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In practice, the frequency of a signal seems to have a significant impact on whether or not it can reasonable be transmitted over a line. As an example, consider the IEEE 802.3ab standard on Gigabit Ethernet (1000BASE-T), which deliberately employs a symbol rate of just \$125\text{ MBaud}\$.

Intuitively, I understand that you cannot use a conventional twisted pair cable to transmit a \$100\text{ GHz}\$ signal. But what exactly is it that keeps us from doing so? In real-world scenarios, what are the equations that when being populated with these frequencies would yield infeasible results?

Here are my thoughts so far:

  1. With increasing signal frequency, the parasitic inductance as well as the parasitic capacitance will probably lead to a significant damping. Is this effect captured by the real part of the characteristic impedance $$\sqrt{\frac{R'+j\omega L'}{G'+j\omega C'}}$$ in sufficient detail? And if this is the case: Wouldn't this mean that lossless transmission lines (with \$R'=G'=0\$) are unaffected by this frequency-dependent effect?
  2. According to these slides, higher signal frequencies "tend to radiate better." What exactly does this mean? Is this related to the damping that the resulting EM wave experiences during its propagation?
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  • \$\begingroup\$ ad 1) L and C won't give any dumping, since they are reactive. With increasing R and G the signal gets weakened or damped as you say. Yes without G and R, the transmission line is not affected VS. frequency. \$\endgroup\$ Nov 28, 2020 at 20:36
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    \$\begingroup\$ The problem is almost 100 percent caused by losses in the dielectric (which, in most cases is some type of polymer). Transmission lines for very high frequencies use special dielectrics or cavity type waveguides. The loss in the dielectric is a strong function of frequency. \$\endgroup\$
    – user57037
    Nov 28, 2020 at 22:49
  • \$\begingroup\$ Doesn't the parasitic RC structure, for instance, lead to some kind of low-pass behavior that becomes visible as damping as well? In general, I would have thought that the attenuation constant exhibits some dependence on L' and C' (see also my question below). \$\endgroup\$
    – user269746
    Nov 29, 2020 at 8:33

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Skin effect in metal conductors contribute losses. This loss mechanism can be compensated to some extent by using large diameter conductors (which could be hollow for weight savings).

Most of the loss, however, is caused by losses in the dielectric material between conductors. All practical dielectrics that can be used with coax, for example, have increased losses at high frequencies. This loss mechanism can be compensated somewhat by using dielectrics that have voids in them, with the goal of approximating an air dielectric.

Why do dielectrics have increased loss at higher frequencies? This is caused at the molecular level by repetitive motion of polar molecules. When exposed to electric fields, the molecules in these dielectrics rotate to align their polarities with the field. Because the field is changing, these rotations occur at the same frequency as the signal in the cable. As the frequency increases, these molecules rotate faster and faster, generating more and more heat in the dielectric. This means that power is required to cause this rotation. That power comes at the expense of the signal propagating in the line.

Incidentally, waveguides can also be used for RF transmission. For normal waveguides, air is the dielectric. In some cases, waveguides are actually evacuated so that there is no dielectric (other than the vacuum inside the sealed waveguide). This allows transmission of high power RF with low losses (for example for radar on naval vessels).

Technically, coax is also a waveguide. But what I mean above is higher order waveguides like this: https://www.fairviewmicrowave.com/popup.aspx?src=/images/Product/large/SMF112S-12.jpg

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  • \$\begingroup\$ I see, thank you very much! \$\endgroup\$
    – user269746
    Nov 29, 2020 at 8:06
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    \$\begingroup\$ I have a side question about dialectric losses. I have read (it could be wrong!) that the G parameter does not change with frequency. So, are dialectric losses modeled by putting a resistance in series with the capacitance (i.e. no longer the standard Telegrapher's Equation model)? Or can this loss be modelled by G changing with frequency? I omitted discussion of dialectric losses in my discussion, because I don't yet know the answer to that question. \$\endgroup\$ Nov 29, 2020 at 15:12
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    \$\begingroup\$ @MathKeepsMeBusy I am not sure either. I am not approaching this theoretically. It is more based on knowledge about what kinds of cables are used for which purposes. In order to support microwave frequencies and beyond in coax, special cables are needed to avoid very high losses (many dB per foot). What is special about those cables is the dielectric. G may be a function of frequency, or maybe when you analyze it correctly, constant G gives rise to frequency dependent losses. \$\endgroup\$
    – user57037
    Nov 29, 2020 at 18:50
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It is mainly the skin effect that limits high frequency signal transmission in transmission lines. The skin effect leads to high damping (R') with increasing frequency.

