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Richard Clark wrote:
This is an amusing irony of where you have found power where you have always posited there is none, and where you have rejected there is power when it has been shown to exist. Each example exposes your lack of experience in the scale of the error and its relation to the equipment measuring it. In exactly the same way that forward power can exceed generated power in a 1/4WL matching section of transmission line, so can the forward irradiance in 1/4WL of thin film exceed the forward irradiance in the air before incidence. That you are still so terribly confused about such a simple fact of light physics is sad. You obviously don't understand the information on the Melles-Groit web page. http://www.mellesgriot.com/products/optics/oc_2_1.htm Please study it until you understand it. -- 73, Cecil http://www.qsl.net/w5dxp |
Richard Clark wrote:
As the last "thin-film" comment reveals, it has always been about how one mind can encompass two contradictory positions (total cancellation - non total cancellation) about the same mechanism (a quarterwave matching section). In other words: A Troll. If the incident irradiance is a single frequency coherent signal, the requirement for TOTAL CANCELLATION OF REFLECTIONS is still that the index of refraction of the 1/4WL thin-film be the square root of the medium upon which it is deposited. QED -- 73, Cecil http://www.qsl.net/w5dxp |
On Fri, 14 Oct 2005 16:22:39 GMT, Cecil Moore wrote:
It was previously reported that the path through the Bird is 40mm. The path through the Bird is actually about five Cecil, If you are referring to my statement in another thread, it was that the sampling element is about 40mm inside the 50 ohm Thruline section (ie that there is about 40mm of 50 ohm transmission line between the Bird 43 terminals and the sampling element), not as you have stated above. Owen -- |
Jim Kelley wrote:
I think it might also be interesting to discuss the instance in which the Bird is interfaced with a real halfwave section. If the Bird is inserted at a point where the net voltage divided by the net current is equal to 50, apparently a 50 ohm Z0-match is achieved at that point and any length of lossless 50 ohm coax can be inserted without altering the forward/reflected conditions in the adjacent transmission lines. -- 73, Cecil http://www.qsl.net/w5dxp |
Owen Duffy wrote:
On Fri, 14 Oct 2005 16:22:39 GMT, Cecil Moore wrote: It was previously reported that the path through the Bird is 40mm. The path through the Bird is actually about five If you are referring to my statement in another thread, it was that the sampling element is about 40mm inside the 50 ohm Thruline section (ie that there is about 40mm of 50 ohm transmission line between the Bird 43 terminals and the sampling element), not as you have stated above. I'm sorry, Owen, I took your statement to mean that the total length of path through the Bird is 40mm (1.5") which is not enough length to force a 50 ohm environment. I downloaded the Bird manual this morning and discovered that, unlike an MFJ, the path through the Bird is about 5 inches and coaxial. I apologize to Bird and you Bird experts for that bad assumption. -- 73, Cecil http://www.qsl.net/w5dxp |
That explains why my new tuner failed to work. I have a rotary switch that
selects 40mm of 50, 75,300, 450, and 600 ohm coax. I thought it should match anything! Now I see I have to go to 5" lengths. I should know better than to build anything before all the errata sheets are in. I can use some additional diversion as almost all the soldering iron burns have healed. "Cecil Moore" wrote in message m... Owen Duffy wrote: On Fri, 14 Oct 2005 16:22:39 GMT, Cecil Moore wrote: It was previously reported that the path through the Bird is 40mm. The path through the Bird is actually about five If you are referring to my statement in another thread, it was that the sampling element is about 40mm inside the 50 ohm Thruline section (ie that there is about 40mm of 50 ohm transmission line between the Bird 43 terminals and the sampling element), not as you have stated above. I'm sorry, Owen, I took your statement to mean that the total length of path through the Bird is 40mm (1.5") which is not enough length to force a 50 ohm environment. I downloaded the Bird manual this morning and discovered that, unlike an MFJ, the path through the Bird is about 5 inches and coaxial. I apologize to Bird and you Bird experts for that bad assumption. -- 73, Cecil http://www.qsl.net/w5dxp |
