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[edit] Part 1: Lumped Acoustical Systems

[edit] Simple Oscillation

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Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11

Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3

Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24

[edit] Solving for the Position Equation

For a simple oscillator consisting of a mass m attached to one end of a spring with a spring constant s, the restoring force, f, can be expressed by the equation

$f = -sx\,$

where x is the displacement of the mass from its rest position. Substituting the expression for f into the linear momentum equation,

$f = ma = m{d^2x \over dt^2}\,$

where a is the acceleration of the mass, we can get

$m\frac{d^2 x}{d t^2 }= -sx$

or,

$\frac{d^2 x}{d t^2} + \frac{s}{m}x = 0$

Note that

$\omega_0^2 = {s \over m}\,$

To solve the equation, we can assume

$x(t)=A e^{\lambda t} \,$

The force equation then becomes

$(\lambda^2+\omega_0^2)A e^{\lambda t} = 0,$

Giving the equation

$\lambda^2+\omega_0^2 = 0,$

Solving for $λ$

$\lambda = \pm j\omega_0\,$

This gives the equation of x to be

$x = C_1e^{j\omega_0 t}+C_2e^{-j\omega_0 t}\,$

Note that

$j = (-1)^{1/2}\,$

and that C₁ and C₂ are constants given by the initial conditions of the system

If the position of the mass at t = 0 is denoted as x₀, then

$C_1 + C_2 = x_0\,$

and if the velocity of the mass at t = 0 is denoted as u₀, then

$-j(u_0/\omega_0) = C_1 - C_2\,$

Solving the two boundary condition equations gives

$C_1 = \frac{1}{2}( x_0 - j( u_0 / \omega_0 ))$
$C_2 = \frac{1}{2}( x_0 + j( u_0 / \omega_0 ))$

The position is then given by

$x(t) = x_0 cos(\omega_0 t) + (u_0 /\omega_0 )sin(\omega_0 t)\,$

This equation can also be found by assuming that x is of the form

$x(t)=A_1 cos(\omega_0 t) + A_2 sin(\omega_0 t)\,$

And by applying the same initial conditions,

$A_1 = x_0\,$
$A_2 = \frac{u_0}{\omega_0}\,$

This gives rise to the same position equation

$x(t) = x_0 cos(\omega_0 t) + (u_0 /\omega_0 )sin(\omega_0 t)\,$

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[edit] Alternate Position Equation Forms

If A₁ and A₂ are of the form

$A_1 = A cos( \phi)\,$ $A_2 = A sin( \phi)\,$

Then the position equation can be written

$x(t) = Acos( \omega_0 t - \phi )\,$

By applying the initial conditions (x(0)=x₀, u(0)=u₀) it is found that

$x_0 = A cos(\phi)\,$
$\frac{u_0}{\omega_0} = A sin(\phi)\,$

If these two equations are squared and summed, then it is found that

$A = \sqrt{x_0^2 + (\frac{u_0}{\omega_0})^2}\,$

And if the difference of the same two equations is found, the result is that

$\phi = tan^{-1}(\frac{u_0}{x_0 \omega_0})\,$

The position equation can also be written as the Real part of the imaginary position equation

$\mathbf{Re} [x(t)] = x(t) = A cos(\omega_0 t - \phi)\,$

Due to euler's rule (e^jφ = cosφ + jsinφ), x(t) is of the form

$x(t) = A e^{j(\omega_0 t - \phi)}\,$

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[edit] Forced Oscillations(Simple Spring-Mass System)

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Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11

Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3

[edit] Recap of Section 1.3

In the previous section, we discussed how adding a damping component (e. g. a dashpot) to an unforced, simple spring-mass system would affect the response of the system. In particular, we learned that adding the dashpot to the system changed the natural frequency of the system from to a new damped natural frequency , and how this change made the response of the system change from a constant sinusoidal response to an exponentially-decaying sinusoid in which the system either had an under-damped, over-damped, or critically-damped response.

