# Simple but Fundamental Antenna Question

Discussion in 'Electronic Basics' started by JLD, Jul 22, 2004.

1. ### JLDGuest

Hello all.

I have never studied RF communications. With that said I have a
antennas (simple straight wires) like on a car.
How can an AC current exist in a wire that is connected to some
circuitry at one end but is unconnected, open, floating at the other
end? How can the current alternate back and forth in an open wire?
I understand about how alternating E fields induce alternating U
fields and vice versa, and I understand that an alternating EM field
introduced to a condcutor causes an AC in the conductor.
Take a look at this circuit (I took this link from another post):
http://www.zen22142.zen.co.uk/Circuits/rf/2bjttx.htm

From that schematic it just seems puzzling to me how an AC signal from
the amplified voice signal from the Mic can exist in the antenna in
order to produce EM waves that will transmit?
What I figured was that cap labled/valued at 4p7 "pushes and pulls"
current as a result from that LC network, but again how can a current
exist through the unconnected antenna?

tia
JLD

2. ### andyGuest

i think it's because of the combined capacitance and inductance of the
antenna - even a straight metal rod has both of these. i.e. you can push
current into one end of the wire for a short time, and the energy is
stored in the wire's electric and/or magnetic field. but someone else
probably knows (a lot) more than me on this.

or how about this - imagine it as a short ditch connected at one end to a
canal or river. you can't get any long lasting flow into it, because
there's nowhere for the water to go. but if you make waves in the river,
they will push and pull water into and out of the ditch.

3. ### Animesh MauryaGuest

Good imagination Andy.

Best regards,
AM

4. ### John JardineGuest

New to electronics, I once built a simple 27MHz radio control transmitter.
The author wrote that the transmitter should be tuned for maximum power by
looking at the light output from a bulb in series with the aerial. Not sure
I'd read it correctly, so gave it a try. It worked!. I just couldn't get my
head round why this 1watt bulb would burn so brightly using just a single
wire and no return circuit. It was obvious that all the basic text book
stuff I'd read upto that point was bollocks. (Subsequent perusal of wave
propogation theory text books, ensured I had no wish to investigate any
further. .
I found the simplest way to look at it, was that the single wire (the
antenna) has some capacitance to earth, so the ac currents have a return
path to ground and then via capacitance back to the battery in the
transmitter.
The capacitance thing allowed me to get on with my life but is useless for
explaining why a space probe can launch EM energy back to earth or even why
we see sunshine.
The EM radiation can't be mentally visualised as it's running through 4
dimensions and nobody really know what the stuff is anyway. Only a maths
description can offer any insight but then it's not really electronics
anymore.

5. ### John PopelishGuest

(snip)

You have run up against one of the limitations of viewing electric
circuits as consisting of lumped components (resistors, capacitors,
inductors, etc.), only. Since electric energy gets from place to
place as a wave, traveling at the speed of light, an understanding of
things like transmission lines, antennas and free traveling waves have
to take this property of space, matter and time into effect if it is
going to make any sense at all. A whip antenna may be visualized from
the energy's point of view (and as a mental transition from lumped
circuits) as a long chain of little inductors with small capacitors
connected, all along the line between inductors into nearby space
(essentially one plate capacitors). If the antenna were surrounded by
a tube of conductor, it would be easier to picture where the far end
of those capacitors connected, but that would be a tansmission line,
rather than anatenna.
The property that makes a whip an antenna instead of a transmission
line is that the magnetic fields of those tiny inductances and the
electric fields of those tiny capacitances do not terminate into a
conductive container, but into waves of energy that propagate away
from the antenna, always leaving the space next to the antenna
available to soak up magnetic and electric fields from the next part
of the cycle. Those waves consist of globs of electric and magnetic
fields that build each other in the direction they are moving and
extinguish each other where they have been.