On Wed, 12 Oct 2005 22:09:51 GMT, Owen Duffy wrote:
The myth: Measurements with a Bird 43 of the conditions on the Thruline section are invalid unless it has some minimum length of 50 ohm line on both sides of itself. All, Can I offer the suggestion that the key to understanding why this is so, it to understand the sampler. I don't have the detail of the physical and electrical parameters of the Bird sampler, however I suspect that like so many other reflectometers, it comes down to a device that samples independently the net current and the net voltage associated with any travelling waves, and those RF samples are in proportion and phase relationship that when algebraicly added and rectified, they produce a DC voltage that is proportional to the power flow in one direction only (provided that Zo is real). The proportions calibrate the instrument for a specific V/I ratio. Apart from mentioning that Zo must be real, and I will address that in another thread, Zo in the region where the sample is unimportant to the "proportion and phase relationship..." bit. Zo of the through line is important only to the extent that you would generally: - not want a significant transformation of impedance between the load terminals and the calibrated sensor at that calibrated V/I ratio. - not want a significant transformation of impedance between the generator and load terminals at that calibrated V/I ratio to minimise disruption to the system being measured. Everyday we use instruments to measure something, somewhere and apply that knowledge to infer something else, somewhere else using appropriate other knowledge. For example, you might measure the voltage drop across a cathode resistor and make some reasonable inference about anode current using appropriate other knowledge. Making measurements with a Bird 43 in one place and inferring the situation somewhere else using appropriate other knowledge is reasonable. For example, I may have an antenna system (say a loop) that uses a transmission line transformer (TLT) to transform the loop terminal Z to 50 ohms. I can attach the generator end of the TLT to the Bird 43, attach the Bird 43 via 50 ohm cable to a transceiver that is rated for a nominal 50+j0 ohm load, and proceed to adjust the antenna / TLT for zero reflected power indication on the Bird 43 knowing that I can reasonably infer that the load presented to the transceiver will be approximately 50+j0 ohms using the knowledge that Bird readings indicate Z at that point is 50+j0 and there will be insignificant transformation on the 50 ohm cable to the transceiver. This is a proper and sound application of the instrument. Did I get that wrong? Did I need to mention environments? Owen -- |
THERE ARE ABOUT 4.17 WATTS OF REFLECTED ENERGY FLOWING BACK THROUGH THE BIRD AND THE BIRD COMPLETELY IGNORES IT. So the Bird is not even yielding valid readings for forward and reflected power through itself. That's exactly what I have been saying all along. If it were calibrated for 75 ohms, it would indicate the correct values. -- 73, Cecil http://www.qsl.net/w5dxp Cecil, With a 1/2 wl of 75 ohm line terminated with 50 ohms at both the source and load end, the 4.17 watts reflected energy does not flow back thru the Bird. With 100 watts incident on the source (Bird), 100 watts will be either radiated, or consumed as heat at the load end. 4.17 will be reflected back to the source. When it reaches the source, a portion will be reflected back towards the load, and a portion consumed as heat. ect. The small portion consumed by heat might account for the 1/2 needle deflection Owen observed on reflected power. 4.17 watts does not flow back thru the Bird as reflected power, and the Bird, of course acknowledges. I don't think the Bird ignores that. I acknowledge that the Bird does not report the actual forward/reverse power in this example Gary N4AST |
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On Fri, 14 Oct 2005 21:23:25 GMT, Owen Duffy wrote:
that when algebraicly added and rectified, they produce a DC voltage that is proportional to the power flow in one direction only (provided that Zo is real). The proportions calibrate the instrument for a Sorry, that should be: "that when algebraicly added and rectified, they produce a DC voltage that is proportional to the square root of power flow in one direction only (provided that Zo is real). The proportions calibrate the instrument for a..." Owen -- |
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