Timeline for Why is 50 Ω often chosen as the input impedance of antennas, whereas the free space impedance is 377 Ω?
Current License: CC BY-SA 4.0
41 events
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| Aug 30, 2021 at 22:07 | history | edited | Null♦ | CC BY-SA 4.0 |
don't use "edit" in the text since we can see the edit history
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| Aug 22, 2021 at 21:41 | comment | added | schnedan | en.wikipedia.org/wiki/Nominal_impedance#50_%CE%A9_and_75_%CE%A9 | |
| Feb 15, 2019 at 17:06 | answer | added | SH12 | timeline score: 1 | |
| Dec 2, 2018 at 7:48 | comment | added | Glenn W9IQ | " free space. The argument goes: no power from the feed line is reflected and must go to the antenna. The antenna can be assumed to be resonant and therefore radiates all its power into free space (disregarding heat losses etc)." - if you disregard resistive losses, an antenna radiates all of the applied power, regardless of its complex impedance (i.e. resonant or not). | |
| Oct 22, 2018 at 18:03 | comment | added | richard1941 | So what if there is a mismatch? Enough signal usually leaks through to do the job. If you happen to have a 377 ohm load, you can use a transformer. There are two kinds of transformers in common use: coils with shared flux, and transmission line arrangements (that are limited to a single frequency). You will be able to design these if you learn to work with Smith charts. | |
| Oct 22, 2018 at 17:59 | comment | added | richard1941 | Ah, free space has no reflections. I guess you don't believe in radar. | |
| Oct 15, 2018 at 12:14 | comment | added | crateane | See my new answer: the free-space impedance is for fields and has no relation to impedances defined for currents and voltages (such as line impedance, resistance). It is only "by accident" that it has the same unit (ohms) | |
| Oct 15, 2018 at 12:11 | answer | added | crateane | timeline score: 22 | |
| Oct 14, 2018 at 12:56 | vote | accept | DK2AX | ||
| Oct 13, 2018 at 16:30 | answer | added | Laurin Cavender WB4IVG | timeline score: 0 | |
| Oct 13, 2018 at 1:07 | answer | added | Phil Frost | timeline score: 8 | |
| Oct 12, 2018 at 22:33 | answer | added | Learned_so_long_ago | timeline score: 2 | |
| Oct 12, 2018 at 22:14 | comment | added | floppy380 | Yes and as far as I understand that Rrad is part of tuning proccess I mean included in impedance matching | |
| Oct 12, 2018 at 22:06 | comment | added | DK2AX | @user1234 Sure, I know where \$Z_0\$ and radiation resistance come from, that's not the issue. As you said yourself, \$R_{rad}\$ depends on the antenna length. Expanded it is something like \$R_{rad} = 197.3 \Omega \cdot (l/\lambda)^2 (1 + 1.316 \cdot (l/\lambda)^2 + ...)\$. Plugging in \$l = 0.5\$ for the half wave dipole approaches 73 instead of 377 Ohm. | |
| Oct 12, 2018 at 21:51 | comment | added | floppy380 | 377 comes from 120 pi | |
| Oct 12, 2018 at 21:50 | comment | added | floppy380 | Antenna current take it I. Far field is pure resl so Prad=1/2 I^2 Rrad. Rrad= 2 pi/3 nu (L/lambda)^2 where nu is 120 pi. You need to read Friis transmission equation. It is a model assuming far field is not reactive derived from radiated power | |
| Oct 12, 2018 at 21:47 | comment | added | DK2AX | @user1234 That about the far field might actually be an important point. Since the phase angle between fields in the near field is different from the far field, it seems also the impedance would be different, and maybe "transforms itself" while moving into the far field. | |
| Oct 12, 2018 at 21:44 | comment | added | DK2AX | @user1234 "The 37 ohms is the radiation resistance and the antenna transforms the natural impedance of free space (377 ohms) to 37 ohms [...]". It seems it's just reformulating roughly what was mentioned in the answers here already. The radiation resistance is not 377 Ohm; the radiation resistance is one part of the input impedance (the rest being losses), so it's something like 73 Ohm for a half wave dipole or 37 for a quarter wave dipole | |
| Oct 12, 2018 at 21:02 | comment | added | floppy380 | Btw it is resistive only if it is in far field | |