6. ### John JardineGuest

What I've never really been able to rationalise, is that we know with
certainty that real energy is being radiated from the aerial yet we're told
that E=MC^2 no longer applies, as the EM 'energy' travels at light speed
therefore can't have mass. Photon thingies are then invoked. I'd feel easier
if I knew the power supply was losing mass all the time the transmitter was
switched on. (like Newtonian mechanics and powering a rocket by throwing
mass out of the back).
regards
john

7. ### John MillerGuest

So THAT's why the batteries are lighter after they're exhausted!

--
John Miller

Real computer scientists like having a computer on their desk, else how

8. ### andyGuest

i think it is, isn't it - not by losing particles, but by losing the
E=MC^2 mass-energy of their chemical energy (if it's a battery)

also, the thing photons don't have is rest-mass, but they do have a
mass-energy related to the wavelength of light they are carrying. i think,
from my somewhat flaky memory of this stuff.

9. ### Rich GriseGuest

I _really_ hope you people are joking.

10. ### andyGuest

I'm /really/ not. If you wind up a clockwork alarm clock, it will be
heavier by a tiny amount, according to relativity theory. People have
measured stuff like this.

A very rough calculation - say it takes 100 turns to wind it, the winder
is 2 cm long (1cm on each side), and the force needed at each end is the
same as a weight of 500g.

force in N = 500*10=5000N
distance = 1cm*2*pi*100=6 m
energy=5000*6=30000J

corresponding mass-energy:

E=m*c^2, so m=E/c^2 =30000/(3*10^8)^2
= 3 * 10^-13 grams.
= 0.0000000000003 grams.

11. ### Bob MyersGuest

I've tried to explain this in the past to various people, with
varying degrees of success, and here's the best I've been able
to come up with so far.

Imagine a circuit consisting of a 1V source - a battery, let's say,
and a 1 ohm resistor, connected by whatever length of wire you
find convenient and with a switch at the battery. You close the
switch, and what happens? Well, by Ohm's Law, we have
1V/1 ohms, and we see 1A of current flowing, right?

But what happens if the resistor is one mile away from the battery,
or ten miles, or a hundred miles, or...? Ignoring the resistance of
the wire for the moment, we've still got a problem, because we
know that nothing, not even electricity, travels at truly infinite
speed. It will take longer for the electricty to "get to" the resistor
if it's a hundred miles away than if it's ten feet away, right? Not
a whole lot longer, on a human time scale, but still definitely long
enough to measure. (Work it out - even things travelling at the
speed of light take about a nanosecond to go one foot, so for
100 miles...) But if that's true, then how much current is flowing
when we first close the switch? (There should be SOME - it
has to get to the resistor eventually, right? - but the battery clearly
can't just "know" that there's a 1 ohm resistor and magically send
out 1A of current right from the start, so what DOES happen?)

This is where we get the idea of "characteristic impedance." Even
before the source "sees" the final load, there IS an impedance
presented by the wire itself (even if it has no resistance at all, as we
are imagining here), and it comes from the physical configuration of
the wires - it is due to the capacitance and inductance presented by
the conductive path itself, the "distributed" capacitance and inductance
that is the same, for a given configuration, all along the path (and so
is given in terms of units like "picofarads per meter" or "microhenries
per foot" or some such). Without getting into the math of how it's
derived, there IS a "load" presented by this impedance to the source,
immediately upon the closure of the switch, and that set the voltage
and current relationship.

So now we've got the electricity going toward the resistor - what
happens when it gets there? What if the value of the resistance does
NOT match the value of the characteristic impedance of the line?
The proper resistance between voltage and current, per Ohm's Law,
MUST always be maintained, and this is what leads to "reflections"
occuring on the line. Reflected voltage and current signals are
created such that (1) the proper relationship exists at the load, and
(2) the relationship of the reflected voltage and current signals are
per the characteristic impedance of the line.