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    \$\begingroup\$ And dielectric losses. \$\endgroup\$
    – user57037
    Nov 28, 2020 at 22:50
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The formula you give, \$Z=\sqrt{\frac{R+j\omega L}{G + j\omega C}}\$ relates the voltage of a signal to it's current, not the change of the signal over distance. Note that Z is frequency dependent, but there is a fairly broad band in which it is fairly constant, and approximately equal to \$\sqrt{\frac{L}{C}}\$. Thus, we speak of a 50\$\Omega\$ coaxial cable, or a 100\$\Omega\$ twisted pair.

Note that impedance does not directly prevent us from using a transmission line to transmit signals. A 50\$\Omega\$ line is not "better than", a 100\$\Omega\$ line or even a 100\$k\Omega\$ line. Each will have different current for a given sinusoidal voltage, but if the line is "lossless", that same current and voltage will appear in the far end of the line as was fed into the near end of the line. What prevents us from using long transmission lines at high frequencies is attenuation, the change in the signal over distance.

The change of a sinusoidal signal over the length of a transmission line is described by the propagation parameter (often called propagation constant although it is frequency dependent). \$\gamma = \sqrt{(R+j\omega L)(G + j\omega C)}\$

The real part of \$\gamma\$ is called the attenuation parameter, and gives the exponential rate of decay over distance. The imaginary part of \$\gamma\$ gives the phase angle difference over distance (from which we can compute the phase velocity).

At low frequencies, the attenuation factor rises with frequency. It then reaches a relative plateau. As the frequency increases further, it begins to rise again, due to changes in the RGLC parameters, primarily R increasing due to the skin effect. At very high frequencies, the Telegrapher's Equation model breaks down because it assumes the electric field across the insulator propagates instantaneously, when in fact it propagates at the speed of light. The model assumes signal propagation in the form of what are called TEM waves, but as the wavelength approaches the scale of the distance between conductors, the propagation becomes TM waves. At that point, a wave guide will have less attenuation than parallel conductors separated by insulation.

Returning to the characteristic impedance. As mentioned, it is not what prevents us from using a transmission line above certain frequencies. However, it is frequncy dependent, (though there is a large plateau), and you ask in a comment whether a transmission line behaves like a filter. Yes, in a way. If one had a long length of transmission line, one could use this as a filter, by choosing to refer to the voltage as the input, and the current as the output. (Why one would want to implement a filter in this way is another question!). The output current could be taken at either end. However, there is another factor that complicates things. Transmission lines need to be properly terminated to prevent signal reflections. If the signal has only one frequency, that is simple, terminate with an impedance \$Z\$ according to the formula. \$Z=\sqrt{\frac{L}{C}}\$ will terminate properly for a wide band. But if the frequency changes enough for the impedance to change, well, termination becomes problematic.

I hope this expanded explanation helps.

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  • \$\begingroup\$ Thank you, I seem to have mixed them up. Allow me to ask a follow-up question: Intuitively, I would expect the transmission line to behave like some kind of low-pass filter (because of the RC structure, for instance). Is this effect also captured by the attenuation constant? Or maybe by the phase constant? \$\endgroup\$
    – user269746
    Nov 29, 2020 at 8:12
  • \$\begingroup\$ @bosonic I significantly expanded my answer to address your question. I hope it helps. \$\endgroup\$ Nov 29, 2020 at 14:10
  • \$\begingroup\$ It did, thank you very much! \$\endgroup\$
    – user269746
    Dec 3, 2020 at 21:41
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Additionally, the physical size of any connectors and other wire terminations practical with a given cable, especially if these connectors are not designed to be perfectly matched, will impose limits here - for example, termination of Cat.5 cable to a patch panel will often mean a cm or a few of the twisted pair being untwisted and treated as "just wire" - while marginally acceptable with 125MHz, this would look like an enormous mess of massive lumped components at 100GHz.

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