Influence of the Analysis Window on the Metrological Performance of the Grid Method

Abstract

This paper deals with the grid method in experimental mechanics. It is one of the full-field methods available for estimating in-plane displacement and strain components of a specimen submitted to a load producing slight local deformation. This method consists in, first, depositing a regular grid on the surface of a specimen, and, second, comparing images of the grid before and after deformation. A possibility is to perform windowed Fourier analysis to measure these deformations as changes of the local grid aspect. The aim of the present study is to investigate the choice of the analysis window and its influence on the metrological performances of the grid method. Two aspects are taken into account, namely the reduction of the harmonics of the grid line profile, which are not pure sine because of manufacturing constraints, and the transfer of the digital noise from the imaged grid to the mechanical measurements. A theoretical study and a numerical assessment are presented. In addition, the interested reader can find in this paper a calculation of the Wigner–Ville transform of a triangular function which, to the best of the present authors’ knowledge, is not available in the existing literature.

Appendices

Proof of Proposition 1

Proof

Let E be the expectation of any random variable. Since n is a white noise of variance v, we have $E(n(x_i,y_j)n(x_k,y_l)) = 0$ if $x_i\ne x_k$ or $y_j\ne y_l$, and $=v$ otherwise.

By expanding the real and imaginary parts of $\widehat{n}$ and replacing the discrete Riemann sums by integrals:

$$\begin{aligned}&\text {Cov}(\text {Re}(\widehat{n}(\xi ,\eta )),\text {Re}(\widehat{n}(\xi ',\eta ')))\nonumber \\&\quad = v \sum _{i,j} w_{\varvec{\sigma }}(x_i-\xi ,y_j-\eta ) w_{\varvec{\sigma }}(x_i-\xi ',y_j-\eta ')\nonumber \\&\qquad \cdot \cos ^2(2\pi f x_i) (\varDelta _x\varDelta _y)^2\nonumber \\&\quad \simeq v \varDelta _x \varDelta _y \iint w_{\varvec{\sigma }}(x-\xi ,y-\eta ) w_{\varvec{\sigma }}(x-\xi ',y-\eta ') \nonumber \\&\quad \cdot \cos ^2(2\pi f x) \;\text {d}x\;\text {d}y \end{aligned}$$

(98)

Let us define:

$$\begin{aligned} I(\xi ,\eta ,\xi ',\eta ')= & {} \iint w_{\varvec{\sigma }}(x-\xi ,y-\eta ) w_{\varvec{\sigma }}(x-\xi ',y-\eta ') \nonumber \\&\cos ^2(2\pi f x) \;\text {d}x \;\text {d}y \end{aligned}$$

(99)

We obtain:

$$\begin{aligned}&I(\xi ,\eta ,\xi ',\eta ')=\frac{1}{2} \biggl ( \int w^{\mathbf {x}}_{\sigma _x}(x-\xi ) w^{\mathbf {x}}_{\sigma _x}(x-\xi ') \biggr .\nonumber \\&\qquad \biggl . (1+\cos (4\pi f x)) \;\text {d}x \biggr ) \left( \int w^{\mathbf {y}}_{\sigma _y}(y-\eta ) w^{\mathbf {y}}_{\sigma _y}(y-\eta ') \;\text {d}y \right) \nonumber \\&\quad = \frac{1}{2} \biggl ( w^{\mathbf {x}}_{\sigma _x}*w^{\mathbf {x}}_{\sigma _x} (\xi -\xi ') +\biggr .\nonumber \\&\qquad \biggl . \text {Re}\left( e^{-2i\pi f (\xi +\xi ')}\int w^{\mathbf {x}}_{\sigma _x}(x-\alpha )w^{\mathbf {x}}_{\sigma _x}(x+\alpha ) e^{-4i\pi f x} \;\text {d}x\right) \biggr )\nonumber \\&\qquad w^{\mathbf {y}}_{\sigma _y}*w^{\mathbf {y}}_{\sigma _y} (\eta -\eta ') \end{aligned}$$

(100)

where $\alpha =(\xi -\xi ')/2$.

Since the analysis window is symmetric, we have $w^{\mathbf {x}}_{\sigma _x}(x-\alpha )=w^{\mathbf {x}}_{\sigma _x}(\alpha -x)$. In addition note that

$$\begin{aligned} \int w^{\mathbf {x}}_{\sigma _x}(x-\alpha )w^{\mathbf {x}}_{\sigma _x}(x+\alpha ) e^{-4i\pi f x} \;\text {d}x = \frac{1}{2} {{\mathcal {W}}} w^{\mathbf {x}}_{\sigma _x}(\alpha , 2\pi f) \end{aligned}$$

(101)

where ${{\mathcal {W}}}t_\sigma$ is the Wigner–Ville transform of the 1D window $w^{\mathbf {x}}_{\sigma _x}$. Equation (52) follows.

