In a transmission line, the condition for minimal distortion (caused by the line) are when (R/L = G/C) in the line, right? However, I don't see any reference to that in any application notes or reference guides that I read of microstrips or striplines or high frequency designs. The only condition that I read for the signal integrity is for reflections (impedance matching). Is there any reason for that? Is it just for simplification of the model or I'm just looking at bad quality info?

There is distortion in a line when you can't write the signal in the output as:

V2(t) = K * V1(t-tau)

That means that the signals are not same (without counting the attenuation and displacement of the signal).

There is going to be distortion in the line unless:

a-) Z_0 doesn't depend of the freq. b-) alpha doesn't depend of the freq. c-) V_p doesn't depend of the freq.

All of those conditions are ok when R/L = G/C

If you think about a,b,c : you are changing the characteristics in the line trough the frequency spectrum, so if you are going to change more some frequency components of the signal than others, so that is distortion in the line.

You can see more of the theory of distortion in t.line in this book in section 4.2.

  • \$\begingroup\$ Dave, that's not correct. \$\endgroup\$ – JAMS88 Mar 29 '14 at 20:57
  • \$\begingroup\$ Are you talking about distortion as a result of the spectrum being altered because of either or all of (a) termination impedances being wrong or (b) spectral changes due to not matching things properly? (the two are usually related but can look like different issues). \$\endgroup\$ – Andy aka Mar 29 '14 at 21:34
  • \$\begingroup\$ ... that probably doesn't sound like helpful but i know what I mean.... how does this distortion manifest itself on the signals you are considering? \$\endgroup\$ – Andy aka Mar 29 '14 at 21:35
  • \$\begingroup\$ Thank you for expanding on what you mean by distortion. The gist of your question seems to be that you're not satisfied with the information you're getting, but you need to give us some specific examples. \$\endgroup\$ – Dave Tweed Mar 29 '14 at 21:44
  • \$\begingroup\$ @Andy aka You have a input singal (a pulse) let say that you are not obeying the minimal distortion conditions, so may be the high freq. signals will have an alpha bigger than low freq. signals, so your are going to attenute the high freq signals, so you can imagine that the pulse is going to look a lot different in the load. \$\endgroup\$ – JAMS88 Mar 29 '14 at 22:09

The Heaviside Condition

\$\frac{R}{L}\$ = \$\frac{G}{C}\$ or \$R C\$ = \$G L\$ or \$\frac{R}{G}\$ = \$\frac{L}{C}\$

Why isn't the Heaviside Condition more widely applied? Heaviside getting no respect even after all these years. More likely just the practicalities (impracticalities) of making the equality true.

Heaviside came up with this idea as a response to problems with telephone signal quality (in 1885ish). He found that for lossless lines and for lossy lines that met the above condition propagation (phase) velocity was independent of signal frequency. So, you would think we would be using it everywhere all the time.

For real lines though \$\frac{R}{L}\$ > \$\frac{G}{C}\$ which means you need to make R or C less, or L or G more. Decreasing R or C make the cable larger and heavier (more copper). Increasing G means exponential attenuation along the line length. Not something desirable for telephonic or long distance cables. So, for telephone lines, they inductively loaded the lines every quarter wavelength. They got good results for audio stuff, but the line in now a low pass filter. It wasn't too good for code.

Really the idea is kind of too simple. It is based on the parameters R, C, G, and L being constants over frequency. That isn't true. There is skin effect for the conductors. Materials have frequency dependent dielectric constants. If the line is loaded with a magnetic material to increase inductance, that will be frequency dependent too. If the relation works, it will only be over a restricted frequency band.

That's why all the work has been on wave compensation approaches, like solitons.

Heaviside Condition hasn't been completely abandoned though, see "Approaching Speed-of-light Distortionless Communication for On-chip Interconnect" as an example. Of course these researchers weren't trying to go across the country, just across a chip, so increasing G was OK in their case.


Is this helpful? I'll delete this if it isn't etc...

In all circumstances of signal the characteristic impedance for a transmission line is: -

\$Z_0 = \sqrt{\dfrac{R + j\omega L}{G + j\omega C}}\$

At low frequencies (below 100 kHz) where R and C dominate characteristic impedance is: -

\$\sqrt{\dfrac{R}{j\omega C}}\$ and this basically means the input impedance is X\$\angle -45\$

In other words at low frequencies you cannot terminate with a simple resistor but need a complex impedance. At high frequencies it becomes \$\sqrt{\frac{L}{C}}\$ which is a resistor.

  • \$\begingroup\$ I'm not talking about reflections, so terminations is not important here. \$\endgroup\$ – JAMS88 Mar 29 '14 at 23:11
  • \$\begingroup\$ Basically then, you are talking about a complex signal that at the low end of its spectrum borders (or includes) audio; and at the hi end encompasses 1 MHz and beyond. If this is true then if the line is properly terminated in a complex impedance that "suits" the line throughout all of the signal's spectrum you won't see the distortion you allude to? \$\endgroup\$ – Andy aka Mar 29 '14 at 23:29
  • \$\begingroup\$ Of course if the line losses vary with frequency then you'll reap a distortion that can only be corrected by a compensator at the receiving end or pre-emphasis at the transmission end (or both). Still do-able and hasn't really got anything to do with R:L ratio. \$\endgroup\$ – Andy aka Mar 29 '14 at 23:33
  • \$\begingroup\$ you are wrong, you are missing theory of t. lines, see 4.2 in book \$\endgroup\$ – JAMS88 Mar 30 '14 at 0:31
  • \$\begingroup\$ It's possible I am wrong but I'm not playing guessing games any more so maybe you can copy n paste the section of the google book into your question because, my browser won't open it. \$\endgroup\$ – Andy aka Mar 30 '14 at 6:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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