In this section, we will digress a bit by going back to the simple (undamped) oscillator system of the previous section, but this time, a constant force will be applied to this system, and we will investigate this system’s performance at low and high frequencies as well as at resonance. In particular, this section will start by introducing the characteristics of the spring and mass elements of a spring-mass system, introduce electrical analogs for both the spring and mass elements, learn how these elements combine to form the mechanical impedance system, and reveal how the impedance can describe the mechanical system’s overall response characteristics. Next, power dissipation of the forced, simple spring-mass system will be discussed in order to corroborate our use of electrical circuit analogs for the forced, simple spring-mass system. Finally, the characteristic responses of this system will be discussed, and a parameter called the amplification ratio (AR) will be introduced that will help in plotting the resonance of the forced, simple spring-mass system.

[edit] Forced Spring Element

Taking note of Figs. 1, we see that the equation of motion for a spring that has some constant, external force being exerted on it is...

$\hat{F} = s_M \Delta \hat{x}\qquad (1.4.1) \,$

where $s_M \,$ is the mechanical stiffness of the spring.

Note that in Fig. 1(c), force $\hat{F}$ flows constantly (i.e. without decreasing) throughout a spring, but the velocity $\hat{u}$ of the spring decrease from $\hat{u_1}$ to $\hat{u_2}$ as the force flows through the spring. This concept is important to know because it will be used in subsequent sections.

In practice, the stiffness of the spring $s_M \,$ , also called the spring constant, is usually expressed as $C_M = \frac {1}{s_M}\,$ , or the mechanical compliance of the spring. Therefore, the spring is very stiff if $s_M \,$ is large $\Rightarrow \; C_M$ is small. Similarly, the spring is very loose or “bouncy” if $s_M \,$ is small $\Rightarrow \; C_M$ is large. Noting that force and velocity are analogous to voltage and current, respectively, in electrical systems, it turns out that the characteristics of a spring are analogous to the characteristics of a capacitor in relation to, and, so we can model the “reactiveness” of a spring similar to the reactance of a capacitor if we let $C = C_M \,$ as shown in Fig. 2 below.

$Reactance\ of\ Capacitor:\ X_C = \frac{1}{j\omega C} \qquad (1.4.2a)$

$Reactance\ of\ Spring:\ X_{MS} = \frac{1}{j\omega C_M} \qquad(1.4.2b)$

[edit] Forced Mass Element

Taking note of Fig. 3, the equation for a mass that has constant, external force being exerted on it is...

$\hat{F} = M_M \hat{a} = M_M \hat{\dot{u}} = M_M \hat{\ddot{x}} \qquad (1.4.3)$

If the mass $M_M \,$ can vary its value and is oscillating in a mechanical system at max amplitude $A_M \,$ such that the input the system receives is constant at frequency $\omega \,$ , as $M_M \,$ increases, the harder it will be for the system to move the mass at $\omega \,$ at $A_M \,$ until, eventually, the mass doesn’t oscillate at all . Another equivalently way to look at it is to let $\omega \,$ vary and hold $M_M \,$ constant. Similarly, as $\omega \,$ increases, the harder it will be to get $M_M \,$ to oscillate at $\omega \,$ and keep the same amplitude $A_M \,$ until, eventually, the mass doesn’t oscillate at all. Therefore, as $\omega \,$ increases, the “reactiveness” of mass $M_M \,$ decreases (i.e. $M_M \,$ starts to move less and less). Recalling the analogous relationship of force/voltage and velocity/current, it turns out that the characteristics of a mass are analogous to an inductor. Therefore, we can model the “reactiveness” of a mass similar to the reactance of an inductor if we let $L = M_M \,$ as shown in Fig. 4.

