I can't even begin to tell you how helpful your videos are. Your three videos in this series are so very understandable. I just spent over a week and many hours trying to understand what was said in my lecture and my book and was getting so frustrated. Why it couldn't have been made this clear in class is beyond me. Thank you sooooo very much!!!
you're a genius , not because of the knowledge of smith chart but the way you explain it. you did it in 5 mn what teacher in school does it in 6 months.
Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh My dear Teacher, u r the best I've ever seen in youtube. You explain hard thing as an easy thing. Congratulationssssss and Thank u very much
when my college sir taught this topic..i was wondering if i'l ever understand such things.. and after watching your video for just one time..i am so confident that i can teach my professor..thanks a lot Professor Carl!
VSWR is always greater than 1(or equal), since the maximum of the wave can't be lower than it's minimum. So you only take numbers to the left of the center point.
while making the circle for vswr from centre , the central line cuts the circle at two points..then why the right side one point is taken as vswr? Why not the left side point? what's the rule for selection of one intersecting point from the available two points?
+Abhishek Poudel , here's a simple way to look at this... VSWR = Vmax / Vmin Therefore, the smallest value VSWR can have is, 1 . Values on the left of the center of VSWR circle are less than, 1 and thus invalid for the purpose of determining VSWR.
How do you physically measure the reactive component of an antenna? I would imagine that is essential before you can use the chart to figure stuff out with.
Hi, thank you for the great video's on smith chart. I can see how to create a matching network between generator and load, but when this is swept over a frequency range (HF) in mind, the match at the design frequency goes out the window. Perhaps you could look at matching over wider frequency ranges in a future video? many thanks, Joe
the resistive line will pass through 2 points of a circle, left and right side.. so which one is the vswr? does it always have to be on the right side of the circle, where resistive line passes through?
It always has to be the right side. The easiest way to remember is this: the range of possible values for VSWR is 1 (perfect match, no reflections) up to infinity (open or short circuit, 100% reflected power). This means that it cannot be less than 1, ruling out all the values on the left side.
it is very helpful video to a student who study about electronic engineering which is me.!!! Awesome video thank you for video. I subscribe your channel.
Please help me. You're given a line's characteristic impedance, but you're given just the magnitude of the load impedance, not the complex expression of the impedance. How do we solve VSWR for this case? Should we assume that the load is purely resistive?
Is the VSWR ratio always to 1? I understand the left part of the ratio is determined from the VSWR circle, but just curious about the 1. Thanks in advance.
I wish i can find a practical tutorial on S parameters.I am thinking like a block showing the network port connections and the explanations - the maths will come but just how you connect and what measurements you make.
For Sxy, the signal is measured at port x, and it came from port y. Simple as that. So for S21, the signal is transmitted out of port 1, and you measure it at port 2. Since it has to travel through a device or TX line to get there, it is insertion loss. For S11, the signal is transmitted out of port 1 and is measured at port 1. Wait! These are the same port! So it must be REFLECTED power that is being measured: return loss. Check out the video on insertion and return loss for more on that.
The Smith Chart can be thought of as a plot of complex reflection coefficient Gamma in polar form (hence the circular boundary). The magnitude of Gamma is linearly scaled from the horizontal axis (the X'=0 line which has the R' values marked on it), and the angle (or argument) is plotted per a scale marked around the outer circle (which hasn't been mentioned in the video as it's not pertinent to finding VSWR easily). Accordingly, the circle centred on the R'=1 point and including the plotted Z' point will have a radius directly proportional to the magnitude of Gamma. (Edit: this is addressed in the next video ruclips.net/video/9KlIgae0ad8/видео.html) VSWR = (1+|Gamma|)/(1-|Gamma|), where |Gamma| is the magnitude of the reflection coefficient. It turns out that when the magnitude of Gamma is plotted on the right-hand segment of X'=0 line, the R' value equates to the VSWR value for the given magnitude of Gamma. To understand this fully, I recommend you read up on how the Smith Chart is mathematically derived from R, jX, and reflection coefficient. Any decent electromagnetics text dealing with transmission lines should include such a derivation. The maths isn't particularly difficult: some complex number manipulation and understanding the equation for a circle would suffice.
Wow, that's really easy and clear, my professor uses more than 6 hours to tell what you said in your video. You are an expert.
Most of engineering is actually not that difficult to explain, but professor often make it harder than it needs to be.
