What does dbi mean, in context of antenna performance?
Frank Robey
>
>The textbooks I have pretty well steer away from dBd, but I've never
>heard anyone (before now) state that it was anything but 2.15 dBi.
>This is the gain of a dipole in free space.
>
I also think that dBi is preferable over dBd. The problem is that
there are unstated assumptions when the gain of a dipole is assumed
to be 2.15 dBi.
There are many types of dipoles, and 2.15 dBi is the free-space gain of a
half wave-length zero-diameter zero-wire-resistance dipole. In fact, in
many antenna books, (e.g. Antenna Theory, Balanis) the first dipole that is
analyzed is the infinitesimal dipole with a gain of 1.76 dBi. There is
also the magnetic dipole with a gain of 1.76 dBi.
If the length of a half wave-length (zero ...) dipole is lengthened, then the
directivity increases. For instance, a dipole that is 1.25 wavelengths
long will have a free-space gain of about 5 dBi.
Anyway, this is just one more reason why I prefer dBi over some dBd
with an unstated gain reference.
Frank Robey V73FR/N1PKT Kwajalein Systems Engineering
Kwajalein, RMI MIT Lincoln Laboratory
Latitude: 8 deg 43' 24", Longitude: 167 deg 44' 12" E
P.O. Box 224
APO AP 96555
(617) 981-0465 (shared)
fax: (617) 981-2473
Alan Bloom
: From: ro...@tekgp4.cse.tek.com (Roy W Lewallen)
: Date: 27 Jan 1995 21:47:45 GMT
: Message-Id: <3gbpm1$m...@tekadm1.cse.tek.com>
: 7ygb/la...@sr.hp.com (Alan Bloom):
: >"dBd" means "gain relative to a dipole in free space" only if the antenna
: >under test is also in free space.
: ...Likewise, if I know that dBd is gain relative to a dipole in free space,
: I can calculate the same information.
Let's say an antenna manufacturer determines (through measurement and
calculation) the free-space gain of his antenna over a dipole in free
space. So you're saying he can in good conscience add 6 dB to that
number and advertise that as the "gain of his antenna"?
After all, that's the "gain" you'd get if you mounted the antenna at
a great height over an infinite, perfectly-conducting ground plane.
If he wants to be a little more honest, he could only add 4 dB, since
that's a typical number you'd get at a reasonable height over a good
real ground.
I'll say it one last time: It makes no sense to quote "antenna gain"
with the antenna under test over earth and the reference antenna
in free space.
AL N1AL
David Stockton
******************** What is antenna gain? **************************
I think that there are serious flaws in the concept of antenna gain.
I've had the job of explaining antennae to a variety of people, including
non-mathematical beginners. It seems far simpler to go right back to
absolute first principles, and start again. In explaining to a beginner,
you cannot have assumptions, you have to be able to justify everything.
Let's forget the word "Gain" for a while. Let's use the words Energy
and Power in their proper meanings.
Transmission:
=============
We can break the performance of a transmitting antenna down into a few
simple stages:
1) Matching - what fraction of the power reaching the antenna is NOT
reflected back down the feeder?
2) Efficiency - what fraction of the power that made it past the
matching factor, above, gets turned into radio waves? (The rest turns
into heat)
3) Directivity - A complicated 3-dimensional expression of how the
power that got this far is distributed over different directions. We
could express this as power fraction density (power fraction per steradian)
Simple and easily visualised, n'est-ce pas?
It is nice and easy to put this in terms of "How much more (or less)
power would we need to put into a perfect 3-D omnidirectional antenna
(an isotropic radiator) to get the same power density IN THE TARGET
DIRECTION ?" This just happens to be the same as is normally meant by
antenna gain compared to an isotropic radiator (dBi).
Reception
=========
Think of a bit of free space, where we will later place an antenna.
