Modal Data Processing for High Resolution Deflectometry

Abstract

In this paper, we present a modal data processing methodology, for reconstructing high resolution surfaces from measured slope data, over rectangular apertures. One of the primary goals is the ability to effectively reconstruct deflectometry measurement data for high resolution and freeform surfaces, such as telescope mirrors. We start by developing a gradient polynomial basis set which can quickly generate a very high number of polynomial terms. This vector basis set, called the G polynomials set, is based on gradients of the Chebyshev polynomials of the first kind. The proposed polynomials represent vector fields that are defined as the gradients of scalar functions. This method yields reconstructions that fit the measured data more closely than those obtained using conventional methods, especially in the presence of defects in the mirror surface and physical blockers/markers such as fiducials used during deflectometry measurements. We demonstrate the strengths of our method using simulations and real metrology data from the Daniel K. Inouye Solar Telescope (DKIST) primary mirror.

Appendix

1.1 Appendix 1

See Fig. 12.

Fig. 12

Quiver plots of a few gradient polynomials

1.2 Appendix 2

1.2.1 Orthonormality of the vector gradient polynomials

G polynomial orthogonality can be observed by noting that the dot product of two G polynomials is orthogonal:

$$\mathop {\iint }\limits_{ - 1}^{1} \left[ {\vec{G}_{j} \left( {x,y} \right) \cdot \vec{G}_{{j^{\prime}}} \left( {x,y} \right)} \right]\sqrt {1 - x^{2} } \sqrt {1 - y^{2} } dxdy = \left\{ {\begin{array}{l} {0, j \ne j^{{\prime }} } \\ {N_{G} , j = j^{{\prime }} } \\ \end{array} .} \right.$$

(A.1)

Next, we demonstrate the G polynomials’ orthonormality using the expanded form of the polynomials (from Eq. (7)):

$$\frac{1}{{N_{G} }}\mathop {\iint }\limits_{ - 1}^{1} \left[ \begin{aligned} &\left\{ {T_{n} \left( y \right)T_{m}^{{\prime }} (x)\hat{i} + T_{m} \left( x \right)T_{n}^{{\prime }} (y)\hat{j}} \right\} \cdot \hfill \\ & \left\{ {T_{{n{\prime }}} \left( y \right)T_{m{\prime}}^{{\prime }}(x)\hat{i} + T_{{m{\prime }}} \left( x \right)T_{n{\prime}}^{{\prime }} (y)\hat{j}} \right\} \hfill \\ & \sqrt {1 - x^{2} } \sqrt {(1 - y^{2} } \hfill \\ \end{aligned} \right]dxdy = \delta_{{mm{\prime }}} \delta_{{nn{\prime }}} ,$$

(A.2)

where $m,\text{ }m^{{\prime }} ,\text{ }n,$ and $n^{{\prime }}$ are integers used for indexing and $N_{G}$ is the normalization factor, which is given by,

$$N_{G} = \left\{ \begin{aligned}& \frac{{\pi^{2} n^{2} }}{4},\;m = 0 \hfill \\ & \frac{{\pi^{2} m^{2} }}{4},\;n = 0 \hfill \\ & \frac{{\pi^{2} }}{8}\left( {m^{2} + n^{2} } \right) ,\;{\text{otherwise}} \hfill \\ \end{aligned} \right..$$

(A.3)

The proof of this derivation and details on how to obtain the normalization factor are as follows:

$$\begin{aligned} I = & \mathop {\iint }\limits_{ - 1}^{1} (G_{j} \cdot G_{{j^{\prime}}} )\sqrt {1 - x^{2} } \sqrt {1 - y^{2} } dxdy \\ = & \mathop {\iint }\limits_{ - 1}^{1} T_{n} \left( y \right)T_{{n^{{\prime }} }} \left( y \right)T_{m}^{{\prime }} \left( x \right)T_{{m^{{\prime }} }}^{{\prime }} \left( x \right)\sqrt {1 - x^{2} } \sqrt {1 - y^{2} } dxdy \\ & + \mathop {\iint }\limits_{ - 1}^{1} T_{m} \left( x \right)T_{{m^{{\prime }} }} \left( x \right)T_{n}^{{\prime }} \left( y \right)T_{{n^{{\prime }} }}^{{\prime }} \left( y \right)\sqrt {1 - x^{2} } \sqrt {1 - y^{2} } dxdy. \\ \end{aligned}$$

(A.4)

