Understanding Standing Wave Ratio: SWR & VSWR

VSWR or voltage standing wave ratio is a phenomenon that occurs on radio frequency feeders.
VSWR, voltage standing wave ratio is an important parameter in many areas, and especially where transmitters are used.
When looking at VSWR it is necessary to look at feeder and load impedances.
It is found that to transfer the maximum amount of power from a feeder to a load, the two must have the same impedance.
Coaxial feeder, twin-line feeder, waveguide and strip-line feeder all have a characteristic impedance. If the load has a different impedance then not all the power can be transferred and the remaining power is reflected back along the feeder.
The voltages and currents from the forward and reverse power combine together and set up standing waves along the feeder. It is found that peaks and troughs occur as a result of the standing waves. These peaks and troughs can rise to twice the voltage of current for a complete mismatch, i.e. infinity to 1 VSWR.
For lesser values of VSWR, the plot of voltage against current along the feeder is less marked.
Accordingly it is possible to defined the voltage standing wave ratio, VSWR as the maximum voltage divided by the minimum voltage.
The reflection coefficient is also a very useful quantity when looking at VSWR. This is the voltage of the forward waveform divided by the voltage of the reflected waveform. It is also possible to obtain the VSWR from a knowledge of the forward and reflected powers. These are all described in the video.
More information about Voltage Standing Wave Ratio, VSWR can be found at: www.electronics-notes.com/art...
Useful mathematical summary of standing waves & it includes the reason why peaks are half a wavelength apart: eng.libretexts.org/Bookshelve...
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[@gianfrancomior7391]

Very good explanations! Quick & clear!

[@ElectronicsNotes]

Glad you liked the video.

[@jsmith5443]

Great job and a simple and quick explanation. Thank you

[@ElectronicsNotes]

I’m really glad you found the video useful - thanks for your comment.

[@robetereo]

Thank you so much! Excellent explanation and video!

[@Hino_55]

ELI & ICE Current lags Voltage by 90 degrees in an ac circuit with a pure inductive reactance. In a circuit with inductive reactance and resistance, the current lags the voltage by less than 90 degrees. Current leads Voltage by 90 degrees in an ac circuit with a pure capacitive reactance. In a circuit with capacitive reactance and resistance, the current leads the voltage by less than 90 degrees.

[@curtpiazza1688]

Great explanation! 🎉

[@ElectronicsNotes]

I’m glad you appreciated the video. Thanks for your comment.

[@adonikam1]

Excellent video, THANK YOU!!!

[@ElectronicsNotes]

Glad we could help.

[@advanceddesignsystemadsrfi449]

thank you for sharing your knowledge

[@ElectronicsNotes]

Our pleasure

[@OxTongue0]

Wonderful explanation thanks for the video

[@ElectronicsNotes]

Glad you liked it. Thanks for the comment

[@AbhishekKumar1902]

Excellent video, I do not hail from RF background but this clearly helped me in the understanding!

[@ElectronicsNotes]

I am really pleased that you found the video useful. We took a while to get the video done and we tried to explain the concept in a clear and concise fashion, so it is great to hear it has been helpful. Thanks.

[@deepakdavid8355]

excellent video

[@ElectronicsNotes]

Thanks for your comment - much appreciated.

[@blakely1317]

Huge thank you! it is my gold i found while preparing for an upcoming exam

[@ElectronicsNotes]

Thank you very much for your comment. Glad the video helped.

[@blakely1317]

@@ElectronicsNotes yes, it did help for sure, I got my A haha :)

[@ElectronicsNotes]

@@blakely1317 Great news!! Well done.

[@BenjaminGSlade]

Good video, but he doesn’t say that, for a short circuit, the reflected voltage is the inverse of the forward voltage. Also the voltage source has an orange sine wave in it, but the forward voltage wave in the graph is gray and the reflected voltage wave is orange. Very confusing for someone who’s trying to figure this out for the first time

[@ElectronicsNotes]

Glad you thought the overall video as good, but sorry you didn't like that fact we did not mention the voltage was the inverse of the forward voltage. I had hoped that would come out of the animation, but obviously not so. We need to keep he videos short and concise and that means missing some details out.

[@glenmartin2437]

Thank you.

[@ElectronicsNotes]

You're welcome!

[@mahendraprasadyadav806]

Excellent💯

[@ElectronicsNotes]

Thank you! Cheers!

[@sharonjuniorchess]

Surely the wavelength is measured as the gap between peak-to-peak. So instead of labelling your intervals as λ/4 they should actually be labelled as λ/2? see 4:10.

[@mr.niceguy724]

@Heads Mess I still don't understand what you want to say :D how can it be, that peak to peak is just λ/4?

[@mr.niceguy724]

@Heads Mess now I get it! thank you very much! have a nice day :)

[@LiHao8377]

1:23 is it all the power or half of the power?

