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Near-field vs Far-field radiation pattern

 
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gelunmak
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Joined: 05 Jun 2009
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Location: Hong Kong

PostPosted: Mon Aug 24, 2009 9:42 am    Post subject: Near-field vs Far-field radiation pattern Reply with quote

Hi All

As far as i know, the relationship between NearField Radiation Pattern and FarField Radiation Pattern is bounded by Fourier Transform.

i.e. NearField --> (FFT) ---> FarField

But why FFT?

Thanks
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admin
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Joined: 03 Jan 2007
Posts: 205

PostPosted: Tue Aug 25, 2009 4:32 am    Post subject: Fourier Transform Reply with quote

At a very low-level intuitive idea, consider that a very short pulse in time would have a very large frequency span. Similarly, a very large antenna (many wavelengths long) is typically very directive; very short antennas are often more omnidirectional. So at first glance it makes a little bit of sense.

To understand the relationship, think about a simple aperture in an infinite metallic sheet. Let the fields within the aperture (the source fields), be Ea. This, in effect, are the near fields. How do you find the far fields from the aperture fields?

You can integrate across the aperture: suppose you want the fields at some large position R away from the aperture. This can be found from:
integral[ Ea (dot) e^(-j*k (dot) R )/R ]

The exponent term comes from the phase propagation, the 1/R is the power falloff of the fields due to distance. This integral is basically a Fourier transform - note that R is fixed. The fourier transform in this case maps the near field domain (the aperture locations) to the far field domain (the angles theta and phi in the far field).

Hence, its just math. If you work out the equations that take the near fields out to the far fields, it resembles a Fourier Transform if you factor it right.
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kviksand81
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Joined: 18 Dec 2014
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PostPosted: Sun Jan 04, 2015 3:52 pm    Post subject: Re: Fourier Transform Reply with quote

admin wrote:
At a very low-level intuitive idea, consider that a very short pulse in time would have a very large frequency span. Similarly, a very large antenna (many wavelengths long) is typically very directive; very short antennas are often more omnidirectional. So at first glance it makes a little bit of sense.

To understand the relationship, think about a simple aperture in an infinite metallic sheet. Let the fields within the aperture (the source fields), be Ea. This, in effect, are the near fields. How do you find the far fields from the aperture fields?

You can integrate across the aperture: suppose you want the fields at some large position R away from the aperture. This can be found from:
integral[ Ea (dot) e^(-j*k (dot) R )/R ]

The exponent term comes from the phase propagation, the 1/R is the power falloff of the fields due to distance. This integral is basically a Fourier transform - note that R is fixed. The fourier transform in this case maps the near field domain (the aperture locations) to the far field domain (the angles theta and phi in the far field).

Hence, its just math. If you work out the equations that take the near fields out to the far fields, it resembles a Fourier Transform if you factor it right.


Say, if I have an aperture field from a horn antenna and just want to "brute force" compute the far-field, doing as few approximations as possible, I would just do a double integral across the limits of the aperture multiplied with the complex exponential e^(-j*k (dot) R)/R ?

In this case, what would the "k" and "R" in the exponent look like? I guess I would have to do a coordinate transformation at some point along the process, since I have my horn aperture field in rectangular coordinates, would this be before or after the computation of the far-field?

Best Regards,
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admin
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PostPosted: Tue Jan 06, 2015 12:28 am    Post subject: Reply with quote

You can check out some of the math on this page:
http://www.thefouriertransform.com/applications/radiation.php

Basically the math is the same whether you're integrating E or Js (surface current). You can see how the dot product gets factored in, and you basically end up with the Fourier Transform of the aperture distribution.
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kviksand81
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Joined: 18 Dec 2014
Posts: 3

PostPosted: Thu Jan 22, 2015 12:16 pm    Post subject: Reply with quote

My thought exactly!

I've designed a standard pyramidal horn using the design procedure outlined in Balanis Antenna Theory - Analysis and Design, 3rd Ed. and used the dimensions from this design to calculate the aperture field of the horn and stored the field in a matrix by using the expression for the tangential E-field component on each location in the discretized aperture, i.e. the expression shown below:



Then, as per my understanding, it should be possible to propagate the aperture field to the far-field, viewed in spherical coordinates by implementing the following equations, where a1 and b1 are the dimensions of the horn:





The conversion to spherical coordinates is done using



and setting Ex and Ez to zero and the integrals are solved for every point defined by r, theta, phi. I simply let Matlab crunch these equations numerically.

The total E-field in the far-field should then just be given by farfield = sqrt(|Er|^2+|Etheta|^2+|Ephi|^2), right?

Am I getting it right here or not? The reason I ask is that when I compare with a simulation of the far-field for the exact same horn simulated in a EM-solver and monitoring only the E-field, I get a far-field result that differs from the Matlab result. So I'm trying to figure out where I am doing it wrong. And the first very helpful thing would be to know if I at least got the process right?

Thanks for your reply!

Hope to hear from you again!

Best Regards,


Last edited by kviksand81 on Fri Feb 27, 2015 10:33 am; edited 1 time in total
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admin
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PostPosted: Fri Jan 23, 2015 9:53 pm    Post subject: Reply with quote

It looks about right, except:

- why do you need to convert anything from (x,y,z) to (R, theta,phi) coordinates?

- your equation for Ey'....what is pho_2 and pho_1? Usually you assume the dominant mode for Ey which is just the cos() term. I don't recall needing that exponential there
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