I am confused on when you would use the telegraphers equations vs when you would use the transmission line wave equation.

Telegraphers equations: enter image description here

enter image description here

Transmission line wave equations:

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  • 1
    \$\begingroup\$ Can you indicate exactly which formulation you are referring to in "transmission line wave equation"? The set of Telegrapher's Equations includes lossy media in their formulation while depending on your course/text your transmission line equation may not. \$\endgroup\$
    – nanofarad
    Commented Mar 12, 2019 at 21:11
  • \$\begingroup\$ Asking about the difference between two formulas would greatly benefit from adding both formulas to your question! In fact, that's the only way an answerer could use your notation of the equations to explain differences, if any. \$\endgroup\$ Commented Mar 12, 2019 at 21:25
  • \$\begingroup\$ Sorry I will provide the formulas. \$\endgroup\$ Commented Mar 12, 2019 at 21:29
  • \$\begingroup\$ So, what you've just added is what, according to you? \$\endgroup\$ Commented Mar 12, 2019 at 21:30
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    \$\begingroup\$ Transmission line formulas are SOLUTIONS to the Telegraphers Equations. What could be confusing here? \$\endgroup\$ Commented Mar 12, 2019 at 21:47

1 Answer 1


As you can see, the telegrapher's equations are coupled to one another, that is, the voltage equation contains a current term, and the current equation contains a voltage term.

That is why you then see the wave equation, which decouples those (that is, differentiate the telegrapher's voltage equation and plug in your current equation into it) and you end up with:

$$ \dfrac{d^2V(z)}{dz^2}=\gamma ^2V(z)$$ $$ \dfrac{d^2I(z)}{dz^2}=\gamma ^2I(z)$$

With \$\gamma=\sqrt{(R+jwL)(G+jwC)}\$

Those are the differential wave equations.

This is a lot simpler to solve than the telegrapher's equations. Solving those yield exactly what you have for the wave equations.

So what should you use? I'd say most of the time you will use the solutions to the wave equations, because those are the solutions to the telegrapher's equations as well.

What happens is that you start with telegrapher's equations, you then decouple them (differential wave equations) and you finally solve the differential wave equations. The solutions to the differential wave equations is what you probably want to use. All this process just tries to find the solution to telegrapher's equations with the wave equations being an intermediate step along the way.

  • \$\begingroup\$ Thanks @Big6 and forgive my ignorance. \$\endgroup\$ Commented Mar 12, 2019 at 21:54
  • \$\begingroup\$ @GeraltofRivia no worries. These are questions we could all have, especially when seeing it for the first time \$\endgroup\$
    – Big6
    Commented Mar 12, 2019 at 21:56

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