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Thread: Impedance in Free Space 377 Ohms

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1. Ham Member
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Feb 2003
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Just wondering what is the theory behind this

= [(1.257 x 10-6 H/m)/(8.85 x 10-12 F/m)]1/2

= 377 ohms (approximately)

Thanks

2. Sounds like a good value to me.

TOM K8ERV Montrose Colo

3. Oh, BTW, free space isn&#39;t free anymore, nothing is.

TOM K8ERV Montrose Colo

4. Because it&#39;s like that.
Why is water H²O ?

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Heres a good explanation.

Incidentally, that 377 ohms is evidenced in that a whip of some height will produce about that many times as much signal in the far-field as a (balanced or shielded) loop the same height.

Cortland
KA5S

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Originally Posted by [b
Quote[/b] (ka5s @ July 04 2006,08:39)]Heres a good explanation.

Incidentally, that 377 ohms is evidenced in that a whip of some height will produce about that many times as much signal in the far-field as a (balanced or shielded) loop the same height.

Cortland
KA5S
Courtland,

I disagree that the explaination in that link is a good one, and I also disgree that the impedance of freespace has something to do with which antenna works best. Neither is correct.

We have a limited number of ways to measure the force created by charges. The two ways we measure are by things we called electrical fields and by magnetic fields.

Radiation is an entirely different force or action at a distance than a force caused by a difference in charge distribution (electric field) or a force caused by charges moving (magnetic field). Radiation is caused by charge acceleration, and it even decays at a different rate with distance than the other effects do.

Because we measure the strength by either electric or magnetic fields, there is a set ratio between those forces when an electromagnetic wave is in space. The ratio between the ways we measure the force works out to be about 377 if freespace, but the electric and magnetic fields are just different ways we quantify the actions of or on charges at a distance.

An antenna does NOT have to &quot;match&quot; freespace, and an antenna closer to 377 ohms will couple no better to space than one with an impedance of 10 ohms or 10,000 ohms.

All of the losses in an antenna system are ohmic losses caused by current flow, either displacement currents through dielectrics or ohmic losses in conductors. If the ratio of loss resistances to radiation resistance (as determined by the effective current causing radiation vs the net power radiated as an EM wave) is high compared to resistive losses, more power makes it to space.

That is the ONLY ratio that matters, and that ratio should be as high as possible for maximum efficiency.

If I have an antenna that has a real radiation resistance of 20 ohms normalized to one particular spot in the system and it has a loss resistance of 20 ohms normalized to the very same spot, 50% of the applied power will be turned to EM radiation.

If I have an antenna that is 188.5 ohms radiation resistance and has 188.5 ohms of loss normalized to the same spot, 50% of the applied power will be radiated.

Getting closer to or further from 377 ohms does not change a thing.

Maxwell&#39;s equations tutorial

73 Tom

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Originally Posted by [b
Quote[/b] (w8ji @ July 04 2006,14:48)]Courtland,

I disagree that the explaination in that link is a good one, and I also disgree that the impedance of freespace has something to do with which antenna works best. Neither is correct.

We have a limited number of ways to measure the force created by charges. The two ways we measure are by
No I guess it isn&#39;t.

&quot;Time has stopped for everyone but you,&quot; said Sweeper patiently. &quot;Actually that sentence is wrong in every particular, but it&#39;s quite a useful lie.&quot;
Lu Tze (Sweeper) in Night Watch, by Terry Pratchett

As to the rest... I certainly agree that 377 ohms has nothing to do with making an efficient antenna. However, it does describe the circumstance in which 1 microampere per meter is also 377 microvolts per meter.

Cortland
KA5S

8. Ham Member
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So is it correct to say that the impendance is the capacitance and inductance playing off each other in free space?

Can you tell me how the numbers are derived in the formula?

Here.

10. Guest
Originally Posted by [b
Quote[/b] (KB1IDH @ July 04 2006,18:13)]So is it correct to say that the impendance is the capacitance and inductance playing off each other in free space?

Can you tell me how the numbers are derived in the formula?
An electromagnetic wave consists of an electric field E and a magnetic field H. Maxwell&#39;s equations describe the physics of how E and H are coupled, how they depend on charges and currents etc... Tom, W8JI&#39;s, link to the Maxwell equations is a good one. Otherwise, this stuff is covered in advanced undergraduate physics courses.

In free space; i.e., vacuum - no dielectric material - no charges, no currents - the propagation of an electromagnetic wave works out such that the ratio of E/H is equal to the square root of the ratio of the &quot;free-space permeability&quot; (a fundamental constant of physics), to the &quot;free-space permittivity&quot; (another fundamental constant). Which if one chooses to express in units of Ohms, is 377 Ohms, which, on the simplest level, sounds like an impedance so it&#39;s called the free-space impedance.

So the numbers themselves in the in formula are fundamental constants of physics, not &quot;derived&quot; from anything else.

I&#39;d avoid trying to make too much of an analogy to transmission line inductances and capacitances. The characteristic impedence of a transmission line depends on geometry of the conductors as well as the permeability and permittivity of the dielectric.

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