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
Electric devices ; R , L , C , e
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
Cm [m/N]∴ 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
Velocity v ↔ Current : i
f
Cm
f
Cm=
< Fig 1.3 >
m
f
v
B) Mass ; m
Mech. ; f = m ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.4)
L
e
i
< Fig 1.4 >
Elect. ; e = L ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.5)
< Fig 1.5 >
m
e
v
Analogy ; Mass m inductance L
m
f
< Fig 1.6 >
C) Mechanical Resistance ; Rm
fluid
v
f
→ frichtion , viscosity
v
m
f
Mech. ; f = Rm v ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.6)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
sound velocity is given by,
δV
c= ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.12)
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
B) inertance ;
f = Sp = m ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.16)
u = Sv < Fig 1.13 >
p
u
P = = e = L ∙∙∙∙∙∙∙∙∙∙∙∙∙(1.17)
Where , =
< 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 ;
