Final answer:
The progressive phase excitation needed between the elements for a six-element end-fire array with a spacing of d = 0.4λ is 0.8π rad or 144°. The angle at which the first null occurs is approximately 99.47°.
Step-by-step explanation:
To design a six-element ordinary end-fire array with elements along the z-axis and directed towards p = 0° with a spacing of d = 0.4λ, we must first calculate the progressive phase excitation between the elements. The phase difference φ required for constructive interference in an end-fire array is given by φ = 2πd/λ, where d is the spacing and λ is the wavelength. For a spacing of d = 0.4λ, the phase excitation between the elements is φ = 2π(0.4) = 0.8π rad or 144°.
To determine the angle where the nulls of the array factor occur, we can use the formula θ = cos-1[(mλ)/Nd - 1], where m is an integer representing the order of the null, λ is the wavelength, N is the number of elements, and d is the element spacing. Since we're looking at a six-element array with d = 0.4λ, the first null occurs at θ = cos-1[(1λ)/(6*0.4λ) - 1], which yields θ = cos-1(-0.1667). This resolves to an angle where the null occurs, approximately at θ = 99.47°. Subsequent nulls can be calculated by substituting higher values of m.