Lumped Elements

Lecture 7 Acoustics of Speech & Hearing 6.551 - HST 714J Lecture 7: Lumped Elements I. What is a lumped element? Lumped elements are physical structures that act and move as a unit when subjected to controlled forces. Imagine a two-dimensional block of lead on a onedimensional frictionless surface. mass=M Acceleration FORCE x When a force is imposed on the block, the block moves as a unit in a direction described by the difference in force acting on its two surfaces, or analytically: dV Net Force = (5.1) dt Mass The key features is that a gradient of a physical parameter produces a uniform physical response throughout the lump. Another example of a lumped element is an electrical resistor where a difference in the Voltage (E) across the resistive element produces a current (I) that is uniform throughout the resistor: I E1 E2 a resistor of value R where: 30 Sept -2004 I = (E1 − E2 )/ R (5.2) page 1 Lecture 7 Acoustics of Speech & Hearing 6.551 - HST 714J II. Lumped Acoustic Elements A. Elements: A lumped element is a representation of a structure by one or two physical quantities that are homogenous or varying linearly throughout the structure Standing Waves in P and V in a long tube with a rigid termination at x=0. The spatial variation in the sound pressure magnitude and phase P(x) is defined by a cosine function. The spatial variation in particle velocity magnitude and phase V(x) is defined by a sine function. The region where the tube can act as a lumped element is the region where the pressure amplitude is nearly constant and the ‘volume velocity’ (v x tube crosssection) varies linear with x. B. An example of a lumped acoustic element is a short open tube of moderate diameter, where length l and radius a are <0.1 λ. length l u(t) p (t) 1 p (t) A SHORT CIRCULAR TUBE OF RADIUS a 2 Under these circumstances particle velocity V and the sound pressures are simply related by: dV (P1 − P2 ) = (5.3) dt ρ0 l 30 Sept -2004 page 2 Lecture 7 Acoustics of Speech & Hearing 6.551 - HST 714J where Eqn. 5.3 is the specific acoustic equivalent of Eqn. 5.1. (Hint: you can describe the forces acting on the lump by multiplying the pressures by the cross-sectional area of the tube πa2. C. Volume Velocity and Acoustic Impedance In discussing lumped acoustic elements, it is convenient to think about velocity in terms of a new variable Volume Velocity U where in the case of the tube above, length l P1 U P2 2xa the volume velocity is defined by the product of the particle velocity and the crosssectional area of the tube, i.e. U = πa 2 V = SV . The relationship between volume velocity and the pressure difference in the open tube above can be obtained by multiplying both sides of Eqn, 5.3 by S=πa2, i.e. dV (P 1 − P 2 ) S = S dt ρ0 l (5.4) P1 − P 2 ) 2 dU (P1 − P 2 ) ( = S= S ,where Sl = Tube Volume. dt ρ0 l ρ0 Sl P − P2 The Acoustic Impedance of the tube is : 1 U III. Separation into ‘Through’ and ‘Across’ Variables , where power (t ) = through (t ) across( t ) ‘Across’ variable ‘Through’ variable Electrics voltage e(t) current i(t) Mechanics: Impedance analogy force f(t) velocity v(t) Mechanics: Mobility analogy velocity v(t) force f(t) Acoustics: Impedance sound pressure p(t) volume velocity u(t) 30 Sept -2004 page 3 Lecture 7 Acoustics of Speech & Hearing 6.551 - HST 714J analogy Acoustics: Mobility analogy volume velocity u(t) sound pressure p(t) In all of the above analogies, power(t) = through (t) across(t) has units of watts. IV. Two Terminal Elements A. Electrical Elements Resistor Capacitor + Inductor + i(t) + i(t) v(t) R= 1 G v(t) i(t) = C or i(t) = Gv(t) or Units of R are ohms (Ω) Units of G are siemens (S) L - v(t) + - v0(t) v(t) = v0(t) independent of i(t) - dv(t) dt v(t) = L or di(t) dt 1 v(t) - v(0) = i(t )dt C m0 i(t) - i(0) = Units of C are farads (F) Units of L are henries (H) Ideal Independent Voltage Source i(t) v(t) C - v(t) = Ri(t) + i(t) t 1 v(t )dt L m0 t Figure 5.1 Simple linear 2-terminal lumped electrical elements and their constitutive relations. The orientation of the arrow and the +/- signs identifies the positive reference direction for each element. In this figure the variable i is current and v is voltage. (From Siebert “Circuits, Signals and System, 1986). Ideal Independent Current Source - v(t) + i(t) i0(t) i(t) = i0(t) independent of v(t) Note that R, C and L are the coefficients of the 0th and 1st order differential equations that relate v(t) (or e(t)) to i(t). 30 Sept -2004 page 4 Lecture 7 Acoustics of Speech & Hearing 6.551 - HST 714J B. Analogous Elements m Mass Cm = 1/s Compliance Rm Figure 5.2 Electric elements and their mechanical and acoustic counterparts in the “Impedance analogy” From Kinsler, Frey, Coppens, & Sanders, Fundamentals of Acoustics , 3rd Ed. (1982) L M Inertance Inductance C C Compliance Capacitance R R Resistance Resistance Resistance Mechanical Acoustical Electrical Fig. 10.3. Acoustic, electrical and mechanical analogues. C. Analogous Constitutive Relationships Mechanical V vs F Electrical I vs E Acoustical U vs P Spring Capacitor Compliance Spring df (t ) v(t ) = C M dt Capacitor de(t ) i( t ) = C E dt Compliance dp(t ) u( t ) = C A dt Damper Resistor Resistor Damper 1 v(t) = f (t) RM Resistor 1 i(t) = e(t) RE Resistor 1 u(t) = p(t) RA Mass Inductor Inertance Mass 1 v(t) = ∫ f (t)dt LM Inductor 1 i(t) = ∫ e(t)dt LE Inertance 1 u(t) = ∫ p(t)dt LA 30 Sept -2004 page 5 Lecture 7 Acoustics of Speech & Hearing 6.551 - HST 714J In the Sinusoidal Steady State: Mechanical Electrical Acoustical V vs F I vs E U vs P Spring Capacitor Compliance Spring Capacitor Compliance V (ω) = jωC M F (ω) I (ω) = jωC E E (ω) U (ω) = jωC A P(ω) Damper Resistor Resistor Damper 1 F (ω) V (ω) = RM Resistor 1 E (ω) I (ω) = RE Resistor 1 U (ω) = P (ω) RA Mass Inductor Inertance Mass 1 F (ω ) V (ω) = jωLM Inductor 1 E (ω ) I (ω) = jωLE Inertance 1 P (ω) U (ω) = jωL A { } dp(t) = Real{jω Pe jωt }= −ω P sin (ωt + ∠P ) dt p(t) = Real Pe jωt = P cos(ωt + ∠P ) = ω P cos(ωt + ∠P + π /2) 30 Sept -2004 page 6 Lecture 7 Acoustics of Speech & Hearing 6.551 - HST 714J V. Acoustic Element Values and Physics Element constraints result from physical process and element values are determined by physical properties including the dimensions of structures, e.g. the electrical resistance of a resistor depend on the dimensions and the resistivity of the material from which it’s constructed. Acoustic mass: units of kg/m4 A. An open ended tube with linear dimensions l and a <0.1 λ and S=πa2 circular tube l p2 u(t) p1 LA = ρο l S = ρο Volume S2 p(t) = p1(t) - p2(t) p(t) = LA du(t) dt assumes only inertial forces ρο = equilibrium mass density of medium The Electrical Analog P1-P2=U jω LA . Note that the acoustic mass is equivalent to the mass of the air in the enclosed element divided by the square of the cross-sectional area of the element. Also since some small volume of the medium on either end of the tube is also entrained with the media inside the tube, the “acoustic” length is usually somewhat larger than the physical length of the tube. For a single open end, the difference between the physical length and the acoustic length is ∆l ≈ 0.8 a . This difference is called the end correction. 30 Sept -2004 page 7 Lecture 7 Acoustics of Speech & Hearing 6.551 - HST 714J B. Acoustic Compliance: units of m3/Pa Volume displaced per unit pressure difference (2 examples, both of which assume resistance and inertia are negligible). 