A rectangular aperture composed of four equal squares placed at equal distances from the center. The total matrix of rectangular shape has dimensions 1024 × 1024 pixels. While, the dimensions of each square (*x*_{0}, *y*_{0}) = 128 × 128 pixels and located along the cartesian coordinates (*x*, *y*) at distances ± *x*_{d} and ± *y*_{d} as shown in the Fig. 1a. A central obstruction is governed by the difference between two circles where the width of the annulus = 32 pixels. A uniform illumination emitted from laser beam is incident upon this new aperture, hence the amplitude transmittance is written as follows:

$$\begin{aligned} A\left( {x,y} \right) & & = rect\left( {x - x_{d} ,y} \right) + rect\left( {x + x_{d} ,y} \right) + rect\left( {x,y - y_{d} } \right) \\ & \quad + rect\left( {x,y + y_{d} } \right) + circ\left( {r_{1} } \right) - circ\left( {r_{2} } \right) \\ \end{aligned}$$

(1)

where

$$rect\left( {x,y} \right) = 1; \left| {\frac{x}{{x_{0} }}\left| { \le 1 , } \right|\frac{y}{{y_{0} }}} \right| \le 1,$$

$$circ\left({r}_{1}\right)=1; |\frac{r}{{r}_{01}}|\le 1, \quad {\text{and }} \quad circ\left({r}_{2}\right)=1; |\frac{r}{{r}_{02}}|\le 1$$

The Point Spread Function (PSF) corresponding to the novel aperture described by Eq. (1) is obtained by operating the Fourier transform as follows:

$$\begin{aligned} h\left( {x,y} \right) & = F.T. \left\{ {A\left( {x,y} \right)} \right\} \\ & = F.T. \{ rect\left( {x - x_{d} ,y} \right) + rect\left( {x + x_{d} ,y} \right) + rect\left( {x,y - y_{d} } \right) \\ & \quad + rect\left( {x,y + y_{d} } \right) + circ\left( {r_{1} } \right) - circ\left( {r_{2} } \right)\} \\ & =F.T. \{ rect\left( {x,y} \right) \otimes \delta ( x-x_{d},y)\} + F.T. \{ rect\left( {x,y} \right) \otimes \delta (x+x_{d},y)\} \\ & \quad F.T. \{ rect\left( {x,y} \right) \otimes \delta ( x,y - y_{d} )\} + F.T. \{ rect\left( {x,y} \right) \otimes \delta (x,y + y_{d} )\} \\ & \quad + F.T. \left\{ {circ\left( {r_{1} } \right)} \right\} - F.T. \left\{ {circ\left( {r_{2} } \right)} \right\} \\ \end{aligned}$$

(2)

The 1st transformation is solved taking in consideration that convolution of two functions is equivalent to simple product of the F.T. of each function, hence we write:

\(F.T. \{rect\left(x,y\right)\otimes\) δ(\(x-{x}_{d},y)\}=F.T.\{rect\left(x,y\right)\}. F.T. \{\updelta ( x-{x}_{d},y)\}\)

It is known that the Fourier transform of rect function is written as:

$$F.T.\left\{rect\left(x,y\right)\right\}={x}_{0}{y}_{0}\left[\frac{\mathrm{sin}\left(\frac{\pi {x}_{0}u}{\lambda f}\right)}{\left(\frac{\pi {x}_{0}u}{\lambda f}\right)}\right]\left[ \frac{\mathrm{sin}\left(\frac{\pi {y}_{0}v}{\lambda f}\right)}{\left(\frac{\pi {y}_{0}v}{\lambda f}\right)}\right]={x}_{0}{y}_{0} sinc\left({x}^{^{\prime}}\right)sinc\left({y}^{^{\prime}}\right)$$

(3)

where sinc function is defined as:

sinc(*x*’) = sin (π*x*’)/π*x*’, *x*’ = *x*_{0} *u*/*λf*, and a similar expression for sinc(*y*’).

Also, the Fourier transform of circ function as obtained as follows [11]:

$${\text{F}}.{\text{T}}. \, \left\{ {{\text{circ}}\left( {r_{{1}} } \right)} \right\} = {2}J_{{1}} \left( {Z_{{1}} } \right)/Z_{{1}} ,{\text{ and F}}.{\text{T}}. \, \left\{ {{\text{circ}}\left( {r_{{2}} } \right)} \right\} = {2}J_{{1}} \left( {Z_{{2}} } \right)/Z_{{2}}$$

(4)

where *Z*_{1} = 2π*r*_{01}*w*_{1}/*λf* and *Z*_{2} = 2π*r*_{02}*w*_{2}/*λf* represent the reduced coordinates in the Fourier planes corresponding to the external and internal circles of radii *r*_{01} and *r*_{02.} \({w}_{\mathrm{1,2}}=\sqrt{{u}^{2}+{v}^{2 }}\,is\,the\,radial\,coordinate\,in\,the\,Fourier\,plane\) \(corresponding\,to\,each\,circle.\)