Concerning the imaginary part of the noise:

$$\begin{aligned}&\text {Cov}(\text {Im}(\widehat{n}(\xi ,\eta )),\text {Im}(\widehat{n}(\xi ',\eta ')) \simeq v \varDelta _x \varDelta _y \nonumber \\&\quad \cdot \iint w_{\varvec{\sigma }}(x{-}\xi ,y{-}\eta ) w_{\varvec{\sigma }}(x-\xi ',y{-}\eta ') \sin ^2(2\pi f x) \;\text {d}x\;\text {d}y\nonumber \\ \end{aligned}$$

(102)

Thus, the same route as the preceding one gives (53).

Moreover,

$$\begin{aligned}&\text {Cov}(\text {Re}(\widehat{n}(\xi ,\eta )),\text {Im}(\widehat{n}(\xi ',\eta ')) \simeq v \varDelta _x \varDelta _y \nonumber \\&\quad \cdot \iint w_{\varvec{\sigma }}(x-\xi ,y-\eta ) w_{\varvec{\sigma }}(x-\xi ',y-\eta ')\nonumber \\&\quad \cdot \sin (2\pi f x) \cos (2\pi f x) \;\text {d}x\;\text {d}y \nonumber \\&\quad \simeq \frac{1}{2} v \varDelta _x \varDelta _y \cdot \iint w_{\varvec{\sigma }}(x-\xi ,y-\eta ) w_{\varvec{\sigma }}(x-\xi ',y-\eta ') \nonumber \\&\quad \cdot \sin (4\pi f x) \;\text {d}x\;\text {d}y \end{aligned}$$

(103)

which gives, in a similar fashion, (54). $\square$

Wigner–Ville Transform of Some Windows

The Wigner–Ville transform of the rectangular and Gaussian windows can be found in the literature [7, 22]. However, they sometimes contain typos. For the sake of completeness, we give here the calculation of these transforms. We also give the calculation of the Wigner–Ville transform of the triangular windows, which we have not been able to find in the available literature.

A numerical assessment is provided at the following URL: http://www.loria.fr/~sur/software/VerifWV/

By definition, ${{\mathcal {W}}}f(x,\lambda )=\int f(x+\tau /2)f(x-\tau /2)e^{-i\tau \lambda } \;\text {d}\tau$. Since f is symmetric,

$$\begin{aligned} {{\mathcal {W}}}f(x,\lambda )=2\int f(\tau +x)f(\tau -x)e^{-2i\tau \lambda }\;\text {d}\tau \end{aligned}$$

(104)

Since ${{\mathcal {W}}}f(-x,\lambda )={{\mathcal {W}}}f(x,\lambda )$, we assume, without loss of generality, that $x\ge 0$.

1.1 Rectangular Window

Here, $f=r_a$ in Table 1. For any $u\in [-a,a]$,

$$\begin{aligned} {{\mathcal {W}}}f(x,\lambda )= & {} \frac{1}{2a^2}\int _{x-a}^{a-x} e^{-2i\tau \lambda } \;\text {d}\tau \end{aligned}$$

(105)

$$\begin{aligned}= & {} \frac{1}{2a^2\lambda }\sin (2\xi (a-x)) \end{aligned}$$

(106)

Thus, for any $x\in {\mathbb {R}}$,

$$\begin{aligned} {{\mathcal {W}}}f(x,\lambda ) = \frac{1}{2a^2\lambda }\sin (2\xi (a-|x|)) {\mathbbm {1}}_{[-a,a]}(x) \end{aligned}$$

(107)

The value for $\lambda =0$ is given by a Taylor expansion, that is,

$$\begin{aligned} {{\mathcal {W}}}f(x,0) = \frac{a-|x|}{a^2} {\mathbbm {1}}_{[-a,a]}(x) \end{aligned}$$

(108)

1.2 Triangular Window

Here, $f=t_b$ in Table 1:

$$\begin{aligned} t_b(x) = \frac{b-|x|}{b^2} {\mathbbm {1}}_{[-b,b]}(x) \end{aligned}$$

(109)

Moreover, if $x\ge b$, ${{\mathcal {W}}}f(x,\lambda )=0$.

First case. $b/2\le x\le b$. We calculate successively

$$\begin{aligned} {{\mathcal {W}}}f(x,\lambda )= & {} 2\int _{x-b}^{b-x} \frac{b-x-\tau }{b^2}\frac{b-x+\tau }{b^2} e^{-2i\tau \lambda } \;\text {d}\tau \end{aligned}$$

(110)

$$\begin{aligned}= & {} \frac{2}{b^4}\int _{x-b}^{b-x} \left( (b-x)^2-\tau ^2\right) e^{-2i\tau \lambda }\;\text {d}\tau \end{aligned}$$

(111)

$$\begin{aligned}= & {} \frac{2(b-x)^2}{b^4} \int ^{b-x}_{x-b}e^{-2i\tau \lambda }\;\text {d}\tau \nonumber \\&-\frac{2}{b^4} \int _{x-b}^{b-x} \tau ^2e^{-2i\tau \lambda }\;\text {d}\tau \end{aligned}$$

(112)

On the one hand,

$$\begin{aligned} \int _{x-b}^{b-x} e^{-2i\tau \lambda }\;\text {d}\tau = \sin (2\xi (b-x))/\lambda \end{aligned}$$

(113)

On the other hand,

$$\begin{aligned}&\int ^{b-x}_{x-b} \tau ^2e^{-2i\tau \lambda }\;\text {d}\tau = \frac{(b-x)^2}{\lambda }\sin (2\lambda (b-x)) \nonumber \\&\qquad + \frac{1}{i\xi }\int ^{b-x}_{x-b} \tau e^{-2i\tau \lambda }\;\text {d}\tau \end{aligned}$$

(114)

$$\begin{aligned}&\quad =\frac{(b-x)^2}{\lambda }\sin (2\lambda (b-x)) \nonumber \\&\qquad + \frac{1}{i\lambda } \left( \frac{x-b}{i\lambda } \cos (2\lambda (b-x)) + \frac{1}{2i\lambda ^2}\sin (2\lambda (b-x)) \right) \nonumber \\ \end{aligned}$$

(115)

$$\begin{aligned}&\quad =\frac{(b-x)^2}{\lambda }\sin (2\xi (b-x)) \nonumber \\&\qquad + \frac{b-x}{\lambda ^2} \cos (2\lambda (b-x)) - \frac{1}{2\lambda ^3}\sin (2\lambda (b-x)) \end{aligned}$$

(116)

Thus,

$$\begin{aligned} {{\mathcal {W}}}f(x,\lambda )= & {} \frac{2}{b^4\lambda ^2}\Biggl ( -(b-x)\cos (2\lambda (b-x)) \Biggr .\nonumber \\&\Biggl . +\frac{1}{2\lambda }\sin (2\lambda (b-x)) \Biggr ) \end{aligned}$$

(117)

Second case. $0\le x \le b/2$. We calculate successively

$$\begin{aligned} {{\mathcal {W}}}f(x,\lambda )= & {} 2\int _{x-b}^{-x} \frac{b+x+\tau }{b^2}\frac{b-x+\tau }{b^2} e^{-2i\tau \lambda } \;\text {d}\tau \nonumber \\&+2\int _{-x}^{x} \frac{b-x-\tau }{b^2}\frac{b-x+\tau }{b^2} e^{-2i\tau \lambda } \;\text {d}\tau \nonumber \\&+2\int _{x}^{b-x} \frac{b-x-\tau }{b^2}\frac{b+x-\tau }{b^2} e^{-2i\tau \lambda } \;\text {d}\tau \nonumber \\= & {} \frac{2}{b^4}\int _{-x}^{x} \left( (b-x)^2-\tau ^2\right) e^{-2i\tau \lambda } \;\text {d}\tau \nonumber \\&+\frac{4}{b^4}\int _{x}^{b-x}\left( (b-\tau )^2-x^2\right) \cos (2\tau \lambda )\;\text {d}\tau \nonumber \\ \end{aligned}$$

(118)

$$\begin{aligned}= & {} \frac{2(b-x)^2}{b^4}\int _{-x}^{x}e^{-2i\tau \lambda } \;\text {d}\tau \nonumber \\&- \frac{2}{b^4}\int _{-x}^x \tau ^2e^{-2i\tau \lambda } \;\text {d}\tau \nonumber \\&+ \frac{4}{b^4}\int _{x}^{b-x}(b-\tau )^2\cos (2\tau \lambda )\;\text {d}\tau \nonumber \\&-\frac{4x^2}{b^4}\int _{x}^{b-x}\cos (2\tau \lambda )\;\text {d}\tau \end{aligned}$$

(119)

Now,

$$\begin{aligned}&\int _{-x}^{x}e^{-2i\tau \lambda } \;\text {d}\tau = \sin (2\lambda x)/\lambda \end{aligned}$$

(120)

$$\begin{aligned}&\int _{-x}^{x}\tau ^2 e^{-2i\tau \lambda } \;\text {d}\tau = \frac{x^2}{\lambda }\sin (2\lambda x) + \frac{x}{\lambda ^2} \cos (2\lambda x)\nonumber \\&\qquad - \frac{1}{2\lambda ^3}\sin (2\lambda x) \end{aligned}$$

(121)

$$\begin{aligned}&\int _{x}^{b{-}x}\cos (2\tau \lambda )\;\text {d}\tau = \frac{1}{2\lambda }\left( \sin (2\lambda (b{-}x)){-}\sin (2\lambda x)\right) \end{aligned}$$

(122)

$$\begin{aligned}&\int _{x}^{b-x}(b-\tau )^2\cos (2\tau \lambda )\;\text {d}\tau = \frac{x^2}{2\lambda } \sin (2(b{-}x)\lambda ) \nonumber \\&\qquad - \frac{(b{-}x)^2}{2\lambda }\sin (2\lambda x) +\frac{1}{\lambda }\int _x^{b{-}x} (b{-}\tau ) \sin (2\tau \lambda )\;\text {d}\tau \nonumber \\&\quad =\frac{x^2}{2\lambda } \sin (2(b-x)\lambda ) - \frac{(b-x)^2}{2\lambda }\sin (2\lambda x) \nonumber \\&\qquad -\frac{x}{2\lambda ^2}\cos (2\lambda (b-x)) + \frac{(b-x)}{2\lambda ^2}\cos (2\lambda x)\nonumber \\&\qquad - \frac{1}{2\lambda ^2}\int _x^{b-x} \cos (2\tau \lambda )\;\text {d}\tau \nonumber \\&\quad = \frac{x^2}{2\lambda } \sin (2(b-x)\lambda ) \nonumber \\&\quad - \frac{(b-x)^2}{2\lambda }\sin (2\lambda x) -\frac{x}{2\lambda ^2}\cos (2\lambda (b-x)) \nonumber \\&\qquad + \frac{(b-x)}{2\lambda ^2}\cos (2\lambda x) - \frac{1}{4\lambda ^3}\left( \sin (2\lambda (b{-}x)){-}\sin (2\lambda x)\right) \nonumber \\ \end{aligned}$$

(123)

Consequently,

$$\begin{aligned} {{\mathcal {W}}}f(x,\lambda )= & {} \frac{2}{b^4\lambda ^2}\Bigl ( (b-2x)\cos (2\lambda x) - x \cos (2\lambda (b-x)) \Bigr .\nonumber \\&\Bigl . - \frac{1}{2\lambda }\sin (2\lambda (b-x)) + \frac{1}{\lambda }\sin (2\lambda x)\Bigr ) \end{aligned}$$

(124)

Conclusion. For any $x\in {\mathbb {R}}$, the Wigner–Ville transform of the triangle function writes

$$\begin{aligned}&{{\mathcal {W}}}f(x,\lambda ) = \frac{2}{b^4\lambda ^2} \Biggl [ \Bigr ( -(b-|x|)\cos (2\lambda (b-|x|)) \Bigr . \Bigr . \nonumber \\&\qquad \Bigl . \Bigl . +\frac{1}{2\lambda }\sin (2\lambda (b-|x|)) \Bigr ) {\mathbbm {1}}_{[-b,b]}(x) \Bigr .\nonumber \\&\qquad + \Bigl . 2\cos (\lambda b) \left( (b-2|x|)\cos (\lambda (b-2|x|)) - \frac{1}{\lambda }\sin (\lambda (b-2|x|)) \right) \nonumber \\&\qquad \cdot {\mathbbm {1}}_{[-b/2,b/2]}(x) \Biggr ] \end{aligned}$$

(125)

The value for $\lambda =0$ is given by a Taylor expansion:

$$\begin{aligned} {{\mathcal {W}}}f(x,0)= & {} \frac{8(b-|x|)^3}{3b^4}{\mathbbm {1}}_{[-b,b]}(x) \nonumber \\&\quad - \frac{4(b-2|x|)^3}{3b^4} {\mathbbm {1}}_{[-b/2,b/2]}(x) \end{aligned}$$

(126)

1.3 Gaussian Window

Note that generalized Gaussian functions give positive Wigner–Ville transform, and only them [10]. The Wigner–Ville transform of a Gaussian function writes as follows:

$$\begin{aligned} {{\mathcal {W}}}f(x,\lambda )= & {} \frac{1}{\pi \sigma ^2}\int e^{{-}((\tau {+}x)^2{-}(\tau -x)^2)/(2\sigma ^2)} e^{-2i\tau \lambda } \;\text {d}\tau \end{aligned}$$

(127)

$$\begin{aligned}= & {} \frac{e^{-x^2/\sigma ^2}}{\pi \sigma ^2}\int e^{-\tau ^2/\sigma ^2} e^{-2i\tau \lambda } \;\text {d}\tau \end{aligned}$$

(128)

$$\begin{aligned}= & {} \frac{1}{\sqrt{\pi }\sigma }e^{-x^2/\sigma ^2-\lambda ^2\sigma ^2} \end{aligned}$$

(129)

since the Fourier transform of $e^{-x^2/\sigma ^2}$ is $\sqrt{\pi }\sigma e^{-f^2\sigma ^2/4}$.

Noise Derivative with a Birectangular or Triangular–Rectangular Window

Let X(t) be a stationary random process of covariance function C(u). For any $\varepsilon >0$, the covariance of $\frac{X(t+\varepsilon )-X(t)}{\varepsilon }$ writes as follows.

$$\begin{aligned}&\text {Cov}\left( \frac{X(t+\varepsilon ) - X(t)}{\varepsilon },\frac{X(t'+\varepsilon ) - X(t')}{\varepsilon } \right) \nonumber \\&\quad = \frac{1}{\varepsilon ^2} \Bigl ( \text {Cov}(X(t+\varepsilon ),X(t'+\varepsilon )) \Bigr . \nonumber \\&\qquad \bigl . + \text {Cov}(X(t),X(t')) - \text {Cov}(X(t+\varepsilon ),X(t')) \nonumber \\&\qquad \Bigl .- \text {Cov}(X(t),X(t'-\varepsilon )) \Bigr ) \nonumber \\&\quad = \frac{1}{\varepsilon ^2} \left( C(u)-C(u+\varepsilon )-C(u-\varepsilon )) \right) \end{aligned}$$

(130)

with $u=t-t'$.

When the second derivative of the covariance function C exists, this latest term converges to $-C''(u)$ as $\varepsilon$ tends to 0. This makes it possible to define the mean-square derivative $X'(t)$ of the random process X at t as soon as the second derivative exists; the covariance function of X(t) being $-C''(t)$.

This obviously generalizes to stationary random fields (see, e.g., [1]). In the case of rectangular windows, the second derivatives do not exist. However, actual computations do not depend on the mean-square derivatives, but instead on a finite difference scheme.

If a central difference scheme of step $\varDelta _x=1$ pixel is used to estimate the derivatives,

$$\begin{aligned}&\text {Cov}\left( \frac{X(t+\varDelta _x) - X(t-\varDelta _x)}{2\varDelta _x},\frac{X(t'+\varDelta _x) - X(t'-\varDelta _x)}{2\varDelta _x} \right) \nonumber \\&\quad = \frac{1}{4\varDelta _x^2} \left( C(u+2\varDelta _x) - 2C(u) + C(u+2\varDelta _x) \right) \end{aligned}$$

(131)

which is the opposite of the second derivative of C (in the sense of the central difference scheme), denoted here by $-\varDelta ^2C$. The problem is to compute the noise covariance in the case of birectangular and triangular–rectangular windows, thus to compute the $-\varDelta ^2 r_a*r_a(u) = -\frac{1}{2} \varDelta ^2 {{\mathcal {W}}}r_a(u/2,0)$.

A straightforward calculation gives

$$\begin{aligned}&-\varDelta ^2 r_a*r_a(u) \nonumber \\&\quad = \frac{-1}{8a^2\varDelta _x^2} \Bigl ( (\varDelta _x-|2a+t|/2){\mathbbm {1}}_{[-2a-2\varDelta _x,-2a+2\varDelta _x]}(t) \Bigr .\nonumber \\&\qquad \Bigl .- 2(\varDelta _x-|t|/2){\mathbbm {1}}_{[-2\varDelta _x,2\varDelta _x]}(t) + \Bigr . \nonumber \\&\quad \Bigl . (\varDelta _x-|2a-t|/2){\mathbbm {1}}_{[2a-2\varDelta _x,2a+2\varDelta _x]}(t)\Bigr ) \end{aligned}$$

(132)