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#11
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To the gentleman who posted the original to this thread: QST had an
article on how to convert a particular tv antenna to a vhf t/r antenna. I can't recall how long ago, perhaps in the fall or summer. KF5DE Cecil Moore wrote: Henry Kolesnik wrote: Cecil Good thought. As I understand it a receiver doesn't really need any power, so if the antenna is 50 ohms and the Rx input is say 1000 ohms most of the power has to be relflected to the antenna. Make it one megohm and virtually all receive power in a perfectly matched antenna is lost to re-radiation? So by astute choice of feedline length we can choose constructive interference or destructive interference in the re-radiated signals? :-) |
#12
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Henty Kolesnik wrote:
"How about a little more explanation on the 50% reradiation?" The antenna doesn`t care where the power comes from. It can be from connected drive or induced. Induced power has two loads, radiation resistance and connected load (receiver input). If they are matched (equal), 50% is absorbed in each. If they are mismatched, paximum power transfer can`t occur. A higher percentage of the available energy will be rejected or reradiated because the receiver input is nearer to an open circuit (rejection) or a short circuit (reradiation) than it is when it is matched to the antenna`s radiation resistance. Best regards, Richard Harrison, KBWZI |
#13
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Cecil, W5DXP wrote:
"If we simply amplify the receiver voltage through a near infinite impedance FET with virtually no load on the signal, do we get the signal information?" Not as much as possible. Just as the reactance must be tuned out of an antenna to receive maximum available power, to get maximum received carrier power, the resistance of the load in the antenna must be reduced to only that matcing the radiation resistance. Walter Maxwell, W2DU has had it right all along. Maximum power transfer requires a conjugate match. Best regards, Richard Harrison, KB5WZI |
#14
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OK....I think I uderstand the transmitter side, perhaps a little better. If
a 50 ohm transmitter is matched to a 50 ohm antenna there is no reflection and all the power is radiated. If the antenna is not 50 ohms then there is reflected power from the antenna to the transmitter and since the transmitter cannot absorb power it reflects it to the antenna ad infinitum. I also know that electric power generating station don't like a match because a 50/50 deal is not good for profits. I understand the transmitter and power generator however I'm having trouble getting my mind around whats happening with the receiver and antenna. Can it be put another way so I can try to understand it? tnx -- 73 Hank WD5JFR "Richard Harrison" wrote in message ... Henty Kolesnik wrote: "How about a little more explanation on the 50% reradiation?" The antenna doesn`t care where the power comes from. It can be from connected drive or induced. Induced power has two loads, radiation resistance and connected load (receiver input). If they are matched (equal), 50% is absorbed in each. If they are mismatched, paximum power transfer can`t occur. A higher percentage of the available energy will be rejected or reradiated because the receiver input is nearer to an open circuit (rejection) or a short circuit (reradiation) than it is when it is matched to the antenna`s radiation resistance. Best regards, Richard Harrison, KBWZI |
#15
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![]() Richard Harrison wrote: Cecil, W5DXP wrote: "If we simply amplify the receiver voltage through a near infinite impedance FET with virtually no load on the signal, do we get the signal information?" Not as much as possible. Just as the reactance must be tuned out of an antenna to receive maximum available power, to get maximum received carrier power, the resistance of the load in the antenna must be reduced to only that matcing the radiation resistance. Walter Maxwell, W2DU has had it right all along. Maximum power transfer requires a conjugate match. I happend across this just the other day. On page 1426, "Physics: Volume Two Electricity, Magnetism, and Light", Ronald Blum/Duane E. Roller it reads "Note that the optimum power transfer occurs when p=0, R=Z0, and t=1. [p and t are reflection and transmission coefficients] In general, for lossy lines, this takes place when the load impedance in the complex conjugate of the characteristic impedance, just as with lumped circuits." Seems to be the only way for p = (R-Z0)/(R+Z0) = 0 and t = 2R/R+Z0 = 1 to hold true for any Z0. ac6xg ac6xg Best regards, Richard Harrison, KB5WZI |
#16
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On Wed, 06 Apr 2005 21:16:53 GMT, "Henry Kolesnik"
wrote: If the antenna is not 50 ohms then there is reflected power from the antenna to the transmitter and since the transmitter cannot absorb power it reflects it to the antenna ad infinitum. Hi Hank, This again points out the fallacy of treating different matches as having equal outcomes. A simple, intuitive test of your statement about would ask the question: "Who needs a tuner for mismatches then?" No doubt the thread will enlarge with discourse on the theory of Thevenin (Norton by implication), Conjugation, Efficiency, Max available power, Max power transfer, and any number of other mantras dear to the heart. And yet none of them will explain that if we have: transmitter cannot absorb power it reflects it to the antenna ad infinitum. "Who needs a tuner for mismatches then?" Saving a few hundred rounds of such non-answers, and suspecting if I held feet to the fire, the discussion would then turn to answering "Who needs a tuner for mismatches then?" in terms of psychology: "A tuner makes the transmitter happy." A such, it reduces to the economy of time. If you are in this hobby long enough, the investment in a tuner will yield that same level of transmitter happiness at a lesser expense than would monthly visits to the pharmacist for Prozac. 73's Richard Clark, KB7QHC |
#17
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![]() "ken wood" wrote in message ... can you tx on a tv antenna yep. The tv stations do it all the time. many hams have used those rabbit-ears for 2-meter work |
#18
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Richard Harrison wrote:
Cecil, W5DXP wrote: "If we simply amplify the receiver voltage through a near infinite impedance FET with virtually no load on the signal, do we get the signal information?" Not as much as possible. Just as the reactance must be tuned out of an antenna to receive maximum available power, to get maximum received carrier power, the resistance of the load in the antenna must be reduced to only that matcing the radiation resistance. Walter Maxwell, W2DU has had it right all along. Maximum power transfer requires a conjugate match. But the information in a received signal can be had while consuming very little of the received signal's power. For instance, ten feet of wire into my IC-756PRO results in perfect reception of Neal Boortz on WTAW on 1620 kHz. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
#19
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Cecil, W5DXP wrote:
"But the information in a received signal can be had while consuming very little of the received signal`s power." Very true, but with AM, or even with weak FM signals, below the "FM Improvement Threshold", the modulation recovered varies directly with the strength of the received carrier Receiver alignment instructions often advise using a modulated signal from the generator which is weak to ensure operating in a region where there is a dB to dB relationship between the received carrier strength and the demodulated signal strength. You can probably hear Neal Boortz on Aggieland`s 1620 kHz without a receiver. Best regards, Richard Harrison, KB5WZI |
#20
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"Richard Harrison" wrote
Very true, but with AM, or even with weak FM signals, below the "FM Improvement Threshold", the modulation recovered varies directly with the strength of the received carrier _____________ But -- isn't program audio recovered in FM systems a function of carrier deviation, only? The threshold deals only with the amount of noise present with it. RF |
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