Can someone please explain what's the difference between a symmetric vs unsymmetric transmission line model ?

The normal un-symmetric one has R and L in series with G, C in parallel. While the symmetric model has the R and L divided in two halves and distributed on the two sides of the parallel G and C.

Is there any advantage what so ever of representing the same thing two different ways? We can get to the telegrapher's equations either way, so what's the point? Is it that the transmission line model should represent the line as being same from either point?

## Best Answer

There's absolutely no difference, in fact four different circuits could be used, the L and the T depicted in the question and the Pi and the inverted L.

All of them are equivalent.

You have two constitutive equations, one for the the voltage drop and the other for the shunt current.

Taking for instance this L diagram

^{simulate this circuit – Schematic created using CircuitLab}and writing the current change along the x direction we have $$ i(x)-i(x+\mathrm{d}x) = G\,\mathrm{d}x\;v(x+\mathrm{d}x)+C\,\mathrm{d}x\;\frac{\partial}{\partial t}v(x+\mathrm{d}x)$$

then we substitute for removing v(x+dx) $$v(x+\mathrm{d}x)=v(x)-R\,\mathrm{d}x\;i(x)-L\,\mathrm{d}x\;\frac{\partial}{\partial t}i(x)$$

we get

$$ \begin{align} i(x)-i(x+\mathrm{d}x) &= G\,\mathrm{d}x\left(v(x)-R\,\mathrm{d}x\;i(x)-L\,\mathrm{d}x\;\frac{\partial}{\partial t}i(x)\right)+\\ &+C\,\mathrm{d}x\;\frac{\partial}{\partial t}\left(v(x)-R\,\mathrm{d}x\;i(x)-L\,\mathrm{d}x\;\frac{\partial}{\partial t}i(x)\right) \end{align}$$

$$ \begin{align} i(x)-i(x+\mathrm{d}x) &= G\,\mathrm{d}x \;v(x) -\left(RG\;i(x)+LG\;\frac{\partial}{\partial t}i(x)\right)(\mathrm{d}x)^2+ \\ &+C\,\mathrm{d}x\;\frac{\partial}{\partial t}v(x)-\frac{\partial}{\partial t}\left(RC\;i(x)+LC\;\frac{\partial}{\partial t}i(x)\right)(\mathrm{d}x)^2 \end{align}$$

so, in turn, we have terms proportional to dx while those proportional to dx squared, are more infinitesimal and has then to be neglected.

$$(\mathrm{d}x)^2=\omicron(\mathrm{d}x) $$

and finally

$$i(x)-i(x+\mathrm{d}x)=G\,\mathrm{d}x \;v(x)+C\,\mathrm{d}x\;\frac{\partial}{\partial t}v(x)$$

$$\frac{\partial}{\partial x}i(x,t)=G \,v(x,t)+C\,\frac{\partial}{\partial t}v(x,t)$$

The same applies to the voltage drop and shunt current for thr L and inverted L circuits and, possibly with some more boring algebra, to the T and Pi versions as well.