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

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.