The wave equations

As shown in the figure, a transmission line can be modeled by its resistance and inductance in series, and the conductance and capacitance in parallel, all distributed along its length in $x$ direction. Here $R$, $L$, $G$ and $C$ represent, respectively, the resistance, inductance, conductance, and capacitance per unit length ( $[ohm]/[meter],\;[siemens]/[meter],\;[henry]/[meter],\;[farad]/[meter]$).

The voltage $v(x,t)$ and current $i(x,t)$ along the transmission line are functions of both time variable $t$ and space variable $x$. Across an infinitesimal section $\triangle x$ along the line, the voltage and current changes are

$$\triangle v=-R\triangle x\; i-L\triangle x \;\frac{di}{dt}, \... ...;\; \triangle i=-G\triangle x\; v-C\triangle x\; \frac{dv}{dt}$$

Dividing both sides of these equations by $\triangle x$ and let $\triangle x\rightarrow 0$ we get the telegrapher's equations:

$$\frac{\partial v}{\partial x}+L \frac{\partial i}{\partial t... ...c{\partial i}{\partial x}+C \frac{\partial v}{\partial t}+Gv=0$$

These coupled partial differential equations (PDEs) of two variables $x$ and $t$, called the telegrapher's equations, can be more conveniently solved by the Fourier transform method. We denote the Fourier transforms of the voltage and current with respect of $t$ ($x$ treated as a parameter) as

$${\cal F}[v(x,t)]=V(x,\omega),\;\;\;\;\;\;\;\;\;\;\;{\cal F}[i(x,t)]=I(x,\omega)$$

For convenience we may sometimes denote the voltage $V(x,\omega)$ and current $I(x,\omega)$ in the Fourier domain as $V(x)$ and $I(x)$ or simply $V$ and $I$. Taking the Fourier transform on both sides of the two PDEs we get two ordinary differential equations (ODEs) with respect to a single variable $x$:

$$\frac{dV(x)}{dx}+(R+j\omega L)I(x)=0, \;\;\;\;\;\;\;\;\frac{dI(x)}{dx}+(G+j\omega C)V(x)=0$$

These ODEs can also be obtained when both voltage $v(x,t)$ and current $i(x,t)$ are represented respectively as phasors $V(x)$ and $I(x)$, and the transmission line is represented in terms of the impedances $R$, $G$, $j\omega L$, and $1/j\omega C$. Therefore the variables $V(x)$ and $I(x)$ in the ODEs can be considered as either the Fourier transform or the phasor representations of the voltage $v(x,t)$ and current $i(x,t)$.

Combining the two equations we get

$$\frac{d^2V}{dx^2}=(R+j\omega L)(G+j\omega C)V=s^2V, \;\;\;\;\;\;\;\; \frac{d^2I}{dx^2}=(j\omega C+R)(j\omega C+G)I=s^2I$$

where we have defined

$$s=\sqrt{(j\omega L+R)(j\omega C+G)}=s_R+js_I=\vert s\vert e^{j\angle s}$$

with

$$\vert s\vert=\left[ (RG-\omega^2LC)^2+(\omega LG+\omega CR)^2... ...\tan^{-1}\left(\frac{\omega LG+\omega CR}{RG-\omega^2LC}\right)$$

and

$$s_R=\vert s\vert\cos\angle s,\;\;\;\;\;\;\;s_I=\vert s\vert\sin\angle s$$

The solutions of these two second order ODEs can be found to be:

$$V(x)=V_f e^{-xs}+V_b e^{xs}=V_F(x)+V_B(x), \;\;\;\;\;\;\;\; I(x)=I_f e^{-xs}+I_b e^{xs}=I_F(x)+I_B(x)$$

where we have defined

$$V_F(x)=V_f e^{-st},\;\;\;\;\;V_B(x)=V_b e^{st};\;\;\;\;\; I_F(x)=I_f e^{-st},\;\;\;\;\;I_B(x)=I_b e^{st}$$

Here $e^{sx}$ and $e^{-sx}$ are the two particular solutions, weighted by the arbitrary constants $V_b$ and $V_f$, which are to be determined based on the boundary conditions $V(0)={\cal F}[v(0,t)]$ at the front and $V(l)={\cal F}[v(l,t)]$ at the back end of the transmission line of length $l$. Note that $V_f$ and $V_b$ are constants with respect to variable $x$, but in time domain, they are still functions of variable $t$. Constant $I_f$ and $I_b$ can be found the same way.

