Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-10 Thread John Allen 
IIRC 

 

IIRC, most of the basic principles were explained, with illustrations, in
Albert Smith’s early IEE papers published in the 1980s. 

 

I only had a paper copy of those when I started experimenting with his
swept-frequency site calibration methods in the late 80’s (at HP) and I
assume that the principles have been expounded and expanded many times over
the years – so can anyone point to some decent downloadable documents which
illustrate the issues?

 

John Allen

W.London, UK

 

From: Pawson, James [mailto:james.paw...@echostar.com] 
Sent: 10 August 2015 09:56
To: EMC-PSTC@LISTSERV.IEEE.ORG
Subject: Re: [PSES] Calculating Reflection Angles on OATS/SAC

 

Very helpful, thanks Brent!

 

From: Brent DeWitt [mailto:bdew...@ix.netcom.com] 
Sent: 08 August 2015 01:19
To: Pawson, James; EMC-PSTC@LISTSERV.IEEE.ORG
Subject: RE: [PSES] Calculating Reflection Angles on OATS/SAC

 

Hi James,

 

The image concept again is useful.  By definition, the  ground reference
plane is at zero potential.  For that to be true, charges on the real
antenna and its image must be equal and opposite.  Put a plus on one end of
a dipole and a minus on the other and look at them.  If they are vertical,
and the bottom of the real dipole has the minus sign, the top of the image
must be plus for the charges to cancel.  For the horizontal example, if the
left end is plus the same end of the image must be minus for the same
reason.

 

In the extreme thought experiment, if you lowered the vertical dipole so its
center point were at the ground plane (now a monopole), its image would
complete the dipole.  The same extreme applied to the horizontal dipole
would have the two cancelling each other out entirely.  We can see this in
reality, since the vertical polarization with the antenna at one meter
height is usually the strongest emission at low frequencies where the path
length in wavelengths is small.  The first maximum from the horizontal
dipole occurs when there is a 180 degree path length difference between the
real antenna and its image.

 

Does that help any?

Brent

 

From: Pawson, James [mailto:james.paw...@echostar.com] 
Sent: Thursday, August 06, 2015 5:13 AM
To: EMC-PSTC@LISTSERV.IEEE.ORG
Subject: Re: [PSES] Calculating Reflection Angles on OATS/SAC

 

Many thanks for all of the replies on this topic. The conceptual key I
lacked was the “image” of the receiver below the ground plane which made the
calculations a lot simpler and I’ve now got an up and running spreadsheet.
I’ve also been introduced to things like cotangents and arctangents which
are new to me.

 

The only thing I still remain confused about is the phase of the reflection
from the ground plane.

 

 Gert wrote: “Note that vertical waves invert in polarity on reflection
with the ground plane, where horizontal polarized waves do not.”

 

 Brent wrote: “…and take the difference for phase, remembering that the
horizontally polarized image is 180 degrees out of phase to start with while
the vertical image is in phase.”

 

I might be misunderstanding but these statements seem to contradict each
other. I can kind of see how a vertically polarised wave would be reflected
inverted. If this was the case, could this be compensated for by subtracting
180° from the reflected ground ray to ensure the phases added/subtracted
correctly at the RX antenna?

 

Thanks again

James

 

 

 

_
From: Pawson, James 
Sent: 31 July 2015 15:59
To: EMC-PSTC@LISTSERV.IEEE.ORG
Subject: Calculating Reflection Angles on OATS/SAC

 

 

Hi,

 

I’m trying to calculate the distances/angles at which a maximum (in phase)
or minimum (anti-phase) signal would occur on an OATS/SAC.

 

I can do this simply when the TX and RX antennae are the same height above
the reflecting surface as the point of reflection lies halfway between the
two antennae, Distance_tx = Distance_rx. The direct and reflected paths can
be calculated using simple geometry and the wavelength is given by lambda =
c / f.

 

However when the height of the RX antenna is different to the height of the
TX antenna then the horizontal distance to the reflection point is no longer
equidistant. I can see that the ratio Height_tx / Distance_tx = Height_rx /
Distance_rx remains the same because the angle of reflection is the same.
But I’m left with two unknown Distance terms in the equation.

 

Is there a standard equation for calculating the reflection angle on an
OATS/SAC with a varying height antenna? Or can someone give me some pointers
to help me figure it out myself? I was so distracted thinking about this
that I missed my turnoff whilst cycling home the other day.

 

I’ve tried Googling but maybe I’m not putting in the right search term.

 

Any assistance gratefully received.

Thanks and regards,

James

 

-


This message is from the IEEE Product Safety Engineering Society emc-pstc
discussion list

Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-10 Thread Pawson, James 
Very helpful, thanks Brent!

From: Brent DeWitt [mailto:bdew...@ix.netcom.com]
Sent: 08 August 2015 01:19
To: Pawson, James; EMC-PSTC@LISTSERV.IEEE.ORG
Subject: RE: [PSES] Calculating Reflection Angles on OATS/SAC

Hi James,

The image concept again is useful.  By definition, the  ground reference plane 
is at zero potential.  For that to be true, charges on the real antenna and its 
image must be equal and opposite.  Put a plus on one end of a dipole and a 
minus on the other and look at them.  If they are vertical, and the bottom of 
the real dipole has the minus sign, the top of the image must be plus for the 
charges to cancel.  For the horizontal example, if the left end is plus the 
same end of the image must be minus for the same reason.

