Thanks and welcome

Thanks

Welcome

Thanks

You are correct. To find length of stub if reactance is known you should consider left side as short and right side as open. But in this case susceptance is known so we need to do opposite. Hope it's clear.

@@RFDesignbasics I think the right side is open circuit also for admittance chart ( G=0)

That is the reason we are doing opposite. Admittance chart will be mirror image of impedance chart. So right becomes left and left becomes right.

Thank you 🙏

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Since we know the required input admittance and we have to find Stub length, we should move towards generator from the given SC/OC load.

actually yes. i did the same way with you and found the result same with the video. It does not matter much.

Normally stub 1 is at load, but if the distance is explicitly given then move that distance from the load and consider the new point as load. Lambda calculation is for given frequency. It's c/f

For more clarity watch my recent video for double stub matching 2.0.

@@RFDesignbasics I see, thank you for the answer

No!! The rotated circle should be 1/16 lambda in anticlockwise direction from the original circle that means it should be at 0.25-0.0625= 0.1875lambda by following clockwise scale of the wavelength. Which is the same as 0.3125lambda if we follow anticlockwise scale.

@@RFDesignbasics thanks for the help!

@@RFDesignbasics is it always rotated by anticlockwise direction from the original 1+jb? for double parallel and series stub

If your question is about circle ⭕. Yes it will be rotated anticlockwise always.

@@RFDesignbasics Thank you, Sir.

If you take final point minus initial point and it comes out to be negative then add 0.5 lambda. Otherwise no need to add. Kindly check again 0.068-0.25 is a negative quantity so add 0.5 and it will become same 0.318 lambda. Alternatively you can also take 0.25 from SC to OC clockwise and add 0.068 which is also same as 0.318 lambda.

Thanku so much for the comments.

Arh never mind, the 0.5 lambda just gives the same value when added or subtracted - my mistake.

Admittance may be in bottom half or upper half based on given value of the load. If the load is given as impedance and is in upper half , than admittance will be in lower half and vice versa.

Normalised impedance on impedance chart is equal to normalised admittance on admittance chart. We are working on rotated impedance chart i.e. admittance chart.

Even single series is impractical. If you really want to do double series follow the same method without changing the load impedance to load admittance.

It's not 1.25, it's 0.125

For open circuit stub length of stub calculation will start from left side of smith chart. Also one thumb rule is that the difference between length of open and short stub will b always 0.25 lambda.

Because distance between two stub is given as lambda/8 so the original 1+jb circle should be rotated in anticlockwise by lambda/8 i.e. 0.125lambda from 0.25 lambda.

You are welcome.. Since I coudn't understand anything other than thank you. Can you translate please?

The translation is: Thank you , you are the best one in KZfaq explaining this and your explaining sounds very good

Ohh So nice to hear.. Thanks a lot. :-)

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ليش تكون منافق تحجي شي وترجمله شي !

That means distance between the stub is given as 0.375 lambda.

@@RFDesignbasics In there z=0.25 -j0.5 and distance was lambda/8 itself but still they drew it below

Then it's wrongly drawn

For more clarity send a copy to my email which is available in about section.

If jb1 is positive then you need to add -jb1 and if it is negative then add jb1. Same is applied for any stub.

@@RFDesignbasics thank you for your response. Now what i think i understand - we are looking to get the conjugate of the imaginary part to cancel out the imaginary part of the interceptions, the load admittance moved along the swr circle to the interception with the rotated 1+ jb circle hence why you subtracted -0.4j which is the imaginary part of the load admittance; then I'm assuming the load admittance also moved along the swr circle from the load admittance to the interception with the original 1 + jb circle to get jb2 which is the conjugate of the imaginary part of the interception. Now what i don't understand is for jb1 you added -0.4j but for jb2 you just took the conjugate of the interception, why different?

Ok, now I understand your doubt. To get jb1, you need to follow final point minus initial point. Since we are moving on constant resistance circle. For jb2 also the final point is centre of Smith chart, so final point is 1+j0 and initial point is 1+jb2.

For more clarification, watch my another video on double stub matching 2.0

@@RFDesignbasics Thank you, will do

pasa double stub bhakkar herdai xu yaar hahaha

Welcome

Do the same thing on other side of the rotated 1+jb circle.

I try that method in second solution i found the L1' but how to find L'2 as i have problem in finding L'2.

The rotated 1+jx should be 180 degree rotated circle or 1+jx circle simply as was in the case of solution #1.

@@muhammadafzaalkhan9277 For L2 you will be moving to the lower side of the original 1+jb circle and to match negative value of susceptance is required. You can find the length using same method.

@@RFDesignbasics thanks alot.

Watch recent uploaded video on double stub.. May be you will understand 😉.