Substituting $V(x)=V_fe^{-xs}+V_be^{xs}$ into the equation $dV(x)/dx+(R+j\omega L)I(x)=0$ and solving for $I(x)$, we get

$$I(x)=-\frac{1}{j\omega L+R}\frac{dV(x)}{dx} =\frac{s}{(j\ome... ... e^{-xs}-V_b e^{xs}] =\frac{1}{Z_0}\;[V_f e^{-xs}-V_b e^{xs}]$$

where

$$Z_0=\sqrt{\frac{j\omega L+R}{j\omega C+G}}$$

is the characteristic impedance of the transmission line measured in ohm. Comparing the two expressions of $I(x)$ above, we see that

$$I_f=\frac{V_f}{Z_0},\;\;\;\;\;\;I_b=\frac{V_b}{-Z_0}, \;\;\;\... ...{-xs}+I_b e^{xs} =\frac{V_f}{Z_0}e^{-xs}-\frac{V_b}{Z_0}e^{xs}$$

Lossless transmission line

When the frequency $\omega$ is high, $\omega L$ and $\omega C$ are much greater than $R$ and $G$, we can assume the transmission line is loss-less with $R=G=0$. In this case the two ODEs above become

$$\frac{dV(x)}{dx}+j\omega L\,I(x)=0,\;\;\;\;\;\;\;\;\frac{dI(x)}{dx}+j\omega C\,V(x)=0$$

and we also have

$$Z_0 = \sqrt{L/C},\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; s = \sqrt{-\omega^2 LC}=\pm j\omega\sqrt{LC}=\pm j\omega/\mu$$

where $\mu$ the transmission speed (measued in meter/second) defined as

$$\mu=1/\sqrt{LC}$$

and

$$V(x) = V_f e^{-sx}+V_b e^{sx} = V_f e^{-j\omega x/\mu}+V_b e^{j\omega x/\mu}$$

$$I(x) = I_f e^{-j\omega x/\mu}+I_b e^{j\omega x/\mu} =\frac{1}{Z_0}\left(V_fe^{-j\omega x/\mu}-V_be^{j\omega x/\mu} \right)$$

The voltage and current in time domain can be obtained as the inverse Fourier transforms of $V(x)=V(x,\omega)$ and $I(x)=I(x,\omega)$:

$$v(x,t)={\cal F}^{-1}[V(x,\omega)]={\cal F}^{-1}[V_f e^{-j\omega x/\mu}+V_b e^{j\omega x/\mu}] =v_f(t-x/\mu)+v_b(t+x/\mu)$$

and

$$i(x,t)={\cal F}^{-1}[I(x,\omega)]={\cal F}^{-1}[I_f e^{-j\omega x/\mu}+I_b e^{j\omega x/\mu}] =v_f(t-x/\mu)+v_b(t+x/\mu)$$

where

$$v_f(t)={\cal F}[V_f],\;\;\;v_b(t)={\cal F}[V_b],\;\;\;\;\;\;\; i_f(t)={\cal F}[I_f],\;\;\;i_b(t)={\cal F}[I_b]$$