I can't even begin to tell you how helpful your videos are. Your three videos in this series are so very understandable. I just spent over a week and many hours trying to understand what was said in my lecture and my book and was getting so frustrated. Why it couldn't have been made this clear in class is beyond me. Thank you sooooo very much!!!
you're a genius , not because of the knowledge of smith chart but the way you explain it. you did it in 5 mn what teacher in school does it in 6 months.
thumbs up if you are watching this night before your exam!!
I'm here 27years too late, but still enjoying good teachers. ;-)
You are my hero Carl, please keep on whistling and doing things "easily"
Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh My dear Teacher, u r the best I've ever seen in youtube. You explain hard thing as an easy thing.
Congratulationssssss and Thank u very much
It is the best video, you blow the whistle and I laughed, totally worth it to see this video. Keep making it.
this is the first time in my life I have understood a smith chart :)
Maaan. You explained everything to me in less than 10 minutes. My teacher couldn't explain it in a month. Thanks a lot sir!
Thank you for teaching. I finally know how to find out VSWR.
I didn't have time to study from the book.. you just made my life easier #king
Thank you very much sir, this is an excellent explanation.
great!your video is much more clear than my professor’s explanation saved my life!
when my college sir taught this topic..i was wondering if i'l ever understand such things.. and after watching your video for just one time..i am so confident that i can teach my professor..thanks a lot Professor Carl!
You are a true asset! Thank you for these videos
Thank you so much. It helped a lot. And thank you for responding to comments. I had the same question about the other side of the circle drawn :)
U kno u came a long way when khan academy can no longer be of service
Perfect teaching in simple way
Easy explanation but would like to see if there is 2:1 match ? how to do it . thanks
Sir, I am confidence enough now to teach other about Smith Chart! Thanks a lot Professor.
73
OH6EAC
a very helpful video, easy and to the point and very good explanation, thank you so veryy much!
God bless you, this is really helpful
Greetings from Mexico!
thank you a lot sir.... U just saved my semester :)
Thank u so much...its such an awsme video fr undrstndng Smithcharts...👍👍👍👍👍
but the circles touch the axis in 2 points, what about the other number?
VSWR is always greater than 1(or equal), since the maximum of the wave can't be lower than it's minimum. So you only take numbers to the left of the center point.
Thank you very much.
It helped me to prepare for my test!
while making the circle for vswr from centre , the central line cuts the circle at two points..then why the right side one point is taken as vswr? Why not the left side point? what's the rule for selection of one intersecting point from the available two points?
Abhishek Poudel left side is voltage minimum then right for maximum
+Abhishek Poudel , here's a simple way to look at this...
VSWR = Vmax / Vmin
Therefore, the smallest value VSWR can have is, 1 .
Values on the left of the center of VSWR circle are less than, 1 and thus invalid for the purpose of determining VSWR.
awesome dude bcoz of u i'll get 5 marks 2mrw..thnx
Great stuff. Very blurry video on my end. Anywhere else it can be view? Thanks!
But why we didn't take the other side that passes the line? Since we are moving clock-wise from the negative part(navigating through the angel too).
The first jumper is the impending. How did you derive at -f100
How do you physically measure the reactive component of an antenna? I would imagine that is essential before you can use the chart to figure stuff out with.
Hi, thank you for the great video's on smith chart. I can see how to create a matching network between generator and load, but when this is swept over a frequency range (HF) in mind, the match at the design frequency goes out the window. Perhaps you could look at matching over wider frequency ranges in a future video? many thanks, Joe
You sir are a wonderful help!
Exellent tutorial. Love it
it was really really really helpful and clear :) thanks a lot
the resistive line will pass through 2 points of a circle, left and right side.. so which one is the vswr? does it always have to be on the right side of the circle, where resistive line passes through?
It always has to be the right side. The easiest way to remember is this: the range of possible values for VSWR is 1 (perfect match, no reflections) up to infinity (open or short circuit, 100% reflected power). This means that it cannot be less than 1, ruling out all the values on the left side.
How come you didn't choose the points on the other side of the VSWR circle that crossed center line?
Because it's closer to the load, if you add a 1/4 wavelength stripline it will bring closer to the other point.
You are too good!Thank you so much!
Excellent video. Where can I obtain Smith Chart Paper like you are using?
www.cems.uvm.edu/~keoughst/EE141-142/Smith_Chart.jpg
You are a saint! Thank you!
