For optical systems, is there a way to estimate its point-spread function from its line-spread function?

In particular, I wish to estimate the encircled energy.

  • $\begingroup$ The line spread function is the convolution of the point spread function with a line. It follows that the point spread function is the deconvolution of line spread function and line. It would be helpful if you could let us know if those words mean anything to you... $\endgroup$
    – Floris
    Commented Sep 6, 2015 at 19:50
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    $\begingroup$ Thank you @Floris. That was helpful. I was thinking along a different line altogether, so deconvolution didn't occur to me. Was obvious, wasn't it? $\endgroup$ Commented Sep 8, 2015 at 17:29
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    $\begingroup$ Just for the sake of anyone who stumbles onto this question. "Derivation of the Point Spread Function from the Line Spread Function" E. W. Marchand This paper pretty much answered most of what I wanted to know. It is interesting to note that the problem can be modeled in a such a way. $\endgroup$ Commented Dec 28, 2015 at 7:22
  • $\begingroup$ Link to the above (behind paywall): osapublishing.org/josa/abstract.cfm?URI=josa-54-7-915 $\endgroup$
    – Floris
    Commented Dec 28, 2015 at 16:02
  • $\begingroup$ You could summarize the procedure in an answer to your own question. May be helpful to others in the future. $\endgroup$
    – kaylimekay
    Commented Jan 11, 2021 at 6:16

2 Answers 2


Here is the analytical solution using Abel transform.

Let's have $LSF(x), x\in\mathbb{R}$ so that it is symmetric $LSF(x) = LSF(-x)$ and for transform to work$$\lim_{x \to \infty} \frac{LSF(x)}{x} = 0.$$

Let's assume we are looking for radially symmetric function $PSR(r)$, such that $PSF(x,y) = PSR(\sqrt{x^2+y^2})$ and $$LSF(x) = \int_{-\infty}^{\infty} PSF(x,y) dy.$$

Thus $$LSF(x) = 2 \int_{0}^{\infty} PSR(\sqrt{x^2+y^2}) dy$$ and upon substitution $r=\sqrt{x^2+y^2}$ we have $$dr = \frac{y dy}{\sqrt{x^2+y^2}} = \frac{\sqrt{r^2 - x^2} dy}{r}$$ so that $$dy = \frac{r}{\sqrt{r^2-x^2}} dr $$ and $$LSF(x) = 2 \int_{x}^{\infty} PSR(r) \frac{r}{\sqrt{r^2-x^2}} dr,$$ the last equation has a structure of the Abel transform and thus $$PSR(r) = -\frac{1}{\pi} \int_{r}^{\infty} \frac{d LSF(x)}{dx} \frac{1}{\sqrt{x^2-r^2}} dx.$$

Here is the method using Fourier transform in 2D

Fourier transform of PSF is called optical transfer function $$OTF(\xi_x,\xi_y) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} PSF(x,y) e^{-2 \pi i (x \xi_x + y \xi_y)} dx dy$$ and there is also Fourier transform of LSF $$\mathcal{F}(LSF)(\xi) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} LSF(x) e^{-2 \pi i x \xi} dx $$

now we have $$OTF(\xi_x,0) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} PSF(x,y) dy e^{-2 \pi i (x \xi_x)} dx = \frac{1}{2\pi} \int_{-\infty}^{\infty} LSF(x) e^{-2 \pi i (x \xi_x)} dx$$ and also $$OTF(\xi_x,0) = \frac{1}{\sqrt{2\pi}}\mathcal{F}(LSF)(\xi_x)$$ and if you believe that $OTF(\xi_x,\xi_y)$ is radially symmetric then $$OTF(\xi_x,\xi_y) = \frac{1}{\sqrt{2\pi}}\mathcal{F}(LSF)(\sqrt{\xi_x^2 + \xi_y^2}).$$ Now we can sample OTF using that formula and perform inverse 2D Fourier transform. By this kind of slice theorem we obtain the solution. Note that I might use different scaling of Fourier transform.

Now for the case of non-symmetric functions, Marchand 1965 might be of interest. In Marchand 1964 there can be found somehow more heavy-weight derivation of the first method but the reduction of the second method, which then does not need double integral by means of integration using Bessel function of the first kind. In general first method might be of interest for those having some smooth fit of LSF, the second method might be useful in case of sampled LSF.


quick and dirt answer: take the (minimal) width of the line and build a circle with that as diameter. here is your PSF.

  • $\begingroup$ Yeah, okay. But how I prove that? Moreover, isn't it an approximation to do so? The LSF corresponding to a PSF represented (for example) by a Bessel function is typically a Struve function. Not the same thing. $\endgroup$ Commented Oct 20, 2015 at 3:03

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