Hello, We express the radiation from a source by the expression bellow:

\[ U(x,y)=\frac{e^{jkz}e^{\frac{k}{2z}(x^2+y^2)}}{j\lambda z}\iint_{-\infty}^{+\infty}u(x',y')e^{-j\frac{2\pi}{\lambda z}(x'x+y'y)}dx'dy'\] the fourier transform is \[f_x=\frac{x}{\lambda z}\] and \[f_y=\frac{y}{\lambda z}.\ \]

Our system bellow has two apartures which has distance between them.

each aparture has fourier transform of sinc \[ F(rect(ax)rect(by))=\frac{1}{|ab|}sinc(\frac{f_x}{a})sinc(\frac{f_y}{b}) \]

\[ F(\delta(f_x-\frac{a}{2},f_y-\frac{b}{2}))=e^{[i\pi(ax+by)]} \]

I know that convolution with delta is a shift in space.

how to find using the integral expression shown above the fourier transform of the shape bellow?

Thanks.

\[ U(x,y)=\frac{e^{jkz}e^{\frac{k}{2z}(x^2+y^2)}}{j\lambda z}\iint_{-\infty}^{+\infty}u(x',y')e^{-j\frac{2\pi}{\lambda z}(x'x+y'y)}dx'dy'\] the fourier transform is \[f_x=\frac{x}{\lambda z}\] and \[f_y=\frac{y}{\lambda z}.\ \]

Our system bellow has two apartures which has distance between them.

each aparture has fourier transform of sinc \[ F(rect(ax)rect(by))=\frac{1}{|ab|}sinc(\frac{f_x}{a})sinc(\frac{f_y}{b}) \]

\[ F(\delta(f_x-\frac{a}{2},f_y-\frac{b}{2}))=e^{[i\pi(ax+by)]} \]

I know that convolution with delta is a shift in space.

how to find using the integral expression shown above the fourier transform of the shape bellow?

Thanks.

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