It is said that a transmitted UWB pulse should not have any DC because
of the transmitting antenna:



"Without getting into the details of the physical generation of UWB
waveforms, it is sufficient to note in this regard that the
transmitting antenna has the general effect of differentiating the time
waveform presented to it. As a consequence the transmitted pulse does
not have a DC (direct current) value-the integral of the waveform
over its duration must equal zero."

(page 4,
http://www.antd.nist.gov/wctg/manet/...rt_April03.pdf)

Would someone please explain that for me?

Thanks!

-- Harry

Harry wrote:
Would someone please explain that for me?

DC steady-state does not cause electrons to emit photons.
For RF photons to be emitted from a copper wire dipole, the
free electrons must be accelerated and decelerated. The DC
component cannot accomplish that feat.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:

Harry wrote:

Would someone please explain that for me?


DC steady-state does not cause electrons to emit photons.

Except in flashlights, apparently.

For RF photons to be emitted from a copper wire dipole, the
free electrons must be accelerated and decelerated. The DC
component cannot accomplish that feat.

Quantum mechanics is completely unnecessary here, Cecil. I think
Faraday still provides the best explanation.

ac6xg

On 27 Sep 2005 14:44:43 -0700, "Harry" wrote:

Would someone please explain that for me?

Hi Harry,

Because the media, space, does not have DC continuity, but acts rather
more like an infinite succession of inductances and capacitances that
transforms the original waveform (a) into (b) whose DC average is zip.
It should be said that this is not due strictly to the antenna, but
all paths within the complete communication circuit.

And to anticipate the yahoos who interject that yes there is an R
component, its contribution (short of the æther turning into a plasma)
is of no consequence to the outcome you observe.

This is an answer, but it may fall quite short of an explanation.

73's
Richard Clark, KB7QHC

Jim Kelley wrote:
Cecil Moore wrote:

Harry wrote:

Would someone please explain that for me?


DC steady-state does not cause electrons to emit photons.

Except in flashlights, apparently.

For RF photons to be emitted from a copper wire dipole, the
free electrons must be accelerated and decelerated. The DC
component cannot accomplish that feat.

Quantum mechanics is completely unnecessary here, Cecil. I think
Faraday still provides the best explanation.

ac6xg

With a ns pulse into a zero resistance conductor in free space all
energy should be radiated, so no DC component if you look at the input
power. However if you introduce the same pulse to a circuit that has
magnetic properties, i.e. a ferrite inductor as an example, there is
energy used to align the molecules in the material, not radiated, which
results in a DC component. In a flashlight, most of the energy is
heat, which will result in a big DC component with no radiation rf
wise.
That's really advanced stuff for a Ham Radio Newsgroup.
Gary N4AST

Jim Kelley wrote:
Quantum mechanics is completely unnecessary here, Cecil. I think
Faraday still provides the best explanation.

In this particular case, quantum mechanics offers the
easiest-to-understand explanation. As Feynman put it:
"So now, I present to you the three basic actions, from
which all the phenomena of light (including RF) and
electrons arise.
-Action #1: A photon goes from place to place.
-Action #2: An electron goes from place to place.
-Action #3: An electron emits or absorbs a photon."

QED^2 :-)
--
73, Cecil http://www.qsl.net/w5dxp

Harry,

Forget all the nonsense about photons, continuity of space, and other
blather. This is simply a matter of mathematics.

In general the initial waveform will consist of a constant DC level and
a number of AC components. Any DC component in the original pulse is
lost when the pulse is differentiated by the transmitting antenna,
leaving only the differentiated AC components. Therefore the integral of
the differentiated waveform has a zero DC value. Of course one could
always add in a constant to the integrated result, but that would not
have much physical meaning in the context under discussion.

73,
Gene
W4SZ

Harry wrote:
It is said that a transmitted UWB pulse should not have any DC because
of the transmitting antenna:



"Without getting into the details of the physical generation of UWB
waveforms, it is sufficient to note in this regard that the
transmitting antenna has the general effect of differentiating the time
waveform presented to it. As a consequence the transmitted pulse does
not have a DC (direct current) value-the integral of the waveform
over its duration must equal zero."

(page 4,
http://www.antd.nist.gov/wctg/manet/...rt_April03.pdf)

Would someone please explain that for me?

Thanks!

-- Harry

What does it mean by "when the pulse is *differentiated* by the
transmitting antenna" ?

Can I say that because an antenna can be modeled as a series of
inductors and capacitors like a transmission line, the DC part of the
pulse won't radiate from the antenna?

-- Harry


Gene Fuller wrote:
Harry,

Forget all the nonsense about photons, continuity of space, and other
blather. This is simply a matter of mathematics.

In general the initial waveform will consist of a constant DC level and
a number of AC components. Any DC component in the original pulse is
lost when the pulse is differentiated by the transmitting antenna,
leaving only the differentiated AC components. Therefore the integral of
the differentiated waveform has a zero DC value. Of course one could
always add in a constant to the integrated result, but that would not
have much physical meaning in the context under discussion.

73,
Gene
W4SZ

Harry wrote:
It is said that a transmitted UWB pulse should not have any DC because
of the transmitting antenna:



"Without getting into the details of the physical generation of UWB
waveforms, it is sufficient to note in this regard that the
transmitting antenna has the general effect of differentiating the time
waveform presented to it. As a consequence the transmitted pulse does
not have a DC (direct current) value-the integral of the waveform
over its duration must equal zero."

(page 4,
http://www.antd.nist.gov/wctg/manet/...rt_April03.pdf)

Would someone please explain that for me?

Thanks!

-- Harry

Harry wrote:
What does it mean by "when the pulse is *differentiated* by the
transmitting antenna" ?

It means that steady-state DC is incapable of generating photons.
--
73, Cecil http://www.qsl.net/w5dxp

Jim Kelley, AC6XG wrote:
"I think Farqaday stilll provides the best explanation."

Yes. Faraday showed d-c had nothing to do with wireless. It was the a-c
motivating wireless electrical coupling.

An Englishman, Faraday, constructed a transformer in 1831. He used two
isolateed coils of wire wound on the same spool. He connected a
galvanometer across one coil. He noticed a brief deflection of the
galvanometer each time he connected or disconnected a battery to the
second coil.

Joseph Henry, a professor at Albany Academy in New York was
independently making the same observations at the same time as Faraday.
When Henry got news of Faraday`s discoveries, he made no effort to claim
credit for his own work but often referred to Faraday`s discovery.

It was the change in the magnetic field which induced electricity
without a direct connection, not the value of the steady magnetic field
itself.

Same with antennas. A battery connected to an antenna sends no signal. A
varying field is required to produce a signal in an antenna.

Best regards, Richard Harrison, KB5WZI