Chapter 15: Arrays

1. Provide the definition of the directivity function \(D(\theta,\psi)\), directivity factor \(D_i\), and directivity index \(\text{DI}\). [answer]

   The directivity is defined as \[D(\theta,\psi) = \frac{p(r,\theta,\psi)}{p(r,\theta_m,\psi_m)} = \frac{\hat{u}(k\alpha,k\beta)}{\hat{u}(0,0)},\] where \(\theta_m\) and \(\psi_m\) are the polar and azimuthal angles corresponding the direction of the beam's maximum. Note that \(\alpha = \sin \theta \cos\psi\) and \(\beta = \sin\theta\sin\psi\). The \(z\) axis is aligned such that \(\theta_m =0\).

   The directivity factor is \[D_i = \frac{\langle I_\text{max} \rangle }{\langle I_\text{mono} \rangle},\] where \(\langle I_\text{max} \rangle \) is the maximum intensity in the beam (along the \(z\) axis) and \(\langle I_\text{mono} \rangle \) is the intensity at the same distance due to a monopole radiating the same power \(W\).

   The directivity index is simply \(\text{DI} = 10 \log_{10} D_i\) (dB).

2. ☸ Derive an expression for the directivity factor \(D_i\) in terms of the directivity function \(D(\theta, \psi)\). Note from the definition of directivity in the previous problem that \[|D(\theta,\psi)|^2 = \frac{\langle I(\theta,\psi) \rangle }{\langle I_\text{max} \rangle }.\] Combine \(\langle I_\text{mono} \rangle = W/4\pi r^2\) and \(W = \oint \langle I(\theta,\psi) \rangle dS\) to eliminate \(W\) and invoke \(\langle I_\text{max} \rangle |D(\theta,\psi)|^2 = \langle I(\theta,\psi) \rangle \). [answer]

   Eliminate \(W\): \begin{align*} \langle I_\text{mono} \rangle &= \frac{1}{4\pi r^2} \oint \langle I(\theta,\psi) \rangle \, dS\\ &= \frac{1}{4\pi r^2} \oint \langle I(\theta,\psi) \rangle \, r^2 \,d\Omega\\ &= \frac{1}{4\pi} \oint \langle I(\theta,\psi) \rangle \, d\Omega \end{align*} Invoke \(\langle I(\theta,\psi) \rangle = \langle I_\text{max} \rangle |D(\theta,\psi)|^2\): \begin{align*} \langle I_\text{mono} \rangle &= \frac{1}{4\pi} \oint \langle I_\text{max} \rangle |D(\theta,\psi)|^2 \, d\Omega\\ \langle I_\text{mono} \rangle &= \frac{\langle I_\text{max} \rangle }{4\pi} \oint |D(\theta,\psi)|^2 \, d\Omega \end{align*} Therefore, the directivity factor is \begin{align*} D_i &= \frac{\langle I_\text{max} \rangle }{\langle I_\text{mono} \rangle } = \frac{4\pi}{\oint |D(\theta,\psi)|^2 \, d\Omega} \end{align*}

3. Calculate the directivity \(D(\theta,\psi)\) due to \[u_0(x,y) = Q_0 [\delta(x-d/2) + \delta(x+d/2)]\delta(y)\] i.e., two in-phase monopoles separated by distance \(d\). [answer]

   The directivity is simply given by \begin{align*} D(\theta,\psi) &= \frac{\hat{u}_0(k_x,k_y)}{\hat{u}_0(0,0)} \end{align*} The numerator is evaluated using the sifting property of the \(\delta\) function: \begin{align*} \hat{u}_0(k_x,k_y) &= Q_0 \iint_{-\infty}^\infty [\delta(x-d/2) + \delta(x+d/2)]\delta(y) e^{j(k_xx + k_yy)} dx dy\\ &= Q_0 \int_{-\infty}^{\infty} [\delta(x-d/2) + \delta(x+d/2)] e^{jk_xx} dx \int_{-\infty}^{\infty} \delta(y) e^{jk_yy}dy\\ &= Q_0 (e^{jk_xd/2} + e^{-jk_xd/2}) e^0\\ &=2Q_0 \cos (k_x d/2) \end{align*} Recalling that \(k_x = k\sin\theta\cos\psi\), the numerator is \begin{align*} \hat{u}_0(k_x,k_y) &=2Q_0 \cos [(kd/2)\sin\theta\cos\psi]\,. \end{align*} Meanwhile, the denominator is simply \(\hat{u}_0(k_x,k_y) = 2Q_0\). Thus the directivity is \begin{align*} D(\theta,\psi) &= \cos [(kd/2)\sin\theta\cos\psi]\,. \end{align*} The directivity is plotted below for \(kd\in [0,35]\).

