Section 4.

0 : The Calculation Algorithm

By Martin J. King, 07/05/02
Copyright  2002 by Martin J. King. All Rights Reserved.
Section 4.0 : The Calculation Algorithm

The Velocity and Pressure Equations for an Acoustic Element :

In the previous section, the general equations for the displacement, the velocity,
and the pressure were derived as functions of position and time. These relationships are
restated below as Equations (4.1), (4.2), and (4.3).

Equations (4.1), (4.2), and (4.3)

( ( −α − I β ) x ) ( ( −α + I β ) x ) (I ω t )
ξ ( x , t ) = ( C1 e + C2 e )e

( ( −α − I β ) x ) ( ( −α + I β ) x ) (I ω t )
u( x , t ) = I ω ( C1 e + C2 e )e

2 ( ( −α − I β ) x ) ( ( −α + I β ) x ) ( I ω t )
p( x , t ) = I ρ c ( C 1 ( I α + β ) e − C 2 ( −I α + β ) e )e

If the two constants, C1 and C2, could be determined or eliminated from these
equations, then the values of displacement, velocity, and pressure would be known
everywhere within the transmission line. Having a background in mechanical
engineering and experience performing structural finite element analyses, I decided to
apply some of the methods used to derive one-dimensional truss elements to formulate
a one-dimensional acoustic transmission line element. My one-dimensional acoustic
transmission line element is shown in Figure 4.1 along with a positive sign convention,
the geometry definition, and the variables at end 1 and end 2.

Figure 4.1 : One-Dimensional Acoustic Element

p1 p2
u1 u2
S1 S2

u(x,t) = ∂ξ(x,t)/∂t
positive sign convention

At end 1 : x = 0 (m) At end 2 : x = L (m)

p1 – pressure (Pa) p2 – pressure (Pa)
u1 – velocity (m/sec) u2 – velocity (m/sec)
S1 – area (m2) S2 – area (m2)

Page 1 of 11
 Section 4.0 : The Calculation Algorithm
By Martin J. King, 07/05/02
Copyright  2002 by Martin J. King. All Rights Reserved.

Derivation of an Acoustic Element Transfer Matrix :

Equations (4.2) and (4.3) for velocity and pressure contain two unknowns. If two
values are assigned to any of the four variables (u1, u2, p1, or p2) from the acoustic
element shown in Figure 4.1, then the remaining two variables can be used to eliminate
the constants C1 and C2. By careful selection of the two assigned values, convenient
expressions for the two remaining variables result. In the following derivation the time
(I ω t)
varying term e has been dropped since it is common to every expression.

Case 1 : u1 = 1 m/sec and u2 = 0 m/sec

2 (I L β ) ( −I L β )
I ρ c ( ( −α + I β ) e + (α + I β ) e )
p1 =
( −I L β ) (I L β )
ω (e −e )
2 ( −α L )
ρc β e
p2 = −2
(I L β ) ( −I L β )
ω ( −e +e )

Case 2 : u1 = 0 m/sec and u2 = 1 m/sec

2 (α L)
ρc β e
p1 = 2
(I L β ) ( −I L β )
ω ( −e +e )

2 ( −I L β ) (I L β )
I ρ c ((α − I β ) e − (α + I β ) e )
p2 =
( −I L β ) (I L β )
ω (e −e )

At this point, a demonstration problem would probably help the reader

understand the pressure solutions for Cases 1 and 2. Assume that the length L of the
acoustic transmission line element is 1 meter and that the cross-sectional area is a
constant 0.05 m2. Since an expression for the fiber damping has not yet been derived,
assume that the transmission line is empty. Also, define the excitation velocity as 0.001
m/sec. For these assumptions, the resonances of the 1 m long closed end transmission
line should be approximately :

f = n x c / (2 x L) n = 1, 2, 3, … and c = 342 m/sec

f = 171 Hz for n = 1
= 342 Hz for n = 2
= 513 Hz for n = 3

Figures 4.2 and 4.3 show the pressure results for Cases 1 and 2 after applying these
assumptions.

Page 2 of 11
 Section 4.0 : The Calculation Algorithm
By Martin J. King, 07/05/02
Copyright  2002 by Martin J. King. All Rights Reserved.

