z_c = \sqrt{\frac{L}{C}}$

$A_1 = \frac{Z_c}{Z_s+Z_c} \tilde{U}_0(s) + \frac{Z_s-Z_c}{Z_s+Z_c} B_1$

$B_1 = \frac{Z_L-Z_c}{Z_L+Z_c} A_1 e^{-2\gamma l}$

$\gamma = s\sqrt{LC}$

Laplace Inverse Transform#

$\tilde{U}_0(s) \rightarrow U_0(t)$
$\tilde{U}_0(s)e^{-\gamma x} \rightarrow U_0(t-\sqrt{LC}x)$ (traveling wave)
$\tilde{U}_0(s)e^{+\gamma x} \rightarrow U_0(t+\sqrt{LC}x)$ (reflected wave)

Source End Matching#

$\quad Z_s = Z_c$

$u(t,x) = \frac{1}{2} U_0(t-\frac{x}{v}) + \frac{1}{2}\rho U_0(t-(2l-x)/v)$

$\rho = \frac{Z_L-Z_c}{Z_L+Z_c}$

Instantaneous Matching Example#

$R_s = 100\Omega, \quad Z_c = 150\Omega, \quad Z_L = 1k\Omega, \quad V_{s} = 75V$

Wire transmission to the terminal, reflection:

$I = \frac{V}{Z_c+Z_s}$

At $t=0$ , input terminal, wave $V_i$ :

$V_i = \frac{Z_c}{Z_c+Z_s}V_s = 6V$

$I_i = \frac{1}{Z_c+Z_s}V_s = 40mA$

Time Interval	Source Voltage	Source Current	Time Interval	Terminal Voltage	Terminal Current
0 - 2τ	$V_i = 6V$	$I_i = 40mA$	0 - 2τ	0	0
2τ⁺ - 4τ	$V_i + (V_{r1} + V_{r2}) = 9.5V$	$I_i + (I_{r1} + I_{r2}) = 4mA$	τ⁺ - 3τ	$V_i + V_{r1} = 10.4V$	$I_i + I_{r1} = 10mA$
4τ⁺ - 6τ	$V_i + (V_{r1} + V_{r2}) + (V_{r3} + V_{r4}) = 8.9V$	$I_i + (I_{r1} + I_{r2}) + (I_{r3} + I_{r4}) = 9.3mA$	3τ⁺ - 5τ	$(V_i + V_{r1}) + (V_{r2} + V_{r3}) = 8.9V$	$(I_i + I_{r1}) + (I_{r2} + I_{r3}) = 8.4mA$

$\rho_L = \frac{1000-150}{1000+150} = 0.74$

Summary#

As long as the characteristic impedance and load impedance are not matched, reflection will occur. If both the source and load ends are mismatched, multiple reflections will occur.

Appendix#

Original images:

1755070142369
1755070152228
1755070158564

Last update: 2025-08-18
Created: 2025-08-13