Electronic – Must the characteristic impedance of a transmission line be non-real for it to be lossy

theorytransmission line

In Does conductance in the transmission line model represent a physical quantity? it came up that a non-real characteristic impedance just means the line has some loss. With real transmission lines having a little (but perhaps negligible) loss, we could expect all real transmission lines to have a characteristic impedance with at least a small imaginary component.

But is that really true? Let's say \$Z = Y = 1+1j\$. The characteristic impedance of this line is real:

$$ \sqrt{ 1+1j \over 1+1j } = 1 $$

And the attenuation constant is 1 [corrected by edit]:

$$ \operatorname{Re}\sqrt{(1+1j)(1+1j)} = 1 $$

Which to my understanding, means this line is lossless. But how can this be when it has both a non-zero conductance and resistance?

Best Answer

Phil, the real part of the propagation constant is the attenuation constant and this equals: -

\$Re\sqrt{(R+jwL)(G+jwC)}\$ and not the formula you have in your question.

The formula you have used is for characteristic impedance.

This wiki page should confirm this (right at the bottom): -

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So, if you do the math at low frequencies (to make life easier) you see that the attenuation constant becomes \$\sqrt{RG}\$ and if R=G=1 then you have a constant of 1 and a lousy highly lossy line. A lossless line has a re(propagation constant) of zero.