In the extreme thought experiment, if you lowered the vertical dipole so its 
center point were at the ground plane (now a monopole), its image would 
complete the dipole.  The same extreme applied to the horizontal dipole would 
have the two cancelling each other out entirely.  We can see this in reality, 
since the vertical polarization with the antenna at one meter height is usually 
the strongest emission at low frequencies where the path length in wavelengths 
is small.  The first maximum from the horizontal dipole occurs when there is a 
180 degree path length difference between the real antenna and its image.

Does that help any?
Brent

From: Pawson, James [mailto:james.paw...@echostar.com]
Sent: Thursday, August 06, 2015 5:13 AM
To: EMC-PSTC@LISTSERV.IEEE.ORG<mailto:EMC-PSTC@LISTSERV.IEEE.ORG>
Subject: Re: [PSES] Calculating Reflection Angles on OATS/SAC

Many thanks for all of the replies on this topic. The conceptual key I lacked 
was the "image" of the receiver below the ground plane which made the 
calculations a lot simpler and I've now got an up and running spreadsheet. I've 
also been introduced to things like cotangents and arctangents which are new to 
me.

The only thing I still remain confused about is the phase of the reflection 
from the ground plane.

 Gert wrote: "Note that vertical waves invert in polarity on reflection 
with the ground plane, where horizontal polarized waves do not."

 Brent wrote: "...and take the difference for phase, remembering that the 
horizontally polarized image is 180 degrees out of phase to start with while 
the vertical image is in phase."

I might be misunderstanding but these statements seem to contradict each other. 
I can kind of see how a vertically polarised wave would be reflected inverted. 
If this was the case, could this be compensated for by subtracting 180° from 
the reflected ground ray to ensure the phases added/subtracted correctly at the 
RX antenna?

Thanks again
James



_
From: Pawson, James
Sent: 31 July 2015 15:59
To: EMC-PSTC@LISTSERV.IEEE.ORG<mailto:EMC-PSTC@LISTSERV.IEEE.ORG>
Subject: Calculating Reflection Angles on OATS/SAC


Hi,

I'm trying to calculate the distances/angles at which a maximum (in phase) or 
minimum (anti-phase) signal would occur on an OATS/SAC.

I can do this simply when the TX and RX antennae are the same height above the 
reflecting surface as the point of reflection lies halfway between the two 
antennae, Distance_tx = Distance_rx. The direct and reflected paths can be 
calculated using simple geometry and the wavelength is given by lambda = c / f.

However when the height of the RX antenna is different to the height of the TX 
antenna then the horizontal distance to the reflection point is no longer 
equidistant. I can see that the ratio Height_tx / Distance_tx = Height_rx / 
Distance_rx remains the same because the angle of reflection is the same. But 
I'm left with two unknown Distance terms in the equation.

Is there a standard equation for calculating the reflection angle on an 
OATS/SAC with a varying height antenna? Or can someone give me some pointers to 
help me figure it out myself? I was so distracted thinking about this that I 
missed my turnoff whilst cycling home the other day.

I've tried Googling but maybe I'm not putting in the right search term.

Any assistance gratefully received.
Thanks and regards,
James

-


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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-07 Thread Brent DeWitt 
Hi James,

 

The image concept again is useful.  By definition, the  ground reference
plane is at zero potential.  For that to be true, charges on the real
antenna and its image must be equal and opposite.  Put a plus on one end of
a dipole and a minus on the other and look at them.  If they are vertical,
and the bottom of the real dipole has the minus sign, the top of the image
must be plus for the charges to cancel.  For the horizontal example, if the
left end is plus the same end of the image must be minus for the same
reason.

 

In the extreme thought experiment, if you lowered the vertical dipole so its
center point were at the ground plane (now a monopole), its image would
complete the dipole.  The same extreme applied to the horizontal dipole
would have the two cancelling each other out entirely.  We can see this in
reality, since the vertical polarization with the antenna at one meter
height is usually the strongest emission at low frequencies where the path
length in wavelengths is small.  The first maximum from the horizontal
dipole occurs when there is a 180 degree path length difference between the
real antenna and its image.

 

Does that help any?

Brent

 

From: Pawson, James [mailto:james.paw...@echostar.com] 
Sent: Thursday, August 06, 2015 5:13 AM
To: EMC-PSTC@LISTSERV.IEEE.ORG
Subject: Re: [PSES] Calculating Reflection Angles on OATS/SAC

 

Many thanks for all of the replies on this topic. The conceptual key I
lacked was the “image” of the receiver below the ground plane which made the
calculations a lot simpler and I’ve now got an up and running spreadsheet.
I’ve also been introduced to things like cotangents and arctangents which
are new to me.

 

The only thing I still remain confused about is the phase of the reflection
from the ground plane.

 

 Gert wrote: “Note that vertical waves invert in polarity on reflection
with the ground plane, where horizontal polarized waves do not.”

 

 Brent wrote: “…and take the difference for phase, remembering that the
horizontally polarized image is 180 degrees out of phase to start with while
the vertical image is in phase.”

 

I might be misunderstanding but these statements seem to contradict each
other. I can kind of see how a vertically polarised wave would be reflected
inverted. If this was the case, could this be compensated for by subtracting
180° from the reflected ground ray to ensure the phases added/subtracted
correctly at the RX antenna?