Let the place be far enough from the source of the transmission that the
electric and magnetic fields have settled down into their "mature"
relationship of 377 volts/metre of electric field for every Amp per
metre of magnetic field. (Note, 377 (V/m)/(A/m) cancels out the metres
to 377 V/A in other words, 377 Ohms - this is the easy way to
understand "the impedence of free space")
If we multiply the electric field by the magnetic field, we get
Volts * Amps per square metre which is watts per square metre.
Even if you don't like maths tricks, it's easy to think of the
incoming signal as a power density of so many watts per square metre in
this place.
When we put our antenna in this place, we know the power density it
is exposed to, we can measure the power of its electrical output. We can
say that the electrical power represents a collection of the power from
so many square metres. If we want, we can choose to partition this
factor into terms describing the true capture area, efficiency losses
due to the resistance of conductors etc, and reflections at the feeder
connection sending some power back into space, or we can just pick an
equivalent area that include all these effects.
It is easy to think of an antenna in terms of effective capture area,
like some form of solar cell. If our antenna is directional, then it
presents different effective capture areas to signals approaching from
different directions. Easy.
True isotropic radiators are non-reciprocal, they have no defined gain
factor on receive. On transmit, they behave as an omnidirectional
source. Feed them 1 Watt and they radiate exactly one Watt, spread
evenly over ALL directions. The strength of signal radiated is
independant all physical dimensions of this (fictional) antenna. On
receive, if we cannot tie down the size of an antenna, how can we assign
a capture area? If it were possible for an antenna to have a capture area
independant of its physical dimensions, then the door is open to small
antennae with infinite capture area sucking in all the power of the
source, yet leaving space for there to be a second such antenna nearby.
This paradox is a good indication that capture area and physical size
cannot be completely independant.
If we choose to use, say, a halfwave dipole as a reference antenna,
that does have a calculable effective capture area, then we must
remember that halfwave dipoles are different lengths for different
frequencies, and so the capture areas are different.
One of the biggest mistakes in radio engineering was to take a dipole
gain on receive as being a constant, independant of frequency, and then
fudging our way out of the repercussions by assigning a frequency
dependance term to path loss. Paths are most definitely NOT frequency
dependant, it's just radiation spreading out radially from the source
(hence its very name) and hence becoming less dense.
*****************
As far as ground effects go, for real world antennae, there are so
many different combinations of antenna height, ground conductivity (and
varied patterns of ground conductivity) (and seasonal variations of
ground conductivity), nearby objects, and differences in the sensitivity
of different antennae to ground and objects, that there are no truely
useful measurements that will relate closely enough to my station's
conditions that I can rely on them to any extent. At least comparing
antenna X in free space to antenna Y in free space gives verifyable
results. To use these results, I have to consider my own conditions. If
I'd been given results relating to someone else's conditions, I'd have
to know those conditions and try to account for differences. Free space
results aren't perfect, nor are necessarily relevant, but at least they
form a uniform starting point.
Cheers
David GM4ZNX
Alan Bloom
I agreed with everything you said except:
: True isotropic radiators are non-reciprocal, they have no defined gain
: factor on receive. On transmit, they behave as an omnidirectional
: source.
Which means they have an aperture of 1/(4*PI) square wavelengths.
: ... If it were possible for an antenna to have a capture area
: independant of its physical dimensions, then the door is open to small
: antennae with infinite capture area ...
Consider a small lossless 1-meter-diameter loop. It has a physical area
of less than 1 square meter, but its aperture is around 700 square meters
at 3.5 MHz. (Even a half-wave dipole has an aperture much larger than the
cross-sectional area of the wire.) There is no strong correlation between
physical size and capture area. The ratio of aperture to physical size
can indeed theoretically be infinite.
: As far as ground effects go, for real world antennae, there are so
: many different combinations of antenna height, ground conductivity
: ... nearby objects, and differences in the sensitivity
: of different antennae to ground and objects, that there are no truely
: useful measurements that will relate closely enough to my station's
: conditions that I can rely on them to any extent. At least comparing
: antenna X in free space to antenna Y in free space gives verifyable
: results.