$Reactance\ of\ Inductor:\ X_L = j\omega L \qquad (1.4.4a)$

$Reactance\ of\ Mass:\ X_{MM} = j\omega L_M \qquad(1.4.4b)$

[edit] Mechanical Impedance of Spring-Mass System

As mentioned twice before, force is analogous to voltage and velocity is analogous to current. Because of these relationships, this implies that the mechanical impedance for the forced, simple spring-mass system can be expressed as follows:

$\hat{Z_M} = \frac{\hat{F}}{\hat{u}} \qquad (1.4.5) \,$

In general, an undamped, spring-mass system can either be “spring-like” or “mass-like”. “Spring-like” systems can be characterized as being “bouncy” and they tend to grossly overshoot their target operating level(s) when an input is introduced to the system. These type of systems relatively take a long time to reach steady-state status. Conversely, “mass-like” can be characterized as being “lethargic” and they tend to not reach their desired operating level(s) for a given input to the system...even at steady-state! In terms of complex force and velocity, we say that “ force LEADS velocity” in mass-like systems and “velocity LEADS force” in spring-like systems (or equivalently “ force LAGS velocity” in mass-like systems and “velocity LAGS force” in spring-like systems). Figs. 5 shows this relationship graphically.

[edit] Power Transfer of a Simple Spring-Mass System

From electrical circuit theory, the average complex power $P_E \,$ dissipated in a system is expressed as ...

$P_E = \frac{1}{2} \mathbf{Re}\left\{ \hat{V}\hat{I^*} \right\} \qquad (1.4.6)\,$

where $\hat{V}\,$ and $\hat{I^*}\;$ represent the (time-invariant) complex voltage and complex conjugate current, respectively. Analogously, we can express the net power dissipation of the mechanical system $\hat{P}_E\,$ in general along with the power dissipation of a spring-like system $\hat{P}_{MS}\,$ or mass-like system $\hat{P}_{MM}\,$ as...

$\hat{P}_E = \frac{1}{2} \mathbf{Re}\left\{ \hat{F}\hat{u^*} \right\} \qquad \qquad \qquad (1.4.7a)\,$

$\hat{P}_{MS} = \frac{1}{2} \mathbf{Re}\left\{\hat{F} \left( \frac{j\hat{F}\omega}{s_M} \right)^* \right\} \qquad (1.4.7b)$

$\hat{P}_{MM} = \frac{1}{2} \mathbf{Re}\left\{\hat{F} \left( \frac{\hat{F}}{j \omega M_M} \right)^* \right\} \qquad (1.4.7c)$

In equations 1.4.7, we see that the product of complex force and velocity are purely imaginary. Since reactive elements, or commonly called, lossless elements, cannot dissipate energy, this implies that the net power dissipation of the system is zero. This means that in our simple spring-mass system, power can only be (fully) transferred back and forth between the spring and the mass. But this is precisely what a simple spring-mass system does. Therefore, by evaluating the power dissipation, this corroborates the notion of using electrical circuit elements to model mechanical elements in our spring-mass system.

[edit] Responses For Forced, Simple Spring-Mass System

Fig. 6 below illustrates a simple spring-mass system with a force exerted on the mass.

This system has response characteristics similar to that of the undamped oscillator system, with the only difference being that at steady-state, the system oscillates at the constant force magnitude and frequency versus exponentially decaying to zero in the unforced case. Recalling equations 1.4.2b and 1.4.4b, letting be the natural (resonant) frequency of the spring-mass system, and letting $\omega_n \,$ be frequency of the input received by the system, the characteristic responses of the forced spring-mass systems are presented graphically in Figs. 7 below.

$\mathbf{Figs.7}\,$

[edit] Amplification Ratio

The amplification ratio is a useful parameter that allows us to plot the frequency of the spring-mass system with the purports of revealing the resonant freq of the system solely based on the force experienced by each, the spring and mass elements of the system. In particular, AR is the magnitude of the ratio of the complex force experienced by the spring and the complex force experienced by the mass, i.e.

