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UWB pulse signal has no DC? Why?
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 |
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Harry wrote:
What does it mean by "when the pulse is *differentiated* by the transmitting antenna" ? Mathematically it means to take the derivitive, or evaluate the slope of a time varying quantity. Electronically it means the antenna acts as a high pass filter - effectively blocking DC and attenuating low frequency signals. jk Cecil Moore wrote: It means that steady-state DC is incapable of generating photons. You forgot about the flashlight again, didn't ya Cecil. Photons are generated when the steady state DC source is applied to the light bulb inside. They are allowed to escape through the clear part in the front of the flashlight. :-) By the way, my nephew just graduated from UCSD with a bachelors in Photonics Engineering. He told me that the photonics engineering discipline is virtually all applied wave mechanics. ac6xg |
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. Slight correction: it is the change in the magnetic flux that gives induced EMF. You can get an induced current from a constant magnetic field if the circuit loop moves in or out of the field, changing the flux throught the loop. Tor N4OGW |
Cecil Moore wrote: I didn't realize that a UWB pulse involved flashlights but I learn strange new things from you all the time. You didn't learn that from me. What you should have learned is that DC can generate photons, contrary to your repeated assertions. ac6xg |
Jim Kelley wrote:
You forgot about the flashlight again, didn't ya Cecil. I didn't realize that a UWB pulse involved flashlights but I learn strange new things from you all the time. -- 73, Cecil http://www.qsl.net/w5dxp |
Jim Kelley wrote:
Cecil Moore wrote: I didn't realize that a UWB pulse involved flashlights but I learn strange new things from you all the time. You didn't learn that from me. What you should have learned is that DC can generate photons, contrary to your repeated assertions. My repeated assertions are about the *SUBJECT* of this thread, i.e. UWB pulses. Do I need to quote the meaning of the word, "context", for you? -- 73, Cecil http://www.qsl.net/w5dxp |
On Wed, 28 Sep 2005 20:54:39 GMT, Cecil Moore wrote:
My repeated assertions are about the *SUBJECT* of this thread, i.e. UWB pulses. Do I need to quote the meaning of the word, "context", for you? On Wed, 28 Sep 2005 14:36:28 GMT, Cecil Moore wrote: It means that steady-state DC is incapable of generating photons. We can guess which dictionary changes the meanings. |
Richard Clark wrote:
On Wed, 28 Sep 2005 14:36:28 GMT, Cecil Moore wrote: It means that steady-state DC is incapable of generating photons. We can guess which dictionary changes the meanings. I'm obviously, in context, referring to coherent photons, Richard. Would you care to explain the physics behind coherent DC photons? -- 73, Cecil http://www.qsl.net/w5dxp |
On Wed, 28 Sep 2005 23:45:13 GMT, Cecil Moore wrote:
Would you care to explain the physics behind coherent DC photons? DC gozinta Photon comesoutta |
Jim Kelley wrote:
Cecil Moore wrote: I didn't realize that a UWB pulse involved flashlights but I learn strange new things from you all the time. You didn't learn that from me. What you should have learned is that DC can generate photons, contrary to your repeated assertions. Just in case someone doesn't understand your silly humor, Jim, here's the original posting: "It is said that a transmitted UWB pulse should not have any DC because of the transmitting antenna:" Antennas emit coherent photons. As far as I know, there's no such thing as a DC coherent photon, but if I'm wrong, please enlighten us. It's obvious that I was talking about coherent photons but just for you I will repeat with the context included this time: Steady-state DC is incapable of generating coherent photons from an antenna. -- 73, Cecil http://www.qsl.net/w5dxp |
Iok so i am confused on a few things
