Editing Direction finding (section)

=== Two-channel DF ===
[[File:Two-port_DF,_polar_plot.png|thumb|Two-port DF, polar plot (normalized)]]
[[File:Two-port_DF,_log_scale.png|thumb|Two-port DF, log scale (normalized)]]
[[File:Power_Dif._v._Bearing.png|thumb|Power Diff. (dB) v. Bearing]]

Two-channel DF, using two adjacent antennas of a circular array, is achieved by comparing the signal power of the largest signal with that of the second largest signal. The direction of an incoming signal, within the arc described by two antennas with a squint angle of Φ, may be obtained by comparing the relative powers of the signals received. When the signal is on the boresight of one of the antennas, the signal at the other antenna will be about 12&nbsp;dB lower. When the signal direction is halfway between the two antennas, signal levels will be equal and approximately 3&nbsp;dB lower than the boresight value. At other bearing angles, φ, some intermediate ratio of the signal levels will give the direction.

If the antenna main lobe patterns have a Gaussian characteristic, and the signal powers are described in logarithmic terms (e.g. [[decibels]] (dB) relative to the boresight value), then there is a linear relationship between the bearing angle φ and the power level difference, i.e. φ ∝ (P1(dB) - P2(dB)), where P1(dB) and P2(dB) are the outputs of two adjacent channels. The thumbnail shows a typical plot.

To give 360° coverage, antennas of a circular array are chosen, in pairs, according to the signal levels received at each antenna. If there are N antennas in the array, at angular spacing (squint angle) Φ, then Φ = 2π/N radians (= 360/N degrees).

==== Basic equations for two-port DF ====
If the main lobes of the antennas have a Gausian characteristic, then the output P<sub>1</sub>(φ), as a function of bearing angle φ, is given by<ref name = Lipsky />{{rp|238}}

:<math> P_1(\phi)= G_0.\exp \Bigr [ -A. \Big ( \frac{\phi}{\Psi_0} \Big )^2 \Bigr ] </math>

where

: G<sub>0</sub> is the [[antenna boresight]] gain  (i.e. when  ø = 0),
: Ψ<sub>0</sub> is one half the half-power [[beamwidth]]
: A = -\ln(0.5), so that P<sub>1</sub>(ø)/P1<sub>0</sub> = 0.5 when ø = Ψ<sub>0</sub>
: and angles are in radians.
The second antenna, squinted at Phi and with the same boresight gain G<sub>0</sub> gives an output

:<math> P_2 = G_0 .\exp \Bigr [ -A. \Big ( \frac{\Phi - \phi}{\Psi_0} \Big )^2 \Bigr ] </math>

Comparing signal levels,

:<math> \frac{P_1}{P_2} = \frac{\exp \big [-A.(\phi/\Psi_0)^2 \big ]}{\exp \Big [-A \big [ (\Phi - \phi)/ \Psi_0 \big ]^2 \Big ]}  =  \exp \Big [  \frac{A}{\Psi_0^2}.(\Phi^2 - 2 \Phi \phi) \Big ] </math>

The natural logarithm of the ratio is

:<math>\ln \Big ( \frac{P_1}{P_2} \Big ) = \ln(P_1) - \ln(P_2) = \frac{A}{\Psi_0^2}.(\Phi^2 - 2 \Phi \phi) </math>

Rearranging
:<math> \phi = \frac{\Psi_0^2}{2A.\Phi}. \big [ \ln(P_2) -\ln(P_1) \big ] + \frac{\Phi}{2} </math>

This shows the linear relationship between the output level difference, expressed logarithmically, and the bearing angle ø.

Natural logarithms can be converted to [[decibels]] (dBs) (where dBs are referred to boresight gain) by using ln(X) = X(dB)/(10.\log<sub>10</sub>(e)), so the equation can be written

:<math> \phi = \frac{\Psi_0^2}{6.0202 \Phi} . \big [ P_2(dB) - P_1(dB) \big ] +\frac{\Phi}{2} </math>