Figure 4.2 : Results for Case 1

Excitation at x = 0 : u(0) = 1, u(L) = 0

 (−α + j⋅ β r)⋅ exp(j⋅ β r⋅ L) + (α + j ⋅ β r)⋅ exp(−j⋅ β r⋅ L)

p 0 := j⋅ ρ ⋅ c ⋅ 
2
r ( ( ) ( ))
r⋅ dω⋅ exp −j⋅ β r⋅ L − exp j⋅ β r⋅ L

−2⋅ ρ ⋅ c ⋅ β r⋅ exp(−α ⋅ L)
2
p L :=
r ( ( )
r⋅ dω⋅ exp −j ⋅ β r⋅ L − exp j⋅ β r⋅ L( ))

100

p0 10
r
Pressure (Pa)

1000⋅ Pa
1
pL
r
1000⋅ Pa
0.1

0.01
1 .10
3
1 10 100
−1
r⋅ dω⋅ Hz

100
80

( )
arg p 0
r
60
40
Phase (deg)

deg 20

( )
0
arg p L
r 20
deg 40
60
80
100
1 .10
3
1 10 100
−1
r⋅ dω⋅ Hz
Frequency (Hz)

Acoustic Finite Element Model | Lumped Parameter Model

|
m | S0⋅ L
p 0 ⋅ 0.001⋅ = 22.522Pa | Cab :=
1 sec 2
| ρ ⋅c

1 ( )
arg p 0 = −90.000deg |
| S0⋅ 0.001⋅
m
sec
m | p :=
p L ⋅ 0.001⋅ = 22.526Pa | j ⋅ 1⋅ Hz⋅ Cab
1 sec
|
arg p L ( ) = −90.000deg
1
| p = 22.525Pa arg ( p ) = −90.000deg

Page 3 of 11
 Section 4.0 : The Calculation Algorithm
By Martin J. King, 07/05/02
Copyright  2002 by Martin J. King. All Rights Reserved.

Figure 4.3 : Results for Case 2

Excitation at x = L : u(0) = 0, u(L) = 1

 (α − j ⋅ β r)⋅ exp(−j⋅ β r⋅ L) − (α + j⋅ β r)⋅ exp(j⋅ β r⋅ L)

p L := j ⋅ ρ ⋅ c ⋅ 
2
r ( ( ) ( ))
r⋅ dω⋅ exp −j⋅ β r⋅ L − exp j ⋅ β r⋅ L

2⋅ ρ ⋅ c ⋅ β r⋅ exp(α ⋅ L)
2
p 0 :=
r ( ( )
r⋅ dω⋅ exp −j⋅ β r⋅ L − exp j⋅ β r⋅ L ( ))

100

p0 10
r
Pressure (Pa)

1000⋅ Pa
1
pL
r
1000⋅ Pa
0.1

0.01
1 .10
3
1 10 100
−1
r⋅ dω⋅ Hz

100
80

( )
arg p 0
r
60
40
Phase (deg)

deg 20

( )
0
arg p L
r 20
deg 40
60
80
100
1 .10
3
1 10 100
−1
r⋅ dω⋅ Hz
Frequency (Hz)

Acoustic Finite Element Model | Lumped Parameter Model

|
m | S0⋅ L
p 0 ⋅ 0.001⋅ = 22.526Pa | Cab :=
1 sec 2
| ρ ⋅c

1 ( )
arg p 0 = 90.000deg |
| S0⋅ 0.001⋅
m
sec
m | p :=
p L ⋅ 0.001⋅ = 22.522Pa | j ⋅ 1⋅ Hz⋅ Cab
1 sec
|
arg p L ( ) = 90.000deg
1
| p = 22.525Pa arg ( p ) = −90.000deg

Page 4 of 11
 Section 4.0 : The Calculation Algorithm
By Martin J. King, 07/05/02
Copyright  2002 by Martin J. King. All Rights Reserved.
A lot of information is contained in Figures 4.2 and 4.3 that warrants some further
discussion and explanation. Lets start by looking closely at Figure 4.2, the same
observations will apply to Figure 4.3 unless otherwise noted. The top plot is the pressure
magnitude where it can be seen that sharp pressure peaks occur at the expected half
wavelength frequencies. The pressure at both ends of the line spikes to extreme
magnitudes at these frequencies.

There are also some deep nulls in Figure 4.2 but only in the pressure at the
driven end. I would not classify these nulls as quarter wavelength resonances even
thought the frequencies are consistent with quarter wavelength frequencies for the
assumed length. At these frequencies, the length of the line is equivalent to a quarter
wavelength and a pressure distribution exists that tends to unload the driven end. The
pressure at the closed end is set by the velocity magnitude at the driven end and does
not spike in a similar manner to the half wavelength modes. This is a velocity-controlled
standing wave and therefore not what I would define as a true resonance.

