# Derivation of Characteristic Impedance?

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 $\frac{V_o^+e^{-\gamma z}+V_o^-e^{\gamma z}}{I(z)}=\frac{V_o^+e^{-\gamma z}-V_o^-e^{\gamma z}}{I_o^+e^{-\gamma z}+I_o^-e^{\gamma z}}$

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\$$): -

• 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 or 0.2 m/ns.

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): -

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.

$$V_x=-LI_t-RI$$

$$I_x=-CV_t-GV$$

$$Z_x=0$$

Taking Laplace Transforms with respect to time:

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

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

Rearranging:

$$V(s)=\frac{I_x(s)}{-(G+sC)}$$

$$I(s)=\frac{V_x(s)}{-(R+sL)}$$

Combining:

$$Z_0=\frac{V(s)}{I(s)}=\frac{I_x(s)}{V_x(s)}\cdot\frac{R+sL}{G+sC}$$

Recalling that $$\Z_x=0\$$:

$$\frac{I_x(s)}{V_x(s)}=\frac{I_x(s)}{I_x(s)Z_0}=\frac{1}{Z_0}$$

Combining and rearranging, we have

$$Z_0^2=\frac{R+sL}{G+sC}$$

$$Z_0=\sqrt{\frac{R+sL}{G+sC}}$$