I agree completely. Free-space gain is the most useful figure of merit.
And if someone insists on measuring antenna gain over earth, then the
reference antenna (dipole or isotropic) must be over the same earth.
AL N1AL
w7...@teleport.com
> I agree completely. Free-space gain is the most useful figure of merit.
> And if someone insists on measuring antenna gain over earth, then the
> reference antenna (dipole or isotropic) must be over the same earth.
>
> AL N1AL
>
I don't want to beat this subject to death, but I do have a problem with isotropic antennas over
real ground. The effect of real ground on reflections is strongly affected by the polarization of
the signal. What's the polarization of an isotropic source? If we don't know this, we don't know
how its radiation would react when striking real ground.
I've never seen a textbook which even considers an isotropic source in an environment other than
free space. Can you provide a reference, please? I'd like to know how the author handles the
polarization problem.
73,
Roy Lewallen, W7EL
w7...@teleport.com
w7...@teleport.com
>. . . .
> I agree completely. Free-space gain is the most useful figure of merit.
> And if someone insists on measuring antenna gain over earth, then the
> reference antenna (dipole or isotropic) must be over the same earth.
>
> AL N1AL
>
>
I don't want to beat this subject to death, but I have a problem with isotropic antennas over
ground. Aside from the fact that dBi is universally (in all texts and papers I've ever seen) as
gain over an isotropic source in free space, there's a problem with polarization. The effect of
real (i.e., not perfectly-conducting) ground on an incident wave is strongly determined by the
polarization of the wave. I don't know how to figure out the polarization of an isotropic source,
so can't guess what its gain or pattern would be over real ground.
Would you please provide a reference where an isotropic source over real ground is mentioned? I'd
like to see how the author handles the polarization problem.
Alan Bloom
: >
: > I agree completely. Free-space gain is the most useful figure of merit.
: > And if someone insists on measuring antenna gain over earth, then the
: > reference antenna (dipole or isotropic) must be over the same earth.
: >
: > AL N1AL
: I don't want to beat this subject to death, but
: I do have a problem with isotropic antennas over
: real ground. The effect of real ground on reflections
: is strongly affected by the polarization of
: the signal. What's the polarization of an isotropic source? ...
The same as the antenna being measured.
However, you raise an interesting point. Many antennas have different
polarizations in different directions. For example, a helix is circularly
polarized on-axis but is linearly polarized in a perpendicular direction.
That's another good reason to use free-space gain as the proper figure of
merit for antennas. In general, the "gain" of an antenna over earth
depends too much on local conditions, even when using a reference antenna
over the same earth. And when using a reference antenna in free space,
the phrase "gain of an antenna over earth" becomes almost meaningless.
AL N1AL
Alan Bloom
Gary Coffman (ga...@ke4zv.atl.ga.us) wrote:
: Since the E and H fields are 90 degrees out of phase in the EM wave,
: the power factor is zero,
No, Gary. The E and H fields are in phase. You are confused by the fact
that the E and H fields are perpendicular -- that is, oriented 90 degrees
from each other in *position*.
: The capture area of an isotropic antenna is theoretically lambda/(4*pi).
Actually lambda^2 / (4*pi)
AL N1AL
Gary Coffman
> OK, so now for the major-league flamebait:
>
> ******************** What is antenna gain? **************************
> I think that there are serious flaws in the concept of antenna gain.
>I've had the job of explaining antennae to a variety of people, including
>non-mathematical beginners. It seems far simpler to go right back to
>absolute first principles, and start again. In explaining to a beginner,
>you cannot have assumptions, you have to be able to justify everything.
>
> Let's forget the word "Gain" for a while. Let's use the words Energy
>and Power in their proper meanings.
Yes, lets do, see below.
> Reception
> =========
> Think of a bit of free space, where we will later place an antenna.