 

Thanks again

James

 

 

 

_
From: Pawson, James 
Sent: 31 July 2015 15:59
To: EMC-PSTC@LISTSERV.IEEE.ORG <mailto:EMC-PSTC@LISTSERV.IEEE.ORG> 
Subject: Calculating Reflection Angles on OATS/SAC

 

 

Hi,

 

I’m trying to calculate the distances/angles at which a maximum (in phase)
or minimum (anti-phase) signal would occur on an OATS/SAC.

 

I can do this simply when the TX and RX antennae are the same height above
the reflecting surface as the point of reflection lies halfway between the
two antennae, Distance_tx = Distance_rx. The direct and reflected paths can
be calculated using simple geometry and the wavelength is given by lambda =
c / f.

 

However when the height of the RX antenna is different to the height of the
TX antenna then the horizontal distance to the reflection point is no longer
equidistant. I can see that the ratio Height_tx / Distance_tx = Height_rx /
Distance_rx remains the same because the angle of reflection is the same.
But I’m left with two unknown Distance terms in the equation.

 

Is there a standard equation for calculating the reflection angle on an
OATS/SAC with a varying height antenna? Or can someone give me some pointers
to help me figure it out myself? I was so distracted thinking about this
that I missed my turnoff whilst cycling home the other day.

 

I’ve tried Googling but maybe I’m not putting in the right search term.

 

Any assistance gratefully received.

Thanks and regards,

James

 

-


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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-06 Thread John Woodgate 
In message , 
dated Thu, 6 Aug 2015, "ce-test, qualified testing bv - Gert Gremmen" 
 writes:


Yes, you may (at your choice) add or subtract 180 degrees for the 
reflected vertically polarised wave only!


..or add a half wavelength to the physical path length.
--
OOO - Own Opinions Only. With best wishes. See www.jmwa.demon.co.uk
When I turn my back on the sun, it's to look for a rainbow
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-06 Thread ce-test, qualified testing bv - Gert Gremmen 
Yes, you may (at your choice) add or subtract 180 degrees for the reflected 
vertically polarised wave only!

I think Brent meant to say the same, but with another viewpoint

 

Gert

 

 

From: Pawson, James [mailto:james.paw...@echostar.com] 
Sent: Thursday 6 August 2015 11:13
To: EMC-PSTC@LISTSERV.IEEE.ORG
Subject: Re: [PSES] Calculating Reflection Angles on OATS/SAC

 

Many thanks for all of the replies on this topic. The conceptual key I lacked 
was the "image" of the receiver below the ground plane which made the 
calculations a lot simpler and I've now got an up and running spreadsheet. I've 
also been introduced to things like cotangents and arctangents which are new to 
me.

 

The only thing I still remain confused about is the phase of the reflection 
from the ground plane.

 

 Gert wrote: "Note that vertical waves invert in polarity on reflection 
with the ground plane, where horizontal polarized waves do not."

 

 Brent wrote: "...and take the difference for phase, remembering that the 
horizontally polarized image is 180 degrees out of phase to start with while 
the vertical image is in phase."

 

I might be misunderstanding but these statements seem to contradict each other. 
I can kind of see how a vertically polarised wave would be reflected inverted. 
If this was the case, could this be compensated for by subtracting 180° from 
the reflected ground ray to ensure the phases added/subtracted correctly at the 
RX antenna?

 

Thanks again

James

 

 

 

_
From: Pawson, James 
Sent: 31 July 2015 15:59
To: EMC-PSTC@LISTSERV.IEEE.ORG
Subject: Calculating Reflection Angles on OATS/SAC

 

 

Hi,

 

I'm trying to calculate the distances/angles at which a maximum (in phase) or 
minimum (anti-phase) signal would occur on an OATS/SAC.

 

I can do this simply when the TX and RX antennae are the same height above the 
reflecting surface as the point of reflection lies halfway between the two 
antennae, Distance_tx = Distance_rx. The direct and reflected paths can be 
calculated using simple geometry and the wavelength is given by lambda = c / f.

 

However when the height of the RX antenna is different to the height of the TX 
antenna then the horizontal distance to the reflection point is no longer 
equidistant. I can see that the ratio Height_tx / Distance_tx = Height_rx / 
Distance_rx remains the same because the angle of reflection is the same. But 
I'm left with two unknown Distance terms in the equation.

 

Is there a standard equation for calculating the reflection angle on an 
OATS/SAC with a varying height antenna? Or can someone give me some pointers to 
help me figure it out myself? I was so distracted thinking about this that I 
missed my turnoff whilst cycling home the other day.

 

I've tried Googling but maybe I'm not putting in the right search term.

 

Any assistance gratefully received.

Thanks and regards,

James

 

-


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discussion list. To post a message to the list, send your e-mail to 


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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-06 Thread Pawson, James 
Many thanks for all of the replies on this topic. The conceptual key I lacked 
was the "image" of the receiver below the ground plane which made the 
calculations a lot simpler and I've now got an up and running spreadsheet. I've 
also been introduced to things like cotangents and arctangents which are new to 
me.

The only thing I still remain confused about is the phase of the reflection 
from the ground plane.

 Gert wrote: "Note that vertical waves invert in polarity on reflection 
with the ground plane, where horizontal polarized waves do not."

 Brent wrote: "...and take the difference for phase, remembering that the 
horizontally polarized image is 180 degrees out of phase to start with while 
the vertical image is in phase."

I might be misunderstanding but these statements seem to contradict each other. 
I can kind of see how a vertically polarised wave would be reflected inverted. 
If this was the case, could this be compensated for by subtracting 180° from 
the reflected ground ray to ensure the phases added/subtracted correctly at the 
RX antenna?

