BOX_AIR.HTM --- Part of Manual for Driver Parameter Calculator --- by Claus Futtrup.
Created 10. December 2000, last revised 22. May 2001. Ported to XHTML 1.0 on 2. October 2004. Last modified 25. October 2004.

## Introduction

When a driver is mounted in a box the air will not provide a completely linear spring/suspension, as a function of cone position. It is the purpose of this document to prove that this nonlinearity can be neglected under normal circumstances (Kooky Kyle's homepage has been the primary source of inspiration for this document, see bottom of this file for a reference). A closed box is assumed, which is like a traditional measuring box. Other kinds of cabinets are not considered in this document.

## Fundamentals

This analysis is based on the changes of pressure inside a box, and it is then analysed what happens to the pistonic motion of a loudspeaker diaphragm, which is assumed to be a rigid piston. Phase relations between air pressure and diaphragm motion are not considered, ie. the changes are quasi-static, which implies that Mms = 0 and Cms = infinity (the suspension is infinitely soft) or something close. The quasi-static process also implies that the pistonic motion is (much) slower than the speed of sound, which is always the case with (conventional) loudspeakers.

The compression inside the box is adiabatic (ie. energy constant). We get the equality:

        p1 * V1^gamma = p2 * V2^gamma                                (1)

p is the gas pressure, in Pa (Pascal)
V is the volume of the gas, in m3 (cubic meter)
gamma is the adiabatic compression constant


Both sides of the equation represents the gas inside the box, but in two different states. In this analysis p1 and V1 is the average state, where the piston has not moved from its rest position, whereas p2 and V2 represents the maximum change from rest.

The gamma constant is set to 1.40, which has been discussed in AIR.HTM, see equation 7 in section Speed of Sound.

Notice that the ideal gas equation p * V = n * R * T cannot be used directly in this case. This is so because neither p, V or T is constant, ie. the process is not isobar, isochor or isoterm, but isentrop (constant entropi), which means the process is adiabatic, ie. energy-constant, and this is the needed information. It is implied that the process is reversible, ie. after changing the situation from 1 to 2 the process can be returned to situation 1. This also implies that damping material is not present in the box.

The change in volume can be expressed as:

        V2 = V1 + dV
= V1 * (1 + dV/V1)
= V1 * (1 + k)                                            (2)

dV is delta-V, the volume change as an absolute value
k is the volume change (as a ratio measure)


The volume change is achieved by moving a driver cone/piston:

        dV = Sd * x

Sd is the driver diaphragm area, in m2 (square meter)
x is the driver excursion, in m (meter)


Equation 1 can be changed to:

        p1 * V1^gamma = p2 * (V1 * (1 + k))^gamma                    (3)


Dividing both sides by V1^gamma gives us the relation:

                                                          p1
p1 = p2 * (1 + k)^gamma      <===>      p2 = -------------   (4)
(1 + k)^gamma

p1 is the average pressure inside the box
p2 is the max or min pressure (peak values)


p1 is set to be the standard pressure 101325 Pa, which can be measured outside the box because this is the pressure inside the box when the piston is at rest position, and p2 is set to represent the pressure inside the box. The volume change k is represented by the ratio of volume displacement from the cone movement and the volume inside the box:

            dV    Sd * x
k = -- = --------                                            (5)
V1      Vb

Vb = V1 is the box volume, in m3 (cubic meter)


At the same time we need the relation between applied force and pressure:

                                            F
F = Sd * (p1 - p2)      <===>      ---- = p1 - p2            (6)
Sd

F is the motional force, ie. applied force to the cone, in N (Newton)


The motional force is the force required to move the cone from its rest position to the x position. Here the internal pressure (the reaction force) gives us the following force balance ä F(x) = F + Sd * (p2 - p1) = 0. The reaction force is Sd * (p2 - p1) and this is why F is the applied force.

The situation we are looking at can be pictured as follows:

              0 = x --> (positive direction outwards)
|
|-----------
| p   |
| V   |-------- F --> (positive direction outwards)
| T   |
|-----------


Here p, V, T represent the state and F is the applied force on the cone.

