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Near field to near field gain calculation - pyramidal horn - RF
Cafe Forums
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| confused_fella
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Post subject: Near field to near field gain calculation - pyramidal
horn
Posted: Fri Feb 20, 2009 7:20 pm
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Joined: Fri Feb 20, 2009
7:14 pm Posts: 2 |
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Hey guys I was wondering if someone could help me
out with the near field to near field gain calculations.
Let's say I took data for two gain horns, one as
a receiver and one as a transmitter, at one foot.
From this data, can I use some sort of calculation
and get the gain information at 2ft, 3ft, or at
any distance given that I am still inside the near
field region?
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nubbage |
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Post subject: Re: Near field to near field gain calculation - pyramidal
horn
Posted: Sat Feb 28, 2009 4:51 am
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| General |
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Joined: Fri Feb 17, 2006
12:07 pm Posts: 236 Location: London UK
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Hi In the near-field the concept of "gain" in
the usual sense, breaks down. Gain comes about by
focussing or vector addition along one space-vector
of voltages having different phase along that vector.
As a rule of thumb in the near-field I think I am
right in saying that the loss from horn A to horn
B will be 22dB at one wave-length separation and
will then decrease as an inverse cubic law. Thus
doubling to two wavelengths drops the level to -22dB
- 10*LOG(8) In the vicinity of the horns the
polar pattern in the near-field will just be a splurge
of the sum of vectors, without the "main beam and
sidelobe" structure of a far-field pattern.
_________________ At bottom, life is all
about Sucking in and blowing out.
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confused_fella |
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Post subject: Re: Near field to near field gain calculation - pyramidal
horn
Posted: Tue Mar 03, 2009 5:45 pm
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Joined: Fri Feb 20, 2009
7:14 pm Posts: 2 |
Oh hey thanks for answering! I'm still not quite
understanding what exactly is going on here. I believe
you are explaining in full detail the characteristics
of the gain horns and calculating the loss through
vector addition. You'll understand more of what
I mean once I explain. The way I've been
doing the experiments is that I've been using a
network analyzer to capture the data using one pyramidal
horn as transmit and one as a receive. The distance
is set at one meter. I then assume that the gain
of each horn is free space loss minus half of the
value (at each frequency) taken from the network
analyzer. Then by using the equations from Chu and
Semplak " Gain of Electromagnetic Horns" I calculate
the correction factors and project the near field
gain I just calculated (at one meter) to far field
gain. Chu and Semplak's equations for correction
works only if you are correcting to far field from
near field. If i were in near field and I want to
get the gain at another near field position, what
process would I use? Let's say I was at one foot,
and using the simple method of extracting the free
space loss, I obtained the gain. Is there a way
I can use this information to get the near field
gain at two feet? I'm guessing you were
trying to answer my question with the previous post,
but I'm not exactly as educated as you take me to
be. So if you can elaborate a little or give a few
more examples, that would be great. I know in the
near field there's an array of vectors coming out
from the horn, but I don't know how to sum them
up. I'm assuming whatever the network analyzer picks
up on the receiver horn does this... Can
you please help me?
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nubbage |
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Post subject: Re: Near field to near field gain calculation - pyramidal
horn
Posted: Wed Mar 18, 2009 6:49 am
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| General |
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Joined: Fri Feb 17, 2006
12:07 pm Posts: 236 Location: London UK
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Hi The VNA does perform the vector summation
on its receive port, and will present the summation
as a decibel level compared to the transmit port,
if you select this presentation. On mine (HP8410B)
I only have the S parameter set, so I would choose
S21 display in dB. This level will show (assuming
S11 is better than -10dB) the gain of the send and
receive horns, together with the coupling loss.
All the textbooks I have looked at say the near
field energy falls off as the inverse cube of distance.
Thus going from one foot to two feet will decrease
the level by 9 dB (10*LOG[2 cubed]). You would
have to screen the area against reflections using
RAM absorbed. I am not familiar with Chu and
Semplak's paper/method, so I cannot comment. Hope
that helps some.
_________________ At bottom, life is all
about Sucking in and blowing out.
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Posted 11/12/2012
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