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3.16a Electrical model of the acoustical circuit in Fig. 35 of posting 3.15
Keywords: acoustic resonator, coupling iris, modeling, Q_{0}, coupling Q
Topics covered in this posting
- This posting is the first step to mathematically analyze the first of the two resonant scattering animations in an earlier posting.
- In this posting we develop an electrical circuit analog to the acoustical circuit in the original animation.
- We also derive equations for the unloaded Q, i.e. Q_{0} , and coupling Q, i.e. Q_{C} .
Contents of this posting
1. Modeling the coupling hole |
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Fig. 1. Details of the resonator and coupling hole from Fig. 35 of a previous posting. The high velocity region in and near the coupling hole is similar to that of a Helmholtz resonator and has the Helmholtz or breathing mode. On the other hand, because the properties of the Helmholtz mode (resonant frequency, Q, etc.) are so dependent on the details of the hole for the resonance itself (as opposed to just the coupling), we focus on the next mode, the λ/2 mode (which is the first closed pipe mode) that can exist with or without the coupling hole and the hole for this mode mostly affects only the coupling. At the site of the resonator in Fig. 35 of the previous posting, incident waves travel upwards, hit a wall, and reflect back downwards. A small coupling hole in that wall allows the waves to excite the resonator and after the resonator is excited, for the resonator to radiate waves back out the coupling hole into the waveguide. The incident wave reflection process doubles the acoustic pressure at the wall and thereby doubles the excitation of the resonator. Waves traveling through the coupling hole see their acoustic velocity concentrated and increased over what they would have had in the waveguide. While the normal parameters to use with a plane wave in an open space are acoustic pressure p and acoustic velocity v, for an acoustic waveguide one uses acoustic pressure and acoustic volumetric velocity u = Av in units of m^{3}/sec and where A is the cross sectional area of the acoustic waveguide. As discussed on Wikipedia the appropriate impedance to use with a waveguide is the characteristic impedance Z = p/Av = p/u . A coupling hole which is short compared with a wavelength acts primarily as a mass element in this acoustical circuit. At first glance, this may seem odd, since there is very little mass of air in the coupling hole compared with that in the wider waveguide. However the coupling hole acts similar to a pinch point (a constriction of lanes) in a traffic jam: it is the sluggishness of the cars in the pinch point, as few as they may be, that impedes the whole traffic flow for miles behind it and determines the throughput of the all these miles of traffic. We can derive the effective mass (i.e. inductance in the circuit analog) with some simple algebra. Starting with Newton's second law:
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Thus we model the acoustic circuit in Fig. 35 by the electrical circuit in Fig. 2 below where a coupling inductor L_{C} plays the role that the coupling hole does in Fig. 35. Elmore and Heald (Amazon, Google books) on page 149 explains that the length len of the coupling hole should be extended by a factor 1.64a (where a is the hole's radius) to account for the high velocity region extending slightly past the length of the hole itself. Thus, the inductance to use is: where K = voltage/(acoustic pressure) is the conversion constant used to convert acoustical pressure into voltage. This conversion constant, although it is arbitrarily chosen, once it is chosen, the one value needs to be consistently used throughout a calculation or modeling. |
2. Modeling the acoustic circuit in Fig. 35 and the Q's of this circuit | ||
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At the right we see the electrical equivalent of the acoustical circuit of Fig. 35. We see three "legs" protruding out from a circulator in the center. Each leg is a transmission line of characteristic impedance Z_{0} = R_{0}. The top left leg is connected to the wave source. The waves from this source propagate down the transmission line to port 1 of the circulator which sends them out port 2. At port 2 they travel to the right in the right most transmission line to the resonator. There they are partially absorbed and partially reflected. The reflected waves travel to the left, to the circulator's port 2 and are sent out port 3 to the third leg. In the third leg they travel down and to the left to the terminating absorber having resistance R_{0}, i.e. equal to the characteristic impedance of the transmission lines. When viewed from the segment of the transmission line between points A and B, Fig. 2 above is similar to Fig. 3 below. The circulator mostly has the function of making sure there is a clean signal from the only source propagating to the right in the section between points A and B. The circuit to the right of point B labelled "load" (consisting of L_{C} and the resonator) has a a certain impedance, which presents a load to the waveguide to the left of point B. For the moment we will label this impedance Z_{L}. In an upcoming posting (3.21) we will derive an equation for the voltage of a complex load at the end of a transmission line. Using Equation 6 of 3.21, the voltage at point B is given by: where we have replaced Z_{0} with R_{0} because we are assuming the characteristic impedance of the transmission lines is real and equal to R_{0}. We can now draw a simpler equivalent circuit for the waves at point B to replace the more complicated circuit shown in Fig. 2. This simpler circuit is shown in Fig. 3 below. Following (3), we have a voltage source given by V = 2V_{source} with a resistor of value R_{0} in series with V and in series with the "load" from Fig. 2. We do this because the term in parentheses in (3) is just a voltage divider equation, giving the voltage across Z_{L} when it is in series with a resistor R_{0}. Note that Fig. 3 makes sense in several limits.
