Understanding VSWR and Return Loss

[@WietseBuwalda]

“The definition of genius is taking the complex and making it simple.” - A. Einstein. Brilliant channel. Credit to all involved.

[@pauldenisowski]

Thank you!

[@josehenriquemaia136]

Gênio!!! Parabéns!!!​@@pauldenisowski

[@rodericksibelius8472]

I am now 66 retired and way back in 1985, I took a 9 month course of Microwave electronics at Microwave Training Institute in Mountain View, California, a school privately owned by Alan Scott. Learned the basics, Operation of Klystrons, Reflex Klystrons, low frequency circuits, microwave circuits and laboratory experiments using microwave instrumentation and USE OF THE SMITH CHART. This video VISUALIZATIONS are awesome and I HAD A HARD TIME visualizing what those PHYSICISTS and Working Engineers teaching us those concepts, I finished the course knowing all these magic microwave skills and went to work at a few companies building microstrip gallium arsenide microwave amplifiers 4 - 18 Ghz at the time tuning and cascading them, with a lot of microscope work. Thank You for MAKING this VIDEO, the time and dedication you have explained these concepts in simple clear manner. God Bless You.

[@richardbrumant2684]

excellent delivery and I am in the Telecommunication field for the past 40 years and I am impressed.

[@pauldenisowski]

Thanks Richard - really appreciate the feedback!

[@onderdincel]

excellent video, thank you, those 5 dislikes probably from keysight

[@ethendixon4612]

Lollll prolly from Keysight

[@Jota_VA]

3 from keysight, 2 from anritsu XD

[@annguyendang8388]

Why did you think of that?

[@tonofgx531]

@Alter Kater Yeah that kind of bothers me as well. The video assumed that the source is perfectly matched to the tline

[@pranaysharma1255]

😂😂😂

[@nisargtrivedi3314]

very nice and clear explanation. thank you for not just this but the whole video series. very informative on basics. just want to point out a possible typo at page 13 where Return Loss and VSWR are related via an equation. here the numerator and denominator are switched somehow i.e. it should have been Return Loss = 20log10(VSWR-1/VSWR+1).

[@jb3757]

I just found the Gold mine, and the fact it comes from my favorite Test&Measurement company is icing on the cake.

[@pauldenisowski]

We're happy you're here - and more is on the way!

[@nlo114]

At 63, I find this the best explanation so far. The explanation is exactly the same as everyone else's, (so it would be, wouldn't it!), but the waveform animation is slower and easier to understand, with the explanation at a well measured steady pace. Thank you R&S, this video should be played in college courses.

[@Rohde-Schwarz]

Thanks a lot for the positive feedback - that motivates us!

[@pauldenisowski]

Thank you!

[@Tom-dn5de]

@@pauldenisowski Thanks for the great video! and I have some things to ask. The frequency of the reflected signal is the same as the frequency of the forward signal or not? And how about the relative phase between them?

[@pauldenisowski]

​@@Tom-dn5de Generally speaking, the signals appearing at all ports of a network will have the same frequency as the input signal but different phases and amplitudes. It is possible to make S-parameter measurements on frequency-converting devices, but this is a more advanced topic that needs its own video :)

[@Tom-dn5de]

@@pauldenisowski Thanks a lot for your explanation. I am taking a VSWR measurement of an RF source, but I don't know why an auxiliary generator is used which transmits a wave with a slightly offset carrier frequency into the DUT. Could you explain it to me?

[@albertkleyn111]

Paul..... in my 76 years I have seen a "FEW" VIDEOS. THIS ONE IS BY FAR THE BEST and clearest I have seen. on this subject. Coming from your company, i am not surprised, Rhode & Schwartz are well known for the top quality of their instruments. Thank you for taking the time and effort in producing this. Please rest assured that it is VERY MUCH appreciated. IGNORE the dislikes... Even if God himself would have done this video... he too would get dislikes... and probably more of 'em ! Albert EI7II.

[@pauldenisowski]

Thanks for the support! There are always things that I could do better in these videos, so I appreciate any and all comments. 73, Paul KO4LZ

[@deepeshshiwakoti7249]

What a wonderful comment.

[@euanwells6968]

MAN this helped so much. Im final year doing electrical engineering and my teacher speaks no English and i was FREAKING out until i found your video. Thanks mate

[@shayanfatima2522]

One of the best lectures I have ever come across. Thank you so much sir!

