Beamwidth, near/far field

07mar19

Some conversion factors:

r=radian,rd=radian,sr=steraidan,th=theta

radian	2*pi radians in a circle	circum=2pi*radius 360deg in circle rd=360/(2pi)
steradian	AreaOfShere/radius^2 4pi steradians in complete sphere.	4pir^2 = area of sphere with r=1 1sr = 4*pi.
deg->radian	57.29578 deg/rd	360./(2*pi)
deg^2->steradian	1sr=3282.81 d^2 sphere=41253 sq deg.	1deg=(pi/180) radian 1deg^2=(pi/180)^2 ster
solid angle	sa=2pi(1 -cos(th0) sr th0=1/2 angle of opening (in rd) 1sr when th0 = .572rd (32.8 deg) for gausian beam sr=pifwhmfwhm/(4ln(2))=1.133 fwhm^2 (fwhm in rd)	th0=1/2 angle areaSr=2pi*integral(dthsin(th)) 0 to th0 areaSr= 2pi(1 - cos(th0) 1sr when th0=.572 rd

Near, far field and the sky ovverlap at 100 km. (top)

The illuminated area on the dish (and the edge taper) determine the half power beam width (HPBW) at a given frequency.The relationship is:

Diameter=(K*lambda)/hpbw (where k depends on the edge taper .. typically k=1.2)

In the far field, rays from opposite sides of the dish to a point in the far field are parallel. For points in the near field, the rays are not parallel. We can define parallel by the path difference being less than a fraction of a wavelength.

The path difference for rays from the opposite edges of the dish to a point P is:

point P phase diff to center and
edge of dish

We can define the near, far field transition as the height were dl is a fraction of a wavelength:

dl=lambda/n = D^2/(8*H)
H=D^2/(8*lambda/n)

Some values are:

wavelength difference at P	h
lambda/4	D^2/(2*lambda)
lambda/8	D^2/lambda
lambda/16	2*D^2/lambda

The table below shows the transition region for some arecibo frequencies for different values of dl

Htrans=D**2/(8*lamba/N)
freq (Mhz)	Lambda (m)	Transition Region (Km)
		diam=305m		diam=225m
		dl=lambda/8	dl=lambda/16	dl=lambda/8	dl=lambda/16
327	0.917	101	202	55	110
430	0.697	133	267	72	145
1300	0.230	403	806	219	438
2380	0.126	738	1476	401	803
5000	0.060	1550	3100	843	1687
7000	0.043	2171	4341	1181	2362
9000	0.033	2791	5581	1518	3037

Using the hpbw to define the illuminated area (d=k*lambda/hpbw) and solving for H gives:

H=(k*lambda/hpbw)^2/ (8.*dl)
Setting dl=lamba/216 gives: H=(k*lambda/hpbw)^2/lambda
Substituting back for hpbw gives the familiar= D^2/lambda relationship.

see also: ../430yagi/yagi/deltaPhase.xfig, ./nearfarfield.pro

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