1. A Diaphragm of a < 0.1 λ p1 (t) diaphragm at rest diaphragm displaced p2 (t) vol = (p1 (t ) − p 2 (t ))C A = p(t)C A d (vol ) dt dp(t ) u(t) = C A dt u(t) = volume displacement = vol The Electrical Analog U = jω C A ( P1 − P2 ) For a round, flat, “simply mounted” plate CA = πa 6 (7 + v)(1 − v) 16 Et 3 , where: a is the radius of the plate, v = 0.3 is Poisson’s ratio, E is the elastic constant (Young’s modulus) of the material, and t is the thickness (Roark and Young, 1975, p. 362-3, Case 10a). 30 Sept -2004 page 9 Lecture 7 Acoustics of Speech & Hearing 6.551 - HST 714J 2. Enclosed volume of air with linear dimensions <0.1 λ P(jω) U(jω) u(t) p(t) Another structure that may be well approximated by an acoustic compliance. Volume C= Adiabatic Bulk modulus U = jω C A P The variations in sound pressure within an enclosed air volume generally occur about the steady-state atmospheric pressure, the ground potential in acoustics. Therefore, one terminal of an electrical-analog of a volume-determined acoustic compliance should always be grounded. C. Acoustic Resistance: units of Acoustic Ohms (Pa-s/m3) 1. A narrow tube or radius a << 0.001 λ circular (radius = a) rigid tube -filled with acoustic medium l u(t) p1 (t) p2 (t) p(t ) = p1 (t ) − p2 (t) = R Au(t ) ← assumption; only viscous forces 8ηl p (t) − p2 (t) RA = 4 ⇒ 1 u(t) πa η = viscosity of medium Because of the viscous forces, relative motions of fluid at one radial position with respect to an adjacent position exerts a force opposing the motion that is proportional to the spatial derivative of the velocity and the fluid’s coefficient of shear viscosity η. 30 Sept -2004 page 10 Lecture 7 Acoustics of Speech & Hearing 6.551 - HST 714J The action of these forces results in a proportionality of pressure difference and volume velocity that is analogous to an electric resistance. In the sinusoidal steady state, then: U1(jω) P1(jω) RA P2(jω) , The consequence of the viscosity is that the velocity at the stationary where P1-P2=U1 RA . Relative velocity amplitude 1.0 walls is zero, and is maximum in the center of the tube (see Fig. 5.3). 0.5 The viscous forces produce energy loss near the walls where the velocity changes with position. 0.1 0.05 0 0.05 0.1 Radial position, r, in cm; a = 0.1 Fig. 5.3 Relative Particle velocity amplitude as a function of radial position in a small pipe of radius a = 0.1 cm, at frequency f = 200 Hz. After Kinsler and Frey, 1950; p. 238 ⎛ −(0.1−r)⎞ ⎟ ⎜ δ ⎠ ⎝ The velocity profile in Figure 5.3 varies as v(r) =1− e , where the “space 1/ 2 constant” δ = [η /(ρ0ω )] , with η, the coefficient of shear viscosity = 1.86x10-5 N-sm-2 for air at STP, ρ0, density of air = 1.2 kg-m3, and ω, radian frequency = 2πf. At 200 Hz δ=1.1x10-4 m = 0.011 cm (Figure 5.3) At 20 Hz δ=3.5x10-5 m = 0.035 cm The effect of the viscous forces is insignificant when the radius of the tube is an order of magnitude or more larger than the space constant and therefore we can ignore viscosity for short tubes of moderate to large radius, 0.01λ < a <0.1λ. 30 Sept -2004 page 11 Lecture 7 Acoustics of Speech & Hearing 6.551 - HST 714J 2. An infinitely long tube The action of an acoustic resistor is to absorb sound power. The viscous forces within a narrow tube convert the sound power into heat that dissipates away. A second type of acoustic resistance can be constructed from a long tube of moderate crosssectional dimensions (0.01 λ < a < 0.2 λ). Such a construction can conduct sound power away from a system and can be treated as an acoustic resistance where: R= 30 Sept -2004 ρ0c . πa 2 page 12 Lecture 7 Acoustics of Speech & Hearing 6.551 - HST 714J l is effectively infinite There is a catch, however, in that p(t) this lumped element always has P0 u(t) one end coupled to ground and therefore can only be used to U1(jω) either terminate acoustic circuits P1(jω) RA or be placed in