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13.15. The far-field pressure distribution as a spatial frequency spectrum of the source velocity distribution

Two-dimensional system

In a two-dimensional system with a planar source, the far-field pressure distribution is given by a generalized version of Eq. (13.279), where

{\tilde{u}}_{0} (x_{0})

${\tilde{u}}_{0} (x_{0})$

is the source velocity distribution:

\begin{matrix} \tilde{p} (r, θ) = ρ_{0} c \sqrt{\frac{k}{2 π r}} e^{- j (k r - \frac{π}{4})} \int_{- \infty}^{\infty} {\tilde{u}}_{0} (x) e^{j k x_{0} \sin θ} d x_{0} \\ = - j \frac{e^{- j (k r + \frac{π}{4})}}{\sqrt{2 π k r}} \tilde{F} (k_{x}), \end{matrix}

$\begin{matrix} \tilde{p} (r, θ) = ρ_{0} c \sqrt{\frac{k}{2 π r}} e^{- j (k r - \frac{π}{4})} \int_{- \infty}^{\infty} {\tilde{u}}_{0} (x) e^{j k x_{0} \sin θ} d x_{0} \\ = - j \frac{e^{- j (k r + \frac{π}{4})}}{\sqrt{2 π k r}} \tilde{F} (k_{x}), \end{matrix}$

(13.286)

where

\tilde{F} (k_{x}) = \int_{- \infty}^{\infty} \tilde{f} (x_{0}) e^{j k_{x} x_{0}} d x_{0},

$\tilde{F} (k_{x}) = \int_{- \infty}^{\infty} \tilde{f} (x_{0}) e^{j k_{x} x_{0}} d x_{0},$

(13.287)

which is simply the Fourier transform or spatial frequency spectrum of the normal pressure gradient distribution

\tilde{f} (x_{0})

$\tilde{f} (x_{0})$

in the xy plane:

\tilde{f} (x_{0}) = - \frac{\partial}{\partial z_{0}} \tilde{p} {(x_{0}, z_{0}) |}_{z_{0} = 0} = j k ρ_{0} c \tilde{u} (x_{0}),

$\tilde{f} (x_{0}) = - \frac{\partial}{\partial z_{0}} \tilde{p} {(x_{0}, z_{0}) |}_{z_{0} = 0} = j k ρ_{0} c \tilde{u} (x_{0}),$

(13.288)

where k _x is the spatial frequency of the component of a wave in the x direction given by

k_{x} = k \sin θ .

$k_{x} = k \sin θ .$

(13.289)

In the case of a strip of infinite extent in the y direction, the velocity distribution is just a step function in the x direction:

f (x_{0}) = {\begin{matrix} 0, & x < - \frac{d}{2} \\ 1, & - \frac{d}{2} \leq x \leq \frac{d}{2} \\ 0, & x > - \frac{d}{2} . \end{matrix}

$f (x_{0}) = {\begin{matrix} 0, & x < - \frac{d}{2} \\ 1, & - \frac{d}{2} \leq x \leq \frac{d}{2} \\ 0, & x > - \frac{d}{2} . \end{matrix}$

(13.290)

By inspection of Eq. (13.279) we see that

D (k_{x}) = \frac{2 π}{j k d ρ_{0} c {\tilde{u}}_{0}} F (k_{x}) = \frac{\sin (\frac{1}{2} k_{x} d)}{\frac{1}{2} k_{x} d} .

$D (k_{x}) = \frac{2 π}{j k d ρ_{0} c {\tilde{u}}_{0}} F (k_{x}) = \frac{\sin (\frac{1}{2} k_{x} d)}{\frac{1}{2} k_{x} d} .$

(13.291)

Now let us convert from polar coordinates in r and θ to rectangular coordinates in x and z, using

r = \sqrt{z^{2} + x^{2}},

$r = \sqrt{z^{2} + x^{2}},$

(13.292)

\sin θ = \frac{x}{\sqrt{z^{2} + x^{2}}}

$\sin θ = \frac{x}{\sqrt{z^{2} + x^{2}}}$

(13.293)

and project the polar directivity pattern onto a distant parallel screen. Hence, the spatial frequency k _x at a point on the screen a horizontal distance x from the z axis is scaled by

k_{x} = k \sin θ = \frac{k x}{\sqrt{z^{2} + x^{2}}} .