Both $v(x,t)$ and $i(x,t)$ are composed of two components traveling at velocity $\mu$ in opposite directions along the transmission line. The time for the wave to travel the whole length $l$ of the transmission line is $T=l/\mu$. At the front ($x=0$) and back ($x=l$) ends of the line we have

$$V(x)\big\vert _{x=0}=V(0)=V_f+V_b=V_F(0)+V_B(0)$$

$$V(x)\big\vert _{x=l}=V(l)=V_f e^{-sl/\mu}+V_b e^{-sl/\mu} =V_f e^{-j\omega T}+V_b e^{-j\omega T}=V_F(l)+V_B(l)$$

i.e., the coefficients $V_f$ and $V_b$ are just the forward and backward voltages at the front of the line ($x=0$):

$$V_b=V_B(0),\;\;\;\;\;\;\;\;V_f=V_F(0)$$

Lossy transmission line

When the signal frequency is low, $R$ and $G$ can no longer be assumed as zero and the signal is always attenuated due to the resistance $R$ in serie with the inductance $L$ and the leakage conductance $G$ in parallel with the capacitance $C$. The two first order ODEs can be written as

$$\frac{dV(x)}{dx}+(R+j\omega L)I(x)=\frac{dV(x)}{dx}+j\omega L'\,I(x)=0$$

$$\frac{dI(x)}{dx}+(G+j\omega C)V(x)=\frac{dI(x)}{dx}+j\omega C'\,V(x)=0$$

where we have defined

$$L'\stackrel{\triangle}{=}L\left(1+\frac{R}{j\omega L}\right),... ... C'\stackrel{\triangle}{=}C\left(1+\frac{G}{j\omega C}\right)$$

As the above equations take the same form as in the loss-less case (with $L$ and $C$ replaced by $L'$ and $C'$, respectively), it can be solved in the same way. Now we have

$$Z_0=\sqrt{\frac{L'}{C'}}=\sqrt{\frac{L(1+R/j\omega L)}{C(1+G/j\omega C)}} =\sqrt{\frac{j\omega L+R}{j\omega C+G}}$$

and

$$\sqrt{L'C'}=\sqrt{LC}\left[ \left(1+\frac{R}{j\omega L}\right) \left(1+\frac{G}{j\omega C}\right) \right]^{1/2}$$

which can be approximated as below when $R<<\omega L$, $G<<\omega C$:

$$\sqrt{L'C'} \approx \sqrt{LC}\left(1+\frac{R}{j\omega L}+\frac{G}{j\omega C}\right)^{... ...}\left[1+\frac{1}{2}\left(\frac{R}{j\omega L}+\frac{G}{j\omega C}\right)\right]$$

$$\approx \sqrt{LC}+\frac{1}{j\omega}\left( \frac{R}{2}\sqrt{\frac{C}{L}}+\frac{G}{2}\sqrt{\frac{L}{C}} \right)=\frac{1}{\mu}+\frac{\zeta }{j\omega}$$

The second approximation is due to the Taylor series:

$$f(x,y)=(1+x+y)^{1/2}=f(0,0)+f_x(0,0)x+f_y(0,0)y+\cdots =1+\frac{1}{2}(x+y)+\cdots$$

Here we have defined the attenuation constant or damping coefficient as

$$\zeta\stackrel{\triangle}{=}\frac{R}{2}\sqrt{\frac{C}{L}}+ \frac{G}{2}\sqrt{\frac{L}{C}}=\zeta_s+\zeta_p$$

which is simply the sum of the damping coefficient $\zeta_s$ of the series RCL circuit and the damping coefficient $\zeta_p$ of the parallel GCL circuit:

$$\zeta_s=\frac{R}{2}\sqrt{\frac{C}{L}},\;\;\;\;\;\;\;\; \zeta_p=\frac{G}{2}\sqrt{\frac{L}{C}}$$

As before, the solution of the above equations is

$$V(x)=V_be^{j\omega x\sqrt{L'C'}}+V_fe^{-j\omega x\sqrt{L'C'}}... ...e^{\zeta x}e^{j\omega x/\mu}+V_fe^{-\zeta x}e^{-j\omega x/\mu}$$

In the time domain we have

$$v(x,t)={\cal F}^{-1}[V(x)]={\cal F}^{-1}\left[V_f e^{-j\omega... ...\mu}\right] =v_f(t-x/\mu)e^{-\zeta x}+v_b(t+x/\mu) e^{\zeta x}$$

Note that the forward wave $v_f(t)$ attenuates exponentially as $x$ increases from 0 to $l$, while the backward wave $v_b(t)$ attenuates exponentially as $x$ decreases from $l$ to 0.