6
\$\begingroup\$

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}}\$

\$\endgroup\$

3 Answers 3

8
\$\begingroup\$

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.

\$\endgroup\$
1
\$\begingroup\$

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.

\$\endgroup\$
0
\$\begingroup\$

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

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

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

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.