$\mathbf{AR} = \left| \frac{s_M \hat{x}}{M_M \hat{a}} \right| = \left| \frac{s_M \hat{x}}{M_M \hat{\dot{u}}} \right| = \left| \frac{s_M \hat{x}}{M_M \hat{\ddot{x}}} \right| \qquad (1.4.8) \,$

If we let $\zeta = \frac{\omega}{\omega_n}$ , be the frequency ratio, it turns out that AR can also be expressed as...

$\mathbf{AR} = \frac{1}{1 - \zeta^2} \qquad (1.4.9)\,$ .

AR will be at its maximum when $\left| X_{MS} \right| = \left| X_{MM} \right| \,$ . This happens precisely when $\zeta^2 = 1 \,$ . An example of an AR plot is shown below in Fig 8.

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[edit] Mechanical Resistance

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Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11

Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3

[edit] Mechanical Resistance

For most systems, a simple oscillator is not a very accurate model. While a simple oscillator involves a continuous transfer of energy between kinetic and potential form, with the sum of the two remaining constant, real systems involve a loss, or dissipation, of some of this energy, which is never recovered into kinetic nor potential energy. The mechanisms that cause this dissipation are varied and depend on many factors. Some of these mechanisms include drag on bodies moving through the air, thermal losses, and friction, but there are many others. Often, these mechanisms are either difficult or impossible to model, and most are non-linear. However, a simple, linear model that attempts to account for all of these losses in a system has been developed.

[edit] Dashpots

The most common way of representing mechanical resistance in a damped system is through the use of a dashpot. A dashpot acts like a shock absorber in a car. It produces resistance to the system's motion that is proportional to the system's velocity. The faster the motion of the system, the more mechanical resistance is produced.

As seen in the graph above, a linear realationship is assumed between the force of the dashpot and the velocity at which it is moving. The constant that relates these two quantities is $R M$ , the mechanical resistance of the dashpot. This relationship, known as the viscous damping law, can be written as:

$F = R\cdot u$

Also note that the force produced by the dashpot is always in phase with the velocity.

The power dissipated by the dashpot can be derived by looking at the work done as the dashpot resists the motion of the system:

$P_D = \frac{1}{2}\Re\left[\hat{F}\cdot\hat{u^*}\right]= \frac{|\hat{F}|^{2}}{2R_{M}}$

[edit] Modeling the Damped Oscillator

In order to incorporate the mechanical resistance (or damping) into the forced oscillator model, a dashpot is placed next to the spring. It is connected to the mass ( $M M$ ) on one end and attached to the ground on the other end. A new equation describing the forces must be developed:

$F - S_Mx - R_Mu = M_Ma \rightarrow F = S_Mx + R_M\dot{x} + M_M\ddot{x}$

It's phasor form is given by the following:

$\hat{F}e^{j\omega t} = \hat{x}e^{j\omega t}\left[S_M + j\omega R_M + \left(-\omega ^2\right)M_M\right]$

[edit] Mechanical Impedance for Damped Oscillator

Previously, the impedance for a simple oscillator was defined as $\mathbf{\frac{F}{u}}$ . Using the above equations, the impedance of a damped oscillator can be calculated:

$\hat{Z_M} = \frac{\hat{F}}{\hat{u}} = R_M + j\left(\omega M_M - \frac{S_M}{\omega}\right) = |\hat{Z_M}|e^{j\Phi_Z}$

For very low frequencies, the spring term dominates because of the $\frac{1}{\omega}$ relationship. Thus, the phase of the impedance approaches $\frac{-\pi}{2}$ for very low frequencies. This phase causes the velocity to "lag" the force for low frequencies. As the frequency increases, the phase difference increases toward zero. At resonance, the imaginary part of the impedance vanishes, and the phase is zero. The impedance is purely resistive at this point. For very high frequencies, the mass term dominates. Thus, the phase of the impedance approaches $\frac{\pi}{2}$ and the velocity "leads" the force for high frequencies.

Based on the previous equations for dissipated power, we can see that the real part of the impedance is indeed $R M$ . The real part of the impedance can also be defined as the cosine of the phase times its magnitude. Thus, the following equations for the power can be obtained.

