# Fourier optics - Transfer function of free space

In all consulted literature, the transfer function of the free space is given as follows: $$\exp(-i k_z d) = \exp(-i2 \pi d \sqrt{1/\lambda^2 -\nu_x^2-\nu_y^2})$$

When referring to this source, they derive the transfer function from the following equation: $$H(\nu_x,\nu_y) = \frac{f_{in}(x,y)}{f_{out}(x,y)}$$

I'm wondering why they do it this way (note: I've already seen this in other sources). 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).

The source 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 $$\exp(i 2\pi\nu_x x)$$ 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.