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

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?

• How is your second equation equal to 1? Commented Mar 16, 2017 at 18:57
• Because $\sqrt{x^2} = x$, and $\operatorname{Re}(1+1j) = 1$. Commented Mar 16, 2017 at 20:30
• Oh yes I was not paying attention to the Re() operator! Commented Mar 16, 2017 at 20:50
• I don't think the question was answered. The answer is no. You describe a lossy distortionless line which has real characteristic impedance Commented Dec 5, 2022 at 12:04

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): -

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.

• Both the series R and parallel G are resistive and therefore loss components. Commented Mar 16, 2017 at 16:59
• Both the questioner and I know that but, the question is all about the real part of the propagation coefficient. Commented Mar 16, 2017 at 17:19
• Ah whoops, that was a typo. But isn't $\sqrt{(1+1j)(1+1j)} = 1+1j$ still? Commented Mar 16, 2017 at 18:41
• At DC the attenuation constant is $\sqrt{RG}$ and if R=G=1 then the constant is 1 and therefore this is a big loss. It isn't lossless - it may be that you are confusing distortionless for lossless? A lossless line will have a constant of zero. See this: google.com/… Commented Mar 16, 2017 at 21:02