If the pressure inside the box is smaller than the pressure outside, then F is negative and the piston moves inwards.

When x is positive then the piston is moving outwards because when k is positive then V2>V1.

Substituting expressions from equation 4 and 5 into equation 6 gives:

         F               p1                         p1
---- = p1 - ------------- = p1 - -----------------------     (7)
Sd         (1 + k)^gamma        (1 + Sd * x / Vb)^gamma


Solving for the excursion variable x we get:

        (1 + Sd * x / Vb)^gamma * (F / Sd - p1) = - p1

<===>
- p1                1
(1 + Sd * x / Vb)^gamma = ------------- = -----------------
(F / Sd - p1)   1 - F / (Sd * p1)
<===>
/   Sd * p1   \ ^(1/gamma)
1 + Sd * x / Vb = ( ------------- )
\ Sd * p1 - F /
<===>
Vb   Vb    /   Sd * p1   \ ^(1/gamma)
x = - -- + -- * ( ------------- )                            (8)
Sd   Sd    \ Sd * p1 - F /


This final equation is different from what was stated on Kyle Lahnakoski's webpage. Working with this part of the DPC manual documentation I have discussed the derivation with Kyle Lahnakoski, and I believe he has chosen to change his webpage in accordance with the above equation. The error did not change his conclusion.

Looking at equation 8 it can be seen that if F = 0 then the expression in parentheses becomes 1, and no matter what power is used, even 1/gamma, then this becomes 1 as well. Then x = Vb/Sd - Vb/Sd = 0.

Notice that Sd * p1 expresses the static force on both sides of the piston, and then F only represents the difference between the two sides. It can also be seen that when x>0 then F>0 which is in accordance with the definitions.

Given a sine input in F it will be possible to evaluate the distortion in x, based on the above equation.

This is too easy for us, though, since such a numerical approach will not be educational. Primarily it will not be possible to evaluate when the box distortion will be of a significant size.

## Taylor series expansion

The purpose of the following Taylor series expansion is to express x (the excursion) directly as a function of F (the force on the cone), which is proportional to the pressure on the cone.

A Taylor series is an infinite polynomial series with variable terms of the following type:

                 ___
\    x(a)^(k)
x(F) = >   -------- * (F-a)^k                              (9)
/___    k!

a is a point, here chosen to be zero, since F is alternating around
zero, and we therefore want the best possible approximation at zero.
k is a running variable from zero to infinity, and all the contributions
must be added together (to retrieve a perfect approximation).
F is the applied force. In this case, later F is substituted by a
x is the equation to be represented by the Taylor series, it must be a
function with derivatives of all orders in the neighborhood of the
point a. The expression x(a)^(k) means x(a) derived k times.


For a = 0 the Taylor series is sometimes called the Maclaurin series, which is the case because we want x = 0 at the rest position (this is similar to what was given at equation 8). I prefer the general term Taylor series. Be aware that the a value can be chosen to be a non-zero value in case the Taylor series approximation is supposed to be made around a cone position different from the rest position.

The first term, k = 0, is when the function is left unchanged, and when F is set to a = 0, the first term becomes zero. This means that when x = 0 there is no pressure difference on the two sides of the cone, and therefore no force on the cone, which makes sense.

The second term gives us the following derivative of equation 8:

                  Vb    /   Sd * p1   \ ^(1/gamma)
dx   -- * ( ------------- )
x' = -- = Sd    \ Sd * p1 - F /                             (10)
dF   --------------------------------
gamma * (Sd * p1 - F)


Inserting x' into the Taylor series and setting F in equation 10 = a = 0 gives the following total expression:

             Vb    /   Sd * p1   \ ^(1/gamma)
-- * ( ------------- )
Sd    \ Sd * p1 - 0 /              F
x(k=1) = -------------------------------- * -
gamma * (Sd * p1 - 0)         1

Vb
= ----------------- * F                                  (11)
Sd^2 * p1 * gamma


Calculating one further term (the 3. term, x(k=2)) gives us the following Taylor series:

                   Vb
x = ----------------- * F                                   (12)
Sd^2 * gamma * p1

Vb    /          1                   1 / gamma - 1      \
+ -- * ( --------------------- + ------------------------- ) * F^2
Sd    \ gamma * (Sd * p1)^2     2 * gamma * (Sd * p1)^2 /

+ R(F^3)


R is expressing the error from lack of polynomial terms, because the Taylor series is an infinte series. R is called the remainder (representing the residual terms). The first part is called the Taylor polynomium of n'th order, in this case a 2. order because k=0, 1, 2 has been evaluated.