[@GuruGyancenter786]

Good

[@michaelperkins4332]

Coming from an avionics background, this video was beneficial since it was only briefly mentioned, and it was more about hooking equipment up, and it will measure it for you. This will definitely be used to help teach others what VSWR is. Thank you.

[@pauldenisowski]

Thanks!

[@Maceta444]

That graph at min 5 was all that i was needing.

[@pauldenisowski]

Thanks -- sometimes a picture really is worth a thousand words :)

[@butchygra]

Another fantastic video, summarising what my lecturer failed to convey in 2 hours in 10 mins! Love this series!

[@pauldenisowski]

Thanks! There is obviously a LOT more than can be said about VSWR and return loss (not to mention all the underlying theory and math), so we try to focus on the basics :)

[@alanmainwaring1830]

I like this comment and it is spot on. The pacing of the speech and clarity videos is just right in an area of Radio Frequency engineering that is not easy to grasp. No wonder Rhode and Schwartz make the best analysers in the industry. I wish I could afford to buy one.

[@pauldenisowski]

Thanks!

[@ahmedsaad4595]

Amazing and very valuable for someone like me been over 10 yrs in field

[@Jerrythenerdful]

I'd like to make three important points, all of which can be easily proven as fact: 1.) Reflected power is absolutely NOT what damages a PA device or stage. It is the mismatch in optimum impedance at the output port, regardless of standing waves or reflected power on the feedline some distance away from the PA stage. 2.) It is quite possible to have a very high percentage, actually nearly 100%, of transmitter power delivered to a load even with a very high "reflected power" or VSWR. 3.) Many antenna systems, as well as PA matching, splitting, and combining systems, intentionally operate with fairly high standing waves. Collinear antennas, like VHF/UHF multiple bay arrays, commonly use mismatched transmission lines in harnesses to split power and match the multiple elements to the feed impedance. These harnesses and cables often operate well over 2:1 VSWR in the feed cables with negligible loss.

[@selvakumarr6672]

This is one of the best and easy to understand explanation with clear examples I have come across. Well done team. I request you to post more such videos.

[@rangapusuloor5572]

Excellent presentation on understanding the Return loss and importance of VSWR.

[@glenmartin2437]

Thank you. You filled a gap in my knowledge. I repeatedly ran too much power for a GC MS and had a foldback circuit shut the power down. The field engineers could not explain this to me. It did not help that I had no meter to let me know I was approaching the power limit. Thanks again. N0QFT

[@Rohde-Schwarz]

😎 👍

[@Rohde-Schwarz]

That's great to hear Glen! Thank you for the feedback

[@pauldenisowski]

Glad it was helpful. Foldback has saved me from destroying quite a few things in my career :)

[@sunkarasaigoutham]

2:46 example of complex impedance - antennas and that is why there is a range of frequency mentioned. Therefore the level of power reflected will be a function of frequency. There are two ways to quantify these losses 1. VSWR 2. return loss Retrun loss= Forward power - reflected power for example - Forward power is 50dBm and reflected power is 10dBm Return loss = 40dBm The larger the return loss the lesser the reflected power

[@risyamnozawa]

Thank you so much for the explanation and the concept. Your way of your explanation is easy to understand, preety straightforward but still bring the concept. Hope the best for you

[@Alex-M0OOV]

This video is so good that tomorrow I would like to buy some Rohde & Schwarz test gear to give back... if I could afford it :) Thank you!

[@Redhawk03a]

Just getting into amateur radio, this is very helpful.

[@pepe6666]

i swear its james woods doing these lessons. great content though and i really appreciate the education. its been extremely helpful. ooh a piece of candy.

[@nax1807]

this helped me through HF/RF Engineering thanks

[@hunchojet]

Beautifully made presentation/explanation. 5/5!

[@pauldenisowski]

Thank you!

[@alexandrechaillet2504]

most understandable vide I saw on that topic. great job

[@pauldenisowski]

Thanks!

[@ronnie166]

I'm trying to troubleshoot a high voltage power transmitter with valuable freq.. Problem is an over current fault which shuts down the transmitter. This video is helpful, TY

[@pauldenisowski]

Thanks and good luck!

[@ohaya1]

What an excellent video, more like this please!

[@sudeepshetty4530]

Beautifully explained..

