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Transmitting and Receiving Antennas (CP, EP, LP)

 
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gelunmak
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Joined: 05 Jun 2009
Posts: 17
Location: Hong Kong

PostPosted: Mon Jun 08, 2009 10:17 am    Post subject: Transmitting and Receiving Antennas (CP, EP, LP) Reply with quote

Another question related to circular / elliptical polarization:

What i understand is if i've a Linearly Polarized Antenna, say a vertical dipole, transmitting 1W of signal. Theoretically, I can receive all 1W of power using another vertical receiving dipole or 0W if a horizontal receiving dipole is uesd. (Assuming no path loss). Ok, when i change the receiving antenna to perfect CP antenna (Axial Ratio = 1.0, or 0 dB), the received power will be 0.5W (-3dB of 1W).

With the LP antenna remaining as the Tx antnena transmitting 1W. My question is how much power will be received if an Elliptical Polarized antenna (say with Axial Ratio = 2.0 , or 3.0 dB) is used as the receiving antenna ? [Case-1: with the major axis Vertical, and Case-2: with the minor axis Vertical ]

Is it 0.667W for case 1 and 0.333W for case 2?

For the description above related to Transmitting LP and Receiving CP, the roles can be swapped, i.e. Transmitting CP and Receiving LP, and the argument are keeping the same, meaning that 0.5W is received in LP antenna from 1W CP transmitting antenna no matter it's a horz- or vert- dipole.

So My final question is by keeping a CP transmitting antenna (transmitting 1W of power) , how much power will be received if an Elliptical Polarized antenna (say with Axial Ratio = 2.0 , or 3.0 dB) is used as the receiving antenna? [Case-1: same hand of rotation, Case-2: opposite hand of rotation]

Thanks a lot!
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gelunmak
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Joined: 05 Jun 2009
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Location: Hong Kong

PostPosted: Wed Jun 10, 2009 9:33 am    Post subject: Corrections ... Reply with quote

I made a mistake on the vertival dipole transmitting 1W of power. The power will radiate in all directions, so i can't get back all power even when i use a vertical dipole.

So for the question, and make the problem simple, let's assume all antennas (LP, CP, EP) have very high directivity so that all power will radiate at one direction with very narrow beamwidth. And the Tx and Rx antennas are towards each other.

Looking for somebody reply!
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admin
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PostPosted: Wed Jun 10, 2009 8:43 pm    Post subject: Thoughts Reply with quote

Don't worry about pathloss at this point. You can estimate pathloss using the Friis Transmission Equation:
http://www.antenna-theory.com/basics/friis.php

The formula for the Friis equation in the above link assumes the antennas are in the far-field (not close to each other) and that the antennas are polarization matched. If they are not matched, the above equation would be multiplied by the polarization loss factor.

It sounds like you are interested in determining the polarization mismatch loss or polarization loss factor:
http://www.antenna-theory.com/basics/antennapol.php

The factor is -3 db (0.5) for circular polarization antenna to a linearly polarized antenna. You're other calculations for linear polarization to elliptical polarization mismatch are also correct.

Your final question is in regards to the polarization mismatch loss for an elliptically polarized antenna transmitting to a circularly polarized antenna. This is more difficult, in general. I will attempt to provide the answer here:

Let the unit vector in the x-direction be [x]. Let the y-directed unit vector be [y]. Assume you have a wave propagating in the [+z] direction. Then vertical polarized efield could be written as:
Evert = +[y]

Horizontal polarization would be written as:
Ehorz = +[x]

Circular Polarization (note the magnitudes of the field should be 1.0 - hence the sqrt(2) in the denominator):
ERhcp = ([x] + j[y])/sqrt(2)
ELhcp = ([x] - j[y])/sqrt(2)

Now, the Polarization Loss Factor for these could be determined using the squared magnitude of the dot product (multiplying the vector components, with the second vector being the complex conjugate):

PLF (h-pol to v-pol) = abs( Evert (dot) Ehorz )^2 = abs( [y] (dot) [x] )^2= 0 = - infinity dB