4. A line array of \(N\) point sources radiates a pressure field given by \[p(r,\theta) = A \frac{e^{-jkr}}{r} \sum_{n=0}^{N-1} e^{jnkd\sin\theta},\] where \(\theta = 0\) points in the direction perpendicular to the line of sources. Given that \[\sum_{n=0}^{N-1} e^{j2n\phi} = \frac{1-e^{j2N\phi}}{1-e^{j2\phi}},\] show that the directivity of the line array is \[D(\theta) = \frac{\sin N\phi}{N \sin\phi},\] where \(\phi = \frac{kd}{2}\sin(\theta) \). Evaluate the result for \(\theta= 0^\circ\). What is this type of radiation called? What is the type of radiation for \(\theta =90^\circ\) called? [answer]

   The right-hand side of the given identity is manipulated into the desired form. \begin{align*} \sum_{n=0}^{N-1} e^{jnkd\sin\theta} &= \frac{1-e^{j2N\phi}}{1-e^{j2\phi}}\\ &= \frac{e^{jN\phi}}{e^{j\phi}} \bigg(\frac{e^{-jN\phi}-e^{jN\phi}}{e^{-j\phi}-e^{j\phi}}\bigg)\\ &= \frac{e^{jN\phi}}{e^{j\phi}} \bigg(\frac{e^{jN\phi}-e^{-jN\phi}}{e^{j\phi}-e^{-j\phi}}\bigg)\\ &= e^{j(N-1)\phi} \frac{\sin (N\phi)}{\sin \phi}\\ &= e^{j(N-1)kd\sin(\theta)/2} \frac{\sin [Nkd\sin(\theta)/2]}{\sin [kd\sin(\theta)/2]} \end{align*} The pressure field is therefore \begin{align*} p(r,\theta)= A \frac{e^{-jkr}}{r} e^{j(N-1)kd\sin(\theta)/2} \frac{\sin [Nkd\sin(\theta)/2]}{\sin [kd\sin(\theta)/2]} \end{align*} Meanwhile, plotting reveals that the maximum occurs at \(\theta_m=0\), for which the pressure is \begin{align*} p(r,\theta_m)= A \frac{e^{-jkr}}{r} N \end{align*} The directivity is therefore \begin{align*} D(\theta) = \frac{p(r,\theta)}{p(r,\theta_m)} = \frac{\sin [Nkd\sin(\theta)/2]}{N\sin [kd\sin(\theta)/2]}\,. \end{align*} For \(\theta = 0\), \(D(0) = 1\). This is referred to as ''broadside radiation.'' For \(\theta = 90^\circ\), the radiation is called ''end-fire.''

5. Obtain the directivity \(D(\theta,\psi)\) for a continuous line array whose source condition is \(u_0(x,y) = u_0 w\delta(y)\) from \(-L/2 \leq x \leq L/2\) and \(0\) otherwise. [answer]

