Transmission lines

When a cable stops being a wire

• Two boards are joined by 40 cm of coax carrying a 3 GHz signal. At that frequency the cable is six wavelengths long.
• Voltage is no longer the same along it: waves travel down, and reflect back from any mismatch. This lesson treats the cable as a transmission line.

The telegrapher equations

• Model the line as inductance $L'$ and capacitance $C'$ per metre. The telegrapher equations 电报方程 are $\dfrac{\partial v}{\partial x} = -L'\dfrac{\partial i}{\partial t}$ and $\dfrac{\partial i}{\partial x} = -C'\dfrac{\partial v}{\partial t}$.
• Combine them and a wave equation appears, with speed $v_{p} = \dfrac{1}{\sqrt{L'C'}}$.

Characteristic impedance

• A wave travelling one way has voltage and current in a fixed ratio, the characteristic impedance 特性阻抗 $Z_{0} = \sqrt{\dfrac{L'}{C'}}$.
• $Z_{0}$ is a ratio, not a resistor: a lossless 50 Ω line dissipates nothing.

Worked example: a real coax

• $L' = 250\,\dfrac{\text{nH}}{\text{m}}$, $C' = 100\,\dfrac{\text{pF}}{\text{m}}$: $Z_{0} = \sqrt{\dfrac{250 \times 10^{-9}}{100 \times 10^{-12}}} = 50\,\Omega$.
• $v_{p} = \dfrac{1}{\sqrt{250 \times 10^{-9} \times 100 \times 10^{-12}}} = 2 \times 10^{8}\,\dfrac{\text{m}}{\text{s}}$, or $0.67c$. At 3 GHz, $\lambda = 6.7\,\text{cm}$, so 40 cm is six wavelengths.

Reflections at the load

• Where the line meets a load $Z_{L}$, part of the wave reflects. The reflection coefficient 反射系数 is $\Gamma = \dfrac{Z_{L} - Z_{0}}{Z_{L} + Z_{0}}$.
• A matched load ($Z_{L} = Z_{0}$) gives $\Gamma = 0$: no reflection. A short gives $\Gamma = -1$; an open gives $\Gamma = +1$.

Worked example: a step into an open end

• A 12 V step enters the 50 Ω line: the wave carries 12 V and $\dfrac{12}{50} = 0.24\,\text{A}$. At the open end, $\Gamma = +1$.
• The reflected 12 V adds to make 24 V at the load, and the reflected $-0.24\,\text{A}$ cancels the current. An open end draws no current, as it must.

Three common slips

Recap