problem of open end coaxial cable

Now I have a question.
In the coaxial cable, the TEM wave can propagate.
When the coaxial cable is open circuit, why the wave can not still propagate foward in the free space?
I think the free space can supprot TEM wave propagate.
I read some paper about open coaxial cable radiation. But so many maths and no physical meaning explanations.
And another question is the rverse problem. If there is a plane wave in space, why can not we put a open coaxial cable in the propagate direction and receive the wave, i.e. let the wave propagete along the coaxial cable? Why we need a antenna?
why the length of antenna is lamada/4 ??

Thanks in advance!

in fact, the wave does continue to propagate out of the open end. Maybe 1 to 5% of the power continues to propagate, whle the remaining power reflects back from the "open".

In fact to get an open circuit to really look solely like an open circuit you need to carefully design the outer conductor to be a length of cuttoff circular waveguide that extends beyond the center conductor end.

biff44 said:

in fact, the wave does continue to propagate out of the open end. Maybe 1 to 5% of the power continues to propagate, whle the remaining power reflects back from the "open".

In fact to get an open circuit to really look solely like an open circuit you need to carefully design the outer conductor to be a length of cuttoff circular waveguide that extends beyond the center conductor end.

Click to expand...

Maybe 1 to 5% of the power ---------why , why not the all ?

yes, I know the result, but why? that is the thing I wanna know. I 'd like the physical explanations not the maths.

impedance mismatch!
it takes energy to propel the em wave from a conductor into space.
from a black box perspective, space looks like a very small impedance compared to the 50 or 75 ohm impedance of the coax.
So, only a small amount is delivered to the load and the rest reflects back in the coax.
Antennas are optimized by designs of the correct length.
It's not that they won't work at other lengths, they would simply be less efficient.

---------- Post added at 19:50 ---------- Previous post was at 19:42 ----------

xuexucheng said:

yes, I know the result, but why? that is the thing I wanna know. I 'd like the physical explanations not the maths.

Click to expand...

Inside the conductor, the motion of free electrons help propagate the EM wave. At the end of the conductor, free electrons become unavailable and the EM wave must now propagate via photons emitted from the conductor. The photons require energy. The photons leave with the energy. Inside the conductor, the photons are continuously restoring the energy back to free electrons.

Impedance mismatch between cable and antenna results in reflections, as biff44 wrote. Think about it as a transmission line with no losses and ended with a very small antenna (dipole).
For a open coaxial cable as a single point in space, there is no power in space in a specific point, as voltage is a relative value and only exists between two points.
Peak difference in level for a sine wave is between two points of this wave, with distance lambda/2.
Current occur if you connect these two points of different voltage, with a conductor.
A conductor much shorter then lambda/2 can only measure between two points close to each other, seen on sine wave curve.
Both current and voltage is needed to be able to send or receive power.
It is a similar situation for most waves such as sound or water waves.
These physical limitations are bidirectional, a hay-straw is ineffective to create big water waves with as well as catch them.

One way to think about the open end is that it is a shortened antenna structure--one that is way too short to be an efficient radiator. Lets say as an antenna its radiation resistance is 1000 ohms. Clearly, a wave traveling down a 50 ohm transmission line hitting a 1000 ohm load is going to have most of the power reflected.

DartPlayer170 said:

impedance mismatch!
it takes energy to propel the em wave from a conductor into space.
from a black box perspective, space looks like a very small impedance compared to the 50 or 75 ohm impedance of the coax.
So, only a small amount is delivered to the load and the rest reflects back in the coax.
Antennas are optimized by designs of the correct length.
It's not that they won't work at other lengths, they would simply be less efficient.

---------- Post added at 19:50 ---------- Previous post was at 19:42 ----------



Inside the conductor, the motion of free electrons help propagate the EM wave. At the end of the conductor, free electrons become unavailable and the EM wave must now propagate via photons emitted from the conductor. The photons require energy. The photons leave with the energy. Inside the conductor, the photons are continuously restoring the energy back to free electrons.

Click to expand...

