The relationship between the maximum obtainable directivity and aperture of an antenna is described by:

$$ D = \frac{4\pi A_e}{\lambda^2} $$

Where $D$ is the directivity, $A_e$ the effective aperture and $\lambda$ the used wavelength. The derivation can be found in the following document: pdf.

However, if the antenna does not scan as a sphere but in $x, y$ but not $z$. Following the same path as the above-mentioned document the equation becomes:

$$ D = \frac{2\pi A_e}{\lambda^2} $$

The aperture of a lossless isotropic antenna is then also $A_e = \frac{\lambda²}{2 \pi}$ instead of $\frac{\lambda²}{4 \pi}$


I cannot manage to use that equation for a microphone array using delay-and-sum beamforming.

For example, if I have an array of 25 microphones spaced as a grid and that my setup has the following properties:

  • Smallest distance $d$ between them is $0.01$ meters.
  • Amount of orientations $o$ that it looks at is $64$.
  • Frequency $f$ to be detected is 8 kHz.

The wavelength $\lambda = \frac{v}{f} = \frac{343}{8000} = 0.0429 $

The aperture $A$ is $A = \frac{\lambda^2}{precision} = \frac{0.0429^2}{\frac{2\pi}{64}} = 0.0187 $


$$ D = \frac{2\pi A_e}{\lambda^2} = \frac{2\pi A}{\lambda^2} = \frac{2 \pi \cdot 0.0184}{0.0429^2} = 62 $$

I rounded the numbers for the simplicity of formulating my problem, if you do not round then you get $64$ instead of $62$ which is the number of orientations that it looks at furthermore the results cannot be true because:

  • It does not involve the number of microphones used.
  • It does not involve the smallest spacing between microphones.
  • I probably made a mistake somewhere.

How can I calculate the maximum obtainable directivity for an array of sensors?

  • 1
    $\begingroup$ be careful! TdoA is not the same as an aperture antenna or a phased array. These are antennas or antenna systems where you use the phase to form a wavefront. TdoA just compares time and hence loses all phase information. I'm not even sure I'd apply the term "directivity" to a TdoA system at all – it's not like you actually get a gain in a specific direction, as TdoA doesn't combine the signals – it's just a method of estimating direction from time differences. Are you really only doing TdoA or are your microphones actually a phased array? $\endgroup$ Commented Jul 15, 2018 at 12:31
  • $\begingroup$ @MarcusMüller Hi, thanks for answering. It is true that TdoA does not combine signals by itself. I use delay-and-sum beamforming which uses TdoA to, later on, combine the signals. Thanks for mentioning it, I updated the question. $\endgroup$
    – dll
    Commented Jul 15, 2018 at 12:35
  • $\begingroup$ Ah, but what happens when you delay-and-sum a signal that, let's say, consists of a tone at 1 kHz from mic A and the signal from mic B, which is three-and-a-half a 1 kHz-wavelength further away from the source? That way, your signals would cancel out, and you get no (or: negative infinity in dB) directivity in your desired direction (assuming that'd be the source of the 1 kHz tone). $\endgroup$ Commented Jul 15, 2018 at 12:40
  • $\begingroup$ It will be an infinity dB only for that orientation because the other orientations will not cancel out. $\endgroup$
    – dll
    Commented Jul 15, 2018 at 13:03
  • 1
    $\begingroup$ @MarcusMüller most audio delay and sum beam formers will use fractional delays relative to the sample rate to steer a beam. The term “phased array” is less informative than “frequency domain beamforming” . Acoustics is baseband, one time delays, either directly or in the frequency domain, but one time delays. $\endgroup$
    – user28715
    Commented Jul 15, 2018 at 22:56

2 Answers 2


Because we are dealing with only a single frequency $f,$ the 2-d spatial coordinates are more conveniently represented by 2-d phase coordinates where a distance of $2\pi$ represents one wavelength. The microphones are located on a square grid at phase coordinates $(x\Delta, y\Delta)$ for every combination of $x$ and $y$ each in $\{-N, -N+1, \dots, N-1, N\}$, with minimum mic-to-mic distance of $\Delta,$ and the number of microphones equal to $(2N+1)^2$ where $N$ is an integer.

$$\Delta = \frac{2\pi d}{\frac{v}{f}} \approx 1.465465960$$

in radians, where $v = 343\text{ m/s}$ is the speed of sound, $f = 8000\text{ Hz},$ and $d = 0.01\text{ m}$.