[@jianhaowu7368]

very nice presentation

[@storaman12]

SUPERB. What a great explanation.

[@dhananjayw502]

Excellent video sir..... it helped me understand the thing. Thanks a lot !!!

[@nick1f]

Excellent presentation!

[@pauldenisowski]

Thank you!

[@klam77]

Thank you! Such a beautifully explained video!

[@eightfivezerobraxton5509]

Really helped me understand the topic ! you're a great teacher.

[@primeradianttechnologies3085]

Absolutly great presentation!!! Thank you!

[@pauldenisowski]

Thanks for the feedback!

[@davidharrell9263]

Thanks for the great video! I hope more to follow!

[@pauldenisowski]

Thanks! We just posted a series of videos on oscilloscopes today and there will be many more videos in this series, so please stay tuned!

[@abdulhaquemohammed6478]

To the point and very informative. Thanks a lot for the share!

[@navneetkumaryadav7280]

Thanks for the clear explanation!!

[@pauldenisowski]

My pleasure - thanks for the feedback!

[@adonikam1]

Excellent, Excellent, excellent video! THANK YOU!! I subscribed...

[@Parirash123]

A clear and good presentation. Thank you.

[@quyvuuc140]

great, thank so much for that knowledge you provided

[@JackQuark]

Thanks, this is so pleasant to learn from.

[@pauldenisowski]

Thank you!

[@snoobeagle]

I didn't know actor James Woods was into antenna efficiency! :)

[@amilasamaraweera6209]

Very clear explanation. Thanks!

[@tomaszkluska6419]

Wiki _From a certain perspective 'Return Loss' is a misnomer. The usual function of a transmission line is to convey power from a source to a load with minimal loss. If a transmission line is correctly matched to a load, the reflected power will be zero, no power will be lost due to reflection, and 'Return Loss' will be infinite. Conversely if the line is terminated in an open circuit, the reflected power will be equal to the incident power; all of the incident power will be lost in the sense that none of it will be transferred to a load, and RL will be zero. Thus the numerical values of RL tend in the opposite sense to that expected of a 'loss'._

[@tomaszkluska6419]

RL should have the name unloss :))

[@pauldenisowski]

I agree completely that the term "return loss" is very problematic. As I've mentioned a few times in the comments, it's confusing even to RF engineers, so much so that the editor of an IEEE journal had to publish an article explaining the "proper" way to use the term :)

[@conspiracytheory9396]

Very informative. Thanks.

[@pauldenisowski]

Thanks - appreciate the feedback!

[@WallaceAustin]

Very well presented.

[@yaroslavmuradian5959]

Thank you for your brilliant explanation. One question on terminology though: should we use something like "effectively transmitted power" intead of "return loss"? I doubt "return loss" is the proper term to use in this case because we are discussing the energy, which has been absorbed by the antenna and has been radiated into ester. I know this is not your invention. I wish to know your opinion.

[@pauldenisowski]

Hi Yaroslav - Thanks for your comment. "Return loss" is the standard industry term for this measurement, but I will agree that this term can be a bit problematic. In fact, the editor-in-chief of IEEE Transactions on Antennas and Propagation published a short article in 2009 describing the origins and proper use of the term "return loss" -- it appears close to a third of the people submitting papers to this journal were using the term incorrectly. (Link below, available to IEEE members). From the article: "Turning to present-day usage, return loss is now the most common term used to describe reflection and mismatch." ieeexplore.ieee.org/document/5162049?arnumber=5162049

[@yaroslavmuradian5959]

@@pauldenisowski Thank you Paul. I will look into it.

[@andreslucioaigster3356]

Very helpful!

[@anhtuta2767]

thank you so much

[@THILAKCMBEC]

omg this video was so informative; thank you a tonne

[@sciencelearning2326]

amazing explanation thanks sir

[@pauldenisowski]

Thanks for the feedback!

[@martijndecauter5329]

Wow superb video! Thanks!

[@OxTongue0]

Clearly explained the content. thanks for making Video .

[@sunkarasaigoutham]

amazing video

[@dirindirin3983]

Loved that❤❤❤❤❤❤

[@pauldenisowski]

Thank you!

[@tpmbe]

excellent explanations... thank you

[@gumidellivenkatesh1240]

Excellent video, how you got percentage of return loss ?

[@sidharthks7834]

that was really helpful. Thanks for the content.

[@pauldenisowski]

Thank you!

