Lecture 3, Earth Segment

Presentation / Lecture 3, Earth Segment

Date Submitted: 06 June 2001

Written by RPC Telecommunications. Website: http://www.rpctelecom.com

  Rate This Article

This is the third in the series of general satcom tutorial lectures submitted by RPC Telecommunications.

Comment On This Article

 Printable Version
-Section 1
Types and components of an Earth Station
-Section 2
General Construction
-Section 3
Antenna Theory
-Section 4
Radiation Patterns
-Section 5
-Section 6
-Section 7
Low-noise Amplifiers
-Section 8
Power Amplifiers
-Section 9

Antenna Theory

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.

Next: Section 4 - Radiation Patterns


Home | Satellite School | Presentations | Calculators | Software | Forum | Submit | Books | Contact Us


Copyright 2018 Satcom Online (About This Site)