# Linear Array Design with Specified Nulls Given: - Isotropic elements along the $z$-axis - Zeros of the array factor at $\theta = ^\circ,\ 60^\circ,\ 120^\circ$ - Element spacing $d = \lambda/4$ - Progressive phase shift = $B$ Let's solve step by step. --- ## (a) Required Number of Elements The general form of the array factor for a uniform linear array of $N$ elements is: $$ AF(\theta) = \sum_{n=}^{N-1} a_n e^{jn\psi} $$ with $$ \psi = kd \cos\theta + B $$ where $k = 2\pi/\lambda$, $d$ is the element spacing, $B$ is the progressive phase, and $a_n$ are the excitation coefficients. We want the array factor to have **nulls at three specified angles**. For a polynomial to have $M$ nulls, it must be at least of order $M$, i.e., $(N-1) \geq M$. - **Number of required nulls:** 3 - **Minimum order of array:** 3 - $\Rightarrow N = 4$ elements --- ## (b) Excitation Coefficients ### Step 1: Find $\psi$ values at the nulls Given $d = \lambda/4$, $k = 2\pi/\lambda$: $$ kd = 2\pi/\lambda \cdot \lambda/4 = \frac{\pi}{2} $$ So, $$ \psi(\theta) = \frac{\pi}{2} \cos\theta + B $$ Nulls at: - $\theta_1 = ^\circ \implies \cos ^\circ = 1 \implies \psi_1 = \frac{\pi}{2} + B$ - $\theta_2 = 60^\circ \implies \cos 60^\circ = .5 \implies \psi_2 = \frac{\pi}{4} + B$ - $\theta_3 = 120^\circ \implies \cos 120^\circ = -.5 \implies \psi_3 = -\frac{\pi}{4} + B$ ### Step 2: Array Factor as a Polynomial For $N=4$ elements, the general array factor (if symmetric) is: $$ AF(\psi) = a_ + a_1 e^{j\psi} + a_2 e^{j2\psi} + a_3 e^{j3\psi} $$ Since we want zeros at $\psi_1, \psi_2, \psi_3$, the factor can be written as: $$ AF(\psi) = K (\exp(j\psi) - e^{j\psi_1})(\exp(j\psi) - e^{j\psi_2})(\exp(j\psi) - e^{j\psi_3}) $$ Or, equivalently, expanding: $$ AF(\psi) = a_ + a_1 e^{j\psi} + a_2 e^{j2\psi} + a_3 e^{j3\psi} $$ We can solve for $a_, a_1, a_2, a_3$ by expanding the product: Let $x = e^{j\psi}$, $$ AF(x) = (x - x_1)(x - x_2)(x - x_3) $$ where $x_k = e^{j\psi_k}$. Expanding: $$ (x - x_1)(x - x_2)(x - x_3) = x^3 - (x_1 + x_2 + x_3)x^2 + (x_1x_2 + x_1x_3 + x_2x_3)x - x_1x_2x_3 $$ So, - $a_3 = 1$ - $a_2 = - (x_1 + x_2 + x_3)$ - $a_1 = (x_1x_2 + x_1x_3 + x_2x_3)$ - $a_ = - x_1x_2x_3$ #### Compute $x_1, x_2, x_3$ - $x_1 = e^{j(\frac{\pi}{2} + B)}$ - $x_2 = e^{j(\frac{\pi}{4} + B)}$ - $x_3 = e^{j(-\frac{\pi}{4} + B)}$ --- ## (c) Array Factor ### Final Array Factor Let $x = e^{j\psi}$, and $\psi = \frac{\pi}{2}\cos\theta + B$. The array factor is: $$ AF(\theta) = a_ + a_1 x + a_2 x^2 + a_3 x^3 $$ where - $a_3 = 1$ - $a_2 = - (x_1 + x_2 + x_3)$ - $a_1 = (x_1x_2 + x_1x_3 + x_2x_3)$ - $a_ = - x_1x_2x_3$ with - $x_1 = e^{j(\frac{\pi}{2} + B)}$ - $x_2 = e^{j(\frac{\pi}{4} + B)}$ - $x_3 = e^{j(-\frac{\pi}{4} + B)}$ --- ## **Summary Table** | Part | Answer | |------|--------| | (a) | **Number of elements $N = 4$** | | (b) | **Excitation coefficients:** <br> $a_3 = 1$ <br> $a_2 = - (x_1 + x_2 + x_3)$ <br> $a_1 = (x_1x_2 + x_1x_3 + x_2x_3)$ <br> $a_ = - x_1x_2x_3$ <br> where $x_k = e^{j\psi_k}$ with $\psi_1 = \frac{\pi}{2} + B$, $\psi_2 = \frac{\pi}{4} + B$, $\psi_3 = -\frac{\pi}{4} + B$ | | (c) | **Array factor:** <br> $AF(\theta) = a_ + a_1 e^{j\psi} + a_2 e^{j2\psi} + a_3 e^{j3\psi}$ <br> with $\psi = \frac{\pi}{2}\cos\theta + B$ | --- ## **References** - Balanis, C.A. "Antenna Theory: Analysis and Design", 4th Edition. - [Linear Array Factor](https://en.wikipedia.org/wiki/Antenna_array#Array_factor) --- **Note:** If you want numeric values for the coefficients for a specific $B$ (e.g., $B=$), substitute $B=$ above and compute $x_1, x_2, x_3$ explicitly.