Question 4 of 7: Dipole radiation, line-source and plane-source outdoor propagation

Question 4: Dipole radiation, line-source and plane-source outdoor propagation (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Three separate outdoor-propagation configurations: a 1 kHz dipole built from two 10 mW monopoles 50 mm apart; a 10 m elevated steam pipe radiating 110 dB in the 1 kHz octave band; and a 1 m square facade opening radiating 0.02 W in the 2 kHz octave band across a hard parking lot to a house 150 m away.

Find. Intensity and sound pressure level in two directions from the dipole; the pipe's sound pressure level at two distances and with a reflecting floor; and the full break-out calculation for the opening, ending in the level at the nearest house with one and with two openings.

Approach. Each part is a divergence calculation with a different source model: two-source interference for the dipole, spherical divergence for the pipe once it is shown to be acoustically compact at these distances, and hemispherical divergence plus ground and atmospheric terms for the opening.

Part A — Dipole

1. Establish the field of one monopole and the interference factor. A single monopole radiating $W_1$ into a free field produces, at radius $r$, an intensity $I_1 = W_1/4\pi r^2$ and hence a mean-square pressure $\langle p^2\rangle_1 = \rho_0 c\,W_1/4\pi r^2$. Two equal sources driven in antiphase and separated by $d$ superpose to give$$\langle p^2\rangle = 2\langle p^2\rangle_1\left[1-\cos(kd\cos\theta)\right]$$where $\theta$ is measured from the line joining them and $k=2\pi f/c$. Here $k = 2\pi(1000)/343 = 18.32\ \text{m}^{-1}$ and $kd = 18.32\times0.05 = 0.916$, which is not small — the compact-dipole approximation $\cos^2\theta$ would be in error, so the exact two-source form is retained.
2. Evaluate the intensity at 1.0 m, $45^\circ$ off the axis. At $r=1.0$ m, $\langle p^2\rangle_1 = 413\times0.010/(4\pi\times1.0^2) = 0.3287$ Pa$^2$, and $kd\cos 45^\circ = 0.916\times0.7071 = 0.6475$ rad with $\cos(0.6475) = 0.7976$. Therefore$$\langle p^2\rangle = 2(0.3287)(1-0.7976)=0.1331\ \text{Pa}^2$$$$I=\frac{\langle p^2\rangle}{\rho_0 c}=\frac{0.1331}{413}$$$$\boxed{I_{45^\circ}=3.22\times10^{-4}\ \text{W/m}^2}$$
3. Convert that intensity to a sound pressure level. Using $L_p = 10\log_{10}\!\left(\langle p^2\rangle/p_{\text{ref}}^2\right)$ with $p_{\text{ref}}^2 = 4\times10^{-10}$ Pa$^2$:$$L_p=10\log_{10}\frac{0.1331}{4\times10^{-10}}$$$$\boxed{L_{p,45^\circ}=85.2\ \text{dB}}$$For comparison, one monopole alone at 1.0 m would give 89.1 dB; the antiphase partner has therefore taken away about 4 dB at this angle rather than adding 3 dB, which is the signature of dipole cancellation.
4. Repeat on the axis at 1.5 m. Now $\theta = 0$, so the interference factor takes its largest value: $\cos(kd) = \cos(0.916) = 0.6092$, giving $1-0.6092 = 0.3908$. At $r=1.5$ m, $\langle p^2\rangle_1 = 413\times0.010/(4\pi\times2.25) = 0.1461$ Pa$^2$, so$$\langle p^2\rangle = 2(0.1461)(0.3908)=0.1142\ \text{Pa}^2$$$$\boxed{I_{\text{axis}}=2.77\times10^{-4}\ \text{W/m}^2\qquad L_{p,\text{axis}}=84.6\ \text{dB}}$$The on-axis point is 0.6 dB quieter than the $45^\circ$ point despite lying in the strongest lobe, simply because it is half as far again from the source; the extra divergence loss of $20\log_{10}(1.5) = 3.5$ dB slightly outweighs the 2.9 dB directivity gain.

It is worth stating the limiting case explicitly, because it is the physical point of the question: in the broadside direction, on the perpendicular bisector, the two paths are exactly equal, the antiphase contributions cancel completely, and the radiated intensity is zero. That null is what distinguishes a dipole — an unbaffled loudspeaker cone, a vibrating beam, an oscillating rod — from a monopole such as a pulsating sphere or a small open pipe end.