At this point, you're probably wondering what this has to do with
antennas. But now think about what happens if the source isn't
a battery (a DC source) but instead is AC - something that, in
effect, is "turning on and off" every so often, on a regular basis.
This results in a reflected AC signal at the load, if a mismatch
exists - and the combination of the "forward" and "reflected"
signals means that there will be points on the line where the
peaks coincide (and reinforce) and where the two waves
interfere with each other - this results in a pattern of maximum
and minimum amplitudes along the line, the so-called "standing
wave" pattern. Just two more steps until we're at an antenna.

Now, imagine what happens if the wires happen to be some
multiple of a quarter-wavelength long at the frequency of the source.
This means that there will be one of the peaks or minimums of this
"standing wave" out at the load end of the line. If the wave is exactly
a quarter-wave long, in fact, there's a voltage peak (and a current
minimum) right where the load is. In fact, the current can zero at
this point - if you simply remove the load resistor and leave the line
open! This means that we have a situation where the reflected wave
carries all of the energy of the forward wave.

Last step - since you've already taken off the resistor, there's no real
need to leave the conductors in parallel as you're probably envisioning
them. "Open up" this quarter-wavelength of line, so that you wind up
with a half-wavelength long structure - two conductors, leading out
from the source in opposite directions. (This change in the geometry
does change the characteristic impedance, etc., but the basic idea,
that you've got a voltage maximum out at the ends of the line, and
conversely a current maximum right at the source, still holds true.)
But now look at what we've got - you have these wires sticking "up
and down" from the source, with voltage peaks (of opposite
polarity) at either end, and a current peak in the middle. Currents
make magnetic fields, and a voltages across a space create electric
fields - and here we are with a structure that's arranged to create
these in the right relationship so as to create an EM wave. And that's
exactly what happens - the energy, rather than being reflected back
into the source, gets "launched" as an electromagnetic wave.

This is, of course, from the "transmitting" point of view, but the same
argument works in reverse for receiving. Incoming EM waves
induce voltages and currents in the antenna in the proper relationship
so as to create electrical signals that can be accessed at the center
point (so instead of the source being there, we now see a load there).
And, of course, what we've described is the classic "dipole" antenna,
but this is the same basic idea behind a simple vertical whip. In the
whip antenna, or more formally the quarter-wave vertical, only one of
the two elements is actually there, and is driven, and it "works against"
the ground plane below it. (The analysis of this structure shows that
it works as if the "missing" element were actually mirrored "in" the
ground plane.)

This is a somewhat simplified, and clearly non-mathematical, treatment
of this question, but hopefully you now at least sort of get the idea.

Bob M.

12. ### John JardineGuest

Nice description and done well without diagrams!.
I'm having fun messing with the idea and followed this spiral path ...
Energy will still radiate even if the unzipped wire ends happen to be longer
or shorter than 1/4 wave each. Energy will radiate if the wire ends are not
splayed 180degrees apart. Energy will radiate if the wire ends see some
kind of impedance. Energy will radiate at any frequemcy. The amount of
radiation is only at a max when in the classic dipole arrangement.
Leads me to consider that -any- circuit wire or PCB tracks are there just to
physically guide EM energy as it is processed by the electronics. The
energy being held in the space between the tracks and ground returns, Keep
these 'go' and 'returns' near to one another and the fields preferentially
localise themselves so little radiation. Use a ground plane and cancellation
will be near automatic. Seperate the go and return and the energy starts to
preferentially radiate away. Woe betide any tracking that allows the voltage
and currents to become physically 1/4 wave apart. I'm still musing though as
what's happening inside lumped L and C's and have a problem with the
mechanics of energy wishing to detach itself from the circuitry and then
forming a self supporting structure. (It's as if cause and effect are
failing to notice each other).
regards
john

13. ### Bob MyersGuest

Right, and I should have mentioned that. The dipole configuration I
mentioned IS just the "ideal" (at least in terms of being the optimum way,
in a fairly simpe structure, of taking an electrical signal and making EM
eaves out of it), but there's ALWAYS radiation. You get a current
anywhere, and there's a magnetic field; you have two points at
different potentials, and there's an electrical field. "Stray" EM
radiation happens all the time, often when we don't WANT it to
happen, and conversely our nice "closed" circuits are always picking up
stuff from the outside. But at least this sort of explanation, describing
the
extreme "ideal" case (which, I hasten to add, in the real world never
really exists!), will make it easier to visualize what's at work.

