The far-zone electric-field intensity (array factor) of an
end-fire two-element

array antenna, placed along the z-axis and radiating into
free-space, is given by

E = cos

)π

4

(cos θ − 1)

* e

−jkr

r

, 0 ≤ θ ≤ π

Find the directivity using

(a) Kraus’ approximate formula

(b) the Directivity computer program of this chapter.

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​​​​​​Solution:From the given data of the QuestionE_{\max }=\cos \left[\frac{\pi}{4}(\cos \theta-1)\right]_{\max }=1Put \theta=0^{\circ}\begin{array}{c}\therefore 0.707 E_{\max }=0.707(1) \\=\cos \left[\frac{\pi}{4}(\cos \theta,-1)\right] \\\therefore \frac{\pi}{4}(\cos \theta,-1)= \pm \frac{\pi}{4}\end{array}\begin{array}{l}\Rightarrow \theta_{1}=\left[\begin{array}{l}\cos ^{-1}(2) \\\& \cos ^{-1}(0)=90^{\circ} \Rightarrow \frac{\pi}{2} \mathrm{rad} .\end{array}\right. \\H_{1 r}=H_{2 r}=2\left(\frac{\pi}{x}\right)=\pi \\\therefore H_{1 r}=\pi_{2 r}=\pi \\\end{array}\begin{aligned}\therefore D_{0}=\frac{4 \pi}{H_{1 r} \cdot H_{2 r}} & =\frac{4 \pi}{\pi^{2}}=\frac{4}{\pi} \\& =1.273 \\D_{0} & =1.048 \mathrm{~dB}\end{aligned} ...