>Let the place be far enough from the source of the transmission that the
>electric and magnetic fields have settled down into their "mature"
>relationship of 377 volts/metre of electric field for every Amp per
>metre of magnetic field. (Note, 377 (V/m)/(A/m) cancels out the metres
>to 377 V/A in other words, 377 Ohms - this is the easy way to
>understand "the impedence of free space")
> If we multiply the electric field by the magnetic field, we get
>Volts * Amps per square metre which is watts per square metre.
Since the E and H fields are 90 degrees out of phase in the EM wave,
the power factor is zero, and we're talking about volt*amperes of
*energy*, not power. If we were talking about power, the 377 ohm
"resistor" would be getting hot. What we've got is wattless-watts
per square meter. Of interest here mainly because one of the functions
of an antenna is to bring currents and voltages back in phase at its
feedpoint. Picture the dipole, with voltage maxima at it's tips matched
with current minima, and a current maximum at it's center, matched with
a voltage minimum.
> Even if you don't like maths tricks, it's easy to think of the
>incoming signal as a power density of so many watts per square metre in
>this place.
>
> When we put our antenna in this place, we know the power density it
>is exposed to, we can measure the power of its electrical output. We can
>say that the electrical power represents a collection of the power from
>so many square metres. If we want, we can choose to partition this
>factor into terms describing the true capture area, efficiency losses
>due to the resistance of conductors etc, and reflections at the feeder
>connection sending some power back into space, or we can just pick an
>equivalent area that include all these effects.
> True isotropic radiators are non-reciprocal, they have no defined gain
>factor on receive. On transmit, they behave as an omnidirectional
>source. Feed them 1 Watt and they radiate exactly one Watt, spread
>evenly over ALL directions. The strength of signal radiated is
>independant all physical dimensions of this (fictional) antenna. On
>receive, if we cannot tie down the size of an antenna, how can we assign
>a capture area? If it were possible for an antenna to have a capture area
>independant of its physical dimensions, then the door is open to small
>antennae with infinite capture area sucking in all the power of the
>source, yet leaving space for there to be a second such antenna nearby.
>This paradox is a good indication that capture area and physical size
>cannot be completely independant.
This is where you go astray. Now granted isotropic antennas are
purely theoretical, both in transmit and receive, however your
"paradox" is flawed. Given two antennas with infinite capture
area, their capture areas must *completely* overlap. Since the
power to be captured is finite, it must divide equally between
the two isotropic antennas so that each receives only one half
of the power. There is no violation of fundamental laws involved.
The capture area of an isotropic antenna is theoretically lambda/(4*pi).
As wavelength decreases, so does capture area. This fits with what
you say below.
> If we choose to use, say, a halfwave dipole as a reference antenna,
>that does have a calculable effective capture area, then we must
>remember that halfwave dipoles are different lengths for different
>frequencies, and so the capture areas are different.
Note, however, that different antenna forms can have similar
capture areas while being grossly different in physical form.
I need only mention ferrite rod loaded AM broadcast receiving
antennas as an example. They have a much larger capture area
than their physical size alone would imply. That's because
the ferrite alters the permitivity of the surrounding space,
concentrating the EM field for the coil.
> One of the biggest mistakes in radio engineering was to take a dipole
>gain on receive as being a constant, independant of frequency, and then
>fudging our way out of the repercussions by assigning a frequency
>dependance term to path loss. Paths are most definitely NOT frequency
>dependant, it's just radiation spreading out radially from the source
>(hence its very name) and hence becoming less dense.
It's not a mistake so much as it's merely another way of dividing the
calculation load.
Gary
--
Gary Coffman KE4ZV | You make it, | gatech!wa4mei!ke4zv!gary
Destructive Testing Systems | we break it. | emory!kd4nc!ke4zv!gary
534 Shannon Way | Guaranteed! | ga...@ke4zv.atl.ga.us
Lawrenceville, GA 30244 | |
Gary Coffman
>I hate to get involved in this, but what the heck...
