Transmission lines - Part II: The telegraphists' equations

As said in the previous article, the RLC model is the transmission line theory starting point: it is characterized by a few lumped components modelling the main parameters of a small portion of the line; then is possible to describe locally the behaviour of the line and then it can be applied to the whole channel length. Look at the RLC model below, which describes a small line segment and the related equations.

$$R \left[\frac{\Omega}{m}\right], L \left[\frac{H}{m}\right], C \left[\frac{F}{m}\right], G \left[\frac{S}{m}\right]$$ $$ \begin{cases} V(z,t)-V(z+\Delta z,t)=I(z,t)R\Delta z + \frac{dI(z,t)}{dt} L \Delta z \\ I(z,t)-I(z+\Delta z,t)=V(z+\Delta z,t)G\Delta z + \frac{dV(z+\Delta z,t)}{dt} C \Delta z \end{cases} $$ $$ \begin{cases} \frac{V(z,t)-V(z+\Delta z,t)}{\Delta z}=\frac{I(z,t)R\Delta z + \frac{dI(z,t)}{dt} \Delta z}{\Delta z} \\ \frac{I(z,t)-I(z+\Delta z,t)}{\Delta z}=\frac{V(z+\Delta z,t)G\Delta z + \frac{dV(z+\Delta z,t)}{dt} C \Delta z}{\Delta z} \end{cases} $$ $$ \begin{cases} \frac{dV(z,t)}{dz}=I(z,t)R + \frac{dI(z,t)}{dt} L \\ \frac{dI(z,t)}{dz}=V(z,t)G + \frac{dV(z,t)}{dt} C \end{cases} $$

For a lossless line the resistors can be removed from the circuit so R=0Ohm and G=0S. The equations become: $$ \begin{cases} \frac{dV(z,t)}{dz}=\frac{dI(z,t)}{dt} L \\ \frac{dI(z,t)}{dz}=\frac{dV(z,t)}{dt} C \end{cases} $$

Now we can derive the first equation w.r.t. t and the second one w.r.t. z and combine the two, obtaining two wave equations! $$ \begin{cases} \frac{d^2V(z,t)}{dzdt}=\frac{d^2I(z,t)}{dt^2} L \\ \frac{d^2I(z,t)}{dz^2}=\frac{d^2V(z,t)}{dtdz} C \end{cases} $$ $$ \begin{cases} \frac{d^2V(z,t)}{dz^2}=\frac{d^2V(z,t)}{dt^2} L C\\ \frac{d^2I(z,t)}{dz^2}=\frac{d^2I(z,t)}{dt^2} L C \end{cases} $$

The general solution of a wave equation is: $$V(z,t) = V^+(z-v_ft) + V^-(z+v_ft)$$ $$I(z,t) = I^+(z-v_ft) + I^-(z+v_ft)$$ where v_f if the phase speed defined as $$v_f = \frac{1}{\sqrt{LC}}$$

When the wave in the transmission line is time-invariant (constant frequency, constant amplitude, constant initial phase) it is possible to remove the time parameter from the wave equations and to use directly the frequency domain.

$$R \to R, L \to j \omega L, C \to \frac{1}{j \omega C}, G \to G$$ $$ \begin{cases} V(z)-V(z+\Delta z)=-I(z)*(R\Delta z + j \omega L \Delta z) \\ I(z)-I(z+\Delta z)=-V(z+\Delta z)*(G\Delta z + j\omega C \Delta z) \end{cases} $$ $$ \begin{cases} \frac{V(z)-V(z+\Delta z)}{\Delta z}=-\frac{I(z)*(R\Delta z + j\omega L \Delta z)}{\Delta z} \\ \frac{I(z)-I(z+\Delta z)}{\Delta z}=-\frac{V(z+\Delta z)*(G\Delta z + j\omega C \Delta z)}{\Delta z} \end{cases} $$ $$ \begin{cases} \frac{dV(z)}{dz}=-I(z)*(R + j\omega L) \\ \frac{dI(z)}{dz}=-V(z)*(G + j\omega C) \end{cases} $$

