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Single-Stub Impedance Matching

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imranshoaib

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Hi,

I'm little confused with the stub position in stub tuning process. Let's say we have been given to match a load using single-stub whether short or open. I did both as an exercise and where I got myself stuck was the position of the stub. The query is: whatever the stub is? whether short or open, the stub position is not going to change; only length vary while switching from short to open or open to short stub.

Thanks in anticipation!
 

biff44

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It might help if you play around with a smith chart.

I will assume you know a little about transmission lines, and that we are talking about lossless lines. At a single frequency, there is no difference in reflection coefficient magnitude between an open circuited or short circuited stub. In either case, the magnitude is 1. But the phase angle of two equal length stubs WILL be different if they are terminated first in a short, and then in an open.

So, if you have a single stub matching network set up for a given load, and the stub is a shorted stub, then if you replace it with an open stub
1) the position that the stub joins the main line does not change
2) the length of the stub has to increase by 90 degrees.
If you do 1) and 2), then the new matching network will also work.

Why does the stub grow 90 degrees in length? Because the short circuit at the end of the stub has a 180 degree reflection coefficient angle. If you replace the short with an open, which has a 0 degree reflection coefficient, you have to do something to make the open circuit look like it is a short. So adding 90 degrees to the stub makes the reflection coefficient of the open look like 90+0+90= 180 degrees, and it works the same as the original circuit would.
 

imranshoaib

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Thanks for your help! Few more queries:

1. What if the characteristic impedance of the stub is not equal to that of the line?

2. While working with admittances on the Smith chart, when the author says that "if and inductive stub is required". What does this mean? I mean like while working with impedances on the Smith chart, Inductive stub means a short-circuited stub; and the Capacitive stub an open-circuited. So, while working with admittances does this reverse i.e. inductive=open and capacitive=short?
 

biff44

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It is important to have a good "feel" for how microwaves work. One way to develop that feel is to manually use a smith chart, for a long enough period of time, until you can see the smith chart in your head.

First, download and print out some copies of this one:
 

biff44

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I haven't used one of these in a coon's age, so please--someone check this all out to make sure I did not make a mistake!

A smith chart is a handy thing. It is technically a bilinear transform between the impedance and reflection coefficient plane. If you studied you complex variables, that might be of some interest. But if you have not, just think of it as a wicked nice tool to figure out how impedances, admittances, transmission lines, and reflection coefficients interact.

Lets do an example.
Assume you want to match a load of 25 + J75 ohms to a 50 ohm system. You need to use a mainline characteristic impedance Zc=50 ohms. But you have some odd coaxial line that is ZC'=25 ohm impedance, and want to use that as the stub element. Lets say you also want the stub connection to the mainline to be a SHUNT connection.

Smith Charts are usually normalized to the system characteristic impedance. For this type of a problem, you would want a two color smith chart--one that shows both impedances and admitances. That is because to do a shunt stub match, you have to add admitances, but the load's impedance is given, and we do not want to do any more calculations than necessary.

When you normalize an impedance to a smith chart characteristic impedance of 50 ohms, you simply divide the real world impedance by 50. When you normalize an admitance to a smith chart characteristic impedance of 50 ohms, you multiply the real world admitance by 50. Simple so far!

So, the normalized load impedance is ZLnorm=(25 + J75)/50= 0.5 + J1.5

On the first smith chart, that is shown as the point ZL. You read 0.5 on the horizontal red line and follow the constant resistance arc up. You read +J1.5 off of the red peripheral circle, and follow the constant reactance curve down. Where they intersect is 0.5 + J1.5. Note that on the edge of the chart, the length position reads 0.162 wavelegths toward generator. You use this scale since you are starting at a load, and then moving "Towards" the generator.

Now what we want to do is place a 50 ohm transmission line between the physical load and the stub connection point. We want that 50 ohm Tline to be just the right length so that the ADMITTANCE becomes Ynormjunc= 1 +/- J something. Then if we just cancel out the susceptance with a -/+ J something, we are done and the system is matched (at that one frequency).

So, if we start at the load and move toward the generator 0.278 wavelengths, we cross over the constant conductance circle (look only at the green part now) of 1. So that is the length of the mainline 50 ohm line, 0.278 wavelengths at the single frequency that you want the match centered at.

Now you read off the Admitamce of the load, transformed by a 0.278 wavelength 50 ohm line to be: Yjuncnorm= 1 + J2.2. The un-normalized admittance is Yjunc=(1/50)*(1 + J2.2) = 0.02 +J0.044.

With me so far?
Now whip out a new smitch chart. This one is going to have a characteristic impedance of 25 ohms (same as the stub coaxial line impedance).

Plot on this new smitch chart a dot at the short circuit position "S.C." ( I am assuming that you want to terminate the far end of the stub in a short circuit). You want this stub to generate a purely susceptive shunt admittance that will cancel that of the transformed load. The Load admittance at the junction is still 0.02 + J0.044. To normalize this to our new smith chart, That would become Yjuncnorm25= (1/25)*0.02 +J0.044= 0.5 + J1.1. So we want the stub to provide a susceptance of J1.1.

So starting at the short circuit position "S.C.", move from the short "towards the generator" by 0.119 wavelengths. This is a susceptance (read the green numbers) of Ystubnorm=-J1.1. Un-normalized, that susceptance is, of course, Ystub=0 - J0.44.

Now, if you just connect the stub to the mainline, the total input admittance is going to be Yinput= 0.02 +J0.044 - J0.044 = 0.02.

Sounds like a 50 ohm match to me!

Added after 1 minutes:

Here are the two smith charts:
 

yendori

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If nobody is going to say "thank you" to biff44, I will.
Thank you.

Yendori
 

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