Feb 19th 1997

# Distortion due to Non-Linear Effects of Air

## Problem:

Some people are taking advantage of the fact that the compliance of the air in the box is a linear function of cone position. This is not true. Air does not have a linear restoring force and therefore there is distortion that results from this non-linearity. The real question is how much distortion is there, and is it significant?

## Solution:

For adiabatic compression inside the box we have the equality:

P1*V1^y = P2*V2^y (1)

where P is pressure, V is volume and y is the adiabatic compression constant for our gas (1.4 for air). Now if we find a 'k', called the change in volume, such that (1+k)*V1=V2 we can adjust our formula:

y y

=> P1*V1 = P2*((1+k)*V1) (2)

y

=> P1 = P2*(1+k) (3)

If P1 is the equilibruium pressure (101.5 KPa) and P2 is the pressure inside the box, we can relate the excursion, x to k by k = A*x/Vb. Where A is the area of the cone and Vb is the volume of the box. We can also relate the applied force on the cone, F, by F/A = P1-P2. Sub these two into (3) to get:

F P1

=> --- = P1 - --------------

A (1+A*x/Vb)^y

Vb Vb / A*P1 \1/y

=> x = - ---- + ---- * |----------| (4)

A A \ A*P1 - F /

And we will use the computer to find the Taylor series expansion so we have x as a function of F.

Vb Vb (y + 1) 2 3

x = ------- F + 1/2 ---------- F + O(F ) (5)

2 3 2 2

A y P1 A y P1

If F is an applied force; F = K*sin(w*t)

2 2

Vb K sin(w t) Vb (y + 1) K sin (w t)

x = ------------- + 1/2 ---------------------- (6)

2 3 2 2

A y P1 A y P1

A Fourier analysis gives us the amplitude of the first and second harmonics. This can be done easily if we see a little trig identity involving (sin(x))^2.

Vb K

First harmonic (w): --------- (7)

2 A y P1

2

Vb (y + 1) K

Second Harmonic (2w): 1/4 -------------- (8)

3 2 2

A y P1

The second harmonic over the first results in the ratio between the two:

(y + 1) K

Harmonic distortion: 1/4 ----------- (9)

A y P1

We will replace K. We assume the frequency is well below resonance, so the force on the cone is maximum when the excursion is at its maximum.

A P1

K = A P1 - --------------

(1+A*x/Vb)^y

We can approximate the distortion of a NHT 1259 (A=.05067 m^2) in a 1.5 cubic foot box(.04248 m^3). We will take the worst case scenario where the driver is moving +/- 1 cm at a frequency well below resonance. This way the distortion due to the air is dominant.

The max force needed to bring the cone out(or in) this far is approximatly:

A P1

K = A P1 - -------------- = 5143.0 - 5058.3 = 84.7 (about)

(1+A*x/Vb)^y

We sub this into (9) to get:

Harmonic Distortion: .00705 (0.7% HD or 21dB Sig to HD ratio)

Thanks to Tom Danley; we should mention that this is amplitude distortion, not power distortion which is the square of this figure.

Harmonic Distortion: .00073 (.073% HD or 31.3dB Sig to HD ratio)

Remember, this is the distortion only due to the nonlinear effects of the air. Taylor's theorem says it is only an approximation, but it is a good one.

It is up to the individual to determine if this is a significant amount of distortion.

- September 03 2000 - Note from Ron Ennenga pointed out I forgot to divide by four when calculating the driver area of the 1259. Corrected it.
- March 28 2001 - Note from Claus Futtrup noted a missing term. Added it, no effect on the result.