Thanks again
James



_
From: Pawson, James
Sent: 31 July 2015 15:59
To: EMC-PSTC@LISTSERV.IEEE.ORG
Subject: Calculating Reflection Angles on OATS/SAC


Hi,

I'm trying to calculate the distances/angles at which a maximum (in phase) or 
minimum (anti-phase) signal would occur on an OATS/SAC.

I can do this simply when the TX and RX antennae are the same height above the 
reflecting surface as the point of reflection lies halfway between the two 
antennae, Distance_tx = Distance_rx. The direct and reflected paths can be 
calculated using simple geometry and the wavelength is given by lambda = c / f.

However when the height of the RX antenna is different to the height of the TX 
antenna then the horizontal distance to the reflection point is no longer 
equidistant. I can see that the ratio Height_tx / Distance_tx = Height_rx / 
Distance_rx remains the same because the angle of reflection is the same. But 
I'm left with two unknown Distance terms in the equation.

Is there a standard equation for calculating the reflection angle on an 
OATS/SAC with a varying height antenna? Or can someone give me some pointers to 
help me figure it out myself? I was so distracted thinking about this that I 
missed my turnoff whilst cycling home the other day.

I've tried Googling but maybe I'm not putting in the right search term.

Any assistance gratefully received.
Thanks and regards,
James


-

This message is from the IEEE Product Safety Engineering Society emc-pstc 
discussion list. To post a message to the list, send your e-mail to 


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Website:  http://www.ieee-pses.org/
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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-04 Thread McDiarmid, Ralph 
If it helps, the EMC Handbook Vol. 4 (White) suggests that absorbing 
material (6dB loss) in an SAC reduces the error caused by all reflection 
(reinforced/cancellation) to about 4dB peak to peak.

Single bound path with no absorption seems to result in only about 3dB 
(peak to peak) error, according to White.
___ 


Ralph McDiarmid  |   Schneider Electric   |  Solar Business  |   CANADA  | 
  Regulatory Compliance Engineering 




From:
"Pawson, James" 
To:
EMC-PSTC@LISTSERV.IEEE.ORG, 
Date:
07/31/2015 08:17 AM
Subject:
[PSES] Calculating Reflection Angles on OATS/SAC



Hi,
 
I’m trying to calculate the distances/angles at which a maximum (in phase) 
or minimum (anti-phase) signal would occur on an OATS/SAC.
 
I can do this simply when the TX and RX antennae are the same height above 
the reflecting surface as the point of reflection lies halfway between the 
two antennae, Distance_tx = Distance_rx. The direct and reflected paths 
can be calculated using simple geometry and the wavelength is given by 
lambda = c / f.
 
However when the height of the RX antenna is different to the height of 
the TX antenna then the horizontal distance to the reflection point is no 
longer equidistant. I can see that the ratio Height_tx / Distance_tx = 
Height_rx / Distance_rx remains the same because the angle of reflection 
is the same. But I’m left with two unknown Distance terms in the equation.
 
Is there a standard equation for calculating the reflection angle on an 
OATS/SAC with a varying height antenna? Or can someone give me some 
pointers to help me figure it out myself? I was so distracted thinking 
about this that I missed my turnoff whilst cycling home the other day.
 
I’ve tried Googling but maybe I’m not putting in the right search term.
 
Any assistance gratefully received.
Thanks and regards,
James
 

__
This email has been scanned by the Symantec Email Security.cloud service.
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Mike Cantwell  
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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-02 Thread John Woodgate 
In message <002b01d0cd7c$e1128330$a3378990$@integrity.com>, dated Sun, 2 
Aug 2015, Manny Barron  writes:


Back in 1998 when I wrote a paper for the IEEE EMC Symposium on 
calculating the ?Theoretical Normalized Site Attenuation (NSA) by 
Spreadsheet Analysis?, I needed to do a similar calculation as yours in 
order to determine the grazing angle gamma.  I couldn?t find an 
equation for this anywhere so I derived it myself (simple algebra, I 
believe some on this e-mail thread have already provided similar 
results).  See the attached diagram with associated algebra ? I am 
surprised I still had my old notes!


Thank you. There are indeed numerous ways to terminate a feline, 
excluding that used by dentists.

--
OOO - Own Opinions Only. With best wishes. See www.jmwa.demon.co.uk
When I turn my back on the sun, it's to look for a rainbow
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-02 Thread Manny Barron 
Resending without the attached file.

I think EMC-PSTC rejected the e-mail posting due to the attachment.

So here it is again without the attached file - should go through now.

 

Manny Barron

 

 

From: Manny Barron [mailto:mbar...@integrity.com] 
Sent: Sunday, August 02, 2015 4:42 PM
To: 'EMC-PSTC@LISTSERV.IEEE.ORG'
Cc: 'Pawson, James'; 'John Woodgate'; 'ce-test, qualified testing bv - Gert
Gremmen'; 'Heckrotte, Michael'; 'CR'; 'Brent DeWitt'; 'T.Sato'
Subject: RE: [PSES] Calculating Reflection Angles on OATS/SAC

 

Hi James,

 

Back in 1998 when I wrote a paper for the IEEE EMC Symposium on calculating
the "Theoretical Normalized Site Attenuation (NSA) by Spreadsheet Analysis",
I needed to do a similar calculation as yours in order to determine the
grazing angle gamma.  I couldn't find an equation for this anywhere so I
derived it myself (simple algebra, I believe some on this e-mail thread have
already provided similar results).  See the attached diagram with associated
algebra - I am surprised I still had my old notes!