Since we have only included 2 terms, the error becomes a function of F^3. This error is neglected because we are proving that distortion is negligible, and the first term(s) in a Taylor series is always more important than all the following terms together. In fact, the error can be expressed as:

                x(z)^(k+1)
Rn(F) = ---------- * (F - a)^(k+1)
(k+1)!

z is a value between 0 and F
x is the function to be derived, here it is derived k+1 times
a is zero in this case


What this means is, that z is a value between 0 and F, and the error from the following terms (ie. the remainder) can be expressed if you calculate just one extra term of the Taylor polynomial. Finding the z value between 0 and F which gives you the maximum error will be a conservative evaluation of the error from exclusion of the following terms in the Taylor series.

For your information, the following terms are longer than the previous terms, so calculating further terms becomes a tedious job.

Equation 12 can be simplified, and for the purpose of this analysis we omit the error part:

                   Vb
x = ----------------- * F                                   (13)
Sd^2 * gamma * p1

Vb         1 + gamma
+ -------- * -------------- * F^2
2 * Sd^3   (gamma * p1)^2


Equation 13 is the 2. degree Taylor polynomium of x(F).

The 3. term is reduced by the following calculations:

        Vb    /          1                   1 / gamma - 1      \
-- * ( --------------------- + ------------------------- ) * F^2
Sd    \ gamma * (Sd * p1)^2     2 * gamma * (Sd * p1)^2 /

Vb    /      1      \  /  1     1 / gamma - 1 \
= -- * ( ------------- )( ----- + -------------- ) * F^2
Sd    \ (Sd * p1)^2 /  \gamma     2 * gamma   /

Vb    /      1      \  /  1          1              1     \
= -- * ( ------------- )( ----- + ------------ - ----------- ) * F^2
Sd    \ (Sd * p1)^2 /  \gamma   2 * gamma^2     2 * gamma /

Vb    /      1      \  /     1              1      \
= -- * ( ------------- )( ------------- + ----------- ) * F^2
Sd    \ (Sd * p1)^2 /  \ 2 * gamma^2     2 * gamma /

Vb    /      1      \  1  / 1 + gamma \
= -- * ( ------------- ) - ( ----------- ) * F^2
Sd    \ (Sd * p1)^2 /  2  \  gamma^2  /

Vb ( 1 + gamma )
= 1/2 ---------------------- * F^2
Sd^3 * p1^2 * gamma^2


The next term in the equation, which is omitted together with all the following terms, is (for your information):

                    Vb      1 + 3 * gamma + 2 * gamma^2
x(k=3) = -------- * --------------------------- * F^3       (14)
6 * Sd^4         (gamma * p1)^3


Including the term in equation 14 in the analysis could improve by inclusion of 3. harmonic distortion. But 2. harmonic must be dominating. Second harmonic distortion is characterized by a non-symmetric distortion, kind of clipping in one of the two half periods of a sine-wave. Third harmonic distoriton is characterized by a symmetric distortion, kind of clipping in both half periods of a sine-wave. A loudspeaker box will be dominated by the non-symmetric kind of distortion, ie. even order harmonics, where 2. harmonic distortion will be dominating.

I hope to make this point more clear by looking into what happens at overpressure versus underpressure. If we have a box with a piston, first at normal pressure, and we pull the piston out of the box there will be an underpressure in the box -- it changes from 1 atm to a smaller value, though it cannot drop below vacuum (0 atm). This behaviour is nonlinear. On the other hand, if the piston is pushed into the box an overpressure builds up, and the pressure increase will be increasing more than a linear increase, virtually going toward infinity (in the most insane case). If you picture this behaviour on a piece of paper versus a straight line, then you will see a non-symmetric non-linearity. This is why the even order harmonics will be dominating in the distortion from a closed box.

