안타깝게 도형이안나온다는

Ch.1 Electric Circuit Analogy of Mechanical system and Acoustical system

•Mechanical ;  Vibration velocity (v) & force (f)           Electric current(i)

                                                          &

•Acoustical ;  Volume velocity (u) & pressure (p)          voltage (v)           

∙They have same differential egs? → electric circuit analysis

∙The size of elements  λ → lump circuits

1-1 Analogy of Mechanical system and Electrical system

k

ξ

f

A) Compliance ; Cm                         

Mech. ; Hook’s Law for a spring

                                                                      < Fig 1.1 >

F = kξ                           ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.1)

K ; spring const. → stiffness [N/m]

ξ ; displacement [m]

i

∴ f =  =               ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.2)

v ;  particle velocity (=dξ/dt)

e

C

Elecrt. ; e = ξ =        ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.3)

           g : electric charge [C]      

 C : capacitance [F]  

                                                                          < Fig 1.2 >

Analogy ;  compliance    Cm ↔ Capacitance : C

           Force          f  ↔ Voltage     : e

v

f

Cm

f

Cm=

< Fig 1.3 >

m

f

v

Mech. ; f = m                  ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.4)    

L

e

i

Elect. ; e = L               ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.5)

< Fig 1.5 >

m

e

v

m

f

                                 < Fig 1.6 >

C) Mechanical Resistance ; Rm

 fluid

v

f

v

m

f

        Rm : frictional coust. or viscosity

Elect. ; e = Ri              ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.7)

   Mechanical Resistance Rm <-> Electric Resistance R                  < Fig 1.7 >

v

f

Rm

i

e

R

              < Fig 1.8 >                                       < Fig 1.9 >                   

1-2. Basic Analogy Circuit for Mechanical System

Rm

Cm

m

v

f

v

m

f

Cm

Rm

< Circular vibrational plate >          < mass connected with spring on vough surface >

                                    < Fig 1.10 >

f = Rmv + m  +

                                       ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.8) 

e = Ri + L  +               

e

i

R

L

 C

f

v

Rm

m

Cm

                                        < Fig 1.11 >

For sinusoidal vibration ,

v = Vm  = V                    ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.9)

f = Fm  = F

F  = (Rm+jwt+ )V                    ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.10)

= ZmV                                     

Where ,  Zm= Rm+ j(wm- ) ; mechanical impedance

= Rm +jXm

Xm ; mechanical reactance

1-3    Analogy of Acoustical system and Electric system

A) Acoustic Compliance ;  

When ,  λ  then

P = κ                          ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.11)

κ ; Bulk modulus ( = r ,  ; Static Pressure , r ; Specific heat ratio)

u

p

u

δV

V。, P。

XX  (1.12) → (1.11)   then ,

P =                            ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.13)                    < Fig 1.12 >

Volume displacement  is given by

δV =                      ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.14)

u : volume velocity  [ /s ]

P =                 e =       ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.15)

Analogy ;

            pressur P ↔ voltage e

            volume velocity u ↔ current i

            acoustic capacitance ↔ capacitance C

u

P

                                   < Fig 1.13 >              

F

u

s

v

Mass : m

f = Sp = m                  ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.16)

u = Sv                                               < Fig 1.13 >

p

u

                                                             < Fig 1.15 > 

C ) Acoustic Resistance ;  

SP =  ← f ∝ v

u = Sv

∴ P = =  ↔ e = Ri         ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.18)

u

p

1-4          Basic Analogy circuit for Acoustical System

• Helmholtz  Resonator

Xx m the pipe

           m = Sl

The inertance xx the pipe ;

       =  =                        ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.19)

The Compliance of the cavity ;

       =                             ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.20)

The XXXXX   Resistance ;