$W_R = \frac{1}{2}\Re\left[\hat{F}\hat{u^{*}}\right] = \frac{1}{2}R_M|\hat{u}|^2 = \frac{1}{2}\frac{|\hat{F}|^2}{|\hat{Z_M}|^2}R_M = \frac{1}{2}\frac{|\hat{F}|^2}{|\hat{Z_M}|}cos(\Phi_Z)$

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[edit] Characterizing Damped Mechanical Systems

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Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11

Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3

[edit] Characterizing Damped Mechanical Systems

Characterizing the response of Damped Mechanical Oscillating system can be easily quantified using two parameters. The system parameters are the resonance frequency ( $''' w r e s o n a n c e'''$ and the damping of the system $''' Q (q u a l i t y f a c t o r) o r B (T e m p o r a l A b s o r p t i o n''')$ . In practice, finding these parameters would allow for quantification of unknown systems and allow you to derive other parameters within the system.

Using the mechanical impedance in the following equation, notice that the imaginary part will equal zero at resonance.

( $Z m = F / u = R m + j (w * M m - s / w)$ )

Resonance case:( $w * M m = s / w$ )

[edit] Calculating the Mechanical Resistance

The decay time of the system is related to 1 / B where B is the Temporal Absorption. B is related to the mechancial resistance and to the mass of the system by the following equation.

$B = R m / 2 * M m$

The mechanical resistance can be derived from the equation by knowing the mass and the temporal absorption.

[edit] Critical Damping

The system is said to be critically damped when:

$R c = 2 * M * s q r t (s / M m) = 2 * s q r t (s * M m) = 2 * M m * w n$

A critically damped system is one in which an entire cycle is never completed. The absorbtion coefficient in this type of system equals the natural frequency. The system will begin to oscillate, however the amplitude will decay exponentially to zero within the first oscillation.

[edit] Damping Ratio

$D a m p i n g R a t i o = R m / R c$

The damping ratio is a comparison of the mechanical resistance of a system to the resistance value required for critical damping. Rc is the value of Rm for which the absorbtion coefficient equals the natural frequency (critical damping). A damping ratio equal to 1 therefore is critically damped, because the mechanical resistance value Rm is equal to the value required for critical damping Rc. A damping ratio greater than 1 will be overdamped, and a ratio less than 1 will be underdamped.

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[edit] Quality Factor

The Quality Factor (Q) is way to quickly characterize the shape of the peak in the response. It gives a quantitative representation of power dissipation in an oscillation.

$Q = w r e s o n a n c e / (w u - w l)$

Wu and Wl are called the half power points. When looking at the response of a system, the two places on either side of the peak where the point equals half the power of the peak power defines Wu and Wl. The distance in between the two is called the half-power bandwidth. So, the resonant frequency divided by the half-power bandwidth gives you the quality factor. Mathematically, it takes Q/pi oscillations for the vibration to decay to a factor of 1/e of its original amplitude.

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[edit] Electro-Mechanical Analogies

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Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11

Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3

[edit] Why analogs to circuits?

Since acoustic devices contain both electrical and mechanical components, one needs to be able to combine them in a graphical way that aids the user's intuition. The method that is still used in the transducer industry is the Impedance and Mobility analogies that compare mechanical systems to electric circuits.

[edit] Two possible analogies

i) Impedance analog

ii) Mobility analog

                     Mechanical           Electrical equivalent

i)impedance analog

   Potential         Force     F(t)          Voltage    V(t)
   Flux              Velocity  u(t)          Current    i(t)

ii)Mobility analog

   Potential         Velocity  u(t)          Voltage    V(t)
   Flux              Force     F(t)          Current    i(t)

Impedance analog is often easier to use in most accoustical systems while mobility analog can be found more intuitively for mechanical systems. These are generalities, however, so it is best to use the analogy that allows for the most understanding. A circuit of one analog can be switched to the equivalent circuit of the other analog by using the dual of the circuit. (more on this in the next section).

