Smith chart basics part 2: finding VSWR

[@ati4280]

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.

[@rahulrajaram6891]

Most of engineering is actually not that difficult to explain, but professor often make it harder than it needs to be.

[@barbaraszabo7382]

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!!!

[@ParthJoshi09]

thumbs up if you are watching this night before your exam!!

[@lupuszzz]

I'm here 27years too late, but still enjoying good teachers. ;-)

[@TheDzKid268]

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.

[@martinmariscalescobar1491]

You are my hero Carl, please keep on whistling and doing things "easily"

[@dhrumilchavda2585]

It is the best video, you blow the whistle and I laughed, totally worth it to see this video. Keep making it.

[@stevehall7785]

this is the first time in my life I have understood a smith chart :)

[@naamaarnaldoamericoamerico1045]

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

[@王來吉]

Thank you for teaching. I finally know how to find out VSWR.

[@fares1651]

Maaan. You explained everything to me in less than 10 minutes. My teacher couldn't explain it in a month. Thanks a lot sir!

[@mrjohn6021]

I didn't have time to study from the book.. you just made my life easier #king

[@caoxiangyu6687]

great！your video is much more clear than my professor’s explanation saved my life!

[@user-yx6mq2km3m]

You are a true asset! Thank you for these videos

[@desiyatra2230]

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!

[@sarahsal431]

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 :)

[@seandafny]

U kno u came a long way when khan academy can no longer be of service

[@evanpasero1846]

Thank you very much sir, this is an excellent explanation.

[@pk5432]

Perfect teaching in simple way

[@KIKETMASTERMX]

God bless you, this is really helpful Greetings from Mexico!

[@abhishekpoudel4476]

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?

[@flowersleewan]

Abhishek Poudel left side is voltage minimum then right for maximum

[@photonwaves4827]

+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.

[@PeculiarCortex]

a very helpful video, easy and to the point and very good explanation, thank you so veryy much!

[@IbrahimKhandker]

Sir, I am confidence enough now to teach other about Smith Chart! Thanks a lot Professor. 73 OH6EAC

[@electronic_guy8471]

thank you a lot sir.... U just saved my semester :)

[@anonghosty]

Thank you very much. It helped me to prepare for my test!

[@uriel_zapata]

but the circles touch the axis in 2 points, what about the other number?

[@AbdullahAhmad-xc3xl]

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.

[@joebell3567]

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

[@divyanipatil3082]

Easy explanation but would like to see if there is 2:1 match ? how to do it . thanks

[@ThomasDemeter]

How come you didn't choose the points on the other side of the VSWR circle that crossed center line?

[@rafiassilian4572]

Because it's closer to the load, if you add a 1/4 wavelength stripline it will bring closer to the other point.

[@donnymac575]

Excellent video. Where can I obtain Smith Chart Paper like you are using?

[@carloliver4402]

www.cems.uvm.edu/~keoughst/EE141-142/Smith_Chart.jpg

[@danielpenney9077]

Great stuff. Very blurry video on my end. Anywhere else it can be view? Thanks!

[@googleuser4203]

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).

[@msbdevil]

awesome dude bcoz of u i'll get 5 marks 2mrw..thnx

[@kevinobrien8803]

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.

[@meghajohn6427]

Thank u so much...its such an awsme video fr undrstndng Smithcharts...👍👍👍👍👍

[@davidesteves2438]

You sir are a wonderful help!

[@gangstaaglide4074]

Thanks carl you're my saviour now XD

[@poorva19]

You are too good!Thank you so much!

[@engineeringclass9731]

Exellent tutorial. Love it

[@reubengeorge220]

The first jumper is the impending. How did you derive at -f100

[@Knarcout]

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?

[@carloliver4402]

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.

[@keivanvalamanesh120]

it was really really really helpful and clear :) thanks a lot

[@Ionita.Vlad.]

You are a saint! Thank you!

[@dalekwon3091]

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.

[@DG-zv2tj]

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.

[@harryren383]

Excellent, you are my teacher

[@widyaagsari2725]

very good explanation, thank you so much

[@zaeemmazhar973]

Thanks...very helpful video (Y) may God bless you... Amen

[@charlessamuellasiste2752]

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?

[@tidegoesin_tidegoesout]

Excellent video

[@MrAjaynagadeep]

Thank you, Carl!!!

[@PowKu10]

nice job bro! helps a lot

[@hassanaziz39]

brilliant.I must call you sir now

[@Abrahamsproperties]

Thanks to world of technology have been trying to get this in lecture room but finally I got it here #yoppylaspotech

[@thepirates2518]

Dude you are awesome !!!

[@asatallah]

very useful videos , thanks

[@lalayev]

Thank u so much ! Very helpful!

[@robertocarloschavarriamart7045]

OMG, thank you Bro, your videos are better than the spanish version NICARAGUA lol

[@lhcj21]

This helps so much! Thank you!

[@amgadtaher4059]

Thank you sooooooo much.....My Transmition Line exam is tommorow!! :'(

[@electronics_classes3302]

Nice video

[@psonney]

Big Thank you.

[@edwatts9890]

Minor quibble: Load 3 should have been noted as (50 + j0) Ohms.

[@shahriarmostafiz1292]

you are a life saver

[@saket1380]

Thanks man....helps alot ..👍

[@urjathakur4545]

very good explanation .....

[@rahul122112]

Mind if I ask you what "73 OH6EAC" is ? Just curious, no reason for asking.

[@黄嘉宇-z6l]

sounds effect is as good as usual :)

[@carloliver4402]

Glad you think so!

[@seank9122]

wow, shouldv found these videos earlier when i took the smith chart class.

[@rajcodes100]

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.

[@carloliver4402]

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.

[@edwardwonder9258]

great explaination

[@aniket7743]

Well Explained sir

[@nimiechaylone5481]

Thank you very much!

[@sudeep6716]

Thank you Sir!

[@carloliver4402]

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

[@bakhtiaruddin.2577]

Thank you SiR..

[@Kwatcher100]

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.

[@rahul122112]

Ah wow. Thats news for me. Thanks :D

[@nibzlegend99]

How to find impedance , given VSWR

[@neerajdharmadhikari9424]

Thank u very much!!

[@srinunavya5433]

Super video sir

[@thomaspeterson2568]

Thank you!

[@consuelosanchez1833]

thank you!

[@PauloConstantino167]

But the VSWR is 1+ Gamma / 1 - Gamma. What you did is just 1 + Gamma. Why ?

[@vk2ig]

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 kzbin.info/www/bejne/b3zPepqXmpWUmpo) 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.

[@seandafny]

Bro does this shit ever end

[@MrRefract]

Thanks!

[@MrNemay]

Thanks !!

[@omaralngar330]

really thank u :)

[@leonidfinkel4347]

THANKS

[@abdulatifdhia382]

Thanks alot

[@jessiequinn6180]

You're Canadian, Eh?

[@sanjoybasak9646]

thank u..:)

[@TheCheng1993]

agree

[@jwill6676]

please come and Teach at my school!

[@Xristina8anos]

radio amateur's code-name probably : ) 73 is like "greetings" from what i know

[@wolleausmallo1]

To bad. I can't see any circle

[@BessieNick-q1y]

Tromp Alley

[@ConanTodd-t8h]

Aurelio Rapids

[@EmilySavage-u2t]

Hickle Row

[@MCSGproject]

You have a gift lol

[@RobertCoffey-j4k]

Elise Avenue