 

Note:

The file is attached only to direct e-mail's shown on cc: line.

(because I believe the EMC-PSTC server strips off file attachments)

If others are interested please send me an e-mail and I will send it to you.

(not sure how else to distribute file attachments on this forum)

 

Also, I still have the original Excel spreadsheet used for the paper noted
above.  It's one of the most complicated spreadsheets that I have ever
generated plus it contains a macro to move the hypothetical antenna from 1
to 4 meters in small steps, storing the "Ed" value after each step (or using
a Max Hold technique, I don't remember) and then determining "Ed(max)" after
the antenna reaches 4 meters height.  If anyone is interested in this
spreadsheet please let me know and I will send it to you.  Caution:  It was
created using Office 97 in Windows XP, so I don't know how well it works
with the latest Office and Windows version (I tried it a few times today,
and it appears to work okay).

 

Lastly, there was another similar paper referenced in another e-mail on this
thread:  Barron, M., "3-D Surface Plot of Theoretical Normalized Site
Attenuation Calculation by Spreadsheet Analysis", 2001 IEEE International
Symposium on EMC, Seattle, WA. Pp 133-138.  The updated paper I actually
presented at the symposium never made it into the final symposium record so
IEEExplore only has the old version.  If anyone is interested in the updated
version please send me an e-mail and I will send it to you.

 

Hoping maybe this info may be helpful in some way.

 

Manny Barron

 

 

 

From: Pawson, James [mailto:james.paw...@echostar.com] 
Sent: Friday, July 31, 2015 7:59 AM
To: EMC-PSTC@LISTSERV.IEEE.ORG
Subject: [PSES] Calculating Reflection Angles on OATS/SAC

 

Hi,

 

I'm trying to calculate the distances/angles at which a maximum (in phase)
or minimum (anti-phase) signal would occur on an OATS/SAC.

 

I can do this simply when the TX and RX antennae are the same height above
the reflecting surface as the point of reflection lies halfway between the
two antennae, Distance_tx = Distance_rx. The direct and reflected paths can
be calculated using simple geometry and the wavelength is given by lambda =
c / f.

 

However when the height of the RX antenna is different to the height of the
TX antenna then the horizontal distance to the reflection point is no longer
equidistant. I can see that the ratio Height_tx / Distance_tx = Height_rx /
Distance_rx remains the same because the angle of reflection is the same.
But I'm left with two unknown Distance terms in the equation.

 

Is there a standard equation for calculating the reflection angle on an
OATS/SAC with a varying height antenna? Or can someone give me some pointers
to help me figure it out myself? I was so distracted thinking about this
that I missed my turnoff whilst cycling home the other day.

 

I've tried Googling but maybe I'm not putting in the right search term.

 

Any assistance gratefully received.

Thanks and regards,

James

 

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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-02 Thread John Woodgate 
In message 
.com>, dated Sun, 2 Aug 2015, "Heckrotte, Michael" 
 writes:



Actually, he did, in his first paragraph:

"I'm trying to calculate the distances/angles at which a maximum (in 
phase) or minimum (anti-phase) signal would occur on an OATS/SAC."


Apologies all round. I just didn't take in that first paragraph.
--
OOO - Own Opinions Only. With best wishes. See www.jmwa.demon.co.uk
When I turn my back on the sun, it's to look for a rainbow
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-01 Thread T.Sato 
On Fri, 31 Jul 2015 14:59:27 +,
  "Pawson, James"  wrote:

> I'm trying to calculate the distances/angles at which a maximum (in phase) or 
> minimum (anti-phase) signal would occur on an OATS/SAC.
...
> Is there a standard equation for calculating the reflection angle on an 
> OATS/SAC with a varying height antenna? Or can someone give me some pointers 
> to help me figure it out myself? I was so distracted thinking about this that 
> I missed my turnoff whilst cycling home the other day.

I found Smith's article very helpful:

Calculation of Site Attenuation From Antenna Factors, Smith, A.A. et al.,
http://dx.doi.org/10.1109/TEMC.1982.304041

Calculation of path length is simple: sqrt(R^2 + (h1 - h2)^2) for
direct wave and sqrt(R^2 + (h1 + h2)^2) for reflected wave.
https://en.wikipedia.org/wiki/Two-ray_ground-reflection_model#/media/File:2-Ray_Ground_Reflection.png

I think reflection angle is not important as far as we assume
specular reflection on the conductive ground plane.
However, at least for vertical polarization, we probably need to
consider incident angle to the antennas.

Regards,
Tom

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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-01 Thread Heckrotte, Michael 
>Things can get more interesting at the step of determining the Rx
>heights at which the maximum (in-phase) and minimum (out-of-phase)
>signal levels occur.

>>You mean constructive or destructive interference between the direct Tx-Rx 
>>path and the indirect path >>with one reflection?

Yes.

>>I agree it's a valid topic but the original enquirer didn't ask about it.

Actually, he did, in his first paragraph:

"I'm trying to calculate the distances/angles at which a maximum (in phase) or 
minimum (anti-phase) signal would occur on an OATS/SAC."