Again, I agree completely. We need to keep in mind that all of these
descriptions are models of what's going on. Sometimes, it's helpful to
view the transmission of energy as charges moving through conductors;
sometimes it's more helpful to view it as EM waves or fields being
guided by that same set of conductors. "Reality" is both, and neither
- you use the model that gives you the most useful set of answers in
the situation you're analyzing at the moment. But in electronics (and
for that matter, physics in general) we always need to realize that these
things we do the math with ARE just models, and they're never QUITE
right. There is a host of classic puzzles that, in posing what appears to
be a paradox or contradiction, really just show the limits of a given way
of looking at things.

Bob M.

14. ### andyGuest

Have you heard of circular species? On the west coast of europe, there are
two kinds of gulls - can't remember the names but call them black-headed
and red crested gulls. if you follow the population of blackheaded gulls
eastwards across russia, they gradually start looking more like red
crested gulls, and vice versa the other way if you go west across the
atlantic. Where they meet in the middle, they can interbreed, so are
considered as one species, but on the european coast, they can't so are
two species. i just like things like that.

16. ### L. FiarGuest

It is not really a single wire, the antenna usually has two parts.

For a dipole, this is clearly the two sections. For the straight antenna, it
is a ground. That ground may be rods which stick out the base, a car body,
the Earth... or even just the -ve track on the radio circuit board.
An exception would be a loop antenna, but I am sure you can see how current
can flow when it is one loop of wire.

Consider capacitance between the vertical section and the ground part
of the antenna - and you have a complete circuit for AC.
You have to remember that the voice signals are not transmitted as such,
they are hooked on to a radio frequency signal... and that is transmitted.

For a little about antennas, try these:
http://www.hottconsultants.com/pdf_files/dipoles-1.pdf
http://www.hottconsultants.com/pdf_files/dipoles-2.PDF
http://www.hottconsultants.com/pdf_files/dipoles-3.PDF

Best regards.
LF.

17. ### JLDGuest

away with more understanding. I have printed out several of your posts
for reference and I know that when I go back to school this fall
(junior EE student) I will be better prepared to ask questions.

JLD

18. ### Bob MyersGuest

OK - just one follow-up. As a junior in an EE program, you should
be at the point of having one or more classes in the fundamentals of
transmission line effects, EM wave propagation, and antennas. A good
grasp on the basics of these - especially, to start with, just what makes
"transmission lines" the rather complicated subject that they are - is
really essential to a truly good understanding of just about every aspect
of electronics. In short, pay close attention..

Bob M.

19. ### Ian BellGuest

I know junior year is a USA expression but as a Brit I am not sure which
year of a degree course it refers to. In the UK most degrees run for three
years. The first year of all engineering degrees is very general with only
a few modules on electronics for an EE course. Maxwell's equations and all
that is derived therefrom is not introduced until the second year.

But, whenever they occur, pay close attention is definitely the thing.

IAn

Ian

20. ### Bob MyersGuest

In the U.S., the "junior" year is most commonly the third year
of a four-year program, the years being (in order) freshman,
sophomore, junior, senior. I first ran up against Maxwell's
equations in a physics course in my sophomore year (this was,
of course, back in the days when we were studying EM
transmission by candlelight...), then had my first "transmission
lines and wave basics" in the EE course of study as a junior,
followed by a course which concentrated primarily on antenna
theory and design as a senior.

Bob M.