[@elecronics-sc]

Very good 👍😊

[@richardphillips2405]

Great video. I got stuck when trying to find the ratio of reflected power to forward power by the Vswr. Is there an equation that converts Vswr to the amount of reflected power?

[@Rohde-Schwarz]

Dear Richard, for calculating VSWR using power instead of voltage, you can use the following formula: VSWR = (1 + p) / (1 - p), where p = sqrt(reverse_power / forward_power). Best, Rohde & Schwarz Social Media Team

[@munazzahtaimuri3585]

@@Rohde-Schwarz If above formula is to be used fo this conversion, when should we use the the formula mentioned at 6:12 relating return loss and VSWR?

[@lukbrowncs]

Sir, your idea of vswr is different from what some materials say. They say vswr is ratio of voltage max at the peak and voltage minimum at the trough/bottom. Your explanation of it in the graph appears different. Also, textbooks don't mention that impedance matching means load impedance is the complex conjugate of the source or line impedance. They say it should be exact match. Could you please clarify. Thanks in advance.

[@SteveWithnell]

The complex conjugate provides the exact match. This calculator might be helpful to play around with: www.analog.com/en/design-center/interactive-design-tools/rf-impedance-matching-calculator.html

[@斑鳩-p1j]

8:16 why there a circuit about capacitor and inductor? I thought we are talking about impedance.

[@ernestb.2377]

The terminology of Return Loss for me is very counterintuitive. Instead of difference Forward - Return power I would rather use the ratio of Return / Forward. As we call it Loss how can we strive to maximize the number? Loss should be a low number, or low percentage.

[@IZ0MTW]

It is counter intuitive. But you can think at the power “returning”, the one that has been reflected back by the mismatch, the one we don’t want. So we want to get rid of that power, we want to loose that. The more we loose it the happier we are. So in this case we aim to have something missing. Like having less fat in the blood is good. The less we have the better!

[@clarkrichardson5846]

This is great stufff

[@flymirpark918]

Will the IL,RL characteristic value change according to the reference input to the POGO connector? For example, Will the IL change according referance input 0dBm Vsersus 35dBm?

[@mathy5384]

For the matching network (at 8:14), what would you need to do if the source impedance was smaller than the load impedance? Would the matching network have to have some sort of ADMITTANCE to get the source and load to match? Or is it just not possible to do so?

[@yanxili1553]

so nice！

[@pauldenisowski]

Thank you!

[@thisnotjesus]

There's a math error at 4:40 if 50dbm came from the source and 10 dbm came back 40 dbm didn't go to the load around 49.5 dbm went to the load

[@jianjing7854]

Is it correct ? At 4:32, it shows that Forward power -reflected power= return loss. I assume the return loss = reflected power. Who is wrong?

[@Rohde-Schwarz]

Hi, thank you for your feedback. This is correct, yes.

[@MovieShortCuts]

About impedance matching, if i have a Source impedance of 50ohms, Line RG6 75ohms, and a 25ohms load impedance(microstrip antenna). Is this a matching set with less reflected power? How much VSWR do you think I'll get from this setup? Thanks

[@pauldenisowski]

Unless the load is purely resistive (and your antenna almost certainly isn't), the VSWR will be a function of frequency -- i.e. it will change depending on the frequency of the signal generated by the source. In most cases, the easiest way to minimize reflected power is to have a source, load, and line impedance that are all as similar as possible. Even though a lot of people successfully use RG6 with 50 ohm sources, you might want to consider a different cable type. There's not much you can do about the impedance of the antenna, but keep in mind that this may also change based on how and where the antenna is mounted (i.e. what's next to it). It can be very difficult to reliably and accurately determine VSWR based on the (nominal) values of components in a system. I can do the math and calculate how long a dipole *should* be to have a given VSWR over a given frequency range, but when I actually build it and hang it from a tree or (especially) in my attic, the VSWR is never precisely what I calculated (and is sometimes quite different). The math may give you a good starting point, but the actual value of the assembled system will often be at least somewhat different. This is one of the reasons why instruments like network (or antenna) analyzers exist -- measurement is the only reliable way to know for sure what your VSWR actually is. Hope that helps!

[@surajkulkarni6868]

Is it only me who finds “return loss” name for what’s is used is counter intuitive.

