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How should someone conclude what's going on on a TX line that has a Smith Chart that goes around and around the outside? I've reworked a board and expected to see a curve pass close to (0,0), but what I saw was all the curves going around and around the outside.

The Log plot is pretty flat and the phase plots look like sawtooth waves.

Is this a bad solder joint that I'm not seeing that's causing this reactive action in the line?

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  • \$\begingroup\$ What is the circuit you're measuring? Where does the chart start at low frequencies? \$\endgroup\$ – The Photon Jul 16 '15 at 14:28
  • \$\begingroup\$ I think I'm measuring a 5 element low pass Chebychev filter, but I'm seeing these large perimeter walks around the smith chart from 10MHZ to 6GHZ \$\endgroup\$ – testname123 Jul 16 '15 at 14:31
  • \$\begingroup\$ Did you properly terminate the output of the filter when you measured s11 of the input? \$\endgroup\$ – The Photon Jul 16 '15 at 14:32
  • \$\begingroup\$ It's going into a 50ohm pin on a chip. The chip promises that it's 50ohms, but I have not measured it. I however ,have not seen this behavior with other iterations of this design \$\endgroup\$ – testname123 Jul 16 '15 at 14:34
  • \$\begingroup\$ Either a short or an open could cause what you're describing. I'll write more when I get to a real keyboard. \$\endgroup\$ – The Photon Jul 16 '15 at 14:43
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Either an open or a short could cause what you're describing.

Say you start at the point representing an open, (1, 0) in cartesian coordinates, then if there's any transmission line between your calibration plane and the location of the open, there will be a phase delay in the reflected wave. The phase delay increases at higher frequency because the wavelength decreases while the physical length of the tranmission line segment stays the same. So on the Smith chart you see a rotation around the outside of the chart as frequency increases. Same thing if you start with a short at (-1, 0).

To tell the difference you need to look at the phase plot near 0 Hz. If the phase at 0 Hz is 0, you have an open. If it's 180 you have a short. If your sweep doesn't go down to low enough frequencies to extrapolate to 0 Hz you could probably just measure with an ohmmeter.

the phase plots look like sawtooth waves

Phase is usually measured either in the range of (-180, 180) degrees or (0, 360) degrees. So if the sweep passes through more than a full cycle of phase delay, you'll see that as a sawtooth on the phase plot. It doesn't mean there's a real discontinuity, only that the phase/magnitude plot can't show continuous paths from 180 to -180 the way a polar plot can.

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The poster is likely moved on in 4 years, but I found this (8/2019) looking for something else. So, maybe someone else will find it and need the answer.

The reason he has a saw tooth in the phase plot and a trace that wraps around the smith chart is there is several wavelengths of transmission line before the filter he is trying to measure.

VNAs permits a equivelent length of line as that in the test channel to be placed in the reference channel. The reference line path typical has a variable length line with a knob or crank that can be adjusted using a short and an open as the line is adjusted to obtain a R=0, j=0 with the short. Another method is to normalize out the test cable length using the data from measuring a short and open. The best approach is a combination of both approaches.

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