the photo part so regardless of weither or not there is or isn't ac or dc , when i xmit (ham or other freq's) are photons being generated?? where would they be?? nearfield? around the antenna? how would i measure the presence/qty of such?? the other thing i got fuzzy on static polar field as Mr. clark mentions so why does the battery produce it /how?? are you referring to the fact that is has 'mass' and therefore.... or did i miss the obvious thanks fuzzylogic fuzzynavels fuzzypeaches |
On Thu, 29 Sep 2005 00:09:38 GMT, Cecil Moore wrote:
Steady-state DC is incapable of generating coherent photons from an antenna. My Flashlight does it every time. Maybe you need new batteries. What you DON'T say is what you are using for an antenna for those photons. It would be ridiculous to use a 20M dipole. Maybe you don't have the right antenna. |
Cecil Moore wrote:
[snip] Steady-state DC is incapable of generating coherent photons from an antenna. Cecil, Have you ever heard of a laser diode? You may well use one every day. Billions of them are made every year for CD players and many other gadgets. DC goes in; coherent photons come out. This item is actually a transducer in which electromagnetic fields in a conductor are converted to electromagnetic fields in space. Hmmmm, sounds a lot like the function of an antenna. 73, Gene W4SZ |
Richard Clark wrote:
On Thu, 29 Sep 2005 00:09:38 GMT, Cecil Moore wrote: Steady-state DC is incapable of generating coherent photons from an antenna. My Flashlight does it every time. Maybe you need new batteries. Your flashlight is NOT generating DC coherent photons. -- 73, Cecil http://www.qsl.net/w5dxp |
Gene Fuller wrote:
DC goes in; coherent photons come out. An RF antenna emits photons coherent with the RF supply frequency. An LED does not emit photons coherent with the DC supply frequency and neither do flashlights. This is what I have been trying to say. -- 73, Cecil http://www.qsl.net/w5dxp |
ml wrote:
Iok so i am confused on a few things the photo part so regardless of weither or not there is or isn't ac or dc , when i xmit (ham or other freq's) are photons being generated?? where would they be?? nearfield? around the antenna? how would i measure the presence/qty of such?? the other thing i got fuzzy on static polar field as Mr. clark mentions so why does the battery produce it /how?? are you referring to the fact that is has 'mass' and therefore.... or did i miss the obvious thanks fuzzylogic fuzzynavels fuzzypeaches Don't listen to these guys. They're changing the subject by changing the scale of the argument in order to digladiate with one another. The truth is, that if an antenna really differentiates a signal so that what it radiates is based on the rate of change of the original signal, then the DC part of the original signal won't contribute to the radiation because the rate of change of a constant(the DC part)is zero. It isn't any more complicated than that. 73, Tom Donaly, KA6RUH |
On Thu, 29 Sep 2005 03:26:28 GMT, Cecil Moore wrote:
Your flashlight is NOT generating DC coherent photons. Mine does. Like I said, yours probably needs batteries, or perhaps it is suffering a dimbulb behind the on/off switch. |
On Thu, 29 Sep 2005 03:31:58 GMT, Cecil Moore wrote:
This is what I have been trying to say. If you understood what you were saying, you would probably stop saying it. |
Richard Clark wrote:
On Thu, 29 Sep 2005 03:26:28 GMT, Cecil Moore wrote: Your flashlight is NOT generating DC coherent photons. Mine does. Like I said, yours probably needs batteries, or perhaps it is suffering a dimbulb behind the on/off switch. Assuming your flashlight does generate photons with a frequency of zero Hz, how do you detect them? -- 73, Cecil http://www.qsl.net/w5dxp |
"Cecil Moore" bravely wrote to "All" (27 Sep 05 22:14:43)
