An Isotropic Antenna is a (theoretical) antenna that radiates
energy uniformly in all directions.
Thus if the power radiated by an isotropic antenna is Pt then the
power flux density (PFD) at a distance s metres from the antenna (in free space) is:
PFD = Pt / (4*pi*s2) W/m2
Note that (4*pi*s2) which is the area of the sphere of radius s is called the
"spreading area".
A real receiving antenna will collect power in an effective area Ae m2 and if it is
at a distance s metres from the transmitting antenna then the power received (Pr)
is:
Pr = Ae . PFD = Ae . Pt / (4*pi*s2) W
The relationship between the gain G of an antenna and its effective area (where l is the
wavelength in metres) is:
Ae = (G* λ2) / (4*pi) m2
The effective area Ai of an isotropic antenna, which by definition has unit gain is
therefore:
Ai = λ2 / (4*pi) m2
Free Space Attenuation:
The power Pr received by an isotropic antenna as a distance s from an isotropic
transmitter is therefore:
Pr = Ai . PFD = Ai . Pt / (4*pi*s2)
W
= Pt / L W
Where: L = (4*pi*s2) / Ai = (4*pi*s2) / (λ2 /
4*pi)
i.e. L = (4*pi*s / λ2)
So, L, which is the ratio of the spreading area to the area of an isotropic antenna is the
free-space attenuation between isotropic antennas and is often called the "path
loss".
Received Power Between Antennas:
Now, if the transmitting and receiving antennas have gains Gt and Gr
respectively then the power C received is:
C = (Pt . Gt) . Gr / L = EIRP . (Gr / L) W
Note two ways of finding the power received:
- Using the PFD and effective area of the receiving antenna
- Using the EIRP and gain of the receiving antenna
The effective area Ae of an antenna is related to the physical area
of its aperture Aa by the expression:
Ae = h . Aa
where h is the efficiency of the antenna.
The efficiency is less than 100% because the antenna is not perfect and the main factors are:
-
Spillover past the subreflector and main reflector (rays A and C in the
figure)
-
Blockage of the antenna aperture by the subreflector (B) and supports (not
shown)
-
Losses due to profile and other manufacturing errors
-
Ohmic losses
-
Non-uniform amplitude and phase distribution in the aperture
We have already seen that:
Ae = (G*λ2) / (4*pi*m2)
So: G = (4*pi*Ae) / λ2
= (4*pi*h*Aa) / λ2
i.e. G = h * (pi*D / λ)2
where D is the antenna diameter in metres.
As gain increases, the beamwidth decreases.
The half-power (3 dB) beamwidth is given by:
HPBW = (N*λ) / D degrees
Where N is a constant dependant on the aperture illumination.
For an ideally illuminated aperture (i.e. each point in the aperture is illuminated with an RF
signal of the same amplitude and phase)
N = 58 but this is not achievable in practice.
For an efficient real antenna, N is approximately 65.
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