I know the impedance of free space is 377 ohm.
photons can be emit by antenna?

---------- Post added at 06:01 ---------- Previous post was at 05:58 ----------

E Kafeman said:

Impedance mismatch between cable and antenna results in reflections, as biff44 wrote. Think about it as a transmission line with no losses and ended with a very small antenna (dipole).
For a open coaxial cable as a single point in space, there is no power in space in a specific point, as voltage is a relative value and only exists between two points.
Peak difference in level for a sine wave is between two points of this wave, with distance lambda/2.
Current occur if you connect these two points of different voltage, with a conductor.
A conductor much shorter then lambda/2 can only measure between two points close to each other, seen on sine wave curve.
Both current and voltage is needed to be able to send or receive power.
It is a similar situation for most waves such as sound or water waves.
These physical limitations are bidirectional, a hay-straw is ineffective to create big water waves with as well as catch them.

Click to expand...

Both current and voltage is needed to be able to send or receive power.
In free space, there is no current, I think.

---------- Post added at 06:06 ---------- Previous post was at 06:01 ----------

biff44 said:

One way to think about the open end is that it is a shortened antenna structure--one that is way too short to be an efficient radiator. Lets say as an antenna its radiation resistance is 1000 ohms. Clearly, a wave traveling down a 50 ohm transmission line hitting a 1000 ohm load is going to have most of the power reflected.

Click to expand...

Thank you for your answer. I have understood what you say.
However, how can we get the impedance of a structure? i.e. why the shortened antenna structure is about 1000 ohm?

any sugestion is welcome!

photons can be emit by antenna?

Click to expand...

It is possible with right frequency and power-level. Even a filament can be seen as an antenna.

In free space, there is no current, I think.

Click to expand...

That was what I wrote, no current until a conductor connects two different points in space.

Impedance of structure
Can be measured by apply a voltage and read current for a given frequency. U/I=complex impedance.
A small dipole will have weak coupling to environment due to its size => Low I =>High impedance.

All this is pretty basic and this type of short antenna is called Hertzian dipole. Google it for more info.
The explanations you can get in a a forum is over-simplified, deeper understanding includes a lot of math and RF theory.
Hertzian dipole

Here is a good ref on radiation resistance:
The Short Dipole Antenna

in this example the short "antenna' input impedance is 0.49 +j1695 ohms.
.

xuexucheng said:

I know the impedance of free space is 377 ohm.
photons can be emit by antenna?

Click to expand...

Not what I am referring to. 377 ohms is the intrinsic impedance of free space. It is simply a property of how em fields propagate in empty space.
What I am talking about takes into account the transmitting antenna, the receiving antenna and the space in between. Essentially I am Thevenizing all that into a component of Z impedance. So, sizes, direction, shape of both the receiving and transmitting antennas, the distance between and any obstructions will all affect the effective impedance that the transmitter sees. If this impedance does not match the transmitter impedance then back reflections occurs and loss of efficiency.

Electrons in the antenna emit photons when they change energy levels. Photons is what carries the energy of the EM field through space.

Thank you all nice guys.

I learn lots from you.

However, I can still not understand the situation of open circuit.

all the TEM wave still propagate in the free space, why this is wrong?

That is because you do not understand transmission line theory. Get a good book like foundations for microwave engineering by Collin and study it. Only then will you understand the above.

biff44 said:

That is because you do not understand transmission line theory. Get a good book like foundations for microwave engineering by Collin and study it. Only then will you understand the above.

Click to expand...

I have learn the book by Collin. I understand the transmission line theory. I know at the open end there is no current, so the wave will reflect.

But Maxwell tell us the variaed E will produce H, and variaed H will produce E. There is variaed E at the open end. so why all the wave can not propagate forward?
yes , there is no current in the space, but Maxwell did not suggest wave propagation need current.

thank you.

The transmission line equations allow for propagation of waves in two directions. The boundary conditions of Voltage and Current at the imperfect open ciruit then tell you that the forward and reverse travelling waves have approximately equal amplitudes. Therefore, there is very little power left over to propagate into free space. That is a first order approximation.