--- on the heady topic of " UWB pulse signal has no DC? Why?" CM From: Cecil Moore CM Xref: core-easynews rec.radio.amateur.antenna:217466 CM Harry wrote: Would someone please explain that for me? CM DC steady-state does not cause electrons to emit photons. CM For RF photons to be emitted from a copper wire dipole, the CM free electrons must be accelerated and decelerated. The DC CM component cannot accomplish that feat. Very true on the macro scale but even a DC potential becomes discontinuous at a microscopic level. Remember that there is a distance between atoms, electrons, molecules and are not continuous. How is it done except if not by photons, virtual or otherwise, whether the context is AC or DC? A*s*i*m*o*v .... Pandora's Law: Never open a box you didn't close yourself |
Asimov wrote:
How is it done except if not by photons, virtual or otherwise, whether the context is AC or DC? DC applied to an antenna no doubt creates virtual photons. My statements should be considered to exclude such virtual photons. They are meant to include only detectable photons coherent with the frequency of the driving signal capable of propagation through free space which are emitted by the acceleration/deceleration of free electrons in a copper wire antenna. (Contexts nowadays are sure complicated.) -- 73, Cecil, http://www.qsl.net/w5dxp |
On Thu, 29 Sep 2005 06:56:26 GMT, Cecil Moore wrote:
Assuming your flashlight does generate photons with a frequency of zero Hz, how do you detect them? You must be assuming delectable photons. |
"Cecil Moore" bravely wrote to "All" (29 Sep 05 06:56:26)
--- on the heady topic of " UWB pulse signal has no DC? Why?" CM From: Cecil Moore CM Xref: core-easynews rec.radio.amateur.antenna:217576 CM Richard Clark wrote: On Thu, 29 Sep 2005 03:26:28 GMT, Cecil Moore wrote: Your flashlight is NOT generating DC coherent photons. Mine does. Like I said, yours probably needs batteries, or perhaps it is suffering a dimbulb behind the on/off switch. CM Assuming your flashlight does generate photons with a frequency CM of zero Hz, how do you detect them? With a field mill or a moving coil of wire? A*s*i*m*o*v .... I installed a sky-light; now the folks above me are mad! |
Cecil Moore wrote:
Steady-state DC is incapable of generating coherent photons from an antenna. At last a truism, arguably. ;-) ac6xg |
Jim Kelley wrote:
Cecil Moore wrote: Steady-state DC is incapable of generating coherent photons from an antenna. At last a truism, arguably. ;-) Say Jim, what is the frequency of virtual photons? -- 73, Cecil, http://www.qsl.net/w5dxp |
Cecil Moore wrote: Say Jim, what is the frequency of virtual photons? Are you asking me what's "nu"? :-) I always say it's c over lambda. For photons it's E/h. You can be the local expert on virtual photons if you want, Cecil. I'm only on a need to know basis with them. 73, ac6xg |
Jim Kelley wrote:
You can be the local expert on virtual photons if you want, Cecil. I'm only on a need to know basis with them. I understand coherent photons emitted by free electrons being accelerated and decelerated by RF currents on a wire antenna. I don't know much about virtual photons. I was hoping that a physics guru would know. -- 73, Cecil, http://www.qsl.net/w5dxp |
Cecil Moore wrote: Jim Kelley wrote: You can be the local expert on virtual photons if you want, Cecil. I'm only on a need to know basis with them. I understand coherent photons emitted by free electrons being accelerated and decelerated by RF currents on a wire antenna. I don't know much about virtual photons. I was hoping that a physics guru would know. A physics guru might say that if it's a photon, its frequency is E/h unless one of the virtues of being 'virtual' is that its frequency may or may not be equal to E/h. Perhaps it's greatest virtue might even be that you can attribute any physical characteristic you want to it - not unlike some of the other virtual entities. ;-) 73 de ac6xg |
On Mon, 03 Oct 2005 14:27:07 -0700, Jim Kelley
wrote: Perhaps it's greatest virtue might even be that you can attribute any physical characteristic you want to it Now there's a familiar terrain. Hi Jim, Are we verging on the shadow of a photon? 73's Richard Clark, KB7QHC |
Richard Clark wrote:
Are we verging on the shadow of a photon? Only the shadow knows. It's a quantum effect. ac6xg |
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