Feb 19th 1997

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.