[edit] The equivalent spring

Mechanical spring

$F=K\int\,u dt$

Impedance analogy of the mechanical spring

$V=1/C\int\,i dt$

Mobility analogy of the mechanical spring

$i=1/L\int\,V dt$

[edit] The equivalent Mass

Mechanical mass

$F=Mx^{\prime\prime}=M\frac{du}{dt}$

Impedance analogy of the mechanical mass

$V=L\frac{di}{dt}$

Mobility analogy of the mechanical mass

$i=C\frac{dV}{dt}$

[edit] The equivalent resistance

Mechanical resistance

$F = R u$

Impedance analogy of the mechanical resistance

$V = R i$

Mobility analogy of the mechanical resistance

$i=\frac{1}{R}V$

[edit] Review of Circuit Solving Methods

Kirchkoff's Voltage law

"The sum of the potential drops around a loop must equal zero."

This implies that the total potential drop around a series of elements is equal to the sum of the individual voltage drops in the series.

$e t o t a l = d r o p 1 + d r o p 2 + d r o p 3$

Kirchkoff's Current Law

"The Sum of the currents at a node (junction of more than two elements) must be zero"

Using the pipe flow analogy of circuits, this can be thought of as the continuity equation.

For example if there was a node with three elements connected to it (numbered 1,2 and 3) $i 1 + i 2 + i 3 = 0$ From the current law, their sum would equal zero.

Hints for solving circuits:

-Remember that certain elements can be combined to simplify the circuit (the combination of like elements in series and parallel)

-If solving a ciruit that involves steady-state sources, uses impedances! (This reduces the ciruit down to a bunch of complex domain resistor elements that can be combined to simplify the circuit.)

Examples of Electro-Mechanical Analogies

[edit] Additional Resources for solving linear circuits:

Thomas & Rosa, "The Analysis and Design of Linear Circuits", Wiley, 2001

Hayt, Kemmerly & Durbin, "Engineering Circuit Analysis", 6th ed., McGraw Hill, 2002

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[edit] Methods for checking Electro-Mechanical Analogies

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Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11

Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3

After drawing the electro-mechanical analogy of a mechanical system, it is always safe to check the circuit. There are two methods to accomplish this:

[edit] 1. Low-Frequency Limits:

This method looks at the behavior of the system for very large or very small values of the parameters and compares them with the expected behavior of the mechanical system. The basic formula to spot an error in the electro-mechanical circuit is as follows:

               Very large value:   Very small value:
Capacitor (C)   Short circuit       Open circuit
Resistor (R)    Open circuit        Short circuit
Inductor (L)    Open circuit        Short circuit

[edit] 2. Dot Method: (Valid only for planar network)

This method helps obtain the dual analog (one analog is the dual of the other). The steps for the dot product are as follows: 1) Place one dot within each loop and one outside all the loops. 2) Connect the dots. Make sure that only there is only one line through each element and that no lines cross more than one element. 3) Draw in each line that crosses an element its dual element, including the source. 4) The circuit obtained should have the same configuration as the dual analog of the original electro-mechanical circuit.

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[edit] Examples of Electro-Mechanical Analogies

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Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11

Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3

[edit] Example 1

Draw the mobility analog representation of the mechanical system shown below.

[edit] Example 1 Solution

Using the fact that flux is equivalent to force and potential to velocity, the following is the mobility analog representation of the mechanical system given in example 1.

Using the Low-frequency limits method to check the accuracy of the mobility analog circuit drawn, we have:

i) If we make the Cms (inductor) very small, the Cms becomes a short circuit. This agrees with the mechanical system.

ii) If we make the Mm2 (capacitor) very large, the Mm2 becomes a short circuit. No motion is transmitted to the rest of the system and this agrees by inspecting the mechanical system given.

