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adityabiyani
Antenna-Theory.com Newbie
Joined: 05 Sep 2011
Posts: 1
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 Posted: Mon Sep 05, 2011 3:59 am Post subject: EFFECT OF FINITE SIZED GROUND PLANE ON A MONOPOLE |
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Hi all,
I wanted to know how a finite sized ground plane affects the radiatio pattern of a quarter wave monopole. I read somehwere that the outer edge diffracts the incident radiaition, and this changes the current distribution. But that did not make much sense to me. Can someone please help me with this?
Cheers ! |
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bigSteve
Antenna Wizard
Joined: 14 Mar 2009
Posts: 265
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| Posted: Tue Sep 06, 2011 5:23 am Post subject: |
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The main effect of a finite sized ground plane is that the radiation pattern doesn't look like a donut. Rather, the peak radiation direction is no longer at theta=90 degrees. That is, the elevation angle is greater than zero degrees for peak radiation. Generally, the larger the groundplane is, the closer the peak radiation direction is to theta (polar angle) = 90 degrees (elevation angle equal to zero degrees).
As for the outer edge diffracting the incident radiation:
I don't think so. The monopole is a good radiator, so the current magnitude decreases quickly from the source (feed point). The radiation is not strong at the edges (unless the ground plane is less than a half-wavelength in diameter.)
See:
http://www.antenna-theory.com/antennas/monopole.php |
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R. Fry
Antenna Theory Regular
Joined: 06 Jun 2011
Posts: 49
Location: Illinois USA
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| Posted: Tue Sep 06, 2011 10:01 am Post subject: |
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| bigSteve wrote: | | Generally, the larger the groundplane is, the closer the peak radiation direction is to theta (polar angle) = 90 degrees (elevation angle equal to zero degrees). |
This is a common conclusion based on the shape of the far-field elevation pattern of a monopole over other than a perfect ground plane -- especially when looking at a far-field analysis using MoM software (NEC).
But in reality the maximum relative field "launched" by a vertical monopole that is 225 degrees and less in height always lies in the horizontal plane.
The differences in the h-plane fields of a vertical monopole measured several wavelengths away from it are dependent on the height of the monopole, the operating frequency, and system losses (mainly the loss in the r-f ground connection). At such close distances the actual conductivity of the earth has little effect.
Ground planes using fewer and shorter buried radials around the base of the monopole have higher losses, which means that a larger percentage of the transmitter power is dissipated in the ground plane, rather than being radiated by the monopole.
This is illustrated by the graphic at the link below...
http://i62.photobucket.com/albums/h85/rfry-100/BLandERadials.gif |
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bigSteve
Antenna Wizard
Joined: 14 Mar 2009
Posts: 265
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| Posted: Wed Sep 07, 2011 2:55 am Post subject: |
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R. Fry -
Could you elaborate more on this? How is antenna height in degrees measured?
Thanks |
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R. Fry
Antenna Theory Regular
Joined: 06 Jun 2011
Posts: 49
Location: Illinois USA
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| Posted: Wed Sep 07, 2011 10:19 am Post subject: |
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| bigSteve wrote: | | How is antenna height in degrees measured? |
Height of monopole in electrical degrees = (H * F * 360) / 299.8, where
H = Physical height of monopole in meters
F = Frequency in megahertz
Example: What is the electrical height of a 10.4 meter monopole operating at 7.2 MHz?
Electrical Height = (10.4 * 7.2 * 360) / 299.8 = 89.9 degrees, approx.
This equation cannot be simplified from its above form due to different units of measure for the fixed numbers in the numerator and denominator.
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