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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-01 Thread John Woodgate 
In message 
.com>, dated Sat, 1 Aug 2015, "Heckrotte, Michael" 
 writes:


Things can get more interesting at the step of determining the Rx 
heights at which the maximum (in-phase) and minimum (out-of-phase) 
signal levels occur.



You mean constructive or destructive interference between the direct 
Tx-Rx path and the indirect path with one reflection?


I agree it's a valid topic but the original enquirer didn't ask about 
it.


The numerical approach that I took kept things simple. Step the Rx 
height from 1 to 4 meters (I chose a 0.01 m step size as this is on the 
order of the accuracy to which I could set the antenna height using 
reasonable care; the resulting resolution was quite adequate). 
Calculate Ed for each Rx height and store in an array. Find the maximum 
value of Ed; from the corresponding array index determine the Rx height.


The geometry for the in-phase conditions can then be calculated. The 
same process can be used to find EdMin for the anti-phase conditions.


Agreed.


An analytical solution could conceivably be found by taking the partial 
derivative of Ed with respect to Rx height, setting the derivative to 
zero, and solving for Rx height. At each solution, take the second 
partial derivative to determine which are maxima and which are minima. 
Find the highest maximum and the lowest minimum. The equations are 
going to get messy.


Perhaps not so very messy.


I used LabVIEW, which includes complex arithmetic. I can confidently 
say that it took me far less time to write the program, run it and 
print out graphs, than would be needed for me to derive and solve the 
proposed analytical equations.


Obviously there is a choice. I wouldn't just assume that the analytical 
approach is unwieldy.


A literature search shows that Manny Barron followed the same numerical 
approach, however he first derived analytical expressions (as functions 
only of real numbers) for the absolute values of the complex-number 
factors in the EdHoriz and EdVert equations. This enabled calculations 
to be performed in Excel.


I think complex numbers can be avoided by using trig functions. But I 
don't think I'm going to pursue the topic further, because at present I 
don't expect ever to need the results.

--
OOO - Own Opinions Only. With best wishes. See www.jmwa.demon.co.uk
When I turn my back on the sun, it's to look for a rainbow
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-01 Thread Heckrotte, Michael 
Ed is the received field at a particular geometry. EdMax is the maximum value 
of Ed over a range of Rx heights, with Distance and Tx height held constant.

Regarding the complexity of the issue...

Certainly it is easy enough to calculate the entire geometry for any 
combination of Distance, Tx height and Rx height. The various hints in this 
thread help a great deal to simplify the visualization and calculations.

Things can get more interesting at the step of determining the Rx heights at 
which the maximum (in-phase) and minimum (out-of-phase) signal levels occur.

The numerical approach that I took kept things simple. Step the Rx height from 
1 to 4 meters (I chose a 0.01 m step size as this is on the order of the 
accuracy to which I could set the antenna height using reasonable care; the 
resulting resolution was quite adequate). Calculate Ed for each Rx height and 
store in an array. Find the maximum value of Ed; from the corresponding array 
index determine the Rx height.

The geometry for the in-phase conditions can then be calculated. The same 
process can be used to find EdMin for the anti-phase conditions.

An analytical solution could conceivably be found by taking the partial 
derivative of Ed with respect to Rx height, setting the derivative to zero, and 
solving for Rx height. At each solution, take the second partial derivative to 
determine which are maxima and which are minima. Find the highest maximum and 
the lowest minimum. The equations are going to get messy.

I used LabVIEW, which includes complex arithmetic. I can confidently say that 
it took me far less time to write the program, run it and print out graphs, 
than would be needed for me to derive and solve the proposed analytical 
equations.

A literature search shows that Manny Barron followed the same numerical 
approach, however he first derived analytical expressions (as functions only of 
real numbers) for the absolute values of the complex-number factors in the 
EdHoriz and EdVert equations. This enabled calculations to be performed in 
Excel.

Various references:

Smith, A. A., Jr., German, R., Pate, J., "Standard site method for determining 
antenna factors", IEEE Transactions on Electromagnetic Compatibility, EMC-24, 
pp. 316-322, no. 3, Aug. 1982.

Smith, A. A., Jr., German, R., Pate, J., "Calculation of Site Attenuation from 
Antenna Factors", IEEE Electromagnetic Compatibilty", EMC-25, No. 3, August 
1982, pp 301-316

Barron, M., "3-D Surface Plot of Theoretical Normalized Site Attenuation 
Calculation by Spreadsheet Analysis", 2001 IEEE International Symposium on EMC, 
Seattle, WA. Pp 133-138

Best Regards,
Mike


-Original Message-
From: John Woodgate [mailto:j...@jmwa.demon.co.uk]
Sent: Friday, July 31, 2015 2:34 PM
To: EMC-PSTC@LISTSERV.IEEE.ORG
Subject: Re: [PSES] Calculating Reflection Angles on OATS/SAC

In message
, dated Fri, 31 Jul 2015, "Heckrotte, Michael"
 writes:

>Perhaps someone with a postgraduate degree in mathematics could derive
>an analytical formula, but as an engineer I used numerical methods by
>calculating Ed over the range of Rx height variation.

You seem to be addressing a much more complex issue than I understood the 
enquirer to be asking about.