$k_{x} = k \sin θ = \frac{k x}{\sqrt{z^{2} + x^{2}}} .$

(13.294)

At a given frequency ω = kc, only the spectrum up to spatial frequency k _x = k is displayed on the screen as θ varies from 0 to π/2. As the frequency is increased, more of the spectrum is shown but never the whole spectrum. We also note that the amplitude of the spectrum is scaled by (z ² + x ²)^−1/4 because of the

1 / \sqrt{r}

$1 / \sqrt{r}$

term in Eq. (13.279).

Three-dimensional system

Here we consider a three-dimensional system in rectangular coordinates (x,y,z) with an arbitrary source velocity distribution in an infinite baffle in the xy plane, which radiates into half space. The pressure field is given by the Rayleigh integral of Eq. (13.6) using the Green's function given by Eq. (13.4), which for z ₀ = 0 can be written as

g (x, y, z | x_{0}, y_{0}, 0) = \frac{e^{- j k r_{1}}}{4 π r_{1}},

$g (x, y, z | x_{0}, y_{0}, 0) = \frac{e^{- j k r_{1}}}{4 π r_{1}},$

(13.295)

where

r_{1}^{2} = {(x - x_{0})}^{2} + {(y - y_{0})}^{2} + z^{2} .

$r_{1}^{2} = {(x - x_{0})}^{2} + {(y - y_{0})}^{2} + z^{2} .$

(13.296)

If we let x = r sin θ _x, y = r sin θ _y, and z ² = r ² − x ² − y ², then

\begin{matrix} r_{1}^{2} = r^{2} - 2 r x_{0} \sin θ_{x} - 2 r y_{0} \sin θ_{y} + x_{0}^{2} + y_{0}^{2} \\ \approx {(r - x_{0} \sin θ_{x} - y_{0} \sin θ_{y})}^{2}, r^{2} ≫ x_{0}^{2} + y_{0}^{2}, \end{matrix}

$\begin{matrix} r_{1}^{2} = r^{2} - 2 r x_{0} \sin θ_{x} - 2 r y_{0} \sin θ_{y} + x_{0}^{2} + y_{0}^{2} \\ \approx {(r - x_{0} \sin θ_{x} - y_{0} \sin θ_{y})}^{2}, r^{2} ≫ x_{0}^{2} + y_{0}^{2}, \end{matrix}$

(13.297)

where θ _x is the angle of elevation subtended with the z axis in the x direction, and θ _y is that subtended with the z axis in the y direction as follows:

\sin θ_{x} = \frac{x}{\sqrt{x^{2} + y^{2} + z^{2}}} = \frac{x}{r},

$\sin θ_{x} = \frac{x}{\sqrt{x^{2} + y^{2} + z^{2}}} = \frac{x}{r},$

(13.298)

\sin θ_{y} = \frac{y}{\sqrt{x^{2} + y^{2} + z^{2}}} = \frac{y}{r} .

$\sin θ_{y} = \frac{y}{\sqrt{x^{2} + y^{2} + z^{2}}} = \frac{y}{r} .$

(13.299)

Alternatively, in cylindrical coordinates we have sin θ _x = sin θ cos ϕ and sin θ _y = sin θ sin ϕ. Inserting the Green's function of Eq. (13.295) into the Rayleigh integral of Eq. (13.6), while doubling the source strength due to half-space radiation, yields

\begin{matrix} \tilde{p} (x, y, z) = j k ρ_{0} c \frac{e^{- j k r}}{2 π r} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \tilde{u} (x_{0}, y_{0}) e^{j k (x_{0} \sin θ_{x} + y_{0} \sin θ_{y})} d x_{0} d y_{0} \\ = \frac{e^{- j k r}}{2 π r} \tilde{F} (k_{x}, k_{y}), \end{matrix}