Excellent, you are my teacher
very useful videos , thanks
it is very helpful video to a student who study about electronic engineering which is me.!!! Awesome video thank you for video. I subscribe your channel.
Dude you are awesome !!!
Thanks carl you're my saviour now XD
Nice video
nice job bro! helps a lot
very good explanation, thank you so much
Thank you, Carl!!!
Please help me. You're given a line's characteristic impedance, but you're given just the magnitude of the load impedance, not the complex expression of the impedance. How do we solve VSWR for this case? Should we assume that the load is purely resistive?
Is the VSWR ratio always to 1?
I understand the left part of the ratio is determined from the VSWR circle, but just curious about the 1. Thanks in advance.
Big Thank you.
Thanks...very helpful video (Y) may God bless you... Amen
This helps so much! Thank you!
Excellent video
you are a life saver
Thank u so much ! Very helpful!
Thanks man....helps alot ..👍
Thank you sooooooo much.....My Transmition Line exam is tommorow!! :'(
sounds effect is as good as usual :)
Glad you think so!
Thanks to world of technology have been trying to get this in lecture room but finally I got it here #yoppylaspotech
Thank you SiR..
very good explanation .....
brilliant.I must call you sir now
OMG, thank you Bro, your videos are better than the spanish version
NICARAGUA lol
Thank you Sir!
73 is "best regards." The other letters and numbers are an amateur radio operator's callsign. files.qrz.com/a/ta4a/ham_radio_world_map.jpg
Thank you very much!
great explaination
thank you!
Mind if I ask you what "73 OH6EAC" is ? Just curious, no reason for asking.
Super video sir
Thank you!
Well Explained sir
I wish i can find a practical tutorial on S parameters.I am thinking like a block showing the network port connections and the explanations - the maths will come but just how you connect and what measurements you make.
For Sxy, the signal is measured at port x, and it came from port y. Simple as that. So for S21, the signal is transmitted out of port 1, and you measure it at port 2. Since it has to travel through a device or TX line to get there, it is insertion loss. For S11, the signal is transmitted out of port 1 and is measured at port 1. Wait! These are the same port! So it must be REFLECTED power that is being measured: return loss. Check out the video on insertion and return loss for more on that.
How to find impedance , given VSWR
Ah wow. Thats news for me.
Thanks :D
Thank u very much!!
Thanks !!
Thanks!
Minor quibble: Load 3 should have been noted as (50 + j0) Ohms.
wow, shouldv found these videos earlier when i took the smith chart class.
really thank u :)
THANKS
But the VSWR is 1+ Gamma / 1 - Gamma. What you did is just 1 + Gamma. Why ?
The Smith Chart can be thought of as a plot of complex reflection coefficient Gamma in polar form (hence the circular boundary). The magnitude of Gamma is linearly scaled from the horizontal axis (the X'=0 line which has the R' values marked on it), and the angle (or argument) is plotted per a scale marked around the outer circle (which hasn't been mentioned in the video as it's not pertinent to finding VSWR easily). Accordingly, the circle centred on the R'=1 point and including the plotted Z' point will have a radius directly proportional to the magnitude of Gamma. (Edit: this is addressed in the next video ruclips.net/video/9KlIgae0ad8/видео.html)
VSWR = (1+|Gamma|)/(1-|Gamma|), where |Gamma| is the magnitude of the reflection coefficient. It turns out that when the magnitude of Gamma is plotted on the right-hand segment of X'=0 line, the R' value equates to the VSWR value for the given magnitude of Gamma.
To understand this fully, I recommend you read up on how the Smith Chart is mathematically derived from R, jX, and reflection coefficient. Any decent electromagnetics text dealing with transmission lines should include such a derivation. The maths isn't particularly difficult: some complex number manipulation and understanding the equation for a circle would suffice.
Thanks alot
In response to your question of "cany you see my circle?", the answer for me is no. The image was too blurry to make it out.
thank u..:)
Bro does this shit ever end
You're Canadian, Eh?
agree
radio amateur's code-name probably : )
73 is like "greetings" from what i know
please come and Teach at my school!
EXCELLENT and THANKS!!!!
73,
WA4AOS
You have a gift lol
To bad. I can't see any circle
Aurelio Rapids
Tromp Alley
Elise Avenue