Smith chart matching with series and parallel stub (homework question) - Page 1
Forum section: electronics / beginners

author: Gecko
author_rank: Contributor
timestamp: September 05, 2015, 11:57:11 am
content: Hi, I'm kind of desperate here.I have a circuit like in the attached image.3.75GHz 200 Ohm Generator -> lambda/8 Parallel open stub -> lambda/8 series short stub -> 50 Ohm load(Length is lambda/8 because it should filter out 7.5GHz for which the 10mm are lambda/4)The goal is to find the impedances ZL1 and ZL2 of the transmission lines.I already watched several videos and documents explaining smith chart matching. And I thought I understood it. But apparently not.I only found explanations about1. Matching a real Zload to a real Z by inserting a series Transmission line (not series open/short stub) with ZL unequal to Z0 and finding out ZL2. Inserting series open/short stub with ZL=Z0 and finding out the length3. Inserting parallel Open/short stubs and finding out the length4. Matching with Parallel / series L or CBut no explanation on how to find the ZL if I have a series open/short stub with a given lengthAdditional "boundary condition" to the solution: AFAIK in the exam we are only given a Impedance Smith chart, no admittance Smith Chart. That makes things a bit more complicated.I'm totally lost, Monday is exam and I'm totally crapping my pants hereSo if anyone has a good explanation for a panicking student I'll be incredible grateful.(All this is kind of embarrassing because 5 years back I worked in an RF lab and working with the SC was a common task for me. But that was all L and C matching because it was just for a couple 100 MHz, so no transmission line matching there)

author: KJDS
author_rank: Super Contributor
timestamp: September 05, 2015, 02:24:52 pm
content: A little light reading.http://www.highfrequencyelectronics.com/Mar06/HFE0306_Rhea.pdfhttp://www.highfreqelec.summittechmedia.com/Apr06/HFE0406_Rhea-part2.pdf

author: Gecko
author_rank: Contributor
timestamp: September 05, 2015, 02:57:42 pm
content: Okay, figured it out (kind of). But I'm not 100% satisfied with my solution.What I did is:0. R1 = 50 Ohm, Z0 = 200 Ohm. Start point is 50Ohm/200Ohm = 0.25 (yellow dot)1. I draw the 1/Z0 admittance circle in my impedance Smith chart (the grey circle)2. I walk clockwise along the Real(Z)=const circle until I cut the 1/Z0 circle. This is Zin2 + R1 (green dot)3. Because its length is lambda/8 I know that the shorted series Stub has an input impedance of Zin2 = j*ZL2*tan(2*pi/8) = j*ZL2So I can calculate ZL2 from that.4. Then I mirror the green dot about the center. thats 1/(Zin2 + R1) and read the imaginary part of the admittance which represents 1/ZL2.5. Then I mirror the admittance 1/Zin1 to get Zin2 (or just calculate it)What I still find a bit unsatisfying: Is there any way around drawing that 1/Z0 admittance circle? (Without using a combined admittance/impedance Smith Chart). I feel there has to be a way, because otherwise it would be easier to simply use a combined admittance/impedance Smith Chart)@KJDS thanks for the documents. The table with equivalent lumped/distributed elements was very helpful!

author: KJDS
author_rank: Super Contributor
timestamp: September 05, 2015, 03:28:45 pm
content: To turn an impedance chart to an admittance chart, rotate 180 degrees. (This isn't something I've done in 30 years, so the exact details you'll have to work out for yourself)