next up previous
Next: Normal Reflection and Transmission Up: Traveling Waves Previous: Energy Conservation


Transmission Lines

A transmission line is typically used to carry high frequency electromagnetic signals over long distances: that is, distances sufficiently large that the phase of the signal varies significantly along the line (which implies that the line is much longer than the wavelength of the signal). A common example of a transmission line is an ethernet cable.

Figure 37: A section of a transmission line.
\begin{figure}
\epsfysize =2.25in
\centerline{\epsffile{Chapter06/fig01.eps}}
\end{figure}

In its simplest form, a transmission line consists of two parallel conductors that carry equal and opposite electrical currents $ I(x,t)$ , where $ x$ measures distance along the line. (See Figure 37.) Let $ V(x,t)$ be the instantaneous voltage difference between the two conductors at position $ x$ . Consider a small section of the line lying between $ x$ and $ x+\delta x$ . If $ Q(t)$ is the electric charge on one of the conducting sections, and $ -Q(t)$ the charge on the other, then charge conservation implies that $ dQ/dt = I(x,t)-I(x+\delta x,t)$ . However, according to standard electrical circuit theory (Fitzpatrick 2008), $ Q(t) = {\cal C}\,\delta x\,V(x,t)$ , where $ {\cal C}$ is the capacitance per unit length of the line. Standard circuit theory also yields $ V(x+\delta x,t)-V(x,t) = -{\cal L} \delta x \partial I(x,t)/\partial t$ (ibid.), where $ {\cal L}$ is the inductance per unit length of the line. Taking the limit $ \delta x\rightarrow 0$ , we obtain the so-called Telegrapher's equations (ibid.),

$\displaystyle \frac{\partial V}{\partial t}$ $\displaystyle =-\frac{1}{{\cal C}} \frac{\partial I}{\partial x},$ (412)
$\displaystyle \frac{\partial I}{\partial t}$ $\displaystyle = -\frac{1}{{\cal L}} \frac{\partial V}{\partial x}.$ (413)

(See Exercise 7.) These two equations can be combined to give

$\displaystyle \frac{\partial^2 V}{\partial t^2} = \frac{1}{{\cal L}\,{\cal C}}\,\frac{\partial^2 V}{\partial x^2},$ (414)

together with an analogous equation for $ I$ . In other words, $ V(x,t)$ and $ I(x,t)$ both obey a wave equation of the form (383) in which the associated phase velocity is

$\displaystyle v=\frac{1}{\sqrt{{\cal L} {\cal C}}}.$ (415)

Multiplying (412) by $ {\cal C}\,V$ , (413) by $ {\cal L} I$ , and then adding the two resulting expressions, we obtain the energy conservation equation

$\displaystyle \frac{\partial {\cal E}}{\partial t} + \frac{\partial {\cal I}}{\partial x} = 0,$ (416)

where

$\displaystyle {\cal E} = \frac{1}{2} {\cal L} I^{ 2} + \frac{1}{2} {\cal C} V^{ 2}$ (417)

is the electromagnetic energy density (i.e., energy per unit length) of the line, and

$\displaystyle {\cal I} = I V$ (418)

is the electromagnetic energy flux along the line (i.e., the energy per unit time that passes a given point) in the positive $ x$ -direction (ibid.). Consider a signal propagating along the line, in the positive $ x$ -direction, whose associated current takes the form

$\displaystyle I(x,t) = I_0\,\cos(\omega\,t-k\,x-\phi),$ (419)

where $ \omega $ and $ k$ are related according to the dispersion relation

$\displaystyle \omega = k\,v.$ (420)

It can be demonstrated, from Equation (412), that the corresponding voltage is

$\displaystyle V(x,t) = V_0\,\cos(\omega\,t-k\,x-\phi),$ (421)

where

$\displaystyle V_0 = I_0\,Z.$ (422)

Here,

$\displaystyle Z = \sqrt{\frac{{\cal L}}{{\cal C}}}$ (423)

is the characteristic impedance of the line, and has units of ohms. It follows that the mean energy flux associated with the signal is written

$\displaystyle \langle {\cal I}\rangle =\langle I V\rangle= \frac{1}{2} I_0 V_0= \frac{1}{2} Z I_0^{ 2} = \frac{1}{2} \frac{V_0^{ 2}}{Z}.$ (424)

