I start from the telegrapher's equation: \$-\frac{dV(z)}{dz}=(R'+j\omega L')I(z)\$, where \$V(z)\$ and \$I(z)\$ are the phasors of voltage and current respectively, in the transmission line model. \$R'\$ and \$L'\$ are resistance per unit length and inductance per unit length respectively.

The solution to the wave equation \$\frac{d^2V(z)}{dz^2}-\gamma^2V(z)=0\$ where \$\gamma=\sqrt{(R'+j\omega L')(G'+j\omega C')} \$ has the form \$V(z)=V_o^+e^{-\gamma z}+V_o^-e^{\gamma z} \$. \$G'\$ and \$C'\$ are respectively the conductance per unit length and capacitance per unit length of the transmission line.

From the telegrapher's equation we get: \$-\frac{dV(z)}{dz}=\gamma V_o^+e^{-\gamma z}-\gamma V_o^-e^{\gamma z}=\gamma (V_o^+e^{-\gamma z}-V_o^-e^{\gamma z})=(R'+j\omega L')I(z)\$

\$I(z)=\frac{\gamma}{R'+j\omega L'}(V_o^+e^{-\gamma z}-V_o^-e^{\gamma z})\$

...and I'm stuck here.

Given that characteristic impedance \$\frac{V_o^+}{I_o^+}=Z_o=\frac{V_o^-}{I_o^-}\$, how do I arrive at \$Z_o=\frac{R'+j\omega L'}{\gamma}=\sqrt{\frac{R'+j\omega L'}{G'+j\omega C'}}\$?

I'm not sure how to get \$Z_o\$ from \$\frac{V_o^+e^{-\gamma z}+V_o^-e^{\gamma z}}{I(z)}=\frac{V_o^+e^{-\gamma z}-V_o^-e^{\gamma z}}{I_o^+e^{-\gamma z}+I_o^-e^{\gamma z}}\$


3 Answers 3


This seems the simplest mathematical way to derive characteristic impedance. Consider a "lump" of transmission line connected to the continuation of that transmission line (\$Z_0\$): -

enter image description here

  • R is series resistance of cable for a given length
  • L is series inductance of cable for a given length
  • G is parallel conductance of cable for a given length
  • C is parallel capacitance of cable for a given length
  • \$Z_0\$ to the right is the continuation of the cable

Therefore the impedance looking into the left is: -

$$Z_0 = R + j\omega L + Z_0||\dfrac{1}{G + j\omega C}$$

$$= R + j\omega L + \dfrac{\frac{Z_0}{G+j\omega C}}{Z_0 + \frac{1}{G+j\omega C}}$$

$$= R + j\omega L + \dfrac{Z_0}{1 + Z_0(G+j\omega C)}$$

$$Z_0[1 + Z_0(G+j\omega C)] = [R+j\omega L][1 + Z_0(G+j\omega C)]+Z_0$$

$$Z_0 + Z_0^2(G+j\omega C) = R+j\omega L + Z_0[(R+j\omega L)(G+j\omega C)]+Z_0$$

$$Z_0^2(G+j\omega C) = R+j\omega L + Z_0[(R+j\omega L)(G+j\omega C)]$$

The important thing next is to recognize that \$(R+j\omega L)(G+j\omega C)\$ is insignificant as the "lump" approaches zero length and we are left with: -

$$Z_0^2(G+j\omega C) = R+j\omega L$$

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

Update Nov 28th 2023 - an alternative method

An alternative and elegant way of calculating the characteristic impedance is to use the velocity of propagation, \$V_P\$ and, the cable capacitance. Start by recalling that the velocity of propagation is defined by this equation (where L and C are the inductance and capacitance per metre): -

$$V_P = \dfrac{1}{\sqrt{LC}}$$

So, just as an example, for a typical cable we might have: -

  • L = 250 nH/m and
  • C = 100 pF/m.

This implies a \$V_P\$ of 200 million m/s. Then, as a thought experiment, consider applying a rise-time limited step voltage to one end of the cable. It can be any rise-time you want but, for the sake of this example I'm assuming a 1 volt change in 10 ns (0.1 volts per nano second): -

enter image description here

Because we know the velocity of propagation, we can calculate how much capacitance is “trapped” within the boundaries of the 10 ns step. So, 0.2 m/ns multiplied by 10 ns is a distance of 2 metres. That’s a “trapped” capacitance of 200 pF for this particular cable.

Knowing that \$i=C \frac{dv}{dt}\$ we can calculate the current flow in the segment of cable where the step is occurring: -

$$i = C \frac{dv}{dt} = 200\times 10^{-12}\cdot\dfrac{\text{0.1 volts}}{10^{-9}} = \text{0.02 amps}$$

So, the characteristic impedance is 1 volt divided by 0.02 amps = 50 Ω.

Of course I knew it would be 50 Ω because I chose L and C to have a ratio of 2500 and, when taking the square root, the result is 50 Ω.

I'm just trying to point-out that a little bit of thought (about the problem) can save you a ton of math that nobody ever remembers.


It is very easy. First you put \$I(z)=I_0^+e^{-yz} + I_0^-e^{yz}\$ an then substitute \$V_0^+=I_0^+\cdot Z_0\$ and \$V_0^-=-I_0^-\cdot Z_0\$ in the equation where you got stuck.


The chacteristic impedance for a transmission line follows directly from the Telegrapher's Equations and there assumption that the characteristic impedance doesn't vary along the line.




Taking Laplace Transforms with respect to time:

$$V_x(s)= -sLI(s) -RI(s)$$

$$I_x(s)= -sCV(s) -GV(s)$$






Recalling that \$Z_x=0\$:


Combining and rearranging, we have




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