[edit] Example 2

Draw the mobility analog representation of the following axisymmetric device. Does your circuit make sense if you consider behavior at low-frequency?

[edit] Example 2 Solution

The mobility anolog representation of this system would be as follows:

[edit] Example 3

Draw the mobility analog representation of the mechanical system below. Consider the behavior of the circuit at low frequency to check for validity. Then draw the impedence equivalent circuit using the dot method.

The mobility analog representation of the mechanical system is shown as:

The impedence analog representation of the same mechanical system is shown as:

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[edit] Primary variables of interest

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Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11

Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3

[edit] Basic Assumptions

Consider a piston moving in a tube. The piston starts moving at time t=0 with a velocity u= $u p$ . The piston fits inside the tube smoothly without any friction or gap. The motion of the piston creates a planar sound wave or acoustic disturbance traveling down the tube at a constant speed c>> $u p$ . In a case where the tube is very small, one can neglect the time it takes for acoustic disturbance to travel from the piston to the end of the tube. Hence, one can assume that the acoustic disturbance is uniform throughout the tube domain.

Image:Acousticplanewave1.gif

[edit] Assumptions

1. Although sound can exist in solids or fluid, we will first consider the medium to be a fluid at rest. The ambient, undisturbed state of the fluid will be designated using subscript zero. Recall that a fluid is a substance that deforms continuously under the application of any shear (tangential) stress.

2. Disturbance is a compressional one (as opposed to transverse).

3. Fluid is a continuum: infinitely divisible substance. Each fluid property assumed to have definite value at each point.

4. The disturbance created by the motion of the piston travels at a constant speed. It is a function of the properties of the ambient fluid. Since the properties are assumed to be uniform (the same at every location in the tube) then the speed of the disturbance has to be constant. The speed of the disturbance is the speed of sound, denoted by letter $c 0$ with subscript zero to denote ambient property.

5. The piston is perfectly flat, and there is no leakage flow between the piston and the tube inner wall. Both the piston and the tube walls are perfectly rigid. Tube is infinitely long, and has a constant area of cross section, A.

6. The disturbance is uniform. All deviations in fluid properties are the same across the tube for any location x. Therefore the instantaneous fluid properties are only a function of the Cartesian coordinate x (see sketch). Deviations from the ambient will be denoted by primed variables.

[edit] Variables of interest

[edit] Pressure (force / unit area)

Pressure is defined as the normal force per unit area acting on any control surface within the fluid.

Image:Acousticcontrolsurface.gif

$p = \frac {\tilde{F}.\tilde{n}}{dS}$

For the present case,inside a tube filled with a working fluid, pressure is the ratio of the surface force acting onto the fluid in the control region and the tube area. The pressure is decomposed into two components - a constant equilibrium component, $p 0$ , superimposed with a varying disturbance $p' (x)$ . The deviation $p'$ is also called the acoustic pressure. Note that $p'$ can be positive or negative. Unit: $k g / m s 2$ . Acoustical pressure can be meaured using a microphone.

Image:Acousticpressure1.gif

[edit] Density

Density is mass of fluid per unit volume. The density, ρ, is also decomposed into the sum of ambient value (usually around ρ0= 1.15 kg/m3) and a disturbance ρ’(x). The disturbance can be positive or negative, as for the pressure. Unit: $k g / m 3$

[edit] Acoustic volume velocity

Rate of change of fluid particles position as a funtion of time. Its the well known fluid mechanics term, flow rate.

$U=\int_{s}\tilde{u}.\tilde{n}\, dS$

In most cases, the velocity is assumed constant over the entire cross section (plug flow), which gives acoustic volume velocity as a product of fluid velocity $\tilde{u}$ and cross section S.

$U=\tilde{u}.S$

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[edit] Electro-acoustic analogies

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Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11

Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3

[edit] Electro-acoustical Analogies

[edit] Acoustical Mass

Consider a rigid tube-piston system as following figure.

Piston is moving back and forth sinusoidally with frequency of f. Assuming