IPhO 1985 · Theory problem 1

Aerial array for a radio link

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A young radio amateur maintains a radio link with two girls living in two towns. He positions an aerial array such that when the girl living in town A receives a maximum signal, the girl living in town B receives no signal and vice versa. The array is built from two vertical rod aerials transmitting with equal intensities uniformly in all directions in the horizontal plane.

a) Find the parameters of the array, i.e. the distance between the rods, its orientation and the phase shift between the electrical signals supplied to the rods, such that the distance between the rods is minimum.

b) Find the numerical solution if the boy has a radio station transmitting at 27 MHz and builds up the aerial array at Portorož. Using the map he has found that the angles between the north and the direction of A (Koper) and of B (small town of Buje in Istria) are \(72^\circ\) and \(157^\circ\), respectively.

Official English text of the problem.

Answers

  • a) Let \(\alpha\) be the angular separation of the towns as viewed from the array, and let \(\lambda\) be the wavelength. The minimum spacing is \[ \boxed{d_{\min}=\frac{\lambda}{4\sin(\alpha/2)}}. \] The line joining the rods must be perpendicular to the internal bisector of the two town directions. Label the rods so that rod 2 lies toward \(\hat{\mathbf n}_A-\hat{\mathbf n}_B\) from rod 1, where the unit vectors point toward the towns. Rod 2 must lag rod 1 by \(90^\circ\) for A to receive a maximum and B a null; it must lead by \(90^\circ\) for the reverse.

  • b) The angular separation is \(85^\circ\), giving \[ \boxed{d_{\min}\approx4.1\ \mathrm m}. \] The rod-to-rod line has bearings \[ \boxed{24.5^\circ\ \text{and}\ 204.5^\circ} \] measured clockwise from north. The northeastern rod must lag the southwestern rod by \(90^\circ\) for maximum reception at Koper and zero reception at Buje; reversing the phase relation exchanges the two.

Solution

a)

The signals from the rods are coherent, have the same frequency, and have matching vertical polarizations. We use the far-field approximation: the towns are much farther from the array than the rod spacing, so the waves reaching either town have equal amplitudes and effectively parallel propagation directions.

Let the horizontal displacement from rod 1 to rod 2 be \(\mathbf d\), with magnitude \(d\). Define the electrical phase difference by \[ \delta=\phi_2-\phi_1, \] where the source oscillations are proportional to \(\cos(\omega t+\phi_j)\). Thus positive \(\delta\) means that rod 2 leads rod 1.

For a distant receiver in direction \(\hat{\mathbf n}\), the path lengths satisfy \[ r_2-r_1\simeq-\hat{\mathbf n}\cdot\mathbf d. \] Consequently, the phase difference between the received waves is \[ \Delta(\hat{\mathbf n}) =\delta-k(r_2-r_1) =\delta+k\hat{\mathbf n}\cdot\mathbf d, \qquad k=\frac{2\pi}{\lambda}. \]

If either rod alone produces intensity \(I_0\) at that receiver, superposition gives \[ I=2I_0(1+\cos\Delta) =4I_0\cos^2\!\left(\frac{\Delta}{2}\right). \] A maximum therefore requires \(\Delta=2m\pi\), whereas a null requires \(\Delta=(2n+1)\pi\).

For a maximum at A and a null at B, \[ \delta+k\hat{\mathbf n}_A\cdot\mathbf d=2m\pi, \] \[ \delta+k\hat{\mathbf n}_B\cdot\mathbf d=(2n+1)\pi. \] Subtracting eliminates the adjustable electrical phase: \[ k(\hat{\mathbf n}_A-\hat{\mathbf n}_B)\cdot\mathbf d =\text{an odd multiple of }\pi. \] Thus \[ \left|(\hat{\mathbf n}_A-\hat{\mathbf n}_B)\cdot\mathbf d\right| \geq\frac{\lambda}{2}. \]

Since the angle between the two unit vectors is \(\alpha\), \[ \left|\hat{\mathbf n}_A-\hat{\mathbf n}_B\right| =\sqrt{2-2\cos\alpha} =2\sin\frac{\alpha}{2}. \] The magnitude of a scalar product cannot exceed the product of the vector magnitudes, so \[ 2d\sin\frac{\alpha}{2}\geq\frac{\lambda}{2}. \] It follows that \[ \boxed{d_{\min}=\frac{\lambda}{4\sin(\alpha/2)}}. \]

Equality is attained when \(\mathbf d\) is parallel to \(\hat{\mathbf n}_A-\hat{\mathbf n}_B\). Moreover, \[ (\hat{\mathbf n}_A-\hat{\mathbf n}_B) \cdot(\hat{\mathbf n}_A+\hat{\mathbf n}_B)=0. \] Because the sum points along the internal angular bisector, the required rod-to-rod line is perpendicular to that bisector.

Choose the direction from rod 1 to rod 2 along \(\hat{\mathbf n}_A-\hat{\mathbf n}_B\). At the minimum spacing, \[ \hat{\mathbf n}_A\cdot\mathbf d =d_{\min}\sin\frac{\alpha}{2} =\frac{\lambda}{4}, \qquad \hat{\mathbf n}_B\cdot\mathbf d=-\frac{\lambda}{4}. \] Hence the received phase differences are \[ \Delta_A=\delta+\frac{\pi}{2}, \qquad \Delta_B=\delta-\frac{\pi}{2}. \]

For maximum reception at A, choose \[ \boxed{\delta=-\frac{\pi}{2}\pmod{2\pi}}. \] Then \(\Delta_A=0\) and \(\Delta_B=-\pi\), as required.

For maximum reception at B, choose \[ \boxed{\delta=+\frac{\pi}{2}\pmod{2\pi}}. \] Then \(\Delta_B=0\) and \(\Delta_A=\pi\). Switching between the two operating modes therefore requires a \(180^\circ\) change in the relative electrical phase.

b)

The angular separation of the towns is \[ \alpha=157^\circ-72^\circ=85^\circ. \] Using \(c\simeq3.00\times10^8\ \mathrm{m\,s^{-1}}\), the wavelength is \[ \lambda=\frac{c}{f} =\frac{3.00\times10^8}{27\times10^6} \simeq11.1\ \mathrm m. \] Therefore, \[ d_{\min} =\frac{11.1}{4\sin42.5^\circ}\ \mathrm m \simeq4.11\ \mathrm m, \] or approximately \[ \boxed{d_{\min}=4.1\ \mathrm m}. \]

The internal bisector has bearing \[ \frac{72^\circ+157^\circ}{2}=114.5^\circ. \] The perpendicular rod-to-rod line consequently has bearing \[ 114.5^\circ-90^\circ=24.5^\circ, \] with the opposite direction at \(204.5^\circ\).

Label the southwestern rod 1 and the northeastern rod 2. The projections toward the towns are \[ \hat{\mathbf n}_A\cdot\mathbf d =d\cos(72^\circ-24.5^\circ) =d\sin42.5^\circ=\frac{\lambda}{4}, \] \[ \hat{\mathbf n}_B\cdot\mathbf d =d\cos(157^\circ-24.5^\circ) =-\frac{\lambda}{4}. \] Thus the phase settings are:

Desired reception Phase of northeastern rod relative to southwestern rod
Koper maximum; Buje zero Lags by \(90^\circ\)
Buje maximum; Koper zero Leads by \(90^\circ\)

The solution was checked against the official solution.