In all consulted literature, the transfer function of the free space is given as follows: exp ⁡ ( − i k z d ) = exp ⁡ ( − i 2 π d 1 / λ 2 − ν x 2 − ν y 2 )When referring to Fourier Optics, they derive the transfer function from the following equation: H ( ν x , ν y ) = f i n ( x , y ) f o u t ( x , y ) I'm wondering why they do it this way. I thought the transfer function is defined in the frequency domain rather than in the time domain (frequency and space for the fourier optics respectively).

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In all consulted literature, the transfer function of the free space is given as follows:  exp  ⁡  (  −  i      k    z    d  )  =  exp  ⁡  (  −  i  2  π  d      1          /              λ      2        −          ν      x      2        −          ν      y      2        )When referring to Fourier Optics, they derive the transfer function from the following equation:  H  (      ν    x    ,      ν    y    )  =                    f                  i          n                    (      x      ,      y      )                      f                  o          u          t                    (      x      ,      y      )      I&#039;m wondering why they do it this way. I thought the transfer function is defined in the frequency domain rather than in the time domain (frequency and space for the fourier optics respectively).

Phiplyrhypelw0 · Accepted Answer

The Fourier Optics is specifically referring to the input and output being a plane wave. The key point on that slide is that complex exponentials in the form   exp  ⁡  (  i  2  π      ν    x    x  ) are eigenfunctions of linear, shift-invariant optical systems. Thus, a complex exponential signal goes into the system, and another complex exponential with the same frequency goes out. A plane wave has the same form as this complex exponential. As a plane wave propagates through a linear system, it remains a plane wave. It can only be amplified/attenuated and phase-delayed. New spatial frequencies (plane wave components) cannot appear in the output.

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