Fig. 3. The electrical circuit which is equivalent to the circuit of Fig. 2 above, from the point of view of the resonator. At the left we see the equivalent electrical circuit of the acoustic circuit, from the point of view of the resonator. We model the waveguide and source with an ideal AC voltage source and resistor. The amplitude of the voltage source is twice that of the sound source in Fig. 40 (with whatever conversion factor we wish to use between sound pressure and electronic voltage). That is V_{equivalent} = 2V_{waveguide}. The factor of two allows an open circuit at the end of the waveguide to experience twice the waveguide source voltage (source voltage in an actual waveguide and not this simplified equivalent circuit), while allowing a perfectly matched load resistor at the waveguide end to experience a voltage equal to the actual source voltage. The waveguide "resistance" R_{0} is set equal to the equivalent acoustic impedance of the waveguide in Fig. 40. |
3. Q's of the above circuit |
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To calculate Q_{0} (the intrinsic Q of the resonator alone) and Q_{C} (the coupling Q) of our circuit (Fig. 3) we rearrange equation (125) of an earlier posting: The resonator is modeled as a parallel LRC circuit. The intrinsic Q, i.e. Q_{0} (the quality factor if the LRC resonator were in a circuit all to itself but magically powered or in free decay) is given by: where ω_{0} = 1/√LC and V_{res} is the voltage amplitude across the parallel elements of the resonator. The factor of ½ in the numerator is part of the formula for energy in a capacitor, an energy that shuttles between the capacitor and inductor during resonance. The factor of ½ in the denominator is due to time averaging the power loss in the resistor. This power loss oscillates between a minimum value of 0 and maximum value of I_{Ro}^{2}R_{0}. The coupling Q_{C} is calculated by assuming that the source voltage is turned off (replaced by a short) and the resonator is magically excited by an ideal voltage source (zero output resistance). The coupling Q equals ω_{0} times the energy in the resonator divided by the power loss in the coupling resistor: . (6) Fig. 4. Coupling Q and loaded Q versus the coupling inductor showing the effect the coupling has on these Q's. Both Q's have been divided by Q_{0} (which equaled 20 in our example) and the coupling inductor was multiplied by the factor ω_{0}/R_{0} . As expected, the coupling inductor L_{C} affects the coupling Q_{C}. Larger L_{C} means larger Q_{C}'s and weaker coupling. This corresponds to a smaller coupling hole in the acoustic circuit of Figs. 35 and 1, consistent with (2) remembering that the hole's area A in the denominator of (2) is proportional to the hole's radius squared. Equation (6) is plotted at the right. On the vertical axis, we plot a normalized coupling Q_{C} defined as Q_{C}/Q_{0} , i.e. (6) divided by (5) or Q_{C}/Q_{0} = 1/β = (R_{0}/R)×(1 + x^{2}) where x = ω_{0}L_{C}/R_{0} . On the horizontal axis we plot x, the normalized coupling inductance. For the graph we assumed R_{0}/R = 1. To obtain normalized coupling Q's for other R_{0}/R values, multiple the value from the graph by R_{0}/R . We also plotted the normalized loaded Q i.e. Q_{L}/Q_{0} . Note that a larger coupling inductor mean weaker coupling, a higher coupling Q and higher loaded Q. In a future posting (3.16c), we calculate the steady state equations for this equivalent circuit above and graph the results. There, we shall see that the resonant frequency is given by ω_{0} = 1/√LC for high loaded Q's but at lower Q's the resonant frequency is affected by the coupling inductor L_{C} . Before we get to the entire circuit, we first examine just the resonator part in the next posting. |
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