Likewise, for a signal propagating along the line in the negative $ x$ -direction,

$\displaystyle I(x,t)$ $\displaystyle = I_0\,\cos(\omega\,t+k\,x-\phi),$ (425)
$\displaystyle V(x,t)$ $\displaystyle =-V_0\,\cos(\omega\,t+k\,x-\phi),$ (426)

and the mean energy flux is

$\displaystyle \langle {\cal I}\rangle =-\frac{1}{2} I_0 V_0= -\frac{1}{2} Z I_0^{ 2} =- \frac{1}{2} \frac{V_0^{ 2}}{Z}.$ (427)

As a specific example, consider a transmission line consisting of two uniform parallel conducting strips of width $ w$ and perpendicular distance apart $ d$ , where $ d\ll w$ . It can be demonstrated, using standard electrostatic theory (Grant and Philips 1975), that the capacitance per unit length of the line is

$\displaystyle {\cal C} = \epsilon_0\,\frac{w}{d},$ (428)

where $ \epsilon_0= 8.8542\times 10^{-12}\,{\rm C}^{\,2}\,{\rm N}^{-1}\,{\rm m}^{-2}$ is the electric permittivity of free space. Likewise, according to standard magnetostatic theory (ibid.), the line's inductance per unit length takes the form

$\displaystyle {\cal L} = \mu_0\,\frac{d}{w},$ (429)

where $ \mu_0 = 4\pi\times 10^{-7}\,{\rm N}\,{\rm A}^{-2}$ is the magnetic permeability of free space. Thus, the phase velocity of a signal propagating down the line is

$\displaystyle v = \frac{1}{\sqrt{{\cal L} {\cal C}}} = \frac{1}{\sqrt{\epsilon_0 \mu_0}},$ (430)

which, of course, is the velocity of light in vacuum. [See Equation (479).] This is not a coincidence. In fact, it can be demonstrated that the inductance per unit length and the capacitance per unit length of any (vacuum filled) transmission line satisfy

$\displaystyle \frac{\cal L}{\mu_0} = \frac{\epsilon_0}{\cal C}.$ (431)

Hence, signals always propagate down such lines at the velocity of light in vacuum. The impedance of a parallel strip transmission line is

$\displaystyle Z = \sqrt{\frac{{\cal L}}{{\cal C}}} = \frac{Z_0}{{\cal C}/\epsilon_0}=\frac{d}{w}\,Z_0,$ (432)

where the quantity

$\displaystyle Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} = 376.73\,\Omega$ (433)

is known as the impedance of free space. However, because we have assumed that $ w\ll d$ , it follows that the impedance of the line is impractically high (i.e., $ Z\gg Z_0$ ).

Practical transmission lines generally consist of two parallel wires twisted about one another (for example, twisted-pair ethernet cables), or two concentric cylindrical conductors (for example, co-axial TV cables.) For the case of two parallel wires of radius $ a$ and distance apart $ d$ (where $ d>2\,a$ ), the capacitance per unit length is (Wikipedia contributors 2012)

$\displaystyle {\cal C} = \frac{\pi\,\epsilon_0}{\cosh^{-1}(d/2\,a)}.$ (434)

Thus, the impedance of a parallel wire transmission line becomes

$\displaystyle Z = \frac{Z_0}{{\cal C}/\epsilon_0}= \pi^{-1}\cosh^{-1}\left(\frac{d}{2 a}\right) Z_0.$ (435)

For the case of a co-axial cable in which the radii of the inner and outer conductors are $ a$ and $ b$ , respectively, the capacitance per unit length is (Fitzpatrick 2008)

$\displaystyle {\cal C} = \frac{2\pi \epsilon_0}{\ln(b/a)}.$ (436)

Thus, the impedance of a co-axial transmission line becomes

$\displaystyle Z = (2\pi)^{-1}\ln\left(\frac{b}{a}\right) Z_0.$ (437)

It follows that a parallel wire transmission line of given impedance can be fabricated by choosing the appropriate ratio of the wire spacing to the wire radius. Likewise, a co-axial transmission line of given impedance can be fabricated by choosing the appropriate ratio of the radii of the outer and inner conductors.


next up previous
Next: Normal Reflection and Transmission Up: Traveling Waves Previous: Energy Conservation
Richard Fitzpatrick 2013-04-08