I am trying to understand the following piece of text. I am not used to dealing with sound intensity and power so I'm not familiar with the derivation of the formula $(*)$ below.

Statement: 1.

We assume there exist $K$ point sources in the far-field indexed by the letter $k$. Each propagates in the direction of the unit vector $p_k$. Then the signal coming from direction $p ∈ S$ is given by $x(p, ω, t) = \tilde x(p, ω)e^{jωt}$,where $\tilde x(p, ω)$ is the emitted sound signal by a source located at $p$ and frequency $ω$. The intensity of the sound field is then $$ I(p,\omega) = \mathbb{E}[|x(p, ω, t)|^2] = \sum_{k=1}^K \sigma_k^2(\omega) \delta(p - p_k), \tag{$*$} $$ where $\sigma_k^2(\omega)$ is the power of the $k$-the source and $\delta(p)$ is the Dirac delta function on the unit circle.


How is $(*)$ derived? The formula for intensity on Wikipedia is $$ I(r) = P/A(r), $$ where $P$ is the power and $A(r)$ is the surface area of a sphere of radius $r$. The formula in $(*)$ doesn't seem to make any use of a sphere of radius $r$..and it also features Dirac deltas which aren't in the Wikipedia formula.

So what is the relationship between the Wikipedia formula and the formula $(*)$?


1 Answer 1


I believe the confusion comes from the fact that equation $(*)$ is referring to a received intensity (in units of W/m^2), whereas the second equation is referring to a radiated intensity.

The assumption behind equation $(*)$ is that the received power as a function of angle is the surface integral of the incident intensity over some aperture area $A \subset S$. Because the surface area element $dA$ has units of area, the result of the integral will be in units of power (W).

The Dirac delta distribution $\delta(p - p_k)$ expresses that the power density is coming from exactly one direction $p_k$ and nowhere else. In this context, the Dirac delta distribution possesses the following property: $$ \int_A \delta(p - p_k) dA' = \begin{cases} 0 & p_k \notin A \\ 1 & p_k \in A \end{cases}, $$ which is analogous to a "density" where all the "mass" is concentrated at one point $p_k$.

To calculate the total received power $P$, you integrate the received intensity over the receive aperture surface $A$: $$ P = \int_A I(p,\omega) dA' $$

Each term of the integral will look something like this: $$ \int_A \sigma_k^2(\omega) \delta(p - p_k) dA' = \begin{cases} 0 & p_k \notin A \\ \sigma_k^2(\omega) & p_k \in A \end{cases}, $$ so in other words, you are just adding up the power coming from each source that your receive aperture $A$ captures, and each source contributes power $\sigma^2_k(\omega)$.

Implicit in this formulation is that all the sources are incoherent, i.e., the waves do not have a fixed phase relationship and thus the average power of the superposition is the sum of the contributions from the individual sources.

  • $\begingroup$ How was $I(p,\omega)$ in $(*)$ derived? How does taking the expectation lead to a summation of Dirac deltas? $\endgroup$
    – sonicboom
    Nov 6, 2018 at 10:38
  • $\begingroup$ @sonicboom I don't see any expectation operator. It's just one delta per point source, and the received power intensity is the sum of the contributions from each source $\endgroup$
    – Robert L.
    Nov 6, 2018 at 14:28

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