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A transmission line can be shown on the smith chart as a clockwise rotation, each 180 degrees corresponding to a quarter wavelength. But this is when the characteristic impedance (Z0) of the transmission line is equal to that of the smith chart (for example, everything is in the 50 ohm system and the characteristic impedance of the transmission line is also 50 ohms).

My question is, how is a transmission line with a different characteristic impedance shown on the smith chart? For example, how to show a quarter-wave transmission line with Z0 = 100 ohm on a smith chart in the 50 ohm system?

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  • \$\begingroup\$ A Smith chart natively represents normalized impedances/admittances. In order to apply a mismatched quarter-wave section you'll likely need to convert impedances as you go down the line from load to source. \$\endgroup\$
    – nanofarad
    Jan 13, 2020 at 3:20
  • \$\begingroup\$ @ReinstateMonica-ζ-- Can it be equated to a matched transmission line with a different length? \$\endgroup\$ Jan 13, 2020 at 3:22

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A Smith chart natively represents normalized impedances/admittances relative to the characteristic impedance of the transmission line you're working on. In order to apply a mismatched quarter-wave section you'll need to convert impedances as you go down the line from load to source.

For example, I have a source, 0.4λ 50-ohm segment, 0.25λ 100-ohm segment, and a 30+j40 ohm load. I'm interested in the impedance seen by the source as it drives this transmission line.

I take my 20-ohm load, plot it on a smith chart w.r.t. 100 ohm (i.e. at the point 0.3 + j0.5); I transform it through a quarter-wave to the point 1.2 - j1.6, which corresponds to the complex impedance 120 - j160 ohms.

I'm now moving to the 50-ohm line, so I need to convert this to a normalized impedance w.r.t. 50 ohms: this is 2.4 - j3.2. By rotating through 0.4λ, I obtain the normalized impedance seen at the start of the 50-ohm segment: roughly 0.76 + j1.96. I finally convert this to ohms to obtain 38 + j98 ohms.

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