parallel with other elements. There are ways of dealing with long tubes as a collection of series and parallel elements that have already been discussed in Lecture 2. D. Two Mixed mass-resistance acoustic loads 1. A tube of intermediate radius (neither wide nor narrow) has an impedance determined by the combination of an acoustic mass or inertance (associated with accelerating the fluid mass within the tube) and a resistance (associated with overcoming viscous drag at the stationary walls of the tube). Since the pressure drop across the resistance and the mass elements add, we think of these as an R and L in series. length l p (t) 1 u(t) p (t) An intermediate tube 2 ∆P = P2 − P1 = U ( jωρ o l S + R) 30 Sept -2004 page 13 Lecture 7 Acoustics of Speech & Hearing LE 6.551 - HST 714J RE where S is the cross-sectional area of the tube and R is the resistance. 2. The radiation impedance acts whenever sound radiates from some element and is made up of an acoustic mass associated with accelerating the air particles near the surface of the element and a resistance associated with the transmission of sound energy into the far field. Since the volume velocities associated with these two processes add (some fraction of U goes into accelerating the mass layer, while the rest radiates away from the element), we can think of these as two parallel elements. Radiation from the end of an organ pipe of radius a can be modeled by the following: U + P _ LR RR U 1 1 1 = Y Rad = = + P Z Rad jωL R RR where: 1 1 = + jω 0.8 a ρ0 c 2 πa 2 2 πa 2 Note that the radiation mass is equivalent to the addition of a tube of radius a and length 0.8a to the end of the pipe. This is the end correction!! E. Range of applicability of acoustic circuit theory. 1. Pressure and volume-velocity ranges consistent with “linear acoustics”. 30 Sept -2004 page 14 Lecture 7 Acoustics of Speech & Hearing 6.551 - HST 714J 2. Frequency range limited by the assumption of “lumped” elements, i.e. the dimensions of the structures need to be small compared to a wavelength: a and l < 0.1 λ. 30 Sept -2004 page 15 Lecture 7 Acoustics of Speech & Hearing 6.551 - HST 714J VI. Circuit Descriptions of a Real Acoustic System A Jug or Helmholtz Resonator A. An Acoustic Circuit Description If we are using acoustic volume velocity as a through variable; the flow of volume velocity through the neck suggests a series combination of Acoustic Elements. The volume velocity first flows through an a series combination of an acoustic inertance LA, and an acoustic resistor RA, and then into the acoustic compliance CA of the closed cavity, where: Volume ρ l′ . L A = 0 2 ; R A = g(l, a , frequency); C A = γP0 πa Furthermore if we really treat the neck as an L and R combination than U2 = U1. ⎛ 1 ⎞ ⎟⎟ P1 (ω) =U 1 (ω )⎜⎜ jωL A + RA + jωC A ⎠ ⎝ The ratio of P1/U1 defines the acoustic input impedance of the bottle and in this case it is equal to the series sum of the impedance of the three series elements. In the sinusoidal steady state: 30 Sept -2004 page 16 Lecture 7 Acoustics of Speech & Hearing Z IN (ω ) = jωL A + R A + 6.551 - HST 714J 1 jωC A B. An Electrical Analog of the Acoustic Circuit Description In Electrical circuits the wires that connect the ideal elements are perfect conductors. LE + E1 RE + E2 + E3 CE If the numerical values of LE = LA, RE=RA, and CE=CA, then I1=U1, E1=P1 and E2=P2. C. A Mechanical Analog of the Acoustic Circuit Description In Mechanical circuits the rods that attach ideal mechanical elements are rigid and massless. If the numerical values of LM = LA, RM=RA, CM=CA, then V1=U1=U2, and F=P1, then 1 F = jωL M + RM + jωCM V1 Where the total force acting on the elements equals the sum of the forces acting on each. ⎛ 1 ⎞ F = V 1⎜ jωL M + R M + jωC M ⎠ ⎝ 30 Sept -2004 page 17