Aperture antenna directivity calculation

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baby_1
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Hello
Here is an uniform E filed distribution aperture antenna E and H far filed euations
2204887800_1448824658.png


and here is directivity calculation approach
6234799200_1448824658.png

my problem is why don't we calculate D with both E(theta) and E(phi)? My mean is:
instead of using
.latex?U%3Dr%5E2.P_%7Bave%7D%3Dr%5E2%28%5Cfrac%7B%7CE%28%5Ctheta%29%7C%5E2%7D%7B2%5Ceta%20%7D%29.gif

we should use
3Dr%5E2%28%5Cfrac%7B%7CE%28%5Ctheta%29%7C%5E2+%7CE%28%5Cphi%29%7C%5E2%7D%7B2%5Ceta%20%7D%29.gif

because both of them are not zero in far-filed zone and not zero at theta=pi/2
 
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[itex]E_0[/itex] is the magnitude of the electric field vector E. If you were to compute the sum of the magnitude squared of the components as you suggest, it should come out as [itex]E_0^2[/itex] as written in the text.

BTW, you have written something different. I assume that you mean [itex]E_\theta^2 + E_\phi^2[/itex] rather than [itex]E^2(\theta)+E^2(\phi)[/itex]? The latter is valid but means something else.
 
Dear marcusl
you are right I wanted to write
%3Dr%5E2%28%5Cfrac%7B%7CE_%7B%5Ctheta%7D%7C%5E2+E_%7B%5Cphi%7D%7C%5E2%7D%7B2%5Ceta%20%7D%29.gif

As the writer wrote we have
Bab%7D%7B%5Clambda%20%7D%29%5E2*%20%5Cfrac%7BE_%7B0%7D%5E2Sin%28%5Cphi%29%5E2%7D%7B2%5Ceta%20%7D.gif

and we assume phi=pi/2 that maximizes U.
and with my approach:
2%29%3D%28%5Cfrac%7Bab%7D%7B%5Clambda%20%7D%29%5E2*%20%5Cfrac%7BE_%7B0%7D%5E2%7D%7B2%5Ceta%20%7D.gif

and I didn't know why we assume only
gif.gif
to calculate power intensity ( U) ( as in the book mentioned)?
 
Last edited:
The book didn't say to use [itex]E_\theta[/itex], it said that the field is maximum at [itex]\theta=0[/itex]. Putting that angle into the equations and finding the sum of the squares of the components, as you did, gives the answer.
 
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