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214

8 Acoustic Waveguides

8.4Wavenumber Vectors in Acoustic Waveguides

In the acoustic waveguide shown in the right gure below, the acoustic wave can travel in the z-direction but will reect on all four sides of the waveguide.

Waveguides are elongated rectangular cavities with no closed walls in the axial direction. We can construct a waveguide by removing the two walls in the z-direction of an elongated rectangular cavity. The gure below shows that the two walls in the z-direction of a resonant cavity are removed to allow acoustic waves to propagate in the z-direction in an acoustic waveguide.

After the two walls in the z-direction are removed, the four wavenumber vectors become two wavenumber vectors because there is no reection in the z-direction as shown in the gure below:

Resonant Cavity

Acoustic Waveguide

 

 

 

 

 

 

With the two walls in the z-direction, the wavenumbers in both the y-direction and z-direction are discretized as:

kym ¼ π m kym is DISCRETIZED where m is an INTEGER

Ly

8.4 Wavenumber Vectors in Acoustic Waveguides

215

 

π

kzn is DISCRETIZED where n is an INTEGER

 

kzn ¼

 

n

 

Lz

 

Without the two walls in the z-direction, the wavenumbers in the y-direction are still discretized, but the wavenumbers in the z-direction become continuous numbers as:

 

 

 

π

kym is DISCRETIZED where m is an INTEGER

kym ¼

 

m

Ly

kzn ¼

π

n ¼ kz

kz is CONTINOUS since Lz can be ANY LENGTH

 

Lz

In a waveguide, because Lz can be considered as a very large number, and based on the formula above, kzn is continuous:

Resonant Cavity

Acoustic Waveguide

 

 

 

 

 

 

 

 

 

: Continuous Number

Because the wavenumber kz changes from being discretized numbers to being continuous numbers, a pure tone (single-frequency sound) will propagate along the axial direction with any transverse eigenmode of the waveguide.

Example 8.3: (Wavenumber Vectors in a Waveguide)

Assume a waveguide has a rectangular cross-section of Lx ¼ 13 [m] and Ly ¼ 12 [m].

(a) Calculate all the possible transverse eigenmodes that can carry an 850 [Hz] pure

tone

in the waveguide. State the wavenumber vectors as

!

 

k

lmn ¼ kxlbex þ

kymey þ kz ez

 

b

b

 

 

(b) Calculate the cutoff frequency of the waveguide.

216

8 Acoustic Waveguides

Example 8.3: Solution

Part (a)

Similar to the searching method for calculating the natural frequencies which are lower than a given frequency in the previous chapter, a searching method will be used to calculate all of the possible wavenumber vectors that can carry a pure tone.

Calculations of Wavenumber Vectors by Using a Searching Method

The relationship between the combined wavenumber and the component wavenumber is:

k2 ¼ k2xl þ k2ym þ k2zn

where k is the combined wavenumber klmn and:

π π π

kxl ¼ Lx l; kym ¼ Ly m; kzn ¼ Lz n

Since Lz can be ANY LENGTH, kzn is CONTINOUS and is not discretized anymore. A star symbol *is used in the subscript to indicate that a wavenumber is continuous. Based on this and the relationship above, the wavenumber kz in the axial direction can be calculated using:

k2z ¼ k2 k2xl k2ym

where:

kxl ¼

π

l ¼

π

l

Lx

31 ½m&

π

kym ¼ 12 ½m& m

and k is the wavenumber of the 850 [Hz] pure tone as:

k

ω

 

2πf

 

2π 850½Hz&

 

5π

¼ c

¼

 

c

¼

¼

 

 

340 s

 

 

 

 

 

 

 

 

m

 

 

 

h1i m

The following rule can be used to determine if an eigenmode exists to carry a pure tone in a waveguide:

If k2z is greater than zero, the eigenmode exists. If k2z is smaller than or equal to zero, the eigenmode does not exist.

Use the searching method to nd all of the eigenmodes (l, m) with their

wavenumber k2z being greater than zero as the following:

l

m

kxl

 

1

 

kym

 

1

 

k

2

 

1

 

Note

m

m

z

 

2

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

m

 

0

0

0

 

0

 

25

π

 

Direct wave

 

 

0

1

0

 

 

 

2π 1

21π2

kz2 > 0 (eigenmode exists)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

8.4 Wavenumber Vectors in Acoustic Waveguides

217

0

2

0

2π 2

9π2

 

kz2

> 0 (eigenmode exists)

0

3

0

2π 3

11π

2

 

 

0 0 (eigenmode does not exist)

 

kz2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Stop and increase l by 1

 

 

 

 

 

 

 

 

 

1

 

0

3π 1

 

0

 

 

 

16π2

kz2 > 0

(eigenmode exist)

1

 

1

3π 1

 

2π 1

 

 

12π2

kz2 > 0

(eigenmode exist)

1

 

2

3π 1

 

2π 2

 

 

0

kz2

0

(eigenmode does not exist)

Stop and increase m by 1

 

 

 

 

 

 

 

 

 

2

 

0

3π 2

 

0

 

 

 

11π2

kz2

0

(eigenmode does not exist)

Stop

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A summary of all the possible transverse eigenmodes found in the table is:

!

k 00 ¼ 5πbezðdirect waveÞ

!

p

 

 

 

 

 

k 01 ¼ 2πey þ

 

 

z

 

 

!

b

 

 

b

 

 

 

 

 

21πe

 

 

 

k

4πe

3πe

 

 

 

 

!02 ¼

by þ

 

bz

 

 

 

k 10 ¼

b

 

b

12πe

 

3πex þ

4πez

 

 

 

!