The first integral I₁ can be expressed as:

$$\begin{aligned} I_{1} = & \mathop {\iint }\limits_{ - 1}^{1} T_{n} \left( y \right)T_{{n^{{\prime }} }} \left( y \right)T_{m}^{{\prime }} \left( x \right)T_{{m^{{\prime }} }}^{{\prime }} \left( x \right)\sqrt {1 - x^{2} } \sqrt {1 - y^{2} } dxdy \\ = & \mathop \int \limits_{ - 1}^{1} T_{n} \left( y \right)T_{{n^{{\prime }} }} \left( y \right)\sqrt {1 - y^{2} } \mathop \int \limits_{ - 1}^{1} T_{m}^{{\prime }} \left( x \right)T_{{m^{{\prime }} }}^{{\prime }} \left( x \right)\sqrt {1 - x^{2} } dy dx \\ T_{m} \left( x \right) = & \cos \left( {m\cos^{ - 1} (x)} \right). \\ \end{aligned}$$

(A.5)

Then, the following substitutions and subsequent equations can be used:

$$x = \cos \left( t \right),\quad dx = - \sin \left( t \right)dt,\quad \sin \left( t \right) = \sqrt {1 - x^{2} } ,$$

$$\begin{aligned} I_{1} = & \mathop \int \limits_{ - 1}^{1} T_{n} \left( y \right)T_{{n^{{\prime }} }} \left( y \right)\sqrt {1 - y^{2} } \\ & \mathop \int \limits_{\pi }^{0} \frac{{m \sin \left( {m\cos^{ - 1} (x)} \right) m^{{\prime }} \sin \left( {m^{{\prime }} \cos^{ - 1} (x)} \right)}}{\sin \left( t \right)}\left( { - \sin \left( t \right)} \right)dt dy, \\ \end{aligned}$$

$$I_{1} = \mathop \int \limits_{ - 1}^{1} (T_{n} \left( y \right)T_{{n^{{\prime }} }} \left( y \right)\sqrt {1 - y^{2} } dy \frac{{\left( { - mm^{{\prime }} } \right)\left[ {m\cos \left( {\pi m} \right)\sin \left( {\pi m^{{\prime }} } \right)} \right]}}{{m^{2} - m^{{{\prime }2}} }}.$$

(A.6)

This integral goes to zero for all values of m and m’ (except for when $m = m^{{\prime }}$), since $\sin (\pi m) = 0$ for all integer values of m. However, when $m = m^{{\prime }}$,

$$I_{1} = \mathop \int \limits_{ - 1}^{1} T_{n} \left( y \right)T_{{n^{{\prime }} }} \left( y \right)\sqrt {1 - y^{2} } \mathop \int \limits_{\pi }^{0} \sin \left( {mt} \right)\sin \left( {mt} \right)dt dy.$$

(A.7)

Using the identity,

$$\sin^{2} \left( x \right) = \frac{{1 - \cos \left( {2x} \right)}}{2},$$

$$I_{1} = \mathop \int \limits_{ - 1}^{1} T_{n} \left( y \right)T_{{n^{{\prime }} }} \left( y \right)\sqrt {1 - y^{2} } dy \left( - \frac{{m^{2} }}{2}\left[ {\frac{{\sin \left( {2\pi m} \right)}}{2m} - \pi } \right]\right) \delta_{{mm^{{\prime }} }} .$$

(A.8)

As before, $\sin (2\pi m) = 0$ for all integer values of m. Thus,

$$\begin{aligned} I_{1} = & \frac{{\pi mm^{{\prime }} }}{2}\delta_{{mm^{{\prime }} }} \mathop \int \limits_{ - 1}^{1} T_{n} \left( y \right)T_{{n^{{\prime }} }} \left( y \right)\sqrt {1 - y^{2} } dy = \frac{{\pi m^{2} }}{2}\delta_{{mm^{{\prime }} }} \\ & \mathop \int \limits_{ - 1}^{1} \cos \left( {n\cos^{ - 1} (y)} \right)\cos \left( {n^{{\prime }} \cos^{ - 1} (y)} \right)\sqrt {1 - y^{2} } dy. \\ \end{aligned}$$

(A.9)

Using the same substitutions as before,

$$y = \cos \left( t \right),\quad dy = - \sin \left( t \right)dt,\quad \sin \left( t \right) = \sqrt {1 - y^{2} } ,$$

$$I_{1} = \frac{{\pi m^{2} }}{2}\delta_{{mm^{{\prime }} }} \mathop \int \limits_{\pi }^{0} \cos \left( {nt} \right)\cos \left( {n^{{\prime }} t} \right)\left( { - \sin^{2} \left( t \right)} \right)dt.$$

(A.10)

