ZernikeGenerator numerical accuracy

[lbusoni]

@imapeartree reports that Z4 average value is not null.

See for instance

In [41]: zg=ZernikeGenerator(600); aa=zg.getZernike(4); aa.mean()
Out[41]: 0.0002979032005116371

Is 3e-4 an acceptable accuracy for this test case? Is the error related to the discretization of the 600px diameter mask?

[lbusoni]

I still don't know if the error is acceptable.
But at least it is decreasing with the pupil size.
See below the same Z4.mean() error for pupil diameters in the range [2,1000]

res=np.array([ZernikeGenerator(dia).getZernike(4).mean() for dia in range(2, 1000)])
semilogy(abs(res)); xlabel('pupil diameter [px]'); ylabel('Z4.mean()')

[lbusoni]

The tilt mode mean value is actually zero

res=np.array([ZernikeGenerator(dia).getZernike(2).mean() for dia in range(2, 1000)])
semilogy(abs(res)); xlabel('pupil diameter [px]'); ylabel('Z2.mean()')

[lbusoni]

It looks like the centrosymmetric modes only are affected: below only Z11 (purple) has a non-numerically-zero mean value

[lbusoni]

It looks like that not only the centro symmetric modes are affected.
Also the modes that have a 45° symmetry on the outer edge maybe affected, like Z14 and Z26

[lbusoni]

I believe the error is compatible with the fact that the outer edge of the circular pupil is discretized with an approximation <0.5px.

Assuming the radius is wrong by a factor $\epsilon$ with $| \epsilon |$ <1px, the average of Z4 with is $&lt; Z_4 &gt; = \int_{0}^{2 \pi} \int_{0}^{1+\epsilon} \sqrt{3} (2 \rho^2 -1) d\rho \rho d\theta$

This simplifies into

$&lt; Z_4 &gt; = \sqrt{3} \pi \big( (1+\epsilon)^4 - (1+\epsilon)^2 \big) \simeq 2 \sqrt{3} \pi \epsilon \sim 10 \epsilon$

As an example case, with a nominal radius of 300px and a mask that overestimate/underestimate it by 0.5px, we may have an error in the mean value of Z4 up to $10*0.5/300=0.015$. The measured accuracy is 0.003, consistent with this rough estimate.

[lbusoni]

Is it a good idea to force the generated Zernike map to have null average?

At a first glance, I don't find it very elegant: in the example above, all the points of the centrosymmetric modes would be off by 1e-3, that may be quite confusing.

But I recon that there are applications (like modal decomposition) that requires orthogonality of every mode with the piston mode, and for them having a null-average is required.

This last consideration has to be taken into account for the ModalDecomposer (?)