$\begin{matrix} \tilde{p} (x, y, z) = j k ρ_{0} c \frac{e^{- j k r}}{2 π r} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \tilde{u} (x_{0}, y_{0}) e^{j k (x_{0} \sin θ_{x} + y_{0} \sin θ_{y})} d x_{0} d y_{0} \\ = \frac{e^{- j k r}}{2 π r} \tilde{F} (k_{x}, k_{y}), \end{matrix}$

(13.300)

where

\tilde{F} (k_{x}, k_{y}) = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \tilde{f} (x_{0}, y_{0}) e^{j (k_{x} x_{0} + k_{y} y_{0})} d x_{0} d y_{0},

$\tilde{F} (k_{x}, k_{y}) = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \tilde{f} (x_{0}, y_{0}) e^{j (k_{x} x_{0} + k_{y} y_{0})} d x_{0} d y_{0},$

(13.301)

which is simply the Fourier transform or spatial frequency spectrum of the normal pressure gradient distribution

\tilde{f} (x_{0}, y_{0})

$\tilde{f} (x_{0}, y_{0})$

in the xy plane:

\tilde{f} (x_{0}, y_{0}) = - \frac{\partial}{\partial z_{0}} {\tilde{p} (x_{0}, y_{0}, z_{0}) |}_{z_{0} = 0} = j k ρ_{0} c \tilde{u} (x_{0}, y_{0}),

$\tilde{f} (x_{0}, y_{0}) = - \frac{\partial}{\partial z_{0}} {\tilde{p} (x_{0}, y_{0}, z_{0}) |}_{z_{0} = 0} = j k ρ_{0} c \tilde{u} (x_{0}, y_{0}),$

(13.302)

where k _x and k _y are the spatial frequencies given by

k_{x} = k \sin θ_{x},

$k_{x} = k \sin θ_{x},$

(13.303)

k_{y} = k \sin θ_{y},

$k_{y} = k \sin θ_{y},$

(13.304)

and the amplitude in the distant plane is scaled by (x ² + y ² + z ²)^−1/2.

Axisymmetric three-dimensional system

In an axisymmetric system with a planar source, such as a piston in an infinite baffle, the pressure distribution is given by the Rayleigh integral of Eq. (13.6) using the Green's function given by Eq. (13.70), which for z ₀ = 0 and ϕ = π/2 can be written as

\begin{matrix} p (r, θ) = j k ρ_{0} c \frac{e^{j k r}}{2 π r} \int_{0}^{2 π} \int_{0}^{\infty} \tilde{u} (w_{0}) e^{j k w_{0} \sin θ \sin ϕ_{0}} w_{0} d w_{0} ϕ_{0} \\ = \frac{e^{j k r}}{r} \int_{0}^{\infty} \tilde{f} (w_{0}) J_{0} (k w_{0} \sin θ) w_{0} d w_{0} \\ = \frac{e^{j k r}}{r} \tilde{F} (k_{w}), \end{matrix}

$\begin{matrix} p (r, θ) = j k ρ_{0} c \frac{e^{j k r}}{2 π r} \int_{0}^{2 π} \int_{0}^{\infty} \tilde{u} (w_{0}) e^{j k w_{0} \sin θ \sin ϕ_{0}} w_{0} d w_{0} ϕ_{0} \\ = \frac{e^{j k r}}{r} \int_{0}^{\infty} \tilde{f} (w_{0}) J_{0} (k w_{0} \sin θ) w_{0} d w_{0} \\ = \frac{e^{j k r}}{r} \tilde{F} (k_{w}), \end{matrix}$

(13.305)

where we have used Eq. (76) from Appendix II to solve the integral and

\tilde{F} (k_{w}) = \int_{0}^{\infty} \tilde{f} (w_{0}) J_{0} (k_{w} w_{0}) w_{0} d w_{0},

$\tilde{F} (k_{w}) = \int_{0}^{\infty} \tilde{f} (w_{0}) J_{0} (k_{w} w_{0}) w_{0} d w_{0},$