| Oct 12, 2018 at 20:45 | comment | added | DK2AX | @user1234 It seems that questions asks about matching a feed line to a load (antenna). My question was about matching the antenna to free space. | |
| Oct 12, 2018 at 19:41 | comment | added | floppy380 | Similar wuestion was asked here electronics.stackexchange.com/q/395847/125197 | |
| S Oct 12, 2018 at 18:15 | history | suggested | user64795 | CC BY-SA 4.0 |
typography corrected (there shall be a space between the numerical value and the unit symbol)
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| Oct 12, 2018 at 18:14 | review | Suggested edits | |||
| S Oct 12, 2018 at 18:15 | |||||
| Oct 12, 2018 at 17:32 | comment | added | analogsystemsrf | Historical note: from what I recall, the lowest loss coax cables are in the 63 ohm region; this lowest-loss is important for long coax runs between the top of ships-masts-antennas and the radio room. And these were air-filled coaxes, with glass-bead spacers between center and shield, for low losses. If you fill this same coax-cable with polyethylene, you now has a 50 ohm coax. | |
| Oct 12, 2018 at 16:03 | answer | added | Baruch Atta | timeline score: 0 | |
| Oct 12, 2018 at 14:45 | history | edited | DK2AX | CC BY-SA 4.0 |
added some clarifying question and summary
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| Oct 12, 2018 at 12:48 | comment | added | starblue | microwaves101.com/encyclopedias/why-fifty-ohms | |
| Oct 12, 2018 at 9:12 | answer | added | Chu | timeline score: 0 | |
| Oct 12, 2018 at 9:10 | comment | added | Curd | @ahemmetter: ...because it is just a transmission line. It simply does not have the special property of antennas: efficiently transmitting energy to/picking energy up from space. Just matching impedance is not all you need. | |
| Oct 12, 2018 at 9:00 | history | tweeted | twitter.com/StackElectronix/status/1050672531110154240 | ||
| Oct 12, 2018 at 8:36 | history | edited | DK2AX | CC BY-SA 4.0 |
added image
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| Oct 12, 2018 at 8:28 | comment | added | DK2AX | Ah, I see. Assuming some transmission line with 377Ω works well at some frequency - do we still use a dipole then to 'couple' to free space? In such a case, the antenna doesn't need to match anything then as it seems. | |
| Oct 12, 2018 at 8:25 | comment | added | Curd | Both is true. That's no contradiction. Anennas act as transmores and you can build them in ways to transform to high or low impedance depending on the antenna design. The same is true for amplifiers or transmission lines. | |
| Oct 12, 2018 at 8:19 | comment | added | DK2AX | This seems to be in contradiction to @Curd's answer. Does an antenna act as a impedance transformer (50 to 377Ω) or do we just chose 50Ω because the cables and transmitters are easier to build that way and we just accept the radiation losses? | |
| Oct 12, 2018 at 8:18 | comment | added | Bruce Abbott | "My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377?" - you mean how does the antenna transform from 50 to 377 Ohms? If that is what you want to know then it should be in your question. Otherwise the answer is simply "because that is the impedance of that type of antenna". | |
| S Oct 12, 2018 at 8:13 | history | suggested | Richard the Spacecat | CC BY-SA 4.0 |
Replaced TeXy Omega with Ω, as it looks better.
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| Oct 12, 2018 at 8:10 | review | Suggested edits | |||
| S Oct 12, 2018 at 8:13 | |||||
| Oct 12, 2018 at 7:56 | answer | added | Curd | timeline score: 28 | |
| Oct 12, 2018 at 7:42 | comment | added | DonFusili | In short: 50 ohms is nice compromise between power transmission towards the antenna and dielectric losses inside cables we can make easily. It's nice to be able to make stuff easily. | |
| Oct 12, 2018 at 7:35 | comment | added | PlasmaHH | For TV I see more often 75Ω and you need to consider the impedance of the feedline, and then you look up where the best power transfer lies (wikipedia has a chart) and other parameters and then you find a compromise | |
| Oct 12, 2018 at 7:33 | history | asked | DK2AX | CC BY-SA 4.0 |