Alternatively, if we look at the earlier picture of a piston in a box. If the piston is moved halfway into the box the pressure doubles, but if the piston is moved equally far out the pressure will become 2/3. This does not draw a graph as a linear function. It will be a hyperbolic function. The hyperbolic behaviour will look flat on a graph as long as dV << Vb.

## Convergence of the Taylor series

A Taylor series is a kind of power series. The convergence of a power series can be tested by determining the radius of convergence the following way (a simple ratio test):

                                a_n
|F| < R = 1/rho = lim -------
a_(n+1)


Looking at x(k=1) against x(k=2) as well as x(k=2) against x(k=3) we can assume that this tendency continues for all derivatives. The coefficient a_i becomes smaller and smaller for higher numbers of i. This means that the above series converges for all values of F, i.e. |F| is infinity.

An alternative test could e.g. be the root test:

        rho = lim |a_n|^(1/n)


For a_2 we get:

                       Vb ( 1 + gamma )
a_2 = = 1/2 ----------------------
Sd^3 * p1^2 * gamma^2

2 * Sd^3 * p1^2 * gamma^2
|F| < sqrt -------------------------
Vb ( 1 + gamma )


If the example following later is taken up, then F < +/- 23.5 kN will suffice to achieve convergence, which would be more than Sd * p1 (see equation 8) and therefore be impossible.

## Distortion

A relevant way to look at closed box nonlinearity is to calculate the harmonic distortion. This can be done if the fourier transform is calculated, but first F is chosen to be a harmonic function:

        F = A * sin(omega * t)                                      (15)

A is the force amplitude in N (Newton)
t is the running time in s (seconds)


According to Kyle Lahnakoski the fourier transform becomes easy if a trig identity is used, involving (sin(x))^2. This function has a shape similar to the sine function, but with a period of pi instead of 2*pi, and with values between 0 and 1 instead of -1 and 1.

The trigonometric identity is:

        (sin(w t))^2 = (1 - cos (2 w t)) / 2


The identity comes in handy because of the second term in equation 13 (or 12), where F is squared, which gives us an expression in (sin(x))^2. This does not directly relate to a function of sin(2x)), which would be the representative of the 2. harmonic distortion.

To equally find a convertion from (sin(x))^3 to something, which is related to (sin(3x))^3 is a difficult task. This would be required if equation 14 was to be included. When more terms are included there is a "risk" of getting more 2. harmonic and higher harmonic terms. This possibility is ignored because it is assumed that the 2. term of the Taylor series will be dominant.

### The derivation of the trigonometric identity

Kyle Lahnakoski writes:

The derivation is highly geometrical. I will try to describe it in words here. Make sure you have a pencil and paper.

Draw a circle, mark the center as "C". Draw a horizontal line from the center to the right and intersect with the circle. Mark the intersection "S" for start. Next choose an angle theta. Draw a line from the center of the circle to its perimeter so that it is at theta degrees to the horizontal line. The intersection of the circle and this new line is called point "A". Draw another line, like the first only with the angle 2*theta. Make the intersection of this line and the circumference as "B". Note the line at A bisects the angle B-C-S. Draw the line B-S, note that it intersects at right angles to the line C-A. Label this intersection "D". The length of B-D is sin(theta), therefore the length of B-S is 2*sin(theta). Now draw a vertical line at S. This line may extend as far up and down as you please. The horizontal distance from B to the vertical line is (2*sin(theta))*sin(theta). This can be seen by noting the angle between B-S and the vertical line is theta.

	Therefore cos(2*theta) = 1 - (2*sin(theta))*sin(theta)


Just rearrange to get the trig identity.

### End of derivation

The derivation is valid for any angle, but will be easier for you to draw and to understand, if the angle theta is kept less than 45 degrees (eg. try to draw theta = 30 degrees).

I believe Kyle Lahnakoski used general fourier analysis. I think a simple fourier series would be easy because the analysed function is by choice a harmonic function, and the Taylor series has done most of the required work.

### Explanation of the Fourier Transform

Here I intended to have an explanation of the Fourier Transform (FT) in general, grounded in the example above.

