# Transmission Lines

## Impedance of Transmission Lines

A transmission line with characteristic impedance ${\displaystyle Z_{0}}$, driven by a source with impedance ${\displaystyle Z_{s}}$, and terminated with a load impedance of ${\displaystyle Z_{L}}$

Transmission lines are a bit different than the normal wires we’re used to dealing with. For example, if you measured the resistance of a 10m piece of wire, and found it to be 0.01${\displaystyle \Omega }$, then you might reasonably expect that you’d measure the impedance of a 20m piece of wire to be 0.02${\displaystyle \Omega }$. However, when we say that a coaxial cable (SMA, BNC, or otherwise) has an impedance of ${\displaystyle 50\Omega }$, there is no mention of a length. 50${\displaystyle \Omega }$ coaxial cable is 50${\displaystyle \Omega }$ whether it is 1m or 100m long. How can this be?

A per-length transmission line model consisting of a (small) series resistance ${\displaystyle R}$, a series inductance ${\displaystyle L}$, a (small) parallel resistance ${\displaystyle G}$ caused by dielectric conduction, and a parallel capacitance ${\displaystyle C}$.

It turns out that the impedance of a transmission line, although it is real-valued (i.e. resistive), is not caused by the resistance of the wire (which is typically quite small, and results in signal loss along the wire). Rather, for a lossless transmission line, capacitance and inductance are what give rise to the characteristic impedance. If you’ve ever cut a cable in half and seen the dielectric that sits between the conducting wire and the exterior sheath, you are probably not surprised that capacitance plays a role. The other key to understanding transmission lines is to recognize that they are for carrying signals. You have to launch a signal down a transmission line, so we should really be thinking about the relationship between the voltage and current of the signal that is transmitted.

Adding a differential piece of transmission line to an infinite line.

Here is a cute pedagogical derivation of how this works. Supposing a lossless transmission line, we add a differential piece of line (with two half-inductances and a capacitor, as shown above), and argue that this shouldn’t change the overall impedance. In this configuration, the overall impedance of the line ${\displaystyle Z_{0}}$ is given by

{\displaystyle {\begin{aligned}Z_{0}&={\frac {1}{2}}Z_{L}+{\frac {1}{2}}Z_{C}\parallel ({\frac {1}{2}}Z_{L}+Z_{0})\\Z_{0}^{2}&={\frac {1}{4}}Z_{L}^{2}+Z_{L}Z_{C}\\Z_{0}&={\sqrt {Z_{L}Z_{C}+\left({\frac {1}{2}}Z_{L}\right)^{2}}}\\\end{aligned}}\,\!}

Now if we define ${\displaystyle L}$ to be an inductance per unit length, and ${\displaystyle C}$ to be a capacitance per unit length, we have ${\displaystyle Z_{L}=j\omega L\Delta \ell }$ and ${\displaystyle Z_{C}=1/j\omega C\Delta \ell }$, where ${\displaystyle \Delta \ell }$ is some small unit of length. In this case:

${\displaystyle Z_{0}={\sqrt {{\frac {L}{C}}-\left({\frac {1}{2}}\omega L\Delta \ell \right)^{2}}}.\,\!}$

Notice how the differential length and frequency dependence (and, even the definition of ${\displaystyle L}$ and ${\displaystyle C}$ as being per unit length) fall out of the left-hand term under the root. And, of course, as ${\displaystyle \Delta \ell \to 0}$, we are left with:

${\displaystyle Z_{0}={\sqrt {\frac {L}{C}}}\,\!}$