If you want to do a rigourous multimode field solution...you will end up with pretty much the same conclusions.

Impedance mismatch is an explanation in many cases (e.g. transition between two coax cables of different Zo), but is tricky in this case.
Characteristic impedance of free space is 377 ohms. You can have a coaxial cable with Zo=377 ohms, but if it is cut only a part of the energy is radiated. The other part is reflected.
The reason is that the propagation modes are different in the cable and in free space, because boundary conditions are completely different.
In both cases the mode is TEM because both E and H are perpendicular to the direction of propagation. But, for example, in the cable E is radial; in free space it is totally different.
More than an impedance mismatch, i would say that there is a "mode mismatch".

Let's use spherical coordinates (rho, theta, phi). Say that the coax cable is in the z axis and is cut in the origin of coordinates.
In the far field there is a radiated wave that propagates in rho. The radiation diagram is a revolution figure along z axis.
I would expect that at any point in spae the radiation from the cut cable has E component only on theta (not in phi and obviously not in rho) and H has component only in phi (not in theta nor in rho).
The radiation must be null at any point in theta=0 (the direction of the cable). This can sound strange, but think that by symmetry there is no possibility of a particular direction for E field (nor for H), so they must be 0 at any point in the Z axis (i.e. theta=0).

Regards

Z

zorro said:

The reason is that the propagation modes are different in the cable and in free space, because boundary conditions are completely different.
In both cases the mode is TEM because both E and H are perpendicular to the direction of propagation. But, for example, in the cable E is radial; in free space it is totally different.
More than an impedance mismatch, i would say that there is a "mode mismatch".Z

Click to expand...

thank you for your answer.
if I change the coaxial cable to 2 paralell plates, I think the E and H direction will be same. How to explain this ?

Why it can?
There is no currents element aviable that in a open field.

xuexucheng said:

if I change the coaxial cable to 2 paralell plates, I think the E and H direction will be same. How to explain this ?

Click to expand...

Yes. Two infinite parrallel plates support TEM mode in the space between them (as well as in both outer half-spaces).
They form an infinitely-wide waveguide. If you start from a regtangular waveguide in TE10 mode and increase its width (along the H plane), when it becomes infinite you have the two parallel planes. Its characteristic imedance is 120*pi ohms too.
It is not a so bad radiator. Nevertheless, it is not perfect because ot the transition:

The mode supported by the infinely-wide waveguide is a TEM planar wave confined in space. When this waveguide stops, from this point you have (in far field) a cylindrical wave (not omnidirectional). In directions different to those contained in the plane parallel to the plates, the [/b]E field[/b] has not the same orientation.

Now consider a a finitely-wide rectangular waveguide. I mark in bold the differences:

The mode supported by the rectangular waveguide is a TE10 planar wave confined in space. When this waveguide stops, from this point you have (in far field) a spherical wave (not omnidirectional). In directions different to the axis of the waveguide, the E and/or the H fields have not the same orientation.

In both cases, there are also reactive fields in the proximity of the transition (Fresnel near field).

In fact, a rectangular "open-ended waveguide" (it is so called) is not a so bad radiator.
A part of the incident wave in TE10 mode is radiated in the form of TEM mode in free space.
If wou want to improve the matching (i.e. more power or energy is ratiated and less is reflected) you can use a horn.
A horn matches the waveguide to feee space providing a slow transition between the two medias (confined waveguide TE10 mode and unconfined space TEM mode).
An ideal (perfect) horn would be of exponential shape and infinite. But pyramidal finite horns work well.

Regards

Z

zorro said:

The mode supported by the infinely-wide waveguide is a TEM planar wave confined in space. When this waveguide stops, from this point you have (in far field) a cylindrical wave (not omnidirectional). In directions different to those contained in the plane parallel to the plates, the [/b]E field[/b] has not the same orientation.

Click to expand...

Why? Could you explain much in detail. Thank you very much.

You indeed are a excellent expert!