Acoustics

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2.5. Impedance of a closed tube using the inhomogeneous wave equation

Boundary conditions

We have already found the impedance of a closed tube by taking the solution to the following Helmholtz wave equation

(\frac{\partial^{2}}{\partial x^{2}} + k^{2}) \tilde{p} (x) = 0

$(\frac{\partial^{2}}{\partial x^{2}} + k^{2}) \tilde{p} (x) = 0$

(2.79)

and applying boundary conditions to the solution. It is known as a homogeneous wave equation because there are no sound sources explicit in the equation. These are included in the boundary conditions that are applied to the solution. Here we shall consider the inhomogeneous wave equation

(\frac{\partial^{2}}{\partial x^{2}} + k^{2}) \tilde{p} (x) = - δ (x - l) \frac{\partial}{\partial x} {\tilde{p} (x) |}_{x = l},

$(\frac{\partial^{2}}{\partial x^{2}} + k^{2}) \tilde{p} (x) = - δ (x - l) \frac{\partial}{\partial x} {\tilde{p} (x) |}_{x = l},$

(2.80)

which includes the sound source at x = l on the right-hand side, where δ is the Dirac delta function. This useful function describes a singularity when its argument is zero (in this case when x = l) but returns a zero value for all other arguments. In solving the equation using this alternative method, we shall introduce some useful techniques for approaching acoustical problems in general, which make use of orthogonality and the properties of the Dirac delta function. The solution itself will provide useful identities for trigonometrical functions with numerical advantages over more conventional ones. In Fig. 2.6, the piston at x = l oscillates with velocity

{\tilde{u}}_{0}

${\tilde{u}}_{0}$

. Hence, using the relationship of Eq. (2.4a), gives

(\frac{\partial^{2}}{\partial x^{2}} + k^{2}) \tilde{p} (x) = j k ρ_{0} c δ (x - l) {\tilde{u}}_{0} .

$(\frac{\partial^{2}}{\partial x^{2}} + k^{2}) \tilde{p} (x) = j k ρ_{0} c δ (x - l) {\tilde{u}}_{0} .$

(2.81)

In other words, we are describing the piston as a point source at the end of the tube.

Solution of the inhomogeneous wave equation for a closed tube

Let the solution be in the form of an eigenfunction expansion

\tilde{p} (x) = \sum_{n = 0}^{\infty} {\tilde{A}}_{n} \cos (n π x / l) .

$\tilde{p} (x) = \sum_{n = 0}^{\infty} {\tilde{A}}_{n} \cos (n π x / l) .$

(2.82)

When the piston is stationary, this satisfies the boundary conditions

\frac{\partial}{\partial x} {\tilde{p} (x) |}_{x = 0} = 0, \frac{\partial}{\partial x} {\tilde{p} (x) |}_{x = l} = 0.

$\frac{\partial}{\partial x} {\tilde{p} (x) |}_{x = 0} = 0, \frac{\partial}{\partial x} {\tilde{p} (x) |}_{x = l} = 0.$

(2.83)

Inserting Eq. (2.82) in Eq. (2.81) and multiplying both sides by cos(mπx/l) while integrating over the length of the tube gives

\begin{matrix} \sum_{n = 0}^{\infty} (k^{2} - \frac{n^{2} π^{2}}{l^{2}}) {\tilde{A}}_{n} \int_{0}^{l} \cos (m π x / l) \cos (n π x / l) d x \\ = j k ρ_{0} c {\tilde{u}}_{0} \int_{0}^{l} \cos (m π x / l) δ (x - l) d x . \end{matrix}

$\begin{matrix} \sum_{n = 0}^{\infty} (k^{2} - \frac{n^{2} π^{2}}{l^{2}}) {\tilde{A}}_{n} \int_{0}^{l} \cos (m π x / l) \cos (n π x / l) d x \\ = j k ρ_{0} c {\tilde{u}}_{0} \int_{0}^{l} \cos (m π x / l) δ (x - l) d x . \end{matrix}$

(2.84)

The two integrals have the following identities

\int_{0}^{l} \cos (m π x / l) \cos (n π x / l) d x = {\begin{matrix} 0, & m \neq n \\ l, & m = n = 0 \\ l / 2, & m = n \neq 0 \end{matrix}

$\int_{0}^{l} \cos (m π x / l) \cos (n π x / l) d x = {\begin{matrix} 0, & m \neq n \\ l, & m = n = 0 \\ l / 2, & m = n \neq 0 \end{matrix}$

(2.85)

\int_{0}^{l} \cos (m π x / l) δ (x - l) d x = \cos (m π) = {(- 1)}^{m} .

$\int_{0}^{l} \cos (m π x / l) δ (x - l) d x = \cos (m π) = {(- 1)}^{m} .$

(2.86)

The first is the property of orthogonality and the second is a property of the Dirac delta function

\int_{- \infty}^{\infty} F (x) δ (x - l) d x = F (l) .