                ^          1
FT(x) = f(w) = ---------- INTEG( x(F) * e^(-i*w*F)) dF
sqrt(2*pi)


It should be sufficient to limit the integration over 1 period, ie. from 0 to 2*pi.

My source is: Erwin Kreyszig, Advanced Engineering Mathematics, 7. edition, section 10.11, Fourier Transform, page 611.

Quite frankly I need to understand better what is happening here. Fourier transform was not a big issue at my studies. Neither was Laplace transform, which perhaps could have helped me a bit.

If someone can explain to me what is going on, then please do. For now I have accepted the calculation by Kyle Lahnakoski.

### End of Fourier Transform

The following results, and this is not the whole expression when transformed, but rather just the amplitude/size of the values:

                                    Vb * A
Fundamental (1. harm.) : 1HD = -----------------                    (16)
Sd^2 * gamma * p1

Vb * (1 + gamma) * A^2
2. harmonic distortion : 2HD = -------------------------            (17)
4 * Sd^3 * (gamma * p1)^2


The relation to equation 13 is very easy to see. The force, F, is substituted with the force peak amplitude, A, and further more the expression for second harmonic distortion is divided by 2. The reason is in the conversion with the trig identity, where the expression with cos(omega * t) is divided by 2.

Dividing the 2. harmonic by the fundamental gives you the distortion ratio, which makes it easy to express the 2. harmonic distortion (2HD) as a percentage value or alternatively in dB:

                       Vb         1 + gamma
-------- * -------------- * A^2
2HD   4 * Sd^3   (gamma * p1)^2
2HDr = --- = -------------------------------
1HD               Vb
----------------- * A
Sd^2 * gamma * p1

Sd^2 * gamma * p1 * (1 + gamma)
= ------------------------------- * A
4 * Sd^3 * (gamma * p1)^2

A * (1 + gamma)
= -------------------                                  (18)
4 * Sd * gamma * p1

2HD(%) = 2HDr * 100                                         (19)

Amplitude 2HD(dB) = 20 * log10(2HDr)                        (20)

Power 2HD(dB) = 10 * log10(2HDr) = 20 * log10(2HDr^2)


As can be seen from equation 18, all included parameters are linear scalable, which is very convenient for further analysis. Two parameters can be changed by the speaker builder, A (turn down the volume) and Sd (choose larger cones for woofers, or use multiple woofers). Simple analysis shows that 2. harmonic distortion increases linearly with force amplitude. On the other hand, a larger cone will linearly decrease 2. harmonic distortion.

Probably this does not matter, since distortion inherent in the driver design (from nonlinear Bxl and nonlinear Cms) will dominate over box distortion. The following example will show what range of box distortion is normal in a normal design at fairly high volumes.

Since the Taylor series is converging, and each term in the Taylor polynomum represents its own n'th harmonic distortion, and the next term in the calculation of the Taylor polynomium expresses the maximum possible error, it can be assumed that the total harmonic distortion is less than twice the second harmonic distortion:

        THD < 2 * 2HD


Now it is necessary to calculate the force amplitude required to achieve this kind of motion, from equation 7 we have:

                              Sd * p1
A = Sd * p1 - -----------------------                       (21)
(1 + Sd * x / Vb)^gamma


Inserting equation 21 into equation 18 gives:

               Sd * p1 * (1 + gamma)
2HDr = ---------------------
4 * Sd * gamma * p1

Sd * p1 * (1 + gamma)
- ---------------------------------------------
(1 + Sd * x / Vb)^gamma * 4 * Sd * gamma * p1

1 + gamma               (1 + gamma)
= --------- - -----------------------------------      (22)
4 * gamma   4 * gamma * (1 + Sd * x / Vb)^gamma

1 + gamma
= --------- (1 - (1 + Sd * x / Vb)^-gamma)
4 * gamma


These equations are beyond what Kyle Lahnakoski has made, but very useful for an analytical examination of the connections between the distortion and physical properties of the closed box loudspeaker system.

