# Why is the sound field intensity due to $K$ point sources given by $I(p,\omega) = \sum_{k=1}^K \sigma_k^2(\omega) \delta(p - p_k)$?

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.

Question

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 $$(*)$$?

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.

• How was $I(p,\omega)$ in $(*)$ derived? How does taking the expectation lead to a summation of Dirac deltas? – sonicboom Nov 6 '18 at 10:38
• @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 – Robert L. Nov 6 '18 at 14:28