Question 8.14: Find and plot the array factor for three two-element antenna...
Find and plot the array factor for three two-element antenna arrays, that differ only in the separation distance between the elements. Assume the two elements for each antenna array are isotropic radiators. The antennas are separated by 5, 10, and 20 cm, and each antenna is excited in phase. The frequency of the signal applied to each antenna is 1.5 GHz.
The separation between the elements is normalized by the wavelength via \xi = kd/2 = πd/ λ. The wavelength in the free space is
λ = \frac{c}{f} = \frac{3 × 10^{8}}{1.5 × 10^{9} } = 20 cmor the normalized separation d is λ/4, λ/2, and λ, respectively. This yields the following values for the parameter \xi: (a) π/4, (b) π/2, and (c) π. The phase difference is zero (δ = 0). The corresponding array factor F_{a}(θ) is found from equation (8.59), and it is plotted in the figure below for these three cases. Because the element pattern is uniform (F_{1}(θ) ≡ 1), it follows from (8.58) that the total radiation pattern F(θ) = F_{a}(θ). The number of the lobes that are found in the pattern increases linearly with an increase of the normalized length of the array, and it can be approximated by the nearest integer number to the real parameter 4d/λ.
F(\theta ,\phi ) = F_{1}(\theta ,\phi ) F_{a}(\theta ,\phi ) (8.58)
F_{a}(\theta ,\phi ) = \cos\left\lgroup\frac{kd \cos \theta + \delta }{2}\right\rgroup (8.59)
