Feb 19th 1997
Distortion due to Non-Linear Effects of Air
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?
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:
=> P1*V1 = P2*((1+k)*V1) (2)
=> 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:
=> --- = P1 - --------------
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)
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.
First harmonic (w): --------- (7)
2 A y P1
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.
K = A P1 - --------------
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:
K = A P1 - -------------- = 5143.0 - 5058.3 = 84.7 (about)
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.