Again using the trigonometric identity for the square of the sine function,

$$\begin{aligned} I_{1} = & - \frac{{\pi m^{2} }}{2}\delta_{{mm^{{\prime }} }} \mathop \int \limits_{\pi }^{0} \cos \left( {nt} \right)\cos \left( {n^{{\prime }} t} \right)\left( {\frac{{1 - \cos \left( {2t} \right)}}{2}} \right)dt \\ = & - \frac{{\pi m^{2} }}{2}\delta_{{mm^{{\prime }} }} \\ & \left( \begin{aligned} - \frac{1}{4} \left\{ \begin{aligned} \frac{{\sin \left( {\pi \left( {n - n^{{\prime }} - 2} \right)} \right)}}{{n - n^{{\prime }} - 2}} + \frac{{\sin \left( {\pi \left( {n - n^{{\prime }} + 2} \right)} \right)}}{{n - n^{{\prime }} + 2}} \hfill \\ + \frac{{\sin \left( {\pi \left( {n + n^{{\prime }} - 2} \right)} \right)}}{{n + n^{{\prime }} - 2}} + \frac{{\sin \left( {\pi \left( {n + n^{{\prime }} + 2} \right)} \right)}}{{n + n^{{\prime }} + 2}} \hfill \\ \end{aligned} \right\} \hfill \\ - \left\{ {\frac{{n^{{\prime }} \cos \left( {\pi n} \right)\sin \left( {\pi n^{{\prime }} } \right) - n\sin \left( {\pi n} \right)\cos \left( {\pi n^{{\prime }} } \right)}}{{n^{2} - n^{{{\prime }2}} }}} \right\} \hfill \\ \end{aligned} \right). \\ \end{aligned}$$

(A.11)

This integral equals zero for all values of n and n’, except when $n = n^{{\prime }}$, in which case the integral is,

$$\begin{aligned} I_{1} = - \frac{{\pi m^{2} }}{2}\delta_{{mm^{\prime}}} \mathop \int \limits_{\pi }^{0} \cos^{2} \left( {nt} \right)\left( {\frac{{1 - \cos \left( {2t} \right)}}{2}} \right)dt \hfill \\ = \frac{{\pi m^{2} }}{4}\delta_{{mm^{\prime}}} \left[ {\left\{ {\frac{{2\pi n + \sin \left( {2\pi n} \right)}}{4n}} \right\} + \left\{ {\frac{{n\sin \left( {2\pi n} \right)}}{{4\left( {n^{2} - 1} \right)}}} \right\}} \right]\delta_{{nn^{\prime}}} . \hfill \\ \end{aligned}$$

(A.12)

All of the sine terms become zero since n is an integer, so,

$$I_{1} = \frac{{\pi^{2} m^{2} }}{8}\delta_{{mm^{{\prime }} }} \delta_{{nn^{{\prime }} }} .$$

(A.13)

Similarly for I₂,

$$I_{2} = \frac{{\pi^{2} n^{2} }}{8}\delta_{{mm^{{\prime }} }} \delta_{{nn^{{\prime }} }} .$$

(A.14)

Now the original integral becomes:

$$I = \frac{{\pi^{2} m^{2} }}{8}\delta_{{mm^{{\prime }} }} \delta_{{nn^{{\prime }} + }} \frac{{\pi^{2} n^{2} }}{8}\delta_{{mm^{{\prime }} }} \delta_{{nn^{{\prime }} }} = \frac{{\pi^{2} }}{8}\delta_{{mm^{{\prime }} }} \delta_{{nn^{{\prime }} }} \left( {m^{2} + n^{2} } \right).$$

(A.15)

Next we address the cases in which either m = 0 or n = 0. When $m = 0,\text{ }I_{1} = 0$ since,

$$T_{{m^{{\prime }} }}^{{\prime }} \left( x \right) = T_{m}^{{\prime }} \left( x \right) = 0,$$

so

$$I = I_{2} = \mathop {\iint }\limits_{ - 1}^{1} T_{n}^{{\prime }} \left( y \right)T_{{n^{{\prime }} }}^{{\prime }} \left( y \right)\sqrt {1 - x^{2} } \sqrt {1 - y^{2} } dxdy,$$

(A.16)

where we used the fact that

$$T_{{m^{{\prime }} }} \left( x \right) = T_{m} \left( x \right) = 1.$$

(A.17)

Following the same steps as before,

$$I = \frac{{\pi^{2} n^{2} }}{4}\delta_{{mm^{{\prime }} }} \delta_{{nn^{{\prime }} }} .$$

(A.18)

Similarly, for n = 0,

$$I = \frac{{\pi^{2} m^{2} }}{4}\delta_{{mm^{{\prime }} }} \delta_{{nn^{{\prime }} }} ,$$

(A.19)

where m and n are in the original basis in the last two equations.

It can easily be seen that I = 0 when m = 0 and n = 0.