þ 2πey

 

p

 

 

z

k 11 ¼ 3πex

þ

 

b

b

 

 

 

 

b

The detailed calculations are shown below: (l, m, n) ¼ (0, 0, ): . . . this is the direct wave:

h i k2z ¼ k2 0 0 ¼ 25π2 m1

where k is the wavenumber of an 850 [Hz] pure tone as:

k ω 2πf

2π 850½Hz&

 

5π

1

 

 

 

 

 

 

 

 

 

 

¼ c ¼ c ¼

 

¼

 

hmi

340 ms

 

ðl, m, nÞ ¼ ð0, 1, Þ : k2z ¼ k2 k2x0 k2y1 ¼ ð5πÞ2 0 ð2πÞ2 ¼ 21π2 ðl, m, nÞ ¼ ð0, 2, Þ : k2z ¼ k2 k2x0 k2y2 ¼ ð5πÞ2 0 ð4πÞ2 ¼ 9π2 ðl, m, nÞ ¼ ð0, 3, Þ : k2z ¼ k2 k2x0 k2y3 ¼ ð5πÞ2 0 ð6πÞ2 ¼ 11π2 ðl, m, nÞ ¼ ð1, 0, Þ : k2z ¼ k2 k2x1 k2y0 ¼ ð5πÞ2 ð3πÞ2 0 ¼ 16π2

ðl, m, nÞ ¼ ð1, 1, Þ : k2 ¼ k2 k2 k2 ¼ ð5πÞ2 ð3πÞ2 ð2πÞ2 ¼ 12π2

z x1 y1

218 8 Acoustic Waveguides

ðl, m, nÞ ¼ ð1, 2, Þ : k2z ¼ k2 k2x1 k2y2 ¼ ð5πÞ2 ð3πÞ2 ð4πÞ2 ¼ 0 ðl, m, nÞ ¼ ð2, 0, Þ : k2z ¼ k2 k2x2 k2y0 ¼ ð5πÞ2 ð6πÞ2 0 ¼ 11π2

Summary of Wavenumber Vectors in Waveguides

For the eigenmodes having a mode number l ¼ 0, there are two eigenmodes (l, m, n) ¼ (0, 1, ) and (0, 2, ) in y-direction. These two eigenmodes have the following properties:

 

kxl ¼ 0:

 

 

 

kym is discretized due to the boundary condition: kym ¼

π

m.

Ly

kz is continued since Lz can be any length: kz ¼

π

n.

 

Lz

 

 

 

 

 

 

 

k ¼ !k lmn

is the combined wavenumber : k ¼ ωc .

 

 

 

Because kz is continue, eigenmodes can be constructed by the combination of a discretized kym and an arbitrary kz that satises:

k ¼ k2x0 þ k2ym þ k2z

Therefore, the possible eigenmodes are intersections (black dots) between a circle with a radius k and horizontal lines with a height kym as shown in the gure below:

 

 

 

units:

 

 

8.4 Wavenumber Vectors in Acoustic Waveguides

219

Visualization of Wavenumber Vectors by a Wavenumber Grid

Because:

k2 ¼ k2xl þ k2ym þ k2zn

all the possible eigenmodes (kxl, kym) for positive k2zn can be drawn as black dots inside a circle with a radius of k ¼ 5π in the gure below:

Because

For> 0

Eigen-modes ( ) are inside the circle with a radius

= 0,2

= 1,1

units:

This method provides a quick way to nd all the possible eigenmodes but does not provide the wavenumber vectors. The wavenumber vectors still need to be calculated after identifying the eigenmodes.

Part (b)

The cutoff frequency corresponds to the lowest y-direction. In this example, kx1 ¼ 3π and ky1 wavenumber is:

wavenumber in both the x- and ¼ 2π. Therefore, the lowest

 

 

 

 

 

 

ky1 ¼ 2π

 

 

 

 

 

 

 

 

 

 

The corresponding cutoff frequency is:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2π

 

2π

 

 

2

 

 

m

 

 

 

 

 

!

 

¼

 

¼

 

 

¼

340

 

ms

h

 

i

¼

 

½

 

&

f y10

ωy10

c

ky1

 

1

340

Hz

 

 

 

 

 

2π

 

 

 

 

 

 

 

 

 

 

 

 

 

π