5.7 Acoustical Impedance
131
The acoustical impedance in the mks system of units is often given the unit rayl,
named after the physicist Lord Rayleigh. A rayl carries the fundamental mks units
of kilogram per second per meter-squared. The acoustical impedance of air at room
temperature (23 ◦ C) is about 420 rayl; salt water is about 1.54 × 10 6 rayl and steel
is 47 × 10 6 rayl. Some values of acoustical impedance for biological tissue (in cgs
units) are shown in Table 5.7.
The fraction of the sound energy transmitted into a second material depends on
the relative impedance of the materials. A wave which traverses a smooth boundary
between two materials will be partially reflected and partially transmitted (and
refracted if the incident ray is not normal to the boundary). Across the boundary, the
wave amplitude and the pressure on either side of the boundary much match. For
normal incidence, this makes the intensity of the transmitted wave, I 2 , a fraction of
the intensity of the incident wave I 1 , according to
I 2
I 1
=
4z 1 z 2
(z 1 + z 2 ) 2 .
(5.41)
Proof At the boundary, let initial and reflected waves have amplitude A and B,
and transmitted wave have amplitude C. Although the wavelength of the wave may
change traversing the boundary, the frequency cannot, as the material is being forced
to move at the incident frequency. The wave amplitude must have a single value at
the boundary between the two materials, so that A + B = C. Also, a balancing of
pressures (using (5.23)) makes z 1 (A − B) = z 2 C. These two relations give A =
(1/2)(1 + z 2 /z 1 )C and B = (1/2)(1 − z 2 /z 1 )C, so that B/A = (1 − z 2 /z 1 )/(1 +
z 2 /z 1 ) and C/A = 2/(1 + z 2 /z 1 ). The wave intensities can be found from (5.34),
i.e. I i = (1/2)z i ω 2 A 2
i , so that
I 2 /I 1 = (z 2 /z 1 )C
2 /A
2
= 4(z 2 /z 1 )/(1 + z 2 /z 1 )
2 q.e.d.
As can be seen from Eq. (5.41), the percent of sound energy transmitted as a
function of the impedance ratio peaks when z 1 = z 2 , for which all the energy of the
wave passes into the second material, with no reflected wave.
Consider the electrical analog: An amplifier with a output impedance Z A
transfers power to speakers with Z S . If the amplifier power supply has a voltage V ,
the current through speakers is i = V /(Z A + Z S ), so the power supplied to speakers
is P = i 2 Z S = Z S V 2 /(Z A + Z S ) 2 . For fixed V and Z A , the power transferred is
maximum when the slope of P as a function of Z S vanishes. You can check that this
occurs when Z S = Z A .
In acoustics as well as in electronic circuit design, adjusting the transmitter or
receiver impedances to maximize power transfer is called ‘impedance matching’.
An example of poor impedance matching occurs when you to try to push air back
and forth with the tip of your finger. A better match is to push air back and forth
with a hand fan.
The fraction of energy reflected is (z 1 − z 2 ) 2 /(z 1 + z 2 ) 2 . As the impedance of the
second material grows above the first, less and less of the initial wave is transmitted.
131
The acoustical impedance in the mks system of units is often given the unit rayl,
named after the physicist Lord Rayleigh. A rayl carries the fundamental mks units
of kilogram per second per meter-squared. The acoustical impedance of air at room
temperature (23 ◦ C) is about 420 rayl; salt water is about 1.54 × 10 6 rayl and steel
is 47 × 10 6 rayl. Some values of acoustical impedance for biological tissue (in cgs
units) are shown in Table 5.7.
The fraction of the sound energy transmitted into a second material depends on
the relative impedance of the materials. A wave which traverses a smooth boundary
between two materials will be partially reflected and partially transmitted (and
refracted if the incident ray is not normal to the boundary). Across the boundary, the
wave amplitude and the pressure on either side of the boundary much match. For
normal incidence, this makes the intensity of the transmitted wave, I 2 , a fraction of
the intensity of the incident wave I 1 , according to
I 2
I 1
=
4z 1 z 2
(z 1 + z 2 ) 2 .
(5.41)
Proof At the boundary, let initial and reflected waves have amplitude A and B,
and transmitted wave have amplitude C. Although the wavelength of the wave may
change traversing the boundary, the frequency cannot, as the material is being forced
to move at the incident frequency. The wave amplitude must have a single value at
the boundary between the two materials, so that A + B = C. Also, a balancing of
pressures (using (5.23)) makes z 1 (A − B) = z 2 C. These two relations give A =
(1/2)(1 + z 2 /z 1 )C and B = (1/2)(1 − z 2 /z 1 )C, so that B/A = (1 − z 2 /z 1 )/(1 +
z 2 /z 1 ) and C/A = 2/(1 + z 2 /z 1 ). The wave intensities can be found from (5.34),
i.e. I i = (1/2)z i ω 2 A 2
i , so that
I 2 /I 1 = (z 2 /z 1 )C
2 /A
2
= 4(z 2 /z 1 )/(1 + z 2 /z 1 )
2 q.e.d.
As can be seen from Eq. (5.41), the percent of sound energy transmitted as a
function of the impedance ratio peaks when z 1 = z 2 , for which all the energy of the
wave passes into the second material, with no reflected wave.
Consider the electrical analog: An amplifier with a output impedance Z A
transfers power to speakers with Z S . If the amplifier power supply has a voltage V ,
the current through speakers is i = V /(Z A + Z S ), so the power supplied to speakers
is P = i 2 Z S = Z S V 2 /(Z A + Z S ) 2 . For fixed V and Z A , the power transferred is
maximum when the slope of P as a function of Z S vanishes. You can check that this
occurs when Z S = Z A .
In acoustics as well as in electronic circuit design, adjusting the transmitter or
receiver impedances to maximize power transfer is called ‘impedance matching’.
An example of poor impedance matching occurs when you to try to push air back
and forth with the tip of your finger. A better match is to push air back and forth
with a hand fan.
The fraction of energy reflected is (z 1 − z 2 ) 2 /(z 1 + z 2 ) 2 . As the impedance of the
second material grows above the first, less and less of the initial wave is transmitted.