# Reflecting and standing waves in wires

Discussion in 'Electronic Design' started by Danny, Dec 30, 2003.

1. ### DannyGuest

Hi

I have been told by numerous people of the first idea opposed to the
second here: Radio waves are transmitted by standing waves in the wires,
and not because of current flowing up the wire sue to its capacitance
relative to earth causing electromagnetic waves which then form the

Can someone please explain whether i have just been horribly mislead all
the years or whether even the understanding the i have been persuaded of
is wrong in which case the real explanation would be really nice.

The main flaw that i have the the idea i have been afflicted with is how
can the charge be reflected back down the wire for form this standing
wave. Also, how do standing waves work with electricity and stuff like.

Danny

2. ### John WoodgateGuest

I read in sci.electronics.design that Danny <>
This is true for resonant antennas, such as dipoles.

Standing waves occur because when the current injected at the driving
point reaches the end of the driven element it has nowhere to go but
back along the element.
This is true for non-resonant antennas, much shorter than a wavelength.

So you have been misled by being told that two different explanations
are *always* true, whereas they are true under different conditions.

3. ### MacGuest

I just want to point out that there are varying degrees of standing wave.
If you have a source, transmission line, and antenna that are all
perfectly matched to the same impedance, then there will be no standing
wave at all in the transmission line. But if there is an impedance
mismatch, then there will be a standing wave. If the standing wave gets
too big, then you are not delivering much power to your antenna, and you
could even damage your transmitter. You could also get very high voltages
that could cause arcing in your transmission line.

Mac

4. ### Wayne ShanksGuest

RF energy flowing in a transmission line is ideally a traveling wave,
not a standing wave. A Standing wave is actually the superposition of
two traveling wave, traveling in opposite directions (as you implied),
thus the standing wave component on a transmission line represents a
zero net transmission of power. Standing waves are generally to be
avoided. In resonant antennas a standing wave is used to increase the
oscillating charge, thus increase the launched electromagnetic power.
Power splitters and impedance transformers use standing waves to move
power on Tx lines from one characteristic impedance to another, so
standing waves are useful, but power is transmitted on traveling waves.

Like electromagnetic waves in air, electromagnetic energy in a Tx line
is carried as the energy stored in its electric and magnetic field.
This assumes that the wavelength in the tx line is much smaller than the
Tx line length. If the Tx line length is much shorter than the
wavelength, then it can be treated as a wire conductor that carries current.

Wayne s

5. ### Robert BaerGuest

Actually, the waves are standing for two reasons:
1) they cannot sit down
2) the FTL component was stolen by the idiot selling FTL cables (how
else do you think he does it?)

6. ### rickGuest

It's certainly not true that the source, line and load ALL need be matched
to prevent standing waves. Only if there is a mismatch between the
transmission line and the load will there be a reflection and hence standing
waves - so to prevent standing waves, you only need to match the load to the
line - the source is irrelevant.

7. ### Paul BurridgeGuest

That's what I was brought up to believe, IIRC. The idea being to trim
your dipole (for example) to a resonant length so that the voltage
nodes (or is it current nodes) are at minima at the tips of the dipole
(IIRC it's the *voltage* nodes thust must be minimised) so that all
the available power is radiated and none reflected back towards the
transmitter. I also STR that standiing waves on antennas can cause
radiation from the feeder as well but it's not my field so I'll shut
me cakehole.
x-posted for a better take on the subject.

8. ### MacGuest

You are quite right. My apologies. RF sources typically need to be
designed for a specific load (usually 50 Ohms) but the source doesn't
actually need to have a particular impedance to avoid a standing wave.

On the other hand, if there is a standing wave, the severity of the
problem will be somewhat reduced if the source impedance matches the line
impedance.

This is relevant to 75 Ohm video signals as well. The datasheets for
video drivers always seem to show a 75 Ohm series resistor. The purpose of
the resistor is to make sure that reflections coming back from the load
don't get re-reflected at the source.

Mac

9. ### Reg EdwardsGuest

In any ordinary relatively low loss system, from transmitter to free-space,
it is necessary only to match impedances at one point. When this done an
impedance match occurs automatically, simultaneously, at every other point
in the system. To think in terms of minimising standing waves and

Wait for it - wait for it !

10. ### Paul BurridgeGuest

Take it easy, Reg. This is only a Usenet forum. Don't let your
Pacemaker get unduly excited.

