I have a set of first 25 Zernike polynomials. Below are shown few in Cartesin co-ordinate system.

z2 = 2*x

z3 = 2*y

z4 = sqrt(3)*(2*x^2+2*y^2-1)



z24 = sqrt(14)*(15*(x^2+y^2)^2-20*(x^2+y^2)+6)*(x^2-y^2)

I am not using 1st since it is piston; so I have these 24 two-dim ANALYTICAL functions expressed in X-Y Cartesian co-ordinate system. All are defined over unit circle, as they are orthogonal over unit circle. The problem which I am describing here is relevant to other 2D surfaces also apart from Zernike Polynomials.

Suppose that origin (0,0) of the XY co-ordinate system and the centre of the unit circle are same.

Next, I take linear combination of these 24 polynomials to build a 2D wavefront shape. I use 24 random input coefficients in this combination.

w(x,y) = sum_over_i   a_i*z_i         (i=2,3,4,....24)

a_i = random coefficients
z_i = zernike polynomials

Upto this point, everything is analytical part which can be done on paper.

Now comes the discretization!

I know that when you want to re-construct a signal (1Dim/2Dim), your sampling frequency should be at least twice the maximum frequency present in the signal (Nyquist-Shanon principle).

Here signal is w(x,y) as mentioned above which is nothing but a simple 2Dim function of x & y. I want to represent it on computer now. Obviously I can not take all infinite points from -1 to +1 along x axis and same for y axis. I have to take finite no. of data points (which are called sample points or just samples) on this analytical 2Dim surface w(x,y)

I am measuring x & y in metres, and

 -1 <= x <= +1; -1 <= y <= +1

e.g. If I divide my x-axis from -1 to 1, in 50 sample points then dx = 2/50= 0.04 metre. Same for y axis. Now my sampling frequency is 1/dx i.e. 25 samples per metre. Same for y axis.

But I took 50 samples arbitrarily; I could have taken 10 samples or 1000 samples. That is the crux of the matter here: how many samples points?How will I determine this number?

There is one theorem (Nyquist-Shanon theorem) as mentioned above which says that if I want to re-construct w(x,y) faithfully, I must sample it on both axes so that my sampling frequency (i.e. no. of samples per metre) is at least twice the maximum frequency present in the w(x,y). This is nothing but finding power spectrum of w(x,y). Idea is that any function in space domain can be represented in spatial-frequency domain also, which is nothing but taking Fourier transform of the function! This tells us how many (spatial) frequencies are present in your function w(x,y) and what is the maximum frequency out of these many frequencies.

Now my question is first how to find out this maximum sampling frequency in my case. I can not use MATLAB fft2() or any other tool since it means already I have samples taken across the wavefront!! Obviously remaining option is find it analytically ! But that is time consuming and difficult since I have 24 polynomials & I will have to use then continuous Fourier transform i.e. I will have to go for pen and paper.

Any help will be appreciated.


  • 1
    $\begingroup$ It sounds like you already have a discretization. Your wave front is discretized by your choice of basis (Zernike polynomials) and the coefficients of those basis functions (the weights of the linear combination). I'm guessing there's more to the story. What are you really trying to do here? Why can you not just use the coefficients of the linear combination as your discretization? It seems like you've already chosen a set of basis functions (the Zernike polynomials), but for some reason you suddenly want to shift to a different basis (Fourier basis). More details would help. $\endgroup$ Commented Jun 11, 2015 at 16:37
  • $\begingroup$ @DougLipinski: Please see edited version now; does it make sense now ? $\endgroup$
    – atom
    Commented Jun 12, 2015 at 5:52
  • $\begingroup$ @DougLipinski: Any linear combination of finite no. of zernike polynomials is a CONTINUOS i.e. analytical 2-dimensional function; where is discretization here ? $\endgroup$
    – atom
    Commented Jun 12, 2015 at 6:06
  • $\begingroup$ The edit goes into the right direction but still many points are phrased vaguely and ambiguously. As an example: I guess with "But I took 50 samples arbitrarily" you mean "I can take any number of sampling points" and not "I took 50 arbitrary samples". Moreover, the Shannon sampling theorem would not be helpful. I'll add that to my answer. $\endgroup$
    – Dirk
    Commented Jun 12, 2015 at 8:03
  • $\begingroup$ @Dirk: Why Shanon theorem is not helpful here? $\endgroup$
    – atom
    Commented Jun 12, 2015 at 8:20

1 Answer 1


Without additional assumptions on the structure of the set if sampling points there can't be an upper bound. The reason is that a 2D polynomial can have a 1 dimensional set of zeros. If your sampling point all lie in this set, the polynomial looks like the zero polynomial.

After the edit it becomes a bit more clear what you are after. As I understand it, you want to sample linear combinations of Zernicke polynomials on a (uniformly spaces?) rectangular grid and want to know how fine your grid needs to be such that you can reconstruct the coefficients of the linear combination from the functions.

Unfortunately, Shannon's sampling theorem will not be helpful here.

  • You could view Zernike polynomials as functions on the unit disk, but there is no Shannon theorem for functions on the unit disk.

  • You could extend the Zernicke polynomials to the whole plane by zero extension, making them square integrable functions. Unfortunately, these functions will have infinite bandwidth (as they are not continuous).

  • You could extend the Zernike polynomials to the whole plane but using their formula. This would give smooth functions on the plane but unfortunately, they will not be square integrable (not even bounded), and hence, Shannon does not apply.

What you can do is the following: Let $z_1,\dots,z_k$ be your Zernike polynomials (i.e. $k=24$) and let $(x_i,y_i)_{i=1,\dots,n}$ be your sampling points (I am ignoring the grid structure here). Then the mapping that takes coefficients to sampling values is $$ M: \mathbb{R}^k\to\mathbb{R}^n,\qquad (a_1,\dots,a_k) \mapsto \sum_{j=1}^k a_j z_j(x_i,y_i) $$ and is linear in the coefficients. I.e. you can express that mapping as a matrix and need the see if it has full rank.

  • $\begingroup$ Please see edited version of my question now. $\endgroup$
    – atom
    Commented Jun 12, 2015 at 6:07
  • $\begingroup$ I did not get what you are saying. Will you please elaborate ? $\endgroup$
    – atom
    Commented Jun 12, 2015 at 8:22

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