[@backpackerthrulife8497]

3:39 Aren't those plots of the absolute values of the standing wave V and I ??

[@advanceddesignsystemadsrfi449]

thank you for commenting ,,,,, i have ben confused about that

[@mateoestevez9582]

Nice animations and video, but for the plots mentioned in 3:40 and 4:01, shouldn't the standing waves look like a sinusoidal wave with 4V peak to peak? Just as the ones showed earlier in the video?

[@mateoestevez9582]

I mean like the wave shown in 3:26

[@ElectronicsNotes]

It is because of the wave the interference patters sum together. For small levels of re-elected power a near sinusoidal pattern is generated, for much higher ones it approximates to a half sine wave.

[@randykitchleburger2780]

Thanks

[@ElectronicsNotes]

Glad you found it useful.

[@advanceddesignsystemadsrfi449]

3:39 this is the most important part of the video one must understand Aren't those plots of the absolute values of the standing wave V and I ??

[@ElectronicsNotes]

Yes, they are voltage standing waves, but can be applicable for either dependent upon the phase.

[@jackalakalaka11]

It seems like the minimum voltage would always be zero, making the VSWR ∞. Is there just typically a DC offset?

[@ElectronicsNotes]

The minima and maxima are RF (AC) voltage levels hence no DC offset.

[@jonahansen]

Yeah, you'd think since they're AC, they will always pass through zero and this confused me too. But what is really meant is the maximum AC voltage anywhere along a line at least a half wavelength long divided by the minimum AC voltage along the same line. And since it's a periodic sine wave, one only needs to look at any half-wavelength section; furthermore AC voltages are positive values. So if there's no reflected power, the maximum and minimum are the same at every point and VSWR is 1:1, but once power is reflected they start interfering as the video shows with maxima and minima as described. Check out the Wikipedia entry - it's more precise and cleared it up for me. Boomlakalakalaka....

[@oldcowbb]

you need a ground symbol for the short circuit, confused me for a sec

[@user-hc7mi7ft9x]

For the short circuit, if there was a ground/earth connected, wouldn't the signal just go to ground just like any low-pass filters??

[@ElectronicsNotes]

The signal cannot just go to ground because it would result in power disappearing. It has to go somewhere and therefore it is reflected. Hope this helps.

[@user-hc7mi7ft9x]

@@ElectronicsNotes Yes I get what you mean, but does this happen to signals that gets filtered out through low pass filters for examples? those are also going to ground...

[@DumbledoreMcCracken]

What happens when the feed isn't a multiple of λ/4?

[@ElectronicsNotes]

The feeder does not need to be an exact quarter wavelength. The voltages seen at any point along feeder of any length are those seen on the animation at that point. The only point that is fixed is the termination to the feeder. So you can work back 1.25 wavelengths, for example from the termination, and see what the voltages will be at that point on the animation, etc. I hope that explains the situation. This does assume a perfectly matched source impedance, otherwise you get even more reflections and the situation becomes very complicated.

[@ancientbutstillhere6570]

Secret of the real wizz. Troubleshoot this problem with AM manual dial radio. Set at 1000,IF no radio near xmits there. I walk up n down cables. Listen to sound on radio. Signals of any RF swaves picked up. Almost always by damaged cables or bad connectors. Kids $3 am radio with headphones keeps hands free. Works on radar too. If its 100w xmitted stay 2 wave lengths away but you still hear it. Data from computers do the same. When found note distance. Tdr line. Or oscope. Boss will be impressed. Done dozens of times. Check coax. Ig rg-6 mixed with 11 could be issue.

[@rifqirahman6314]

how matching vswr?

[@ElectronicsNotes]

There are many units for matching the impedance of an antenna to a feeder. You can use an antenna tuner: www.electronics-notes.com/articles/antennas-propagation/antenna-tuning-tuner-unit/what-is-an-atu-basics.php This is mainly used fir HF antennas where frequency changes are needed. Alternatively a matching transformer often called a balun or even an un-un dependent upon the application. If high reactance levels are encountered, then a special matching transformer with opposite reactive elements may be used. Hope this helps.

[@vanxaychanthachack6532]

vanxay

[@yoksel99]

Not very clear explanation. Need to explain that the gray and red waveform at the short circuit end are opposite voltages to get zero in sum. Whereas at the open end, they are same value which gives double voltage in sum. So in both cases the unmatched end generates that red waveform going back. And still, not very clear why all that that happens at the end, especially for why the doubling happens at the open circuit case. Would be much nicer to start from the explanation of how the "step" voltage travels via transmission line and how it reflects off the unmatched end. That step explanation is very visual and easy to comprehend. Then, after that reflection phenomenon is clear, switch to sine waves.

[@ElectronicsNotes]

Sorry it was not very clear for you.

[@Pioneer936]

Also no mention of where the reflected power ends up

[@brucetouzel6484]

teaching pace is a bit fast