What is 'Ed'?
--
OOO - Own Opinions Only. With best wishes. See www.jmwa.demon.co.uk When I turn 
my back on the sun, it's to look for a rainbow John Woodgate, J M Woodgate and 
Associates, Rayleigh, Essex UK

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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-01 Thread John Woodgate 
In message <51qlfiadg5uvf...@jmwa.demon.co.uk>, dated Fri, 31 Jul 2015, 
John Woodgate  writes:


Is there a standard equation for calculating the reflection angle on 
an OATS/SAC with a varying height antenna? Or can someone give me some 
pointers to help me figure it out myself


Make a drawing. The only odd bit is that cotangents occur.

Height of Tx = T, height of Rx = R. Distance between antennas = S

Angle between incident ray and plane = angle between reflected ray and 
plane = alpha.


T*COT[alpha] + R*COT[alpha] = S

COT[alpha] = S/(T+R)

Angle of reflection = 180-2[alpha]


That is the angle between the incident and reflected rays. The 
conventional angle of incidence (and therefore the conventional angle of 
reflection) is half that.


Call that angle beta. Then TAN[beta] = S/(T+R)

The physical path length is (T+R)COSEC[alpha] or (T+R)SEC[beta].
--
OOO - Own Opinions Only. With best wishes. See www.jmwa.demon.co.uk
When I turn my back on the sun, it's to look for a rainbow
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-01 Thread Brent DeWitt 
I haven't followed this thread completely, so this has probably already been
said:

The easiest was for me to think of the problem is with image theory.  The
source antenna has an "image" as far below the ground plane as it is above
the ground plane.  Solve for the hypotenuse of both triangles and take the
difference for phase, remembering that the horizontally polarized image is
180 degrees out of phase to start with while the vertical image is in phase.
For an approximation, use Friis equation to get the amplitude of each path
and add them up.  I say approximation because it assumes far field
propagation, which may not be the case depending on the range and frequency.

Hope that helps,

Brent DeWitt, AB1LF
Milford, MA

-Original Message-
From: CR [mailto:k...@earthlink.net] 
Sent: Saturday, August 01, 2015 8:01 AM
To: EMC-PSTC@LISTSERV.IEEE.ORG
Subject: Re: [PSES] Calculating Reflection Angles on OATS/SAC

On 7/31/2015 10:59 AM, Pawson, James wrote:
> I can do this simply when the TX and RX antennae are the same height 
> above the reflecting surface as the point of reflection lies halfway 
> between the two antennae, Distance_tx = Distance_rx. The direct and 
> reflected paths can be calculated using simple geometry and the 
> wavelength is given by lambda = c / f.
> However when the height of the RX antenna is different to the height 
> of the TX antenna then the horizontal distance to the reflection point 
> is no longer equidistant. I can see that the ratio Height_tx / 
> Distance_tx = Height_rx / Distance_rx remains the same because the 
> angle of reflection is the same. But I'm left with two unknown 
> Distance terms in the equation.

-

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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-08-01 Thread CR 

On 7/31/2015 10:59 AM, Pawson, James wrote:
I can do this simply when the TX and RX antennae are the same height 
above the reflecting surface as the point of reflection lies halfway 
between the two antennae, Distance_tx = Distance_rx. The direct and 
reflected paths can be calculated using simple geometry and the 
wavelength is given by lambda = c / f.
However when the height of the RX antenna is different to the height 
of the TX antenna then the horizontal distance to the reflection point 
is no longer equidistant. I can see that the ratio Height_tx / 
Distance_tx = Height_rx / Distance_rx remains the same because the 
angle of reflection is the same. But I’m left with two unknown 
Distance terms in the equation.


Why not reverse the problem?

You know the distance between antennas, and can set the height. This 
creates a st of triangles consisting of a direct path and a reflected 
path whose angle of reflection is equal to the angle of incidence, where 
each end terminates at a source or an antenna.


Using the optical constraint re angle of reflection, generate a table of 
triangles whose two ends terminate at these points and, with frequency 
as an input variable, calculate for each one the frequencies where the 
paths are either in phase or out by 180 degrees.


A quick way (IMO) is to just measure what happens. Using a swept signal 
source and your normal setup, save the data, look for minima and maxima, 
and filter the resulting data. If you've corrected for AF and cable 
losses you're done.



Cortland Richmond

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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-07-31 Thread John Woodgate 
In message 
.com>, dated Fri, 31 Jul 2015, "Heckrotte, Michael" 
 writes:


Perhaps someone with a postgraduate degree in mathematics could derive 
an analytical formula, but as an engineer I used numerical methods by 
calculating Ed over the range of Rx height variation.


You seem to be addressing a much more complex issue than I understood 
the enquirer to be asking about.


What is 'Ed'?
--
OOO - Own Opinions Only. With best wishes. See www.jmwa.demon.co.uk
When I turn my back on the sun, it's to look for a rainbow
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-07-31 Thread Heckrotte, Michael 
Perhaps someone with a postgraduate degree in mathematics could derive an 
analytical formula, but as an engineer I used numerical methods by calculating 
Ed over the range of Rx height variation. Then it was a simple matter to find 
EdMax and the height at which EdMax occurs. Since I was calibrating antennas 
and needed to get the right values I used complex arithmetic and the original 
Smith, German, Pate formulas. These have been republished in various antenna 
calibration standards.

Not sure how close scalar calculations that only consider phase difference but 
neglect the loss terms will be, however this might be a consideration depending 
on what you are going to use this for.

Seconding Gert's comment, the results are enlightening.