(13.306)

which is simply the Hankel transform or spatial frequency spectrum of the normal pressure gradient distribution

\tilde{f} (w_{0})

$\tilde{f} (w_{0})$

in the w plane:

\tilde{f} (w_{0}) = {- \frac{\partial}{\partial z_{0}} \tilde{p} (w_{0}, z_{0}) |}_{z_{0} = 0} = j k ρ_{0} c \tilde{u} (w_{0}),

$\tilde{f} (w_{0}) = {- \frac{\partial}{\partial z_{0}} \tilde{p} (w_{0}, z_{0}) |}_{z_{0} = 0} = j k ρ_{0} c \tilde{u} (w_{0}),$

(13.307)

where k _w is the spatial frequency given by

k_{w} = k \sin θ .

$k_{w} = k \sin θ .$

(13.308)

In the case of a piston in an infinite baffle, the velocity distribution is just a step function in the w direction:

\tilde{f} (w_{0}) = {\begin{matrix} j k ρ_{0} c {\tilde{u}}_{0}, & 0 \leq w_{0} \leq a \\ 0, & w_{0} > a \end{matrix},

$\tilde{f} (w_{0}) = {\begin{matrix} j k ρ_{0} c {\tilde{u}}_{0}, & 0 \leq w_{0} \leq a \\ 0, & w_{0} > a \end{matrix},$

(13.309)

so that applying the integral solution of Eq. (95) from Appendix II yields

\begin{matrix} \tilde{F} (k_{w}) = j k ρ_{0} c {\tilde{u}}_{0} \int_{0}^{a} J_{0} (k w_{0} \sin θ) w_{0} d w_{0} \\ = j k a^{2} ρ_{0} c {\tilde{u}}_{0} \frac{J_{1} (k a \sin θ)}{k a \sin θ} . \end{matrix}

$\begin{matrix} \tilde{F} (k_{w}) = j k ρ_{0} c {\tilde{u}}_{0} \int_{0}^{a} J_{0} (k w_{0} \sin θ) w_{0} d w_{0} \\ = j k a^{2} ρ_{0} c {\tilde{u}}_{0} \frac{J_{1} (k a \sin θ)}{k a \sin θ} . \end{matrix}$

(13.310)

By inspection of Eq. (13.102), we see that

D (k_{w}) = \frac{2}{j k a^{2} ρ_{0} c {\tilde{u}}_{0}} \tilde{F} (k_{w}) .

$D (k_{w}) = \frac{2}{j k a^{2} ρ_{0} c {\tilde{u}}_{0}} \tilde{F} (k_{w}) .$

(13.311)

Now let us convert from polar coordinates in r and θ to cylindrical coordinates in w and z, using

r = \sqrt{z^{2} + w^{2}},

$r = \sqrt{z^{2} + w^{2}},$

(13.312)

\sin θ = \frac{w}{\sqrt{z^{2} + w^{2}}},

$\sin θ = \frac{w}{\sqrt{z^{2} + w^{2}}},$

(13.313)

and project the polar directivity pattern onto a distant parallel screen. Hence, the spatial frequency k _w at a point on the screen a horizontal distance w from the z axis is scaled by

k_{w} = k \sin θ = \frac{k w}{\sqrt{z^{2} + w^{2}}} .

$k_{w} = k \sin θ = \frac{k w}{\sqrt{z^{2} + w^{2}}} .$

(13.314)

At a given frequency ω = kc, only the spectrum up to spatial frequency k _w = k is displayed on the screen as θ varies from 0 to π/2. As the frequency is increased, more of the spectrum is shown but never the whole spectrum. We also note that the amplitude of the spectrum is scaled by (z ² + w ²)^−1/2 because of the 1/r term in Eq. (13.101).

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13.15. The far-field pressure distribution as a spatial frequency spectrum of the source velocity distribution

Two-dimensional system

Three-dimensional system

Axisymmetric three-dimensional system

Table of Contents for
Acoustics