PLF (h-pol to h-pol) = abs( Ehorz (dot) Ehorz )^2 = abs( [x] (dot) [x] )^2= 1 = 0 dB
PLF (h-pol to RHCP) = abs( Ehorz (dot) ERhcp )^2 = abs( [x] (dot) ([x]+j[y])/sqrt(2) )^2= (1/sqrt(2))^2 = 0.5 = - 3 dB

Using this formulation, we can determine the polarization loss factor for elliptical polarizations. Suppose we have an elliptically polarized wave:

Eellip = (2[x] + j[y])/sqrt(5)

Then we can easily determine all the polarization loss factors:

PLF: Eellip -> ERHcp = abs( (2[x]+j[y])/sqrt(5) (dot) ([x]+j[y])/sqrt(2) )^2 = ( ( 2*1 + 1*1 )/sqrt(10) )^2 = 0.9 = -0.45 dB

PLF: Eellip -> ELhcp = abs( (2[x]+j[y])/sqrt(5) (dot) ([x]-j[y])/sqrt(2) )^2 = ( (2*1 - 1*1 )/sqrt(10) )^2 = 0.1 = -10 dB

PLF: Eellip -> Ev = abs( (2[x]+j[y])/sqrt(5) (dot) [x] )^2 = ( 2/sqrt(5) )^2 = 0.8 = -0.969 dB

So basically, if you write out the E-field (with magnitude of 1) in terms of its two orthogonal components, then dot them together and take the squared magnitude, this will give you the polarization loss factor. Hopefully this was clear. If I have time, I'll add a webpage to the site that contains this information in a clearer form.
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gelunmak
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Joined: 05 Jun 2009
Posts: 17
Location: Hong Kong

PostPosted: Thu Jun 11, 2009 7:17 am    Post subject: Thank you Reply with quote

Many thanks. Yup, Polarization Loss Factor is one of my interested toipc.

Yr explaination is clear for me, and so i was wrong in my guess in the question above. In case one, it's not 0.6667, it should be 0.8 (-0.969 dB) as in yr example. So case two should not be 0.333, it should be 0.2 (-7 dB). I just forgot we're considering Power, not E-field, so should think of E square.

In general, it's better to have the PLF equation between two tilted EP antennas. (i.e. not just [cos(phi)]^2 between two tilted LP antennas as in http://www.antenna-theory.com/basics/antennapol.php)

Bty, I guess there're some typo in yr reply (abt the sense of rotation), because in phasor notation {assume propagating in +z direction}.

[x]-j[y] means
[x]e^(-jkz) - j[y]e^(-jkz) = [x]e^(-jkz) + [y]e^j(-kz-pi/2)

so in instantaneous expression, it become
[x] cos (wt-kz) + [y] cos (wt-kz-pi/2) --- > RHCP

I'm referring a book "Field and Wave Electromagnetics (2nd Edition)"
http://www.amazon.com/Field-Electromagnetics-Addison-Wesley-Electrical-Engineering/dp/0201128195
Page 364 - 366


ERhcp = ([x] + j[y])/sqrt(2)
ELhcp = ([x] - j[y])/sqrt(2)


should be
ELhcp = ([x] + j[y])/sqrt(2)
ERhcp = ([x] - j[y])/sqrt(2)



Look forward for yr PLF page.
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admin
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PostPosted: Thu Jun 11, 2009 4:13 pm    Post subject: You're right Reply with quote

Yes you are right in your calculations, I was a bit sloppy and didn't check.

As for the Right Hand or Left Hand debate, I believe physicists and engineers define it differently. I believe the IEEE definition is to determine the sense of rotation while watching a field propagate away from you, and physicists determine the sense of rotation for fields propagating towards you.

So unfortunately it is a matter of convention.

When I get the sense of polarization wrong (left instead of right), I just say I was using the other definition Cool
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gelunmak
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Joined: 05 Jun 2009
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PostPosted: Fri Jun 12, 2009 2:37 am    Post subject: I see. thks Reply with quote

So i'm using the IEEE definition. My way to memorize is to let the Thumb be Propagation Direction, and the 4 fingers will be the sense of rotation.

Anyway, thanks and it was a very good discussion.
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