$\int_{- \infty}^{\infty} F (x) δ (x - l) d x = F (l) .$

(2.87)

Hence, cos(mπx/l) in this case is the orthogonal (or normalizing) function and we can eliminate the summation to yield

{\tilde{A}}_{n} = j k l ρ_{0} c {\tilde{u}}_{0} \frac{(2 - δ_{0 n}) {(- 1)}^{n}}{k^{2} l^{2} - n^{2} π^{2}},

${\tilde{A}}_{n} = j k l ρ_{0} c {\tilde{u}}_{0} \frac{(2 - δ_{0 n}) {(- 1)}^{n}}{k^{2} l^{2} - n^{2} π^{2}},$

(2.88)

so that the solution of Eqs. (2.80) and (2.81), given by Eq. (2.82), becomes

\tilde{p} (x) = j k l ρ_{0} c {\tilde{u}}_{0} \sum_{n = 0}^{\infty} \frac{(2 - δ_{0 n}) {(- 1)}^{n} \cos (n π x / l)}{k^{2} l^{2} - n^{2} π^{2}},

$\tilde{p} (x) = j k l ρ_{0} c {\tilde{u}}_{0} \sum_{n = 0}^{\infty} \frac{(2 - δ_{0 n}) {(- 1)}^{n} \cos (n π x / l)}{k^{2} l^{2} - n^{2} π^{2}},$

(2.89)

where δ _mn is the Kronecker delta function

δ_{0 n} = {\begin{matrix} 0, & m \neq n \\ 1, & m = n . \end{matrix}

$δ_{0 n} = {\begin{matrix} 0, & m \neq n \\ 1, & m = n . \end{matrix}$

(2.90)

Eq. (2.89) is equivalent to the solution of Eq. (2.79), given by Eq. (2.69)

\tilde{p} (x) = j ρ_{0} c {\tilde{u}}_{0} \frac{\cos k x}{\sin k l} .

$\tilde{p} (x) = j ρ_{0} c {\tilde{u}}_{0} \frac{\cos k x}{\sin k l} .$

(2.91)

Impedance of the closed tube

We saw previously that the impedance of a closed tube is given by

Z_{s} = \frac{\tilde{p} (l)}{- {\tilde{u}}_{0}} = - j ρ_{0} c \cot k l,

$Z_{s} = \frac{\tilde{p} (l)}{- {\tilde{u}}_{0}} = - j ρ_{0} c \cot k l,$

(2.92)

which is equivalent to the following expression obtained using Eq. (2.89)

Z_{s} = \frac{\tilde{p} (l)}{- {\tilde{u}}_{0}} = - j ρ_{0} c \sum_{n = 0}^{\infty} \frac{(2 - δ_{0 n}) k l}{k^{2} l^{2} - n^{2} π^{2}} .

$Z_{s} = \frac{\tilde{p} (l)}{- {\tilde{u}}_{0}} = - j ρ_{0} c \sum_{n = 0}^{\infty} \frac{(2 - δ_{0 n}) k l}{k^{2} l^{2} - n^{2} π^{2}} .$

(2.93)

Because modes occur when kl = nπ, the eigenfrequencies are given by

f = \frac{n c}{2 l} or l = \frac{n}{2} λ, n = 1,2,3 \dots

$f = \frac{n c}{2 l} or l = \frac{n}{2} λ, n = 1,2,3 \dots$

(2.94)

Expansions for cot and csc

We have thus obtained a useful expansion for cot kl, which converges when truncated to a finite number of terms and retains the singularities at kl = nπ, unlike the more conventional formula cot kl = (cos kl)/(sin kl), where

\sin k l = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} {(k l)}^{2 n + 1}}{(2 n + 1)!},

$\sin k l = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} {(k l)}^{2 n + 1}}{(2 n + 1)!},$

(2.95)

\cos k l = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} {(k l)}^{2 n}}{(2 n)!} .

$\cos k l = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} {(k l)}^{2 n}}{(2 n)!} .$

(2.96)

Hence

\cot x = \sum_{n = 0}^{\infty} \frac{(2 - δ_{0 n}) x}{x^{2} - n^{2} π^{2}} .

$\cot x = \sum_{n = 0}^{\infty} \frac{(2 - δ_{0 n}) x}{x^{2} - n^{2} π^{2}} .$

(2.97)

Similarly, by comparing the pressures given by Eqs. (2.89) and (2.91) at x = 0, we obtain

\csc x = \sum_{n = 0}^{\infty} \frac{(2 - δ_{0 n}) {(- 1)}^{n} x}{x^{2} - n^{2} π^{2}} .

$\csc x = \sum_{n = 0}^{\infty} \frac{(2 - δ_{0 n}) {(- 1)}^{n} x}{x^{2} - n^{2} π^{2}} .$

(2.98)

It is also useful to have an expression for the difference between the above

\csc x - \cot x = \tan \frac{x}{2} = \sum_{n = 0}^{\infty} \frac{4 x}{{(2 n + 1)}^{2} π^{2} - x^{2}} .

$\csc x - \cot x = \tan \frac{x}{2} = \sum_{n = 0}^{\infty} \frac{4 x}{{(2 n + 1)}^{2} π^{2} - x^{2}} .$

(2.99)

Hence

\tan x = \sum_{n = 0}^{\infty} \frac{2 x}{{(n + 1 / 2)}^{2} π^{2} - x^{2}} .

$\tan x = \sum_{n = 0}^{\infty} \frac{2 x}{{(n + 1 / 2)}^{2} π^{2} - x^{2}} .$

(2.100)

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Table of Contents for Acoustics

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2.5. Impedance of a closed tube using the inhomogeneous wave equation

Boundary conditions

Solution of the inhomogeneous wave equation for a closed tube

Impedance of the closed tube

Expansions for cot and csc

Table of Contents for
Acoustics