Best Regards,
Mike

From: ce-test, qualified testing bv - Gert Gremmen [mailto:g.grem...@cetest.nl]
Sent: Friday, July 31, 2015 11:27 AM
To: EMC-PSTC@LISTSERV.IEEE.ORG
Subject: Re: [PSES] Calculating Reflection Angles on OATS/SAC

Note that vertical waves invert in polarity on reflection with the ground 
plane, where horizontal polarized waves do not.
I remember that calculations were much easier when the send antenna is seen as 
if it were under the ground plane for the reflected wave
and above for the direct wave.
Also note that conclusions drawn (and they are spectacular) about the 
functioning of the OATS are correct but applicable
to a infinitely small radiate/receive antenna only. For larger antenna one 
might need to integrate over the size of both the antennas.

Gert Gremmen

Van: Pawson, James [mailto:james.paw...@echostar.com]
Verzonden: vrijdag 31 juli 2015 16:59
Aan: EMC-PSTC@LISTSERV.IEEE.ORG<mailto:EMC-PSTC@LISTSERV.IEEE.ORG>
Onderwerp: [PSES] Calculating Reflection Angles on OATS/SAC

Hi,

I'm trying to calculate the distances/angles at which a maximum (in phase) or 
minimum (anti-phase) signal would occur on an OATS/SAC.

I can do this simply when the TX and RX antennae are the same height above the 
reflecting surface as the point of reflection lies halfway between the two 
antennae, Distance_tx = Distance_rx. The direct and reflected paths can be 
calculated using simple geometry and the wavelength is given by lambda = c / f.

However when the height of the RX antenna is different to the height of the TX 
antenna then the horizontal distance to the reflection point is no longer 
equidistant. I can see that the ratio Height_tx / Distance_tx = Height_rx / 
Distance_rx remains the same because the angle of reflection is the same. But 
I'm left with two unknown Distance terms in the equation.

Is there a standard equation for calculating the reflection angle on an 
OATS/SAC with a varying height antenna? Or can someone give me some pointers to 
help me figure it out myself? I was so distracted thinking about this that I 
missed my turnoff whilst cycling home the other day.

I've tried Googling but maybe I'm not putting in the right search term.

Any assistance gratefully received.
Thanks and regards,
James

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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-07-31 Thread John Woodgate 
In message , 
dated Fri, 31 Jul 2015, "ce-test, qualified testing bv - Gert Gremmen" 
 writes:


I remember that calculations were much easier when the send antenna is 
seen as if it were under the ground plane for the reflected wave


and above for the direct wave.


It doesn't matter which is up and which is down, of course. That diagram 
leads to TAN[alpha) = (T+R)/S. Just my formula inverted.

--
OOO - Own Opinions Only. With best wishes. See www.jmwa.demon.co.uk
When I turn my back on the sun, it's to look for a rainbow
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-07-31 Thread ce-test, qualified testing bv - Gert Gremmen 
Note that vertical waves invert in polarity on reflection with the
ground plane, where horizontal polarized waves do not.

I remember that calculations were much easier when the send antenna is
seen as if it were under the ground plane for the reflected wave

and above for the direct wave.

Also note that conclusions drawn (and they are spectacular) about the
functioning of the OATS are correct but applicable

to a infinitely small radiate/receive antenna only. For larger antenna
one might need to integrate over the size of both the antennas.

 

Gert Gremmen

 

Van: Pawson, James [mailto:james.paw...@echostar.com] 
Verzonden: vrijdag 31 juli 2015 16:59
Aan: EMC-PSTC@LISTSERV.IEEE.ORG
Onderwerp: [PSES] Calculating Reflection Angles on OATS/SAC

 

Hi,

 

I'm trying to calculate the distances/angles at which a maximum (in
phase) or minimum (anti-phase) signal would occur on an OATS/SAC.

 

I can do this simply when the TX and RX antennae are the same height
above the reflecting surface as the point of reflection lies halfway
between the two antennae, Distance_tx = Distance_rx. The direct and
reflected paths can be calculated using simple geometry and the
wavelength is given by lambda = c / f.

 

However when the height of the RX antenna is different to the height of
the TX antenna then the horizontal distance to the reflection point is
no longer equidistant. I can see that the ratio Height_tx / Distance_tx
= Height_rx / Distance_rx remains the same because the angle of
reflection is the same. But I'm left with two unknown Distance terms in
the equation.

 

Is there a standard equation for calculating the reflection angle on an
OATS/SAC with a varying height antenna? Or can someone give me some
pointers to help me figure it out myself? I was so distracted thinking
about this that I missed my turnoff whilst cycling home the other day.

 

I've tried Googling but maybe I'm not putting in the right search term.

 

Any assistance gratefully received.

Thanks and regards,

James

 

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Re: [PSES] Calculating Reflection Angles on OATS/SAC

2015-07-31 Thread John Woodgate 
In message 
, dated 
Fri, 31 Jul 2015, "Pawson, James"  writes:



 
Is there a standard equation for calculating the reflection angle on an 
OATS/SAC with a varying height antenna? Or can someone give me some 
pointers to help me figure it out myself


Make a drawing. The only odd bit is that cotangents occur.

Height of Tx = T, height of Rx = R. Distance between antennas = S

Angle between incident ray and plane = angle between reflected ray and 
plane = alpha.


T*COT[alpha] + R*COT[alpha] = S

COT[alpha] = S/(T+R)

Angle of reflection = 180-2[alpha]

I hope!
--
OOO - Own Opinions Only. With best wishes. See www.jmwa.demon.co.uk
When I turn my back on the sun, it's to look for a rainbow
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

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