UHF Double stub matching - Page 1
Forum section: electronics / projects

author: Geunsjl
author_rank: Newbie
timestamp: December 26, 2015, 04:10:10 pm
content: HelloSorry if this is in the wrong section.The wavelength of a signal on a lossless transmission line is 45 cm, SWR = 4.5 and thefirst voltage minimum is at 5 cm from the load. Find the length of two open-circuited seriesstubs used to match the line, respectively at 0.22? and 0.45? from the load.(Note no Z0 is given)I know how to find the stub lengths when the load is given, but I don't know how to start with the given wavelenghts and distances from the load. So if anyone has any idea, please explain it to me.Thanks!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 26, 2015, 05:08:54 pm
content: SWR is given, so everything is in ratios to Zo. Don't need to know Zo at all!You can convert SWR to a ratio of Zo using its equation.I guess one should assume the stubs have the same impedance, being cut from the same cable or whatever.FWIW, if I understand "series stub" correctly, this would be a pretty poor way to do it in real life, because a series element also has a common mode impedance with respect to ground. Arranging it with parallel stubs instead, or using different impedance sections, is a more common method. (Go figure for textbooks giving somewhat less practical problems.)Tim

author: Geunsjl
author_rank: Newbie
timestamp: December 28, 2015, 10:39:13 pm
content: Thank you for your reaction, but unfortunately it isn't really helping me. Any idea how to start with this problem?

author: rfeecs
author_rank: Frequent Contributor
timestamp: December 28, 2015, 11:06:19 pm
content: Quote from: Geunsjl on December 26, 2015, 04:10:10 pmI assume you are at least visualizing this on the Smith chart. You know the load VSWR, so you can calculate the magnitude of the reflection coefficient. This is the radius of a circle around the center of the Smith chart. You know at 5 cm from the load is the first voltage minimum. So this is the point where the circle intersects the real axis, and it is on the left side of the Smith chart.From here you rotate to the location of the first stub. The series stub adds capacitive reactance, so you move along the reactance circle. You then rotate to the location of the second stub. The capacitive reactance of the second stub moves you to the center of the chart.I think from that you can come up with the answer.

author: G0HZU
author_rank: Super Contributor
timestamp: December 29, 2015, 12:10:14 am
content: If the VSWR = 4.5:1 then the mag of the reflection coefficient is (4.5 - 1)/(4.5 + 1) = 0.63636The angle of the reflection coefficient for Vmin will be -180degrees5cm is 40degrees if the wavelength is 45cm as (5/45) * 360 = 40degSo to find the load impedance you would have to rotate twice this angle (= 80 degree rotation counterclockwise) on the smith chart because 360deg on a smith chart is half a wavelengthSo this takes you to a reflection coefficient of angle -100 degreesSo the load impedance (at the load) will be 0.63636/-100deg or (18.3, -j38.5) if Zo = 50R or (0.366, -j0.77) when normalised.Hope I've understood the question and my answer is OK. I've got apple pie and custard waiting for me so this was a bit rushed...You said that this was the only bit you needed help with