12.5 Band-Stop Resonator

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Schematic of a Helmholtz resonator

Step 1: Derivation of the Dynamic Equation of the Resonator

The volume of fluid in the neck area will act as a mass just as the open end of a flanged pipe for λ a, where a is the radius of opening. The effective length of the neck is assumed to be 0.85a longer than the actual length of the pipe if the pipe is terminated with a wide flange at the opening. If the pipe has an un-flanged termination, it can be assumed that the effective length is longer only by 0.6a. Since the

interior end of the neck acts like a flanged termination, the effective length of the neck with flanged and un-flanged termination to the outside medium becomes:

Using the effective length, the mass of fluid moving in the neck is:

M ¼ ρoSL0

The mass in the neck works against the stiffness of the fluid in volume V in addition to resistance from the frictional forces in the neck and the radiation impedance from the open end.

The stiffness of the fluid in volume V is calculated from the change of volume inside the container due to the motion of the mass in the neck. Hence, if ξ is the displacement of the neck mass, the change in volume V is then given by:

Dividing both sides of this equation by the volume V and using the relationship above yield:

The acoustic pressure inside the given volume can thus be expressed as:

p ¼ ρoc2sac ¼ ρoc2 Sξ

V

The force necessary to displace the fluid in the neck is simply Sp ¼ Kξ where K is the effective stiffness of the fluid inside V. Multiplying both sides of the pressure equation above by S yields:

Sp ¼ ρoc2 S2 ξ ¼ Kξ

V

and the effective stiffness is:

K ¼ ρoc2 S2

V

In addition to the inertial and stiffness forces acting on the fluid column in the neck, there are also resistive forces that are proportional to the fluid velocity. These forces are due to the radiation resistance Rr and the viscous resistance Rω. These two constants provide damping to the mechanical system since the force they create is proportional to the velocity of the fluid in the neck. Hence, the total mechanical resistance coefficient is given by:

and:

Rω ¼ 2Mcαω

where αω is the absorption coefficient for wall losses.

The mass-spring system described here is forced by the pressure wave of p impinging on its opening. This force is simply the pressure of the wave multiplied by the neck cross-section area. That is:

f ¼ Sp

Now the differential equation of the resonator can be written in terms of mass M, stiffness K, and the damping coefficient Rm given by the equations above as follows:

Step 2: Mechanical Impedance of the Dynamic Equation

Let ξ ¼ Ae jωt, where its complex conjugate is ξ ¼ A e jωt, and we can then solve the following two equations step-by-step as follows for p ¼ Pe jωt and p ¼ P e jωt. Note that K, M, Rm, S, and ω are all real constants:

Let:

H ¼ K ω2M þ jωRm 1

H ¼ K ω2M jωRm 1

to obtain:

A ¼ K ω2M þ jωRm 1SP ¼ HSP

A ¼ K ω2M jωRm 1SP ¼ H SP

Therefore:

ξ ¼ HSPejωt

Note that the equations above show that all complex values, functions, or equations always have a complex conjugate part associated with them. Therefore, we can always hide the conjugate part, but when we need a real-world solution, we must show them.

Step 3: Transmission Coefficient Induced by a Side Resonator

Neglecting Rω for wall losses, the input mechanical impedance of the resonator is:

For ka 1, (Zi,Z2,Pi real) (Z1,P2 complex):