That is correct. The 50 ohm impedance refers to the "characteristic impedance" of the "transmission line". This comes from electromagnetic theory and is usually applicable to RF and high frequency applications. At DC your trace will still be very low impedance (resistance). If you were to take an ohmmeter to it you would probably measure maybe 1 ohm or 0.5 ohm, and that is only because the probe resistances would dominate over the actual trace resistance (which should be in maybe the tens of milliohms range).

Characteristic impedance has (mostly) to do with the capacitance and inductance per unit length of the transmission line. The capacitance and inductance are not that important at low frequencies, but as your signal frequency increases they will cause effects that can no longer be ignored. This is why you often will see coaxial cables advertised as 50 ohm or 75 ohm. This refers to the characteristic impedance (applicable to higher frequencies, like upper MHz range and GHz), not the DC resistance.

Since the characteristic impedance depends on C and L per unit length, as long as you do not change the distance between trace and ground plane (affects capacitance) nor do you change the trace width (affects inductance) your characteristic impedance should not change, no matter how long the trace is (Note: this is a simplified explanation, but captures the basic idea).

Note that often the word "characteristic" is dropped from the term, and it's just called "impedance". It sounds like this is the case in your layout program.

When we talk about the "impedance" of a pcb trace, we are talking about the *characteristic impedance* of a uniform transmission line.

The characteristic impedance depends mainly on *H*, *W*, and \$\epsilon_r\$ in the figure above. To get a 50 Ohm characteristic impedance, you just have to define your trace width in the proper proportion to the *H* of your dielectric (with some small adjustments for *T*_{met} and dielectric and conductor loss terms). As other answers have said, there's numerous online calculators and high priced tools available to help you do that.

The characteristic impedance doesn't depend on any features that change along the "L" direction. For example the different patterns in your example ("H", "L", and so on) don't change the characteristic impedance of the traces that make up the patterns. However, bends, curves, or branches in the lines do create discontinuities and regions where the characteristic impedance doesn't accurately model the transmission line behavior, as other answers have discussed.

*Now, about that paper*

For the record, the paper you referred to is

P. Sujintanarat, *et al.*, "Analysis of Electromagnetic Emission from PCB by using a Near-Field Probe", International Symposium on Communications and Information Technologies (ISCIT), 2006.

It appears that they are just using the different patterns of microstrip traces to produce fields that they can measure using the probe that they developed and which they are reporting on. It is not obvious that they *had* to use 50 Ohm microstrip, but it may have been convenient to do so because that would minimize back reflection where their coax line feeds the the test patch and maximize the signals seen on the samples.

They would have seen very similar results if they had used 40-Ohm traces, 60-Ohm traces, or even 100-Ohm traces. However, they had to pick some uniform characteristic impedance, and 50 Ohms is convenient, and less likely to confuse the pcb shop that built their test patches.

Incidentally, I was somewhat disappointed in the paper, because they started out saying that their goal is to have a low-cost test method that helps to predict the results of radiated emissions tests for EMC. However, they only compare their measurements to simulations; it doesn't appear that they ever compared their probe results with actual EMC measurements, so there's no demonstration that they accomplished what they set out to do.

Did he just add termination resistors in specific places?

It doesn't appear that they terminated the traces on their test patches, but they didn't need to for their purposes.

The signal comes in from the coax feed. It reflects off the various feature and the unterminated ends of the transmission lines. And it creates a field whether they terminate the traces or not.

Since they're only interested in whether they can measure that field, it's not important to their results to correctly terminate the transmission lines.

## Best Answer

Let's look at the formula and equivalent circuit for a transmission line.

(1) Impedance rather than reactance.

Reactance refers to the opposition to the change in current (of an inductor) or voltage (for a capacitor) - single components. The transmission line has \$R,L\$ and \$C\$ components - impedance is the ratio of voltage phasor to current phasor.

(2) It is \$50\Omega\$ because the ratio of inductance to capacitance per unit length produces that value. As \$R << j\omega L\$ and \$G \to 0\$, these values can be ignored and so the expression reduces to \$\sqrt{L/C}\$ (frequency independent).

(3) Nope, but it's generally a good idea to keep things as standard as possible. You may find it difficult to find a suitable connector for your \$167\Omega\$ transmission line. There's also a lot of information available for designing

standardtransmission lines on PCBs, etc. The magic number in my book is 376.73031... the impedance of free space. Now without that one we'd live in a different universe.(4) Going back to the formula. At low frequencies \$R\$ may be significant as the reactance of the inductor will be small). At very high frequencies the dielectric losses may become significant.