Optical transfer function

Remarks

So $H$ is the Fourier transform of $h$, scaled so that $H(0) = 1$. Since $h \ge 0$, one has $\abs{H(\nu)} \le H(0) = 1$, and since $h$ is real, $H(-\nu) = \overline{H(\nu)}$. If the object intensity is $f$, the image intensity is $h * f$, so the spectrum of the image is the spectrum of $f$ times $H$, up to the constant factor $\int h$.

Example: the Gaussian blur $h(x) = e^{-\pi \abs{x}^2}$ has $\int h = 1$ and Fourier transform $e^{-\pi \abs{\nu}^2}$. So $H(\nu) = e^{-\pi \abs{\nu}^2}$, which is real and positive, and the phase transfer function is $0$.

Non-example: the constant function $h = 1$ is not integrable on $\R^2$, so the formula does not define an optical transfer function.

For an aberration-free system, $H$ is the autocorrelation of the pupil function, in scaled coordinates, normalised to $1$ at the origin.