Sir what if normalized zl is 2-2j which is outside the 1+jx circle do we map it the opposite point on the 1+jx circle or we follow 2-2j till it reaches the point in 1+jb circle ?
@@RFDesignbasics The outside circle can get a little confusing. An extension to this video explaining the case of zL outside 1+jx circle would be great.
@@RFDesignbasics if the normalised load lies outside the unity circle then we will have a series impedance connected to Zl so we won't be working with the unit admittance circle, that will come later
Sir i wanna ask, if the impedance normalization is on outside r=1 , for example 1-0.2 j , is the jx that - 0.2j , and after that we paralel it with the jb ? (Admintace) Thank you sir
We should normalise zl not z0. In case the normalised load is on 1+jx circle than you need either capacitor or inductor to match it. You don't need LC combination.
Suhail Hayat 1+jx circle wo resistance circle hai jo smith chart ke centre se hokar gujarata hai. Isliye load point ya to 1+jx circle ke andar hoga ya bahar hoga.
Sorry for the wrong substitution at 15:05 to 15:30. Kindly consider C=b/(2*pi*f*Z0) as correct substitution.
You are a lifesaver! I am studying for my Microwave Engineering final exam and this helps so much!
Very good explanation! One of the few that is based on just the impedance chart, which seem to be what most students are learning on.
This is the easiest way how to get impedance matching using Smith Chart. Thank you so much Sir
Sir, thank you for using the impedance-only chart. Hopefully your explanation will help me during the exam.
Great video, one concern I encountered was when converting to the 1+jx circle, i kept getting j1.6 instead of j1.45
wonderfull explaination.
Thank you! This is a very useful video.
Easy to understand, thanks sir!
This was a really helpful video, thank you sir!!
What should be the procedure if normalize impedance is outside 1+jx circle.
It is a very good video. Cleared my concepts. Thanks!
Thanku so much :-) don't forget to subscribe
kal sahi to kar, phir pata chalega -"cleared my concepts"
All the best to both :-)
Dude I went through dozens of videos just to pick this right one and share. Credits banta h..
+RF Designs Basic great explanation!
Wanted to tell, kedhar brought us here...thanks to him
Great
Excellent! Thanks a lot
Awesome sir.
Respected sir.. Your explanation is so good.. Is it possible to find all possible combination for LC and T and pi networks??
If you really understand the Smith chart, you can find any combination.
At 13:00 , Why is th value 1+j1.45 instead of 1-j1.45? Aren't you on the negative side of the smith chart?
Continue watching upto 14:30, you will see that I have realised my mistake and corrected the same.
thanks bro
should be 0.44 not 0.48 ,right?
Yes, he made a slight miscalculation
Watch in 1.25x speed.... Thank me later
That is saving of 20% of time watching this video. Good Idea.
Dude it works
watching it on 1.5x dude
Congratulations 🎉 you saved your precious 6 min 20 sec.
nice video sir.Thank you
Thanks for the comment. Spread among your friends.
very very thank. you
Sir, great effort but the explanation is not clear in many places, and please zoom in the circles you are marking.
Following Resistance circle on the Smith chart
Sir what if normalized zl is 2-2j which is outside the 1+jx circle do we map it the opposite point on the 1+jx circle or we follow 2-2j till it reaches the point in 1+jb circle ?
2-2j is not outside 1+jx circle. It's inside 1+jx circle So follow the first circuit. Refer starting of this video.
Sorry Sir i meant 0.2-0.2j
You can use second circuit and procedure will be same.
@@RFDesignbasics The outside circle can get a little confusing. An extension to this video explaining the case of zL outside 1+jx circle would be great.
@@RFDesignbasics if the normalised load lies outside the unity circle then we will have a series impedance connected to Zl so we won't be working with the unit admittance circle, that will come later
Sir Is there any method for PI network using smith chart ?
Sir i wanna ask, if the impedance normalization is on outside r=1 , for example 1-0.2 j , is the jx that - 0.2j , and after that we paralel it with the jb ? (Admintace) Thank you sir
What if the line is outside, for example 1.2+1.5j
At 12:00 the value of b is 0.08 and not 0.88
Adding 0.48 and 0.4 will give 0.88 , so the value of b =0.88 is correct
We don’t add we subtract
Sir, How to do the LC Matching for ZL outside 1+jX circle?
Use the other circuit and try.
what if zl is on the 1+jx circle?
what if the normalised zo lie on the unity circle?
We should normalise zl not z0. In case the normalised load is on 1+jx circle than you need either capacitor or inductor to match it. You don't need LC combination.
whay you uesd 1.2 +1.8j ?
After dividing by 50 ohm, which is called normalisation of load impedance.
12:10
Sir 1+jx kha liya h Samaj nhi reha hai
Suhail Hayat 1+jx circle wo resistance circle hai jo smith chart ke centre se hokar gujarata hai. Isliye load point ya to 1+jx circle ke andar hoga ya bahar hoga.
sir please upload the next topics ppt
Which topic?