   The directivity is calculated using Fourier transforms: \begin{align*} D(\theta,\psi) &= \frac{\hat{u}_0 (k_x,k_y)}{\hat{u}_0 (0,0)}\,. \end{align*} The numerator is given by \begin{align*} \hat{u}_0 (k_x,k_y) &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} u_0 w\delta(y) e^{jk_xx}e^{jk_yy} dxdy\\ &=u_0 w \int_{-L/2}^{L/2} e^{jk_xx} dx\\ &=\frac{u_0 w }{jk_x} e^{jk_xx} \bigg\rvert_{-L/2}^{L/2}\\ &=\frac{u_0 w }{jk_x}[e^{jk_xL/2} - e^{-jk_xL/2}]\\ &=\frac{2 u_0 w }{k_x}\sin{k_xL/2}\\ &={u_0 L w }\frac{\sin{k_xL/2}}{k_x L/2} \end{align*} Meanwhile, the denominator is \begin{align*} \hat{u}_0 (0,0) &= u_0 L w\,. \end{align*} Recall that \(k_x = k\sin\theta \cos \psi\). Thus the directivity is \begin{align*} D(\theta,\psi) &= \frac{\hat{u}_0 (k_x,k_y)}{\hat{u}_0 (0,0)}\\ &= \frac{\sin{[(kL/2) \sin\theta \cos \psi]}}{(kL/2) \sin\theta \cos \psi}\,. \end{align*}

6. Consider a circular piston, whose directivity is given by the jinc fuction, \(D(\theta) = 2J_1(\kappa a)/\kappa a\). The wavenumber is \begin{align*} \kappa &= \sqrt{k_x^2 + k_y^2}\\ k_x &= k\sin\theta \cos\psi\\ k_y &= k\sin\theta \cos\psi \end{align*} Suppose the beam is now steered with \(k_x\) with \(k(\sin \theta \cos\psi - \sin\theta_0)\). What will the new directivity be? [answer]

   \[D(\theta) = \frac{2J_1[ka(\sin\theta -\sin\theta_0)]}{ka(\sin\theta -\sin\theta_0)}\].

7. ☸ A distribution of \(N\) velocity sources, each with a source condition of \(u_1(x-x_n,y-y_n)\), has a total source condition of \[u_0(x,y) = \sum_{n=1}^N u_1(x-x_n, y-y_n)\,.\] Prove that the directivity of the distribution is the product of the directivity of each source and the directivity of the array, where the array is treated as a collection of point sources. Hint: write the above source condition in terms of \(\delta\)-function, \[u_0(x,y) = \sum_{n=1}^N \iint_{-\infty}^\infty u_1(x-x_n, y-y_n) \delta(x-x_n)\delta(y-y_n)\, dx\, dy\,,\] and take the 2D Fourier transform. [answer]

   The Fourier transform of the velocity source condition \[u_0(x,y) = \sum_{n=1}^N \iint_{-\infty}^\infty u_1(x-x_n, y-y_n) \delta(x-x_n)\delta(y-y_n)\, dx\, dy\] is taken by noting that the Fourier transform of \(\delta(x-x_n)\) is \(e^{jk_xx_n}\), and similarly the Fourier transform of \(\delta(y-y_n)\) is \(e^{jk_yy_n}\): \[\hat{u}_0(k_x,k_y) = \hat{u}_1(k_x,k_y) \sum_{n} e^{j(k_xx_n +k_yy_n)}\,.\] Dividing by \(\hat{u}_0(0,0) = \hat{u}_1(0,0) \sum_n e^{j(0x_n + 0y_n)} = \hat{u}_1(0,0)\) gives the directivity (see problem 1 of this section), \[D(\theta,\psi) = D_1(\theta,\psi)D_A(\theta,\psi)\,.\]

8. Suppose there are two circular pistons of radius \(a\) separated by a distance \(d\) from each other (center-to-center). What is the directivity of the two pistons? [answer]

   The product theorem makes the problem easy. The directivity of each piston is \[D_1(\theta) = \frac{2J_1(ka\sin\theta)}{ka\sin\theta}\] while the directivity for the \(N=2\) line array is \[D_A(\theta, \psi) = \cos[(kd/2) \sin\theta\cos\psi]\,.\] The directivity of the two circular pistons of radius \(a\) separated by a distance \(d\) is thus \[D(\theta,\psi) = \frac{2J_1(ka\sin\theta)}{ka\sin\theta}\cos[(kd/2) \sin\theta\cos\psi]\,. \]