I'm pretty sure the left side is short circuit (R=0, X=0) and right side is open circuit (R=infinity, X=infinity), so you have to get length starting from the left side
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.
I don't understand why you use first a unitary circle that have half of it at left and the other half at right in order to calculate the first Stub (jb1) and then use a second circle rotated lamba/8 (90° clockwise) to calculate the second Stub (the one most at the left) (I wrote you via Linkedin). I have seen that all the problems begin with a circle that pass for the short circuit point in the smith chart and the point center of the chart.
in the last step when finding the stub lengths, shouldnt you be looking at the direction towards the load, not towards the generator? since you are trying to go to the short circuit (which is the load)
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.
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.
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.
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.
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.
Hi sir, do you have the explanation video on how to calculate the lambda as you told me at the end of the video? also, if the length of stub 1 to stub 2 is fixed, how do you know the length between stub 1 and the load? could it be the same way as a single stub parallel?
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
@@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 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.
Very easy to learn and was immensely useful for my examination
I'm pretty sure the left side is short circuit (R=0, X=0) and right side is open circuit (R=infinity, X=infinity), so you have to get length starting from the left side
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.
Sir i am confused in 4:02
Very helpful. Thank you so much.
Greetings from Turkey!
Thank you. Easy to understand and apply
Thanks
awesome sir ... i wish why u didn't offered to us in transmission lines
Thank you so much, sir.
I don't understand why you use first a unitary circle that have half of it at left and the other half at right in order to calculate the first Stub (jb1) and then use a second circle rotated lamba/8 (90° clockwise) to calculate the second Stub (the one most at the left) (I wrote you via Linkedin). I have seen that all the problems begin with a circle that pass for the short circuit point in the smith chart and the point center of the chart.
Very helpful, Thank you very much Prof.
Thanks and welcome
Awesome sir! thank you..
God bless you!!!.
Wonderful explanation sir
Thank you 🙏
Very helpful,thank you Sir.
Welcome
well explained Sir thanks keep up the great work
Thanks
in the last step when finding the stub lengths, shouldnt you be looking at the direction towards the load, not towards the generator? since you are trying to go to the short circuit (which is the load)
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.
If you had a spacing of 1/16 lambda, would you have place the rotated 1 + jB circle on the top plane on the left hand side where it is 0.0625 lambda?
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.
When you move from "Short circuit point" to 0.068 lambda, did you not forget to add 0.5 lambda?
Besides from the background noise a great video.
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.
my mind is not supoorting this version. How the radius of 1+jB circle is at 3.0??
very nice tutorial
sir how did the imaginary value of jb1 comes to 0.6. first, you wrote 0.06 then you changed what's the correct value and how did it come?
Great ❤️
Great sir..
Thanks. Do subscribe for more updates.
Is It 0.57 or 0.82
Sir should we take admittance in the bottom half of the chart if given in a question??
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.
Hi sir, as we are plotting it on impedence chart, so at 7:20 you are reading an impedence value and writing as admittance. Is my question right?
Normalised impedance on impedance chart is equal to normalised admittance on admittance chart. We are working on rotated impedance chart i.e. admittance chart.
Why there is no any single example of double series stub matching network on utube. Is it not feasible?
I have searched everywhere but not found.
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.
why did u draw the circle in the upper half.
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.
sir, what happens if we use open circuit stubs
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.
Hi sir, do you have the explanation video on how to calculate the lambda as you told me at the end of the video?
also, if the length of stub 1 to stub 2 is fixed, how do you know the length between stub 1 and the load? could it be the same way as a single stub parallel?
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
How to find second solution as i
Double stub has two solution
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.
Hello , i got question , how lambda/8 get 1,25 at te diagram ?
It's not 1.25, it's 0.125
In some problems i saw them draw the rotated circle below instead of at top in the smith chart,why is that done?
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.
قداش ما صوت ديرمك مستفز الزح ، لكن thank you مع انك عفلقتني لكن افدتني
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 TH-cam explaining this and your explaining sounds very good
Ohh So nice to hear.. Thanks a lot. :-)
.
ليش تكون منافق تحجي شي وترجمله شي !
watch at 1.25x
pasa double stub bhakkar herdai xu yaar hahaha
Welcome
Sir, for jb2 you did not subtract -0.4j like you did with jb1 why?
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
How it's 0.068j
Why -0.57 no 0.57
هسه 😂شدوس
hard to understand. Others teach way better. Plz take this dislike from me.
Watch recent uploaded video on double stub.. May be you will understand 😉.