Suppose we connect two transmission lines with characteristic impedances \$Z_1\$ and \$Z_2\$. The reflection coefficient looking into the transmission line with impedance \$Z_2\$ is $$ \Gamma=\frac{Z_2-Z_1}{Z_1+Z_2}. $$ Now consider the case where both lines are two coaxial conductors. For both lines, the impedance may be calculated by $$ Z=\frac{1}{2\pi}\sqrt{\frac{\mu}{\epsilon}}\ln{\frac{D}{d}}, $$ where \$\mu\$ and \$\epsilon\$ are, respectively, the magnetic permeability and dielectric constant of the dielectric between the two conductors, \$D\$ is the inner diameter of the outer conductor, and \$d\$ is the outer diameter of the inner conductor. Assume the dielectric is the same between the two transmission lines, but for one of the transmission lines, \$D\$ and \$d\$ are both smaller than the corresponding values for the other by the same factor. In this case, \$\frac{D_1}{d_1}=\frac{D_2}{d_2}\$, \$Z_1=Z_2\$, and the reflection coefficient \$\Gamma=0\$.

Based on this result alone, I would expect a voltage wave incident on the interface between the two lines to undergo no reflection. On the other hand, there is clearly a geometrical discontinuity at the interface, which I would intuitively expect to produce a reflection. Is my intuition wrong or am I missing something?

  • \$\begingroup\$ Your law for calculating Z from dielectric constants and diameter only holds for a line on infinite length—i.e. it is a simplification of reality. At the termination of the coax, the relation between geometry and actual spot-impedance is very complex and can be manipulated by the way it is mounted. But there is a reason why these steps are avoided as much as possible... \$\endgroup\$
    – EvertW
    Commented Jan 19, 2021 at 15:56

2 Answers 2


If you connect two transmission lines of different dimensions together, even if they are the same impedance, the physical discontinuity in the inner and outer conductors introduces an additional shunt capacitive loading, which does create a reflection.

This is usually combatted by what is called a Kraus (spelling?) step. In coaxial lines, this means staggering the locations of inner and outer diameter change, so that there is effectively a short length of inductive (higher impedance) line between the two shunt capacitances. This creates a low pass filter which eliminates the reflection, up to a certain frequency.

I take what you're saying to mean that the formula I gave for the impedance is not accurate near the interface between the two transmission lines. Is that correct?

Modifying the coaxial impedance formula 'near to the interface' is not a useful way to handle it. You could probably fit any one instance by doing it, but it lacks predictive power and simplicity. Keep the coaxial impedance formula as is, it is exact after all, and handle the non-uniform part separately.

The way all microwave engineers, and microwave tools like ADS and QUCS handle it, is to lump the effects of the step into an S-parameter box, that's connected between the two lines. A simple model would be a small shunt capacitor. More complex models exist which fit the exact results better.

This 'extra component' approach can be used whenever a transmission line comes to a discontinuity. The fringing capacitance and the radiation loss of an open end, the extra capacitance of a bend, or the connection of a stub to an otherwise uniform line, are all handled by having perfect transmission lines connected to a model of the discontinuity.

There are several ways to arrive at that model. Where the junction has some nice symmetries, it's possible to use transformations to arrive at the extra capacitance analytically. However usually either test pieces are measured, or 3D models made in something like HFSS, and the parameters of the discontinuity extracted from the overall results.

  • \$\begingroup\$ I take what you're saying to mean that the formula I gave for the impedance is not accurate near the interface between the two transmission lines. Is that correct? \$\endgroup\$
    – ashwmk
    Commented Jan 19, 2021 at 3:00
  • \$\begingroup\$ @ashwmk The repsonse is too long for a comment, so I've updated my answer \$\endgroup\$
    – Neil_UK
    Commented Jan 19, 2021 at 10:55
  • \$\begingroup\$ Regardless of how you model it, this shunt capacitance seems to be what was missing in my analysis. Thank you for your detailed response. \$\endgroup\$
    – ashwmk
    Commented Jan 20, 2021 at 0:58

In theory, there is no reflection at the connection of both TX lines.

In practice you would need to connect the outer conductor of the bigger TX line with the outer conductor of the smaller one.

This is where you might get an impedance change and thus reflections.

  • 1
    \$\begingroup\$ Exactly. Although it would be possible to design an adaptor that maintains correct impedance throughout the transition. Afterall, the two coax segments can't really be soldered together. Some kind of barrel or something will be needed. \$\endgroup\$
    – user57037
    Commented Jan 18, 2021 at 17:13
  • \$\begingroup\$ My intuition is that there would be a reflection. Take a somewhat absurd case: D1=1000*D2 and d1=1000*d2. Here, the interface between TX1 and TX2 looks like a wall with a tiny ring-shaped hole in the center. It seems unreasonable to expect there to be no reflection in this case, even in theory. \$\endgroup\$
    – ashwmk
    Commented Jan 19, 2021 at 2:55
  • \$\begingroup\$ I totally agree with your arguments. You could also ask: Does the formula above also apply for really big coax cables? My intuition would be that for coax diameters > wavelength it does not. But I admit I do not get the weakness of the formula for really big coax lines. \$\endgroup\$ Commented Jan 19, 2021 at 5:34
  • \$\begingroup\$ There doesn't need to be reflection. The Z-law for coax assumes an infinite length both to the left and right. At the interface, it is possible to design a converter that keeps the correct impedance, using complex simulations. \$\endgroup\$
    – EvertW
    Commented Jan 19, 2021 at 16:15

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