At the bottom of Figures 4.2 and 4.3, the lumped parameter pressure is
calculated and compared to the 1 Hz results from the acoustic transmission line element.
This is really a check on the element static stiffness and it can be seen that both
methods yield identical pressure magnitudes and phases. Also note the 180-degree
difference in the phase calculated in Figure 4.2 and Figure 4.3 at the 1 Hz frequency.
This is a direct result of the positive sign convention assumed in Figure 4.1. A positive
displacement increases pressure in Figure 4.2 while it decreases pressure in Figure 4.3.

Acoustic impedances can be calculated for the four variables (u1, u2, p1, and p2)
from the acoustic element shown in Figure 4.1. These relationships are shown below for
the two load cases.

Case 1 : u1 = 1 m/sec and u2 = 0 m/sec

p1
Z11 =
S1 u1

p2
Z21 =
S1 u1

Case 2 : u1 = 0 m/sec and u2 = 1 m/sec

p1
Z12 =
S2 u2

p2
Z22 =
S2 u2

Page 5 of 11
 Section 4.0 : The Calculation Algorithm
By Martin J. King, 07/05/02
Copyright  2002 by Martin J. King. All Rights Reserved.
These impedance relationships can be arranged to express the pressures in terms of the
volume velocities and then placed in matrix notation.

 p1  Z11 Z12 S1 u1

    
 =  
 p2  Z21 Z22 S2 u2


 p1  Z11 Z12  U1 
    
 =  
 p2  Z21 Z22  U2 


One more rearranging of the equations yields the transfer matrix for the one-dimensional
acoustic element.

Equation (4.4)
 Z22 1 

 − 
 U1   Z21 Z21   U2 

    
  
 p  =  Z Z  
Z11   p2 
 1   11 22 
− + Z12 
 Z 21 Z21 

Equation (4.4) expresses the pressure and volume velocity at one end of the acoustic
element in terms of the pressure and volume velocity at the other end. This transfer
matrix equation is the basis for the transmission line calculation algorithm I have
programmed into my new MathCad worksheets.

Closed End and Open End Simple Transmission Line Models :

Figure 4.4 shows the acoustic equivalent circuit model, using the impedance
analogy, for a simple transmission line. Figure 4.5 shows the corresponding electrical
equivalent circuit model. From the classic papers by Thiele(1) and Small(2-4), all of the
circuit elements have already been defined with the exception of the transmission line’s
acoustic impedance and electrical impedance. Equation (4.4) can be used to determine
these remaining circuit elements.

Simple Transmission Line – Closed End

Assume a unit velocity excitation at the driven end of the transmission line and a
zero velocity boundary condition at the closed end.

S1 m
U1 =
sec

U2 = 0

Page 6 of 11
 Section 4.0 : The Calculation Algorithm
By Martin J. King, 07/05/02
Copyright  2002 by Martin J. King. All Rights Reserved.
The unit velocity input is converted to the equivalent volume velocity and applied
at end 1. A zero volume velocity condition is applied at end 2. In Equation (4.4) this
leaves two equations and two unknowns, p1 and p2, which can be solved for as
functions of frequency.

Having solved for the pressures resulting from the assumed volume velocity
input, the acoustic impedance and the electrical impedance can be determined.

p1 sec
Zacoustic =
S1 m

2 2
B l
Zelectrical =
S1 2 Zacoustic

The equivalent circuits in Figures 4.4 and 4.5 can now be solved to obtain the
electrical and SPL system performance as functions of frequency for a closed end
simple transmission line. This is the basis for the first MathCad model titled “TL Closed
End.MCD”.

Simple Transmission Line – Open End

Assume a unit velocity excitation at the driven end of the transmission line and a
zero pressure boundary condition at the open end.

S1 m
U1 =
sec

Again, the unit velocity input is converted to the equivalent volume velocity and
applied at end 1. However, at the open end a more complicated boundary condition will
be applied. My first models simply set the pressure p2 to zero but this did not lead to a
good correlation with the measured test line data. To make the model more accurate I
applied an acoustic impedance boundary condition at the open end. The acoustic
impedance of a piston in an infinite baffle was used to model the terminus impedance. A
typical terminus (or mouth) impedance is shown in Figure 4.6.

p2 = Zmouth U2

In Equation (4.4) this leaves two equations and two unknowns, p1 and U2, which can be
solved for as functions of frequency.

