Question Three isotropic sources, with spacing d between them, are placed along the z-axis. The excitation coefficient of each outside element is unity while that of the center element is 2. For a spacing of d = ?/4 between the elements, find the (a) array factor (b) angles (in degrees) where the nulls of the pattern occur (0? = ?? = 180?) (c) angles (in degrees) where the maxima of the pattern occur (0? = ?? = 180?) (d) directivity using the computer program Directivity of Chapter 2.

Problen 1:\Rightarrow Three isotropic sources spacon 'd along z-acis.\Rightarrow Center element is 2d=\lambda / 4Solulion:Frion given, number of elemenis, P=3\begin{aligned}\therefore \quad 2 M+1 & =P \\2 M+1 & =3 \\M & =1\end{aligned}(a) Array factor:\Rightarrow Excitation can be eocpressed as,\begin{array}{l}E_{t}=E_{1}+E_{2}+E_{3} \\E_{t}=2 E_{0} \frac{e^{-j k \gamma}}{\gamma}+E_{0} \frac{e^{-j k r_{1}}}{\gamma_{1}}+E_{0} \frac{e^{-j k r_{2}}}{r_{2}}\end{array}For field \Rightarrow \quad r_{1}=r_{2}=r \rightarrow Amphtude\begin{array}{l}\gamma_{1}=\tau-d \cos \theta \rightarrow \text { phase } \\\gamma_{2}=\gamma+d \cos \theta \rightarrow \text { phase }\end{array}\Rightarrow substirute the respecttve values in excitation equation.\begin{array}{l}E_{t}=2 E_{0} \frac{e^{-j k \gamma}}{\gamma}+E_{0} \frac{e^{-j k(\tau-d \cos \theta)}}{\gamma}+E_{0} \frac{e^{-j k(\gamma+d \cos \theta)}}{\gamma} \\E_{t}=E_{0} e^{-j k \tau}\left[2+e^{j k d \cos \theta}+e^{-j k d \cos \theta}\right] \\E_{t}=E_{0} \frac{e^{-j k r}}{r}\left[2\left(1+\frac{1}{2} e^{j j k d \cos \theta}+\frac{1}{2} e^{-j k d \cos \theta}\right]\right. \\E_{t}=E_{0} \frac{e^{-j k \gamma}}{\gamma}[2+2 \cos (k d \operatorname{coc} \theta)] \\E_{t}=2 E_{0} \frac{e^{-j k \gamma}}{\gamma}(1+\cos (k d \cos \theta)) \\E_{t}=E_{0} \frac{e^{-j k r}}{r}[2(1+\cos (k d \cos 0))] \\\end{array}Array factor.\begin{aligned}A F(\theta) & =2[1+\cos (K d \cos \theta)] \\& =2\left(2 \cos ^{2}\left(\frac{K d}{2} \cos \theta\right)\right) \\& =4 \cos ^{2}\left(\frac{k d}{2} \cos \theta\right)\end{aligned}Normalized form is:\begin{aligned}\operatorname{AF}\left(\theta_{n}\right) & =\frac{A F(\theta)}{2} \\& =2 \cos ^{2}\left(\frac{k d}{2} \cos \theta\right)\end{aligned}(b) Angle in degrees where nulls occur 0^{\circ} \leq \theta \leq 160^{\circ} :\Rightarrow Drray factor: A F(\theta) \Rightarrow 2(1+\cos (k d \cos \theta))=0\begin{array}{l}\therefore 1+\cos \left(k d \cos \theta_{n}\right)=0 \\\cos \left(k d \cos \theta_{n}\right)=-1 \\k d\left(\cos \theta_{n}\right)=\cos ^{-1}(-1) \\k d\left(\cos \theta_{n}\right)=n \pi \quad, n= \pm 1, \pm 3, \cdots \\\left(\frac{2 \pi}{\lambda}\right)\left(\frac{\lambda}{4}\right) \cos \theta_{n}=n \pi \\\frac{\pi}{2} \cos \theta_{n}=n \pi \Rightarrow \cos \theta_{n}=2 n \\\theta_{n}=\cos ^{-1}(2 n) \text { for } n= \pm 1, \pm 3 . \pm 5, \ldots\end{array}\Rightarrow For such condition no null can exists.(c) Angle where maxima of pattern occur 0^{\circ} \leqslant \theta \leqslant 180^{\circ}.Amray factor for maxcimasA F(\theta)=1+\cos (\text { kd } \cos \theta m)=2\cos (k d \cos \theta m)=2-1\text { kd } \begin{aligned}\cos \theta m & =\cos ^{-1}(1) \\\text { kd } \cos \theta m & =2 m \pi, m=0, \pm 1, \pm 2, \ldots \\\left(\frac{2 \pi}{\lambda}\right)\left(\frac{\lambda}{4}\right) \cos \theta m & =2 m \pi \\\left(\frac{\pi}{2}\right) \cos \theta m & =2 m \pi \\\pi(\cos \theta m) & =14 \pi \\\cos \theta m & =4 m \\\theta m & =\cos ^{-1}(14 m) \quad m=0, \pm 1, \pm 0,\end{aligned}\Rightarrow angle can exists at " \theta " of m=0\theta m=\cos ^{-1}(0)Macuma angle, \theta m=90^{\circ} ...