Timeline for Why is 50 Ω often chosen as the input impedance of antennas, whereas the free space impedance is 377 Ω?
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| Dec 3, 2018 at 8:39 | comment | added | Curd | @Glenn W9IQ: I wrote: "voltage (or current) amplitude \$V_0\$ (or \$I_0\$)". Amplitude means peak. | |
| Dec 2, 2018 at 20:40 | comment | added | Glenn W9IQ | So do you use RMS or peak (I and V) in your formula? | |
| Dec 2, 2018 at 19:40 | comment | added | Curd | @Glenn W9IQ: Approximations might come into play when E and B are expressed in terms of current/voltage; but I didn't do that in my posting. The linked Wikipedia article does that (but that's not "my formula"). If you mean that: please discuss it there. I linked it just as an example showing how my ansatz can be used. | |
| Dec 2, 2018 at 19:40 | comment | added | Curd | @Glenn W9IQ: I'm not sure if I know what you mean by "my radiation resistance formula". If you mean \$R = 2P/I_0\$: The factor 1/2 is nothing but the squared factor for converting amplitude (=max. value) to effective value (because average power over one period is asked for) of a sinusoidal signal. It is exact if a sinusoidal signal is fed into the antenna. This formula as well as the Poynting vector formula are exact and valid in general; no approximations, idealizations or restrictions (besides assuming a sinusoidal signal). I don't see any other formulas in my posting. | |
| Dec 2, 2018 at 13:41 | comment | added | Glenn W9IQ | I should also add that your radiation resistance formula is only valid for a Hertzian dipole where it is assumed that the current drops off linearly towards the ends (thus the reason to divide maximum current by 2 to obtain average current). As a counter example, in a 1/2 wave dipole, the current drops off by cos(2πz/λ). Thus your formula would predict a radiation resistance of ~50 ohms for a 1/2 wave dipole while in fact it is 73 ohms. | |
| Dec 2, 2018 at 9:23 | comment | added | Glenn W9IQ | Your radiation resistance description is a little loose. The current in the equation must be the maximum effective current causing the radiation not simply the applied current. The distinction becomes important as the feedpoint is moved on the radiating antenna element. On a half wave antenna the feedpoint current changes but the radiation resistance does not. On other antenna geometries, changing the feedpoint location changes the maximum effective radiating current and thus the radiation resistance. | |
| Oct 15, 2018 at 19:44 | comment | added | crateane | @curd i do not completely agree. In your ex., the permittivity changes and has an effect on both kinds of impedances. But the effects are not related: the relation is not predictable. Changing the permittivity has a defined effect on the wave imp. But the rad. resistance effect is not clear but depends on the antenna geometry. Thus, my statement still is that input/line impedances and wave impedances are not related at all. However, changing the physical conditions can or cannot affect both (see my example of the coacial cable where the wave impedance is not changed by changing the geometry) | |
| Oct 15, 2018 at 18:10 | comment | added | Curd | @Faekynn: I wouldn't say they have relation because: suppose you submerge a 50Ω (air) antenna in water (or an other medium) its radiation resistance would change very well. | |
| Oct 15, 2018 at 12:25 | comment | added | crateane | @Curd Looking at the fields is the way to go, see my answer. wave impedances and resistances have no relation at all, and, thus, neither do 50 Ohm and 377 Ohm. | |
| Oct 14, 2018 at 12:56 | vote | accept | DK2AX | ||
| Oct 12, 2018 at 16:30 | history | edited | Curd | CC BY-SA 4.0 |
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| Oct 12, 2018 at 14:23 | comment | added | Curd | @ahemmetter: yes, I think that properly sums it up. | |
| Oct 12, 2018 at 14:05 | comment | added | DK2AX | Thanks for adding the ansatz. So, to clarify: the input impedance (especially the radiation resistance \$R\$) is the impedance 'seen' by the transmission line, whereas power radiated into free space depends on the free space impedance in the Poynting vector \$S = \frac{E^2}{Z_0}\$. And the antenna just transforms between both impedances. Is that more or less correct? | |
| Oct 12, 2018 at 13:51 | history | edited | Curd | CC BY-SA 4.0 |
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| Oct 12, 2018 at 11:06 | history | edited | Curd | CC BY-SA 4.0 |
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| Oct 12, 2018 at 10:25 | history | edited | Curd | CC BY-SA 4.0 |
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| Oct 12, 2018 at 10:11 | history | edited | Curd | CC BY-SA 4.0 |
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| Oct 12, 2018 at 10:03 | history | edited | Curd | CC BY-SA 4.0 |
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| Oct 12, 2018 at 8:42 | comment | added | Curd | @ahemmetter: I don't have access to it right now but I think Jackson's "Classical Electrodynamics" could have some examples of that kind of problems. | |
| Oct 12, 2018 at 8:40 | comment | added | Curd | @Chris Straton: I made clear that is wrong to assume that a good antenna should have 377Ohm impedance (of course at the feed point; anything else doesn't make sense). I did give an answer (high level) to the "Why?". and even more details in the comment and if I have time I will give expand my answer, maybe with some formulas... (what do you expect within a few minutes?) | |
| Oct 12, 2018 at 8:33 | comment | added | Curd | Yes, it's not trivial to see. Nobody said it would be. Maybe I can find a simple example... not within the next few hours however. | |
| Oct 12, 2018 at 8:31 | comment | added | Chris Stratton | At best you are arguing that the feedpoint impedance is what it is because that is what works. That's not an answer. An answer would explain why the feedpoint impedance is what it is. And no, it isn't too match the feedline, if anything the other way around, the feedline is designed with the antenna impedance as one of the goals. | |
| Oct 12, 2018 at 8:24 | comment | added | DK2AX | I'd appreciate to see at least the ansatz to this calculation for a dipole antenna. It's not trivial to see where the impedances appear in such a case and how Maxwell's equations can be applied to find them. Would it be possible to add this to the answer or link to an article explaining it? | |
| Oct 12, 2018 at 8:17 | comment | added | Curd | "Just" apply Maxwell's equations to your antenna geometry and you will find, that it will turn out that it does (not exactly but about). Your expectation to immediately "see" the 50/377 ratio in wire or wave length ratios is not justified; but you will get the result if you do the integrations etc. | |
| Oct 12, 2018 at 8:09 | history | edited | Curd | CC BY-SA 4.0 |
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| Oct 12, 2018 at 8:00 | history | edited | Curd | CC BY-SA 4.0 |
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| Oct 12, 2018 at 7:59 | comment | added | Puffafish | My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377? There isn't an obvious 2:15 ratio there. | |
| Oct 12, 2018 at 7:56 | history | answered | Curd | CC BY-SA 4.0 |