11. ### Active8Guest

It's certainly not true that the source, line and load ALL need be matched
to prevent standing waves. Only if there is a mismatch between the
transmission line and the load will there be a reflection and hence standing
waves - so to prevent standing waves, you only need to match the load to the
line - the source is irrelevant.[/QUOTE]

You are quite right. My apologies. RF sources typically need to be
designed for a specific load (usually 50 Ohms) but the source doesn't
actually need to have a particular impedance to avoid a standing wave.

On the other hand, if there is a standing wave, the severity of the
problem will be somewhat reduced if the source impedance matches the line
impedance.[/QUOTE]

All true, but just so no one gets the idea that matching the source
to the line is never important, remember that it's done for maximum
power transfer. Also, the transducer gain of the amplifier is
dependent on the match and the input impedance of the amp is
dependent on the output impedance and the reverse transfer
admittance or reverse transmission coefficient depending on what
perameters you're using. That's just the \$20 way of saying that if
you change the design load, you change the gain and input
impedance.

Of course, if you need less gain than is available with a given
device, you can purposely mismatch the output to the load

With power transistors in the final, you'd deal with the large
signal impedance parameters.

12. ### Reg EdwardsGuest

Old Wives' Tales.

The SWR meter located at the output of the transmitter does NOT measure
standing waves on the transmission line to the antenna.

Most of you are fooling yourselves.

The instrument indicates only WHETHER or NOT the transmitter is loaded with
the correct design value of resistance, usually 50 ohms. Which, under
normal operating conditions, is all anybody needs to know anyway.

And the internal impedance of the transmitter has nothing to do with it.
Even the transmitter designer has no interest in what the internal impedance
is. It is NOT the designer's objective to maximise output power by matching
the internal impedance to the load.

The instrument is mis-named. It doesn't measure anything. Its correct
----
Reg.

--
............................................................
Regards from Reg, G4FGQ
http://www.btinternet.com/~g4fgq.regp
............................................................

13. ### Paul BurridgeGuest

It certainly curious that this subject more than any other seems to
generate a huge amount of confusion. There are so many conflicting
views on this issue it's hardly surprising some folk get all mixed up
over it. Perhaps you should write a definitive essay on it and stick

14. ### RickGuest

Unless your SWR meter happens to be a scalar reflectomer.
It's cheaper than paying someone to do it.
So your SWR meter "doesn't measure anything", but at the same time it
tells you all you need to know. Spooky.

15. ### Reg EdwardsGuest

So your SWR meter "doesn't measure anything", but at the same time it
======================================

Measuring is not the same as merely distinguishing between good or bad.

Delete the SWR meter scale and replace it with a coloured band, changing
from green at one end, via brown, to red at the other.

One of these miniature, 1/2" square, units will do fine.

Or just a red and a green LED, both in one package.

Get yourselves into the 21st Century.

But then, it must be admitted, there'd be only the weather to argue about.

16. ### Paul BurridgeGuest

Within my limited knowledge on the subject, the SWR meter simply tells
you the ratio between forward and reflected power at the point in the
line the measurement is made. The smaller the degree of reflection,
the better; ideally 1:1 but in practice ISTR that even at 3:1 most of
the power is still usefully radiated.

17. ### Reg EdwardsGuest

The SWR meter located at the output of the transmitter does NOT measure
===========================

Done! See one-liner above.

18. ### Paul BurridgeGuest

Sorry, Reg. Can you repost the link? Nothing appeared here.

19. ### John CrightonGuest

Hello Paul,
just read carefully what Reg has written, and if you happen to
come across a 100 metre drum of coax cable try this experiment.

Tx-VSWR meter-----------long coax cable------------short/open/50

Use an old cheap VHF taxi radios or whatever you have.
Put the VSWR meter at the transmitter end of the long
coax cable and then mess about with the far end, by attaching
a short circuit, open circuit, 25 ohm load.
Take note of what the VSWR meter reads in all cases.

Next, move the VSWR meter to far end of cable and try this,

open circuit and note what the VSWR meter reads in all
cases and compare these readings with the first experiment.

Read again what Reg has written and what you have
Hint: you get little radiation from coax cable with a dead short
at the far end.
What did the VSWR meter tell you, when it was at the TX end
and an open circuit or dead short was on the coax at the far end?

Regards,
John Crighton
Sydney

20. ### Paul BurridgeGuest

Thanks, John. There are a lot of guys here whose 'advice' just goes in
the recycle bin, but you are *not* one of them. BTW, did you see the
post from the guy who wanted to control a "DC motor" - he could really
use that great circuit you referred me to for PWM speed control!