From various sources of Acoustics (such as "Acoustics - An Introduction to Its Physical Principles and Applications" by Allan D. Pierce and "Fundamentals of General Linear Acoustics" by Finn Jacobsen and Peter Møller Juhl) we know that the for an idealised monopole, infinitesimally small point source the pressure at a distance $r$ from the source (assuming it is situated in the origin) is

$$ p \left( r, t \right) = \frac{j \omega \rho Q}{4 \pi r} e^{j \left( \omega t - k r \right)} \tag{1} \label{1} $$

where $t$ denotes time, $j$ the imaginary unit for which $j^{2} = -1$ is true, $\omega$ the radial frequency which is $\omega = 2 \pi f$ with $f$ being the temporal frequency. $Q$ is the "source strength" which has units of volume velocity (surface times velocity) and $k$ is the wavenumber which is given by $k = \frac{\omega}{c} = \frac{2 \pi}{\lambda}$ where $c$ is the speed of propagation (speed of sound) and $\lambda$ the wavelength.


In general the pressure as given by equation \eqref{1} is a complex quantity. Even more, the constant quantity, which will call "Amplitude" for the purpose of this question, given by $A = \frac{j \omega \rho Q}{4 \pi r}$ is a purely imaginary number.

The question that arises is whether there is any physical meaning associated with the imaginary nature of the amplitude $A$ or even the complex nature of the pressure given by equation \eqref{1}.

Furthermore, in the second textbook cited above ("Fundamentals of General Linear Acoustics") it is stated that we could potentially normalise the product in the numerator of $A$ to equal $1$ such that $j \omega \rho Q = 1$. For this to happen $Q$ would have to be imaginary, since the other two quantities are purely real. Is there any physical meaning to that? Is this somehow related to the general context presented above?


1 Answer 1


You have to view this as a time Fourier transform. In general, if your signal in time domain is real, the Fourier transform is complex. In fact, the Fourier transform acquires an imaginary part when the original signal in time domain has an odd component.

In your case, since I think that the equations you are dealing with are linear, you can directly focus on one mode and then reconstruct any real signal by superposition. Note that $A(\omega)$ satisfies $A(-\omega) = A(\omega)^*$, ie $A$ is a real function of $j\omega$. This guarantees that the inverse Fourier transform will give you a real signal in time domain. Note that if your equation is non linear, even if the Fourier transform is still applicable, it may not be the best approach.

Technically, the definition: $$ Q =\frac{1}{\rho j\omega} $$ is underdetermined due to the pole at $0$. I suspect that in your case, the causal function is more relevant so that is rather: $$ Q = \frac{1}{\rho j(\omega+\epsilon j)} $$ with $\epsilon\to 0^+$. This corresponds, in the time domain, $Q = \frac{2\pi}{\rho} H(t)$ with $H$ the Heaviside step function. In general, depending on how you treat the pole, you'll add an additive constant.

Hope this helps.


Essentially, you need to interpret the formula as a Green's function in frequency domain: $$ \hat G(r,\omega) = \frac{j\omega\rho}{4\pi r}e^{-jkr} $$ which becomes in time domain: $$ \hat G(r,\omega) = \frac{\delta'(t)\rho}{4\pi r}e^{-jkr} $$ Even though $p$ contains the exponential factor which gives time dependence, you should rather read it as the defining convention for the Fourier transform. Therefore, for a general source $Q(r,t)$, in frequency domain: $$ \begin{align} \hat p(r,\omega) &= \int d^3r' \frac{j\omega\rho}{4\pi |r-r'|}e^{-jk|r-r'|}\hat Q(\omega,r') \\ &= \hat G(\omega)\star Q(\omega) \end{align} $$ with the convolution only over space. This gives in time domain: $$ \begin{align} p(r,t) &= \int d^3r' \int \frac{d\omega}{2\pi}\frac{j\omega\rho}{4\pi |r-r'|}e^{j(\omega t-k|r-r'|)}\hat Q(\omega,r') \\ &= G\star Q \end{align} $$ with the convolution over space time. Note that the integrand is precisely your formula.

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    $\begingroup$ Yes, I used the hat to indicate the Fourier transform. In time domain, this gives:$$p(r,t)=\int\frac{d\omega}{2\pi}\frac{j\omega}{4\pi r}e^{j(\omega t-kr)} \star\hat Q(r,\omega)$$ with $\hat Q$ the Fourier transform of $Q(r,t)$ and convolution performed over space. $\endgroup$
    – LPZ
    May 27, 2023 at 11:31
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    $\begingroup$ Essentially, the function $$\hat G= \frac{j\omega}{4\pi r}e^{-jkr}$$ is the Green’s function in frequency domain. By doing the inverse Fourier transform:$$G =\frac{\delta’(t)}{4\pi r}e^{-jkr}$$ you can write the pressure as a space time convolution of the source $Q(t,r)$ $\endgroup$
    – LPZ
    May 27, 2023 at 11:35
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    $\begingroup$ Actually, since the Green’s function is real in time domain, it only depends on $Q$ to be real in time domain or equivalently $\hat Q$ to be conjugate symmetric in frequency domain. $\endgroup$
    – LPZ
    May 27, 2023 at 11:38
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    $\begingroup$ Forgot a minus in front of the $\delta’$ $\endgroup$
    – LPZ
    May 27, 2023 at 11:41
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    $\begingroup$ would it be possible to include the information of the comments in the body of your answer? During my work on implementing the point source pressure calculations (for wideband signals) this information was extremely useful and without it would be quite hard to get results just with the information included in the answer. $\endgroup$
    – ZaellixA
    May 29, 2023 at 12:21

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