Equal steering angle and wave source angle: Let $\alpha$ denote the steering angle such that when $\alpha = 0$ the array is steered to receive an incoming plane wave originating from the direction of the positive $x$ axis of the square grid of microphones. A microphone at $(x\Delta, y\Delta)$ receives the wave advanced in phase by $\chi(x, y, \alpha) = \cos(\alpha)x\Delta + \sin(\alpha)y\Delta$ compared to the phase of the wave at coordinates $(0, 0).$ To compensate, the signal from the microphone is artificially delayed by the same amount, $\chi(x, y, \alpha)$. Because the microphone signals are in phase, the gain $G$ relative to the gain of 1 for a single microphone, is equal to the the number of microphones, $G = (2N+1)^2 = 25$ for $N = 2$.

Different steering angle and wave source angle: When the source angle of the wave is $\beta$ rather than the steered angle $\alpha,$ the complex gain is $G(\alpha, \beta) = \sum_{y=-N}^{+N}\sum_{x=-N}^{+N}e^{i\chi(x, y, \beta)}e^{i\chi(x, y, \alpha)},$ with the first exponential representing the phase difference of the wave at the microphone compared to the same at the reference point $(0, 0),$ and the second exponential representing the artificial delay. As an example let's plot $G(\alpha, \beta)$ for two steering angles:

Figure 1. Gain $G(\alpha, \beta)$ as function of the angle $\beta$ of the origin of the incoming plane wave, when the steering angle is aligned with a grid axis: $\alpha = 0$ (blue) or diagonal with respect to the grid: $\alpha = \frac{\pi}{4}$ (red). The microphone array arrangement and delays have symmetries that constrain $G$ to real numbers.

We can also calculate the directivity $D$ as function of the steering angle:

$$D(\alpha) = \frac{\left|G(\alpha, \alpha)\right|^2}{\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|G(\alpha, \beta)\right|^2d\beta},$$

where $G(\alpha, \alpha) = 25$ and the divisor is the average square of the absolute gain over all source angles of the wave.

Figure 2. Directivity $D(\alpha)$ as function of the steering angle $\alpha$. The best directivity is obtained at $\alpha = 0$ and the worst at $\alpha = \frac{\pi}{4}$ (see Fig. 1).


When using a microphone array for baseband acoustic signals, Array Gain is a preferred performance metric over Directivity Index.

In geophysics, the term “stacking” is often preferred over the equivalent “delay-and-sum beam former. It is certainly more descriptive because you essentially time align signals like a stack and add them coherently.

If the noise is independent at each microphone, for the aligned stack, the upper bound on coherent gain is $10 Log N$.

Directivity is a function of the beam pattern which is a function of a single frequency. For an RF system with an obvious carrier, the pattern at that nominal frequency is an often used way to characterize gain.

The stacked array gain is baseband. It is the upper bound on gain.

  • $\begingroup$ Do you mean $10\log_{10}(N)$ dB? Where does this come from? $\endgroup$ Commented Jul 18, 2018 at 10:35
  • $\begingroup$ Van Trees , vol 4 Array Processing page 66. $\endgroup$
    – user28715
    Commented Jul 18, 2018 at 14:04
  • $\begingroup$ Isn't array gain and directivity the same? For example, directivity for 2D is $ D(\phi) = \frac{U(\phi)}{P_{tot} / (2 \pi) }$ with $ P_{tot} = \int_{\phi = 0}^{\phi = 2 \pi} U d\phi $ while array gain is $G_a = \frac{G_d}{G_n}$ where $G_d$ is the gain in the direction of interest and $G_n$ the average gain in all directions. $P_{tot} / (2 \pi) $ is in fact $G_n$ because $\phi$ is in radian and $P_{tot}$ is devided by $2 \pi$. This results in the average power. Or am I missing something ? $\endgroup$
    – dll
    Commented Jul 18, 2018 at 18:49
  • $\begingroup$ no, DI is purely a geometric function of the beam pattern. Array Gain takes non isotropic noise into account. when the noise is independent at each sensor, they are the same. $\endgroup$
    – user28715
    Commented Jul 18, 2018 at 18:57

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