For a lossless line the resistors can be removed from the circuit so R=0Ohm and G=0S. The equations become: $$ \begin{cases} \frac{dV(z)}{dz}=-I(z)*(j\omega L) \\ \frac{dI(z)}{dz}=-V(z)*(j\omega C) \end{cases} $$

Substituting respectively I(z) and V(z) in the other formula we obtain: $$ \begin{cases} \frac{dV^2(z)}{dz^2}=-V(z)*(\omega^2 LC) \\ \frac{dI^2(z)}{dz^2}=-I(z)*(\omega^2 LC) \end{cases} $$

Now the solution can be found, keeping in mind that the generic solution of a 2nd order differential equation is Ae^bx; k is the so-called wave constant; '0' instead is the starting section of the transmission line. $$ k^2 = \omega^2 L C $$ $$ \begin{cases} V(z)=V_{0}^+e^{-jkz} + V_{0}^-e^{jkz} \\ I(z)=I_{0}^+e^{-jkz} + I_{0}^-e^{jkz} \end{cases} $$ V₀⁺ is the voltage amplitude at the 0 section of the line (typically the end close to the generator) while V₀^- is the voltage amplitude at the end of the line (typically close to the load). The first term of the voltage (current) solution represents the forward travelling wave (from generator to load) while the second term represents the backward travelling wave (from load to generator). The second term is zero in case of perfectly matched impedance (when the load impedance is the same as the line characteristic impedance), so no reflection is generated at the load. This will be clarified later; for now, just imagine a voltage sine wave with constant amplitude, at a very high frequency, going through a waveguide that can be partially reflected to the generator, according to the type of load at the end of the line.

Now, we can substitute the obtained solution in the previous equation: $$ \begin{cases} \frac{dV(z)}{dz}=-I(z)*(j\omega L) \\ \frac{dI(z)}{dz}=-V(z)*(j\omega C) \end{cases} $$ $$ \begin{cases} \frac{dV_{0}^+e^{-jkz}}{dz}=-I_{0}^+e^{-jkz}*(j\omega L) \\ \frac{dI_{0}^+e^{-jkz}}{dz}=-V_{0}^+e^{-jkz}*(j\omega C) \end{cases} $$ $$ \begin{cases} -jkV_{0}^+e^{-jkz}=-I_{0}^+e^{-jkz}*(j\omega L) \\ -jkI_{0}^+e^{-jkz}=-V_{0}^+e^{-jkz}*(j\omega C) \end{cases} $$ $$ \begin{cases} \frac{V_{0}^+}{I_{0}^+} = \sqrt{\frac{L}{C}} = Z_0 \\ \frac{I_{0}^+}{V_{0}^+} = \sqrt{\frac{C}{L}} = Y_0 \end{cases} \qquad \begin{cases} \frac{V_{0}^-}{I_{0}^-} = -\sqrt{\frac{L}{C}} = -Z_0 \\ \frac{I_{0}^-}{V_{0}^-} = -\sqrt{\frac{C}{L}} = -Y_0 \end{cases} $$

At the end of the calculation one can find the so-called 'characteristic impedance' of the line, which is a value strictly related to the geometry of the medium and its material. This value is very important for transmission lines and should be known before starting any calculation or simulation. It is a constant value representing the ratio between the forward (or backwards) voltage and current waves along the line.

From the solution of telegraphists' equations we found two waves: a forward and a backward one. This is exactly what happens in a TL: the generator sends some signals along the line and part of them are reflected at the ends if not properly terminated; when this happens, a backward travelling wave generates, summing to the original one. The result is a travelling wave whose maxima changes in time; in some particular cases, instead, a standing wave is generated along the line: a wave whose amplitude changes in time at well-defined line sections (antinodes) and stays to 0 in correspondence of other sections (nodes) - in the second case the wave "doesn't move" along the line.

A line whose impedance is matched will have a voltage sinusoidally changing in time for every line section, and there will be no reflected wave summing up to the forward travelling wave. A line whose impedance is not matched will have a total voltage wave along the line, a travelling wave, with maxima that change following z, the spatial variable.

We will analyse all these concepts in the next articles in more detail.