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# 5. Antenna Arrays ## 5.1 Introduction * An array is a set of $N$ antennas. * The total field radiated by the array is the vector sum of the fields radiated by the individual antennas. * By properly phasing the elements, the fields from the individual antennas can add constructively in the...

# 5. Antenna Arrays ## 5.1 Introduction * An array is a set of $N$ antennas. * The total field radiated by the array is the vector sum of the fields radiated by the individual antennas. * By properly phasing the elements, the fields from the individual antennas can add constructively in the desired direction and destructively in other directions. * Arrays provide increased gain and directivity compared to single antenna elements. **Array factor**: the radiation pattern of an array of isotropic elements. **Overall array pattern**: array factor $\times$ element pattern ## 5.2 Two-Element Array ### 5.2.1 General Case * Consider two antennas separated by a distance $d$, as shown in the figure. * The total electric field at point $P$ is given by $E = E_1 + E_2 = E_0 \frac{e^{-jkr_1}}{r_1} + E_0 \frac{e^{-jkr_2}}{r_2}e^{j\beta}$ (We have included a phase excitation $\beta$ in the second antenna) In the far field, we have $r_1 \approx r$ and $r_2 \approx r - d\cos\theta$. Then $E = E_0 \frac{e^{-jkr}}{r} [1 + e^{jkd\cos\theta}e^{j\beta}]$ The array factor is therefore $AF = 1 + e^{j(kd\cos\theta + \beta)}$ $|AF| = |1 + e^{j(kd\cos\theta + \beta)}| = |e^{-j\psi/2}||e^{j\psi/2} + e^{-j\psi/2}|$ where $\psi = kd\cos\theta + \beta$. Therefore $|AF| = |2\cos(\psi/2)| = 2|\cos(\frac{kd\cos\theta + \beta}{2})|$ ### 5.2.2 Special Cases 1. **Broadside Array**: $\beta = 0$ $|AF| = 2|\cos(\frac{kd\cos\theta}{2})|$ Maximum radiation occurs at $\theta = \pi/2$. 2. **End-Fire Array**: $\beta = -kd$ $|AF| = 2|\cos(\frac{k d(\cos\theta - 1)}{2})|$ Maximum radiation occurs at $\theta = 0$. 3. **End-Fire Array with Increased Directivity**: $\beta = -kd + \pi/2$ $|AF| = 2|\cos(\frac{kd\cos\theta - kd + \pi/2}{2})|$ Maximum radiation occurs at $\theta = 0$. 4. **Phased Array**: By adjusting the phase $\beta$, the direction of maximum radiation can be steered to any desired angle. ## 5.3 N-Element Array: Uniform Amplitude and Spacing * Consider $N$ antennas placed along the $z$ axis with uniform spacing $d$. The excitation phase between the elements is $\beta$. The array factor is given by: $AF = \sum_{n=1}^{N} e^{j(n-1)(kd\cos\theta + \beta)}$ Let $\psi = kd\cos\theta + \beta$. Then $AF = \sum_{n=1}^{N} e^{j(n-1)\psi} = \frac{1 - e^{jN\psi}}{1 - e^{j\psi}} = \frac{e^{jN\psi/2}}{e^{j\psi/2}}\frac{e^{-jN\psi/2} - e^{jN\psi/2}}{e^{-j\psi/2} - e^{j\psi/2}}$ $AF = e^{j(N-1)\psi/2} \frac{\sin(N\psi/2)}{\sin(\psi/2)}$ $|AF| = |\frac{\sin(N\psi/2)}{\sin(\psi/2)}|$ $|AF| = |\frac{\sin(\frac{N}{2}(kd\cos\theta + \beta))}{\sin(\frac{1}{2}(kd\cos\theta + \beta))}|$ * Maxima occur when $\psi = 0, \pm 2\pi, \pm 4\pi,...$ $\rightarrow kd\cos\theta + \beta = 0, \pm 2\pi, \pm 4\pi,...$ * **Broadside Array**: $\beta = 0$ $\theta = \cos^{-1}(\pm \frac{2n\pi}{kd})$, $n = 0, 1, 2,...$ The maximum radiation occurs at $\theta = \pi/2$. * **End-Fire Array**: $\beta = -kd$ $\theta = \cos^{-1}(1 \pm \frac{2n\pi}{kd})$, $n = 0, 1, 2,...$ The maximum radiation occurs at $\theta = 0$. * **Nulls**: $N\psi/2 = \pm \pi, \pm 2\pi, \pm 3\pi,...$ $\frac{N}{2}(kd\cos\theta + \beta) = \pm \pi, \pm 2\pi, \pm 3\pi,...$ $kd\cos\theta + \beta = \pm \frac{2n\pi}{N}$, $n = 1, 2, 3,...$ * **HPBW**: The half-power beamwidth (HPBW) is the angle between the two points on the main lobe where the power is half of its maximum value. The HPBW can be approximated as: $HPBW \approx \frac{2\sin^{-1}(0.443\lambda/d)}{N\cos\theta_0}$ where $\theta_0$ is the angle of maximum radiation. * **FNBW**: The first null beamwidth (FNBW) is the angle between the two points on the main lobe where the radiation pattern is zero. The FNBW can be approximated as: $FNBW \approx \frac{2\sin^{-1}(\lambda/d)}{N\cos\theta_0}$ * **Directivity**: The directivity of an array of $N$ elements with uniform spacing $d$ is given by: $D \approx \frac{4\pi U_{max}}{P_{rad}} = \frac{2Nd}{\lambda}$