Rendering Deformed Speckle Images with a Boolean Model

Abstract

Rendering speckle images affected by a given deformation field is of primary importance to assess the metrological performance of displacement measurement methods used in experimental mechanics and based on digital image correlation (DIC). This article describes how to render deformed speckle images with a classic model of stochastic geometry, the Boolean model. The advantage of the proposed approach is that it does not depend on any interpolation scheme likely to bias the assessment process, and that it allows the user to render speckle images deformed with any deformation field given by an analytic formula. The proposed algorithm mimics the imaging chain of a digital camera, and its parameters are carefully discussed. A MATLAB software implementation and synthetic ground-truth datasets for assessing DIC software programs are publicly available.

Appendix A: Quantization Error for a Compound Probability Distribution

We assume that each real value inside a quantization box [a, b] (with $a<b$) is assigned to a quantized value. In Monte Carlo estimation, sample means are distributed according to a normal distribution of mean m and standard deviation s. A quantization error occurs when the sample mean is not assigned to the same quantized value as m. If s is small with respect to $b-a$ and m is in the middle of the interval [a, b], the probability of an error is small, while it is larger if m is close to a or b.

In order to estimate an overall error, we assume that the mean m is uniformly distributed over [a, b]. In other words, we assume that the sample mean X is a compound random variable distributed according to a Gaussian distribution of standard deviation s and mean m distributed according to a uniform distribution on interval [a, b]. We estimate the probability that X is not correctly quantized, i.e., that it falls outside the interval [a, b]. In other words, we would like to estimate the probability $\mathcal{E}(s,b-a)=1-\Pr (a\le X \le b)$.

We denote by g the standard normal probability distribution function (that is, $g(x) = e^{-x^2/2}/\sqrt{2\pi }$) and by G the associated cumulative distribution function. Marginalizing out m in X gives:

$$\begin{aligned} \Pr (a\le X \le b)&= \frac{1}{b-a} \int _a^b \Pr (a\le X \le b \,|\,\mu ) \;\text {d}\mu \end{aligned}$$

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$$\begin{aligned}&= \frac{1}{b-a} \int _a^b \left( G\left( \frac{b-\mu }{s}\right) \right. \nonumber \\&\quad -\left. G\left( \frac{a-\mu }{s}\right) \right) \;\text {d}\mu . \end{aligned}$$

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Now, for any two real values $\alpha$ and $\beta \ne 0$, an antiderivative of $G(\alpha +\beta x)$ is given by:

$$\begin{aligned} \int G(\alpha +\beta x)= & {} \frac{1}{\beta }\left( (\alpha +\beta x)G(\alpha +\beta x) \right. \nonumber \\&\quad +\left. g(\alpha +\beta x) \right) . \end{aligned}$$

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We calculate:

$$\begin{aligned} \begin{aligned} \frac{1}{b-a} \int _a^b G\left( \frac{b-\mu }{s}\right) \;\text {d}\mu = G\left( \frac{b-a}{s}\right) \\ + \frac{s}{b-a}g\left( \frac{b-a}{s}\right) -\frac{s}{\sqrt{2\pi }(b-a)} \end{aligned} \end{aligned}$$

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and

$$\begin{aligned} \begin{aligned} \frac{1}{b-a} \int _a^b G\left( \frac{a-\mu }{s}\right) \;\text {d}\mu = G\left( \frac{a-b}{s}\right) \\ - \frac{s}{b-a}g\left( \frac{a-b}{s}\right) +\frac{s}{\sqrt{2\pi }(b-a)} \end{aligned} \end{aligned}$$

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We eventually obtain, since $G(-x)=1-G(x)$ and $g(-x)=g(x)$:

$$\begin{aligned} \Pr (a\le X \le b)= & {} 2G\left( \frac{b-a}{s}\right) -1\nonumber \\&\quad +\frac{2s}{b-a}\left( g\left( \frac{b-a}{s}\right) - \frac{1}{\sqrt{2\pi }}\right) \end{aligned}$$

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We are interested in the quantization error when the random fluctuation of the Monte Carlo estimation is small, that is, when $s/(b-a)$ tends to 0.

Since $G(x)=1- g(x)/x + \mathcal{O}( g(x)/x )$ when $x\rightarrow +\infty$ (where $\mathcal O$ is Landau’s “big-O”; this is a property of Mill’s ratio [46, chapter 2]), we have proved the following proposition.

Proposition 1

If a and b are two real numbers such that $a<b$, and X is a random variable distributed according to a compound Gaussian distribution of variance $s^2$ such that its mean is a random variable distributed according to a uniform distribution on [a, b], the following asymptotic expansion holds:

$$\begin{aligned} \Pr (a\le X \le b) =_{s \rightarrow 0} 1 - \frac{\sqrt{2}s}{\sqrt{\pi }(b-a)} + \mathcal{O}\left( s e^{-(b-a)/(2s^2) }\right) \end{aligned}$$

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If $s/(b-a)<< 1$, the latter expression gives an accurate estimate of the probability $\mathcal E$ of quantization error as:

$$\begin{aligned} \mathcal{E}(s,b-a) = \frac{\sqrt{2}s}{\sqrt{\pi }(b-a)}. \end{aligned}$$

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