The Fourier transform of shifted dirac-delta function gives inclined plane wave as:

$$\begin{aligned}F.T. \left\{\updelta \left( x-x_d,y\right)\right\} & = \mathop {\iint }\limits_{ - \infty }^{\infty }\left\{\updelta \left( x-x_d,y\right)\right\} \exp \left[ { - \frac{j2\pi}{\lambda f}\left( {xu + yv} \right)} \right] {\text{d}}x{\text{d}}y \\ & = \exp \left( { - \frac{{j2\pi x_{d} u}}{\lambda f}} \right)\mathop {\iint }\limits_{ - \infty }^{\infty }\left\{\updelta \left( x-x_d,y\right)\right\} \exp \left[ {- \frac{j2\pi}{\lambda f}(\left( {x - x_{d} )u + yv} \right)} \right] {\text{d}}x{\text{d}}y \\ & = \exp \left( { - \frac{{j2\pi x_{d} u}}{\lambda f}} \right) \\ \end{aligned}$$

(5)

Similar expressions are obtained for the other shift of the dirac-delta function as follows:

$$F.T. \left\{\updelta \left( x+{x}_{d},y\right)\right\}=\mathrm{exp} \left(+\frac{j2\pi {x}_{d}u}{\lambda f}\right)$$

(6)

$$F.T. \left\{\updelta \left( x,y-{y}_{d}\right)\right\}=\mathrm{exp} \left(-\frac{j2\pi {y}_{d}v}{\lambda f}\right)$$

(7)

$$F.T. \left\{\updelta \left( x,y+{y}_{d}\right)\right\}=\mathrm{exp} \left(+\frac{j2\pi {y}_{d}v}{\lambda f}\right)$$

(8)

Plugging Eqs. (3–8) in Eq. (2), we get:

$$\begin{aligned} h\left( {x,y} \right) & = x_{0} y_{0} {\text{sinc}}\left( {x^{\prime}} \right){\text{sinc}}\left( {y^{\prime}} \right) \\ & \quad + \left\{ {\exp \left( { - \frac{{j2x_{d} u}}{f}} \right) + \exp \left( { + \frac{{j2x_{d} u}}{f}} \right) + {\text{exp}} \left( { - \frac{{j2y_{d} v}}{f}} \right) + {\text{exp}} \left( { + \frac{{j2y_{d} v}}{f}} \right)} \right\} \\ & \quad + \left\{ {\frac{{2J_{1} \left( {Z_{1} } \right)}}{{Z_{1} }} - \frac{{2J_{1} \left( {Z_{2} } \right)}}{{Z_{2} }}} \right\} \\ \end{aligned}$$

(9)

Hence, the PSF is finally written as follows:

$$\begin{aligned} h\left( {x,y} \right) & = 2x_{0} y_{0} {\text{sinc}}\left( {x^{\prime}} \right){\text{sinc}}\left( {y^{\prime}} \right)\left\{ {{\text{cos}}\left( {\frac{{2x_{d} u}}{f}} \right) + {\text{cos}} \left( {\frac{{2y_{d} v}}{f}} \right)} \right\} \\ & \quad + \left\{ {\frac{{2J_{1} \left( {Z_{1} } \right)}}{{Z_{1} }} - \frac{{2J_{1} \left( {Z_{2} } \right)}}{{Z_{2} }}} \right\} \\ \end{aligned}$$

(10)

Equation (10) is rewritten as follows:

$$\begin{aligned} h\left( {x,y} \right) & = 4x_{0} y_{0} {\text{sinc}}\left( {x^{\prime}} \right){\text{sinc}}\left( {y^{\prime}} \right)\left\{ {\cos \left( {\frac{{\left( {x_{d} u + y_{d} v} \right)}}{f}} \right){\text{cos}}\left( {\frac{{\left( {x_{d} u - y_{d} v} \right)}}{f}} \right) } \right\} \\ & \quad + \left\{ {\frac{{2J_{1} \left( {Z_{1} } \right)}}{{Z_{1} }} - \frac{{2J_{1} \left( {Z_{2} } \right)}}{{Z_{2} }}} \right\} \\ \end{aligned}$$

(11)

The intensity impulse response is computed from the modulus square of Eq. (11) as follows:

$$\begin{aligned} I\left( {x,y} \right) \\ & = 16 x_{0}^{2} y_{0}^{2} {\text{sinc}}^{2} \left( {x^{\prime}} \right){\text{sinc}}^{2} \left( {y^{\prime}} \right)\left\{ {\cos^{2} \left( {\frac{{\left( {x_{d} u + y_{d} v} \right)}}{f}} \right){\text{cos}}^{2} \left( {\frac{{\left( {x_{d} u - y_{d} v} \right)}}{f}} \right) } \right\} \\ & \quad + 8x_{0} y_{0} {\text{sinc}}\left( {x^{\prime}} \right){\text{sinc}}\left( {y^{\prime}} \right)\left\{ {\cos \left( {\frac{{\left( {x_{d} u + y_{d} v} \right)}}{f}} \right)cos\left( {\frac{{\left( {x_{d} u - y_{d} v} \right)}}{f}} \right) } \right\} \\ & \quad \left\{ {\frac{{2J_{1} \left( {Z_{1} } \right)}}{{Z_{1} }} - \frac{{2J_{1} \left( {Z_{2} } \right)}}{{Z_{2} }}} \right\} + \left\{ {\frac{{2J_{1} \left( {Z_{1} } \right)}}{{Z_{1} }} - \frac{{2J_{1} \left( {Z_{2} } \right)}}{{Z_{2} }}} \right\}^{2} \\ \end{aligned}$$

(12)