Project Cyclops, A Design... - Department of Earth and Planetary ...

whereP is the radiated power. Dividing equation (1) by
(2) we find the antenna gain to be
4rr If UdAJ 2
g = -- (3)
X2 p
The effective area of any antenna is X2g/4rr, so from
equation (3) we have
the dish in the transmission mode. Thus,
Ta = (l-a,)Ts+a,T o (8)
where Ts is the sky temperature, To is the ground
temperature, and at is the fraction of the total power
represented by spillover. Solving for al we find
If U dAI 2
Ta - T s
Aeff = P (4) at (9)
70-7",
Obviously, the gain and effective area are reduced if any
of the radiated power P spills past the dish, for then P is
greater than necessary to produce a given U(p,¢). If
there is no spillover, then all the power is reflected by
the dish so
and
P = f IUI2 dA (5)
If U dAI z
Aeff = f IUI= dA (6)
From equation (6) we see that A mr,'is greatest and equal
_ejj
to the physical projected area A if _/is constant, since of
all functions a constant has the smallest mean square
value for a given mean value. For the greatest effective
area the feed should illuminate the dish uniformly. For
any illumination pattern having no spillover the aperture
efficiency
is
If T s = 4 ° K and To = 300 ° K and we wish Ta < 7° then
1
al _