Having solved for the pressures and the open end’s volume velocity resulting
from the unit velocity input, the acoustic impedance and the electrical impedance can be
determined.

Page 7 of 11
 Section 4.0 : The Calculation Algorithm
By Martin J. King, 07/05/02
Copyright  2002 by Martin J. King. All Rights Reserved.

p1 sec
Zacoustic =
S1 m

2 2
B l
Zelectrical =
S1 2 Zacoustic

One last relationship is needed, the velocity at the open end as a function of the
velocity at the driven end.

S1 U2
ε=
S2 U1

u2
ε=
u1

The open end simple transmission line electrical and SPL system performance
can now be calculated as functions of frequency using the equivalent circuit models
shown in Figures 4.4 and 4.5. This is the basis for the second MathCad model titled “TL
Open End.MCD”.

Summary :

A transfer matrix for a one-dimensional acoustic element has been derived and
boundary conditions applied to represent two simple transmission line configurations.
The only remaining unknown to be discussed is the damping coefficient for the fibrous
tangle typically used to stuffing transmission lines. The following section will present the
empirically derived damping coefficient and the correlation with the test data presented
in Section 2.0, using the simple MathCad transmission line model “TL Open End.MCD”.

Page 8 of 11
 Section 4.0 : The Calculation Algorithm
By Martin J. King, 07/05/02
Copyright  2002 by Martin J. King. All Rights Reserved.

Figure 4.4 : Acoustic Equivalent Circuit for a Simple Transmission Line Speaker

Ratd Mad Cad

Ud

pg Zal

where :

pg = pressure source
= (eg Bl) / (Sd Re)

Rad = driver acoustic resistance

= (Bl2 / Sd2) [Qed / ((Rg + Re) Qmd)]

Ratd = total acoustic resistance

= Rad + (Bl)2 / [Sd2 ((Rg + Re) + jω Lvc)]

Cad = driver acoustic compliance

= Vd / (ρair c2)

Mad = driver acoustic mass

= (fd2 Cad)-1

Zal = transmission line acoustic impedance

Ud = driver volume velocity

= Sd ud

ud = driver cone velocity

then :

uL = terminus air velocity

= ε ud

ε = uL / ud

Page 9 of 11
 Section 4.0 : The Calculation Algorithm
By Martin J. King, 07/05/02
Copyright  2002 by Martin J. King. All Rights Reserved.

Figure 4.5 : Electrical Equivalent Circuit for a Simple Transmission Line Speaker

Rg + Re Lvc

eg Lced Cmed Red Zel ed

where :

eg = voltage source
= 2.8284 volt

Rg+Re = electrical resistance of the amplifier, cables, and voice coil

Lvc = voice coil inductance

Lced = inductance due to the driver suspension compliance

= [Cad (Bl)2] / Sd2

Cmed = capacitance due to the driver mass

= (Mad Sd2) / (Bl)2

Red = resistance due to the driver suspension damping

= Re (Qmd / Qed)

Zel = transmission line equivalent electrical impedance

= (Bl)2 / (Sd2 Zal)

ed = Bl ud

Page 10 of 11
 Section 4.0 : The Calculation Algorithm
By Martin J. King, 07/05/02
Copyright  2002 by Martin J. King. All Rights Reserved.

Figure 4.6 : Transmission Line Open End Acoustic Impedance

Terminus Impedance : Piston in an Infinite Baffle Impedance Model

SL
aL :=
π

 k  x
2⋅ k+ 1 

( −1) ⋅  
25
  2 
J1( x) := ∑  k!⋅ Γ ( k + 2) 
 
k =0

 2⋅ k+ 2 
 ( −1) k⋅  x   25
  2 
H1( x) :=   3  ∑
5
k = 0  Γ  k + ⋅Γ  k +  
  2  2

  r⋅ dω ⋅ a  
2⋅ J1 2⋅
 r⋅ dω ⋅ a  
 L  2⋅ H1 2⋅ L 
⋅  1 −
 c  + j⋅  c
ρ ⋅c 
Zmouth :=  
r SL ⋅ dω ⋅ dω
 2⋅
r
⋅ aL  2⋅
r
⋅ aL 
 c  c 

Terminus Impedance
Real and Imaginary Impedance Comps.

1.25

(
SL⋅ Re Z mouth )
r
1

ρ ⋅c
0.75
(
SL⋅ Im Z mouth )
r
0.5
ρ ⋅c

0.25

0
0 2 4 6 8 10 12 14
r⋅ dω
2⋅ ⋅ aL
c

Page 11 of 11