A point response becomes a contrast transfer rule
Connect an incoherent intensity PSF to a normalized OTF and a measurable sinusoidal contrast.
On this page
The question for today
What you’ll understand
- Define the incoherent PSF and its normalization.
- Derive the OTF as a Fourier transform and pupil autocorrelation.
- Calculate contrast loss for an explicit sampled blur.
An unresolved point is recorded as a distribution of light, not an infinitesimal dot. If the imaging system is linear in intensity and shift invariant over the region of interest, that point response determines every image through convolution. Fourier transformation turns the same rule into a statement about contrast at each spatial frequency. Careful normalization lets us distinguish redistributed light from lost throughput and distinguish contrast magnitude from a spatial shift.
Specify the quantity that adds
For mutually incoherent object points, their time-averaged image intensities add. Let h be the intensity point-spread function after factoring out overall throughput, with its spatial integral equal to one. A displaced object point produces a displaced copy of h in appropriately scaled image coordinates. Summing these copies produces the convolution formula. Shift invariance is an approximation over a limited field: a strongly aberrated wide-field instrument may need a different response at different positions.
The normalized PSF is nonnegative because it represents intensity. Its width measures one aspect of blur, while sidelobes describe light redistributed away from a point. Two PSFs with equal widths can transmit different contrasts. The coherent amplitude impulse response is a different object: it can be complex and can change sign. In scalar quasi-monochromatic imaging the intensity PSF is proportional to its modulus squared, with normalization chosen separately. Squaring a PSF already defined as intensity would count the detection nonlinearity twice.
Frequency reveals contrast and position
Define the normalized optical transfer function as the Fourier transform of the intensity PSF. The zero-frequency value is one, expressing conservation of mean intensity after the throughput factor is removed. Its modulus is the modulation transfer function, which multiplies the contrast of a sinusoidal intensity object. Its argument gives a phase shift of that sinusoid. An asymmetric shifted PSF may therefore have a complex OTF even though the PSF itself is entirely real and nonnegative.
For a coherent pupil function P written in spatial-frequency coordinates, the intensity PSF is the squared modulus of the inverse transform of P. The convolution theorem makes its transform a normalized autocorrelation of P. A clear one-dimensional rectangular pupