## Abstract

The iterative blind deconvolution algorithm proposed by Ayers and Dainty [
Opt. Lett. **13**,
547 (
1988)] and improved on by Davey *et al.* [
Opt. Commun. **69**,
353 (
1989)] is applied to the problem of phase retrieval, which is a special case of the blind deconvolution problem. A close relationship between this algorithm and the error-reduction version of the iterative Fourier-transform phase-retrieval algorithm is shown analytically. The performance of the blind deconvolution algorithm is compared with the error-reduction and hybrid input–output versions of the iterative Fourier-transform algorithm by reconstruction experiments on real-valued, nonnegative images with and without noise.

© 1990 Optical Society of America

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### Equations (23)

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(1)
$$\begin{array}{ll}\hfill c(\overline{x})& ={\mathit{\int}}_{-\infty}^{\infty}f({\overline{x}}^{\prime})g(\overline{x}-{x}^{\prime})\text{d}{x}^{\prime}+n(\overline{x})\hfill \\ \hfill & =f(\overline{x})*g(\overline{x})+n(\overline{x}),\hfill \end{array}$$
(2)
$$C(\u016b)=F(\u016b)G(\u016b)+N(\u016b),$$
(3)
$$\begin{array}{ll}F(\u016b)\hfill & =|F(\u016b)|exp[i\psi (\u016b)]=\mathcal{F}[f(\overline{x})]\hfill \\ \hfill & ={\mathit{\int}}_{-\infty}^{\infty}f(\overline{x})exp[-i2\pi (\u016b\xb7x)]\text{d}x.\hfill \end{array}$$
(4)
$$\begin{array}{ll}\hfill r(\overline{x})& ={\mathit{\int}}_{-\infty}^{\infty}f({\overline{x}}^{\prime})f*({\overline{x}}^{\prime}-x)\text{d}{x}^{\prime}\hfill \\ \hfill & ={\mathcal{F}}^{-1}[F(\u016b)F*(\u016b)]={\mathcal{F}}^{-1}[{|F(\u016b)|}^{2}].\hfill \end{array}$$
(5)
$${F}_{k}(\u016b)={\stackrel{\sim}{F}}_{k}(\u016b);$$
(6)
$${F}_{k}(\u016b)=(1-\beta ){\stackrel{\sim}{F}}_{k}(\u016b)+\beta \frac{C(\u016b)}{{\stackrel{\sim}{G}}_{k}(\u016b)};$$
(7)
$$\frac{1}{{F}_{k}(\u016b)}=\frac{1-\beta}{{\stackrel{\sim}{F}}_{k}(\u016b)}+\beta \frac{{\stackrel{\sim}{G}}_{k}(\u016b)}{C(\u016b)}.$$
(8)
$$c(\overline{x})=s(\overline{x})*f(\overline{x})+n(\overline{x}),$$
(9)
$$C(\u016b)=S(\u016b)F(\u016b)+N(\u016b),$$
(10)
$$\widehat{F}(\u016b)=W(\u016b)C(\u016b),$$
(11)
$$W(\u016b)=\frac{S*(\u016b)}{{|S(\u016b)|}^{2}+\u3008{|N(\u016b)|}^{2}\u3009/\u3008{|F(\u016b)|}^{2}\u3009},$$
(12)
$${F}_{k}(\u016b)=\frac{{\stackrel{\sim}{G}}_{k}^{*}(\u016b)}{{|{\stackrel{\sim}{G}}_{k}(\u016b)|}^{2}+{\sigma}^{2}/{|{\stackrel{\sim}{F}}_{k}(\u016b)|}^{2}}C(\u016b),$$
(13)
$${G}_{k}(\u016b)=\frac{{\stackrel{\sim}{F}}_{k}^{*}(\stackrel{\sim}{u})}{{|{\stackrel{\sim}{F}}_{k}(\u016b)|}^{2}+{\sigma}^{2}/{|{\stackrel{\sim}{G}}_{k-1}(\u016b)|}^{2}}C(\u016b).$$
(14)
$${F}_{k}(\u016b)=\frac{{\stackrel{\sim}{G}}_{k}^{*}(\u016b)}{{|{\stackrel{\sim}{G}}_{k}(\u016b)|}^{2}+\alpha}C(\u016b).$$
(15)
$${G}_{k}(\u016b)=\frac{{\stackrel{\sim}{F}}_{k}^{*}(\u016b)}{{|{\stackrel{\sim}{F}}_{k}(\u016b)|}^{2}+\alpha}C(\u016b).$$
(16)
$$\begin{array}{ll}\hfill {F}_{k}(\u016b)& ={\stackrel{\sim}{G}}_{k}^{*}(\u016b)\hfill \\ \hfill & =\frac{{\stackrel{\sim}{F}}_{k}(\u016b)}{{|{\stackrel{\sim}{F}}_{k}(\u016b)|}^{2}+\alpha}{|F(\u016b)|}^{2}.\hfill \end{array}$$
(17)
$${F}_{k}(\u016b)=\frac{{\stackrel{\sim}{F}}_{k}(\u016b)}{{|{\stackrel{\sim}{F}}_{k}(\u016b)|}^{2}+{\sigma}^{2}/{|{\stackrel{\sim}{F}}_{k}(\u016b)|}^{2}}{|F(\u016b)|}^{2}.$$
(18)
$${F}_{k}(\u016b)=|F(\u016b)|exp[i{\mathrm{\Phi}}_{k}(\u016b)]={\stackrel{\sim}{F}}_{k}(\u016b)\frac{|F(\u016b)|}{|{\stackrel{\sim}{F}}_{k}(\u016b)|}.$$
(19)
$${F}_{k}(\u016b)={\stackrel{\sim}{F}}_{k}(\u016b)\frac{{|F(\u016b)|}^{2}}{{|{\stackrel{\sim}{F}}_{k}(\u016b)|}^{2}}.$$
(20)
$$\text{ABSERR}\equiv {\left[\frac{\text{\u2211}_{\overline{x}}{|\alpha {\stackrel{\sim}{f}}_{k}(\overline{x}-{\overline{x}}_{0})-f(\overline{x})|}^{2}}{\text{\u2211}_{\overline{x}}{|f(\overline{x})|}^{2}}\right]}^{1/2},$$
(21)
$$\alpha =\frac{\text{\u2211}_{\overline{x}}f(\overline{x}){\stackrel{\sim}{f}}_{k}^{*}(\overline{x}-{\overline{x}}_{0})}{\text{\u2211}_{\overline{x}}{|{\stackrel{\sim}{f}}_{k}(\stackrel{\sim}{x})|}^{2}}$$
(22)
$$\text{FME}\equiv {\left\{\frac{\text{\u2211}_{\u016b}{[{|F(\u016b)|}_{n}-|F(\u016b)|]}^{2}}{\text{\u2211}_{\u016b}{|F(\u016b)|}^{2}}\right\}}^{1/2}.$$
(23)
$$|\widehat{F}(\u016b)|={\left[\frac{1}{1+{\sigma}^{2}/{|F(\u016b)|}_{n}^{4}}{|F(\u016b)|}_{n}^{2}\right]}^{1/2},$$