If Sd * x / Vb goes toward 0, then 2HDr becomes 0. Mathematically it can be written as:

            Sd * x
lim ------ -> 0 gives 2HDr -> 0
Vb


This is like concluding either turn down the volume (keep x low) and/or use very large cabinets (Vb is high). Why should Sd be as small as possible -- to keep the pressure influence as low as possible, but this lowers the sound pressure level and is simply equivalent to turning down the volume.

A driver with larger Sd will give you less distortion at a given sound pressure level (not from the box but from the driver itself), all else being equal. The reason is the nonlinear structure of the driver (magnet system as well as supension) when moving. See XMAX.HTM for further information on Xmax.

## Example calculation

I have chosen the same example as Kyle Lahnakoski. A 12 inch NHT 1259 driver is mounted in a 1.5 cuft box (42.5 liter, half the recommended size, which exagerates the distortion level) = 0.0425 m3. Neglecting driver distortion and only including contributions from the cabinet the driver is moving +/- 10mm = 0.01 meter, presumably at a fairly low frequency to keep issues like mass intertia (Mms) out of the game. The driver Sd = 0.0519 m2. This gives us a volume displacement of +/- 0.52 liter.

For the NHT 1259 in the given situation we get:

               1 + 1.4                1 + 1.4
2HDr = ------- - ------------------------------------------
4 * 1.4   4 * 1.4 * (1 + 0.0519 * 0.01 / 0.0425)^1.4

= 2.4 / 5.6 - 2.4 / 5.69597 = 0.007221

2HD(%) = 0.72 %

2HD(dB) = -42.8 dB


The amplitude distortion from a closed box is (in this worst case situation) only 0.7 % whereas the driver would normally contribute with excursion nonlinearities in the range of 10 % power distortion (-10 dB, or amplitude distortion level of -20 dB). To convert from amplitude distortion to power distortion the result must be squared:

        20 * log(2HDr^2) = 10 * log(2HDr) = -21.4 dB


Most woofers, even in appropriate boxes, will provide much higher 2. and 3. harmonic distortion products at such high power levels in the range of 10 % harmonic distortion. For further documentation on this, see XMAX.HTM and the quoted sources.

We can conclude that the loudspeaker driver will produce much higher harmonic distortion than what is produced by the nonlinear air (functioning as a spring) in a closed box. From this point of view it is safe to use a closed box for measuring loudspeaker parameters.

Notice that Sd * p1 = 5000 in this case, but F only alternates +/- 100, so the behaviour is quite linear. Only when F alternates +/- 1000 or more will the nonlinearity be quite visible on a graph plot.

With the above equation one could also "ask" the other way around, eg. how much excursion is demanded to give 1% 2. harmonic distortion. The equation becomes:

                   1      /-------------------------
Vb   -----   / 4 * gamma * (1 + gamma)
x = -- * gamma  /  -----------------------
Sd    \    /      1 + gamma
\  /       --------- - 2HDr
\/        4 * gamma


For the particular case above one gets x = 14 mm.

## Numerical approach

To confirm these calculations I have tried an alternative approach, which is described below:

With a piece of math software I have generated a given time scale and a sine wave force. The force amplitude is specified through the desired excursion, as described with equation 21. The chosen frequency is 1 Hz.

gamma = 1.4
f = 1 Hz
Vol = 42.5 liter = 0.00425 m3
Area = 519 cm2 = 0.0519 m2
p1 = 101325 Pa
x = 10 mm = 0.01 m

tau = 2^p, where p = 5 (tau/f = sample rate per wave, here 32)
t = a vector/array containing integer numbers from 0 to tau-1

Ft = A * sin(2*pi*f * t/tau)


Then equation 8 is utilized to calculate Xt = x(t).

Then the FFT (of a single wave) of x(t) is calculated: T = fft(Xt).

The content in T now represents the frequency content from the given situation above. The input values can be changed anytime to represent another given situation.