Part B — Elevated line source

1. Decide which source model applies. A finite line source behaves as a line (cylindrical divergence, $-3$ dB per distance doubling) only in its near field, $r \lt L/\pi$; beyond that it collapses to a point source with spherical divergence ($-6$ dB per doubling). Here $L/\pi = 10/\pi = 3.18$ m, and the receivers are at 50 m and 120 m, so both are far-field points. Integrating the finite line exactly at 50 m confirms this: it differs from the point-source result by only 0.01 dB.
2. Apply free-field spherical divergence at 50 m. With no reflecting surfaces the power spreads over a full sphere:$$L_p = L_W - 10\log_{10}\!\left(4\pi r^2\right)$$At $r = 50$ m, $4\pi r^2 = 31\,416\ \text{m}^2$ and $10\log_{10}(31\,416) = 44.97$ dB, so$$\boxed{L_p(50\ \text{m}) = 110 - 44.97 = 65.0\ \text{dB}}$$This is the familiar $L_p = L_W - 20\log_{10}r - 11$ written out in full.
3. Repeat at 120 m. Now $4\pi r^2 = 180\,956\ \text{m}^2$ and $10\log_{10}(180\,956)=52.57$ dB, giving$$\boxed{L_p(120\ \text{m}) = 57.4\ \text{dB}}$$The 7.6 dB drop from 50 m to 120 m matches $20\log_{10}(120/50)=7.6$ dB, confirming pure spherical spreading.
4. Add the hard concrete floor. Resting the pipe on a hard, acoustically rigid floor removes half the radiating solid angle; all the power is now forced into a hemisphere, so the divergence area halves and the level rises by 3 dB:$$L_p = L_W - 10\log_{10}\!\left(2\pi r^2\right) = 110 - 49.57$$$$\boxed{L_p(120\ \text{m, on floor}) = 60.4\ \text{dB}}$$Equivalently, the directivity factor has changed from $Q=1$ to $Q=2$.

Part C — Break-out through a facade opening

1. Sound power level of the opening (vii). Direct from the definition,$$L_W = 10\log_{10}\!\left(\frac{W}{10^{-12}}\right)=10\log_{10}\!\left(\frac{0.02}{10^{-12}}\right)=10\log_{10}(2\times10^{10})$$$$\boxed{L_W = 103.0\ \text{dB}}$$
2. Level at 6 m with no excess attenuation (viii). Hemispherical divergence into the half space in front of the facade gives$$L_p = L_W - 10\log_{10}\!\left(2\pi r^2\right)=103.0-10\log_{10}(226.2)=103.0-23.5$$$$\boxed{L_p(6\ \text{m}) = 79.5\ \text{dB}}$$This is the level a person standing just outside the opening would experience — already an intrusive level for a residential setting.
3. Excess attenuation due to ground reflection (ix). Over hard, flat ground the reflected ray arrives essentially in phase with the direct ray, so the two contributions add coherently and the mean-square pressure doubles:$$A_g = -10\log_{10}(2)$$$$\boxed{A_g = -3\ \text{dB}}$$The negative sign is not a bookkeeping accident: a hard ground is a gain, not a loss. Only a soft, porous ground (grass, ploughed soil) produces the familiar positive ground attenuation, and then chiefly in the 250–500 Hz region.
4. Atmospheric absorption to the nearest house (x). Table 5.3 gives $m = 15.5$ dB per 1 000 m at 2 kHz for 25 % relative humidity and $20\ {}^\circ\text{C}$. Over 150 m,$$A_{\text{atm}} = m\,\frac{r}{1000}=15.5\times\frac{150}{1000}$$$$\boxed{A_{\text{atm}} = 2.33\ \text{dB}}$$The 2 kHz band is where molecular relaxation absorption first becomes significant; at 250 Hz the same path would lose only 0.23 dB, and at 4 kHz it would lose 8.1 dB.
5. Total level at the nearest house (xi). Assemble the terms — divergence, ground and atmosphere:$$L_p = L_W - 10\log_{10}\!\left(2\pi r^2\right) - A_g - A_{\text{atm}}$$With $10\log_{10}(2\pi\times150^2)=10\log_{10}(141\,372)=51.5$ dB,$$L_p = 103.0 - 51.5 + 3.0 - 2.33$$$$\boxed{L_p(150\ \text{m}) = 52.2\ \text{dB}}$$For context, a 52 dB level in the 2 kHz band alone would breach a typical Canadian municipal night-time limit of 45 dB(A), so this opening would need treatment.
6. Effect of a second identical opening (xii). Two openings close together but not phase-locked radiate incoherently at the receiver, so their powers — not their pressures — add:$$L_{p,\text{total}} = L_p + 10\log_{10}(2) = 52.2 + 3.0$$$$\boxed{L_{p,\text{2 openings}} = 55.2\ \text{dB}}$$Because they are close together relative to the 150 m propagation distance, the divergence, ground and atmospheric terms are unchanged and only the source power doubles.