[@AnnaSarris]

645 Damaris Stravenue

[@glennwillems9924]

2:39 a yagi with a gain of 12 DECABel? I would expect the good people at R&S to at least use graphics in which the SI units are respected. In this case: 12 dBi.

[@pauldenisowski]

Sorry, but could you clarify what you're referring to? I don't believe I ever said anything about antenna gain in units of dB or dBi. The y-axis of the graph at 2:39 was intentionally labelled with the generic unit "impedance" for the purpose of illustrating that impedance is non-constant for most antennas, at least compared to the dummy load on the previous slide A graph of gain for a directional antenna like the yagi shown would usually be a polar plot showing gain in dBi (relative to an isotropic radiator) as a function of azimuth. You are absolutely correct in that the gain of an antenna is almost always given in dBi -- one exception would be something like front-back ratio, which would be in dB, not dBi. Again, my apologies if I'm misunderstanding you. And I can assure you that the good people at R&S are familiar with the different types of dB: in fact, we have an entire educational note on this very topic :) scdn.rohde-schwarz.com/ur/pws/dl_downloads/dl_application/application_notes/1ma98/1MA98_13e_dB_or_not_dB.pdf

[@alanmainwaring1830]

For a start because decibels is based on log functions there are no units as such in the argument. One can refer to an antenna gain without referring to a theoretical antenna that has a perfect spherical radiation pattern, in practice such an antenna does not exist or you can just use decibels without the isotropic reference. Decibels is based on ratios of things like power, voltage so that the log function argument has no units.

[@43SunSon]

@4:30 40dB or dBm ?

[@pauldenisowski]

If the forward and reflected powers are in units of dBm, the difference between these values (X dBm - Y dBm) will be in units of dB (not dBm).

[@43SunSon]

@@pauldenisowski ?? i thought dBm-dBm=dBm. could you please explain more or give me an example ?

[@pauldenisowski]

@@43SunSon Sure :) One of the more common "rules" when it comes to decibels and logarithms is that reducing a value by 3 dB is the same as decreasing it by one half. The difference between one watt (30 dBm) and one-half watt (27 dBm) is .... 3 dB. The power value 3 dBm is approximately 2 mW, not one-half watt. Rohde and Schwarz actually has a whitepaper that explains this and quite a few other things regarding decibels: scdn.rohde-schwarz.com/ur/pws/dl_downloads/dl_application/application_notes/1ma98/1MA98_13e_dB_or_not_dB.pdf Hope that helps.

[@43SunSon]

@@pauldenisowski oh my man! you did well! I am reading that pdf. How did you know that much? Are you in this area as well?

[@pauldenisowski]

@@43SunSon Thanks. I'm an engineer at Rohde & Schwarz working in radio frequency test and measurement, so I deal with dB all day, every day :) If you're interested in learning more about RF, please check our website for additional whitepapers, application notes, etc.

[@kavithasenthilkumar4533]

How to find sd11 through this concept

[@pauldenisowski]

If you mean S11, please see the video "Understanding S-Parameters"

[@kavithasenthilkumar4533]

@@pauldenisowski I am not referring to S11 I am trying to say that when two ports which are connected differential then the S parameter for that is called Sd11 it is not as same as S11 we will get S11,S22,S12,S21 from that we need to find Sd11 that was my doubt if you can help me.please reply.Thank you sir for your reply

[@pauldenisowski]

​@@kavithasenthilkumar4533 Differential mode S-parameter measurement is probably a bit too complex of a topic to address in a KZbin video comment :) Rohde & Schwarz supports these measurements using our VNAs and we have numerous application notes and presentations on this topic, e.g. cdn.rohde-schwarz.com/pws/dl_downloads/dl_application/application_notes/1ez53/1EZ53_0E.pdf Hope that helps!

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[@SteveWithnell]

There is an ambiguity, in that the presentation can leave the impression that reflected power gets back into the source and destroys the equipment. This is of course not true. It would be useful to link the rising voltage in the voltage charts to the damage caused to the source device as the explanation of the danger of high SWR. The alternate case of course is very high currents.

[@glenmartin2437]

The reflected power can damage the finals in radio transmitters. These finals then need to be replaced. Been there and have witnessed this happen.

[@Pioneer936]

​@glenmartin2437 how does reflected power damage finals when it joins the forward power ?

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And what about transmission line? It's not excellent video made by excellent company😒. It's rather support typical misundarstandings according to reflected power.

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