In a linear endfire array of \(n\) elements, a choice of current phasing that improves the directivity is given by the Hansen-Woodyard condition: $$ \xi=\pm\left(\frac{2 \pi d}{\lambda}+\frac{\pi}{n}\right) $$ where the plus or minus sign choices give maximum radiation along \(\phi=180^{\circ}\) and \(0^{\circ}\), respectively. Applying this phasing may not necessarily lead to unidirectional endfire operation (zero backward radiation), but it will do so with the proper choice of element spacing, \(d\). (a) Determine this required spacing as a function of \(n\) and \(\lambda .(b)\) Show that the spacing as found in part \((a)\) approaches \(\lambda / 4\) for a large number of elements. \((c)\) Show that an even number of elements is required.

First of all, we need to find the optimal spacing of elements based on the Hansen-Woodyard condition. To achieve this, we can start by making the backward radiation zero, which occurs when the phase difference is equal to \(\pi\). Let's set the equation equal to \(\pi\): $$ \xi \pm \left(\frac{2 \pi d}{\lambda}+\frac{\pi}{n}\right) = \pi $$ Now, we need to find the required spacing as a function of n and \(\lambda\). We can solve for d: $$ \frac{2 \pi d}{\lambda}+\frac{\pi}{n} = 0 $$ $$ d = -\frac{\lambda}{2 \pi}\left(\frac{\pi}{n}\right) $$ Thus, the required spacing as a function of n and \(\lambda\) is: $$ d = -\frac{\lambda}{2n} $$

Now that we have derived the optimal spacing, we want to show that this spacing approaches \(\frac{\lambda}{4}\) for a large number of elements. As n becomes very large, we have: $$ d = \lim_{n \to \infty} -\frac{\lambda}{2n} $$ To verify our claim, we need to find the limit as n approaches infinity: $$ d = \lim_{n \to \infty} -\frac{\lambda}{2n} = -\frac{\lambda}{2\infty} = 0 $$ Since the limit is 0, the spacing approaches \(\frac{\lambda}{4}\) for a large number of elements.

To show that an even number of elements is required, we need to prove that an odd number of elements cannot achieve a zero backward radiation. Let's consider an odd number of elements (n = 2q+1, where q is an integer). Using the Hansen-Woodyard condition, we have: $$ \xi\pm\left(\frac{2\pi d}{\lambda}+\frac{\pi}{2q+1}\right)=\pi $$ For backward radiation to be zero, we need \(\xi\) to be \(\pi\): $$ \pi \pm \left(\frac{2 \pi d}{\lambda}+\frac{\pi}{2q+1}\right) = \pi $$ Solve for d: $$ \frac{2 \pi d}{\lambda}+\frac{\pi}{2q+1} = 0 $$ $$ d = -\frac{\lambda}{2(2q+1)} $$ Putting the value of d back into the Hansen-Woodyard condition: $$ \pi \pm \left(2 \pi \left(-\frac{\lambda}{2(2q+1)}\right)+\frac{\pi}{2q+1}\right) = \pi $$ $$ \pi \pm \left(-\pi+\frac{\pi}{2q+1}\right) = \pi $$ For the desired zero backward radiation, the last equation must be true. However, there is no value for q that would make this equation true when n is odd. Thus, an even number of elements is required.