T1 is the first element, which is the fundamental frequency, T2 is the second harmonic and so on. Calculating the relative harmonics is a simple task. Here only the power distortion is calculated:

2HD(dB) = 10 * log(T2/T1) = -21.414 dB
3HD(dB) = 10 * log(T3/T1) = -42.592 dB
4HD(dB) = 10 * log(T4/T1) = -63.658 dB
5HD(dB) = 10 * log(T5/T1) = -84.658 dB


THD = 10 * log((|T2|+|T3|+|T4|+|T5|)/|T1|) = -21.380376 dB (=0.73 %)


As can be seen, the distortion level of higher than 2. harmonic distortion can be totally neglected. The 2. harmonic distortion from the box in this situation is something like 10 times smaller (10 dB smaller) than what can be expected from the driver. In this situation it can be neglected too.

3. harmonic distortion from a loudspeaker driver is normally of about the same size as the 2. harmonic distortion, at low frequencies and when driven within its limits.

The above calculations are quite clearly in agreement with the analytical calculation in the previous sections. The 2. harmonic distortion calculation above gives -21.413615 dB whereas the earlier analytical derivation gives -21.413934 dB.

The difference (how extremely small it ever may seem) is because of either the limitations in the development of the Taylor series. Including all the higher order contributions in the Taylor series would perhaps add up for the last 0.00032 dB difference. Alternatively it is a matter of rounding errors in the calculation routines of the math software (an equally possible reason).

The difference in the calculation of 3. harmonic distortion between the analytical (quick and dirty) approach and the numerical approach is approx. 2.5 dB. Surprisingly close, yet the numerical approach is the most accurate here. This is also valid for the higher order distortion components.

Also, it can be seen that THD < 2 * 2HD since the linear development (on the dB scale) of higher order harmonics means that the total harmonic distortion is almost equal to the 2. harmonic distortion. This is valid whether the excursion is changed to +/- 100 mm or down to +/- 1 mm. I am willing to bet my bottom dollar that this linear decrease of distortion is valid for any closed box operating within the above described limitations (e.g. adiabatic situation).

The difference between 2HD and THD in the given situation is only 0.033 dB. This again states that 2. harmonic distortion is the only one which matters.

When increasing the excursion level, then the air becomes an even better alternative to driver nonlinearities. When reducing the excursion level, eg. to a level of +/- 1 mm, then driver nonlinearities may not be dominating anymore, and box air nonlinearity is in this situation (given the above situation) -31.4 dB. This equals 0.07 % distortion, which cannot be registered by the ear. In other words, whenever the distortion from the air becomes (perhaps) significant, then the distortion contribution is so small it cannot be perceived by the human ear.

At low levels, at least in the driver midband, the distortion level from a driver is typically less than -40 dB (typically -50 dB and down to -60 dB for high quality drivers). This signifies that the distortion from a driver changes more rapidly with excursion than the box distortion.

## Conclusions

The conclusion is, that one should not worry about distortion levels from compression nonlinearites in a closed box. I assume that it is different with reflex cabinets, where the pressure changes inside the box is higher for a given cone excursion. This could give less distortion from the driver combined with more distortion from the compression of the air, if the reflex box is designed appropriately.

Does 2. harmonic have relevance? It probably disturbs measurement of Vas in a box less than 3. harmonic distortion, all else being equal. In this case 3. harmonic distortion is too low to have any significant influence in a closed box design.

Theoretically high levels of 2. harmonic distortion can move the piston resting point. In the case of a closed box the resting point would move outwards, but as can be seen the 2. harmonic distortion from the driver is completely dominating the picture.

The analytical approach has given a quick and easy method (equation 22) for calculating the vast majority of the harmonic distortion in a closed box system.

Distortion at low levels provides a risk of box distortion without masking from the driver. Here the level of distortion is so low, though, that I consider it impossible to hear.

## Litterature

Calculus and Analytic Geometry, 3. edition C. H. Edwards and David E. Penney Prentice-Hall (International Edition) 1990 ISBN 0-13-111006-3

Termodynamik, 2. oplag Erik Both og Gunnar Christensen Den private Ingeniørfond, Danmarks Tekniske Højskole, 1993 ISBN 87-7381-056-8

## Acknowledgments

I would like to acknowledge Kyle Lahnakoski for his impressive derivation, found at:

I have basically just followed in his footsteps, trying to understand it, enhanced on the explanations, and corrected what I believe are minor mistakes or errors.