>
>Gary Coffman (ga...@ke4zv.atl.ga.us) wrote:
>
>: Since the E and H fields are 90 degrees out of phase in the EM wave,
>: the power factor is zero,
>
>No, Gary. The E and H fields are in phase. You are confused by the fact
>that the E and H fields are perpendicular -- that is, oriented 90 degrees
>from each other in *position*.
Are you saying that the E field and the H field both are at a maximum
in a single plane perpendicular to the propagating direction of the
wave at the same time?
As far as I know, the E field and H field maxima are 90 degrees
separated in time with respect to any plane perpendicular to the
direction of propagation. This has nothing to do with their being
oriented at right angles to the direction of propagation. The
energy in an EM field oscillates back and forth between an E field
and a H field, the energy isn't in both at the same time at the
same point.
Alan Bloom
: In article <3hhpns$k...@canyon.sr.hp.com> al...@sr.hp.com (Alan Bloom) writes:
: >I hate to get involved in this, but what the heck...
: >
: >Gary Coffman (ga...@ke4zv.atl.ga.us) wrote:
: >
: >: Since the E and H fields are 90 degrees out of phase in the EM wave,
: >: the power factor is zero,
: >
: >No, Gary. The E and H fields are in phase. You are confused by the fact
: >that the E and H fields are perpendicular -- that is, oriented 90 degrees
: >from each other in *position*.
: Are you saying that the E field and the H field both are at a maximum
: in a single plane perpendicular to the propagating direction of the
: wave at the same time?
Yes.
: As far as I know, the E field and H field maxima are 90 degrees
: separated in time with respect to any plane perpendicular to the
: direction of propagation. This has nothing to do with their being
: oriented at right angles to the direction of propagation.
Nope, they both are in phase, meaning they both reach their maximum
at the same time. It's their direction that is 90 degrees out.
The E and H fields are perpendicular (90 degrees) from each other
and also from the direction of propagation.
: The
: energy in an EM field oscillates back and forth between an E field
: and a H field, the energy isn't in both at the same time at the
: same point.
The above statement is true for a tuned circuit, but not for an EM wave.
Think of it this way: In the transmitting antenna, the current and
voltage are in time phase, if the antenna is resonant. (Again they are
out of phase in position -- the current is maximum at the center of a
half-wave dipole, the voltage is maximum at the ends.) You can think of
the E field as being caused by the voltage and the H field as being
caused by the current, which implies that E and H are also in time phase.
An analagous argument applies to the receiving antenna.
AL N1AL
Anthony R. Gold
> Gary Coffman (ga...@ke4zv.atl.ga.us) wrote:
> : Are you saying that the E field and the H field both are at a maximum
> : in a single plane perpendicular to the propagating direction of the
^^^^^^
> : wave at the same time?
>
> Yes.
> The E and H fields are perpendicular (90 degrees) from each other
> and also from the direction of propagation.
:-(
--
Tony - G3SKR / AA2PM / tg...@microvst.demon.co.uk
David Stockton
: Since the E and H fields are 90 degrees out of phase in the EM wave,
: the power factor is zero, and we're talking about volt*amperes of
: *energy*, not power. If we were talking about power, the 377 ohm
: "resistor" would be getting hot. What we've got is wattless-watts
: per square meter. Of interest here mainly because one of the functions
: of an antenna is to bring currents and voltages back in phase at its
: feedpoint. Picture the dipole, with voltage maxima at it's tips matched
: with current minima, and a current maximum at it's center, matched with
: a voltage minimum.
************** HALT ! *****************
The planes of polarisation of the E and H fields are at 90 degrees to
each other, BUT THE FIELDS ARE IN PHASE. At any point in space, Peak H
occurs at exactly the same time as peak E fields. the vectors of the
fields are orthogonal, but the little blighters are exactly in
synchronism. The "impedance of free space" does not exist as a normal
physical thing that you can stub your toe on, but it is purely real in
value, it has a power factor of 1 it is not reactive. It is just like a
lump of coax cable, it presents say 50 ohms, purely resistive to its
source immediately. Only after a time delay for any signal reflected at
its far end to make the round-trip can the coax present any other Z.
Anything that makes electrical energy go away can look like resistive
Z. It doesn't matter where "away" is. In a resistor it gets turned to
heat. Space and transmission lines move it to a different place. To
create reactive impedences, there has to be some reflection somewhere.
We are talking of a power flow density of very real watts per square
metre. If I stick a sheet of resistive material whose resistivity
matches free space, in the signal, it will absorb all the incident
radiation, and will receive the appropriate number of watts
corresponding to the energy density and its area.
: This is where you go astray. Now granted isotropic antennas are
: purely theoretical, both in transmit and receive, however your
: "paradox" is flawed. Given two antennas with infinite capture
: area, their capture areas must *completely* overlap. Since the
: power to be captured is finite, it must divide equally between
: the two isotropic antennas so that each receives only one half
: of the power. There is no violation of fundamental laws involved.
Each antenna must have a capture area that completely surrounds
the source to collect all possible radiation. Two such antennae must
overlap completely, as you say, but if both are perfectly efficient, the
inner one must grab all the power, leaving none for the other. There is
certainly a paradox.
: The capture area of an isotropic antenna is theoretically lambda/(4*pi).
: As wavelength decreases, so does capture area. This fits with what
: you say below.
No, an isotropic antenna of 100% efficiency has no unique value of
capture area, it can be anything you want without the antenna stopping
being either unidirectional or 100% efficient. The equation you've
quoted is just a bodge to CREATE reciprocity. It assigns a value to an
otherwise indeterminate parameter, and the value it assigns gives it
2.18 dB less capture area than a halfwave dipole. Choosing such a value
is neither more nor less legitimate than choosing any other value for
it.
: Note, however, that different antenna forms can have similar
: capture areas while being grossly different in physical form.
: I need only mention ferrite rod loaded AM broadcast receiving
: antennas as an example. They have a much larger capture area
: than their physical size alone would imply. That's because
: the ferrite alters the permitivity of the surrounding space,
: concentrating the EM field for the coil.
True! if we ever get HF micro-loops in room temp (or cold beer
temp?) superconductor we'll have a nice example of this taken to the
limit.
: > One of the biggest mistakes in radio engineering was to take a dipole
: >gain on receive as being a constant, independant of frequency, and then
: >fudging our way out of the repercussions by assigning a frequency
: >dependance term to path loss. Paths are most definitely NOT frequency
: >dependant, it's just radiation spreading out radially from the source
: >(hence its very name) and hence becoming less dense.
: It's not a mistake so much as it's merely another way of dividing the
: calculation load.
True! My objection to it is that if factors had been fairly
assigned to their appropriate objects, life would be less confusing. As
it is it smacks of accountancy and tax law :-(
We're not really arguing, are we?
David GM4ZNX
Gary Coffman
>
>Think of it this way: In the transmitting antenna, the current and
>voltage are in time phase, if the antenna is resonant. (Again they are
>out of phase in position -- the current is maximum at the center of a
>half-wave dipole, the voltage is maximum at the ends.) You can think of
>the E field as being caused by the voltage and the H field as being
>caused by the current, which implies that E and H are also in time phase.
>
>An analagous argument applies to the receiving antenna.
Okay, Al, I'll accept that the currents and voltages on the antenna
are in phase in *time*, but they aren't in phase in *space*. To sum,
they have to be at the *same point*. To get there on the dipole, one
or the other must move 1/4-wave, and that takes 90 degrees of *time*.
So now they are in phase in space, but out of phase in time. To sum
to become power, they have to be in phase in space *and* time
simultaneously. Space and time, two sides of the same coin. Power
factor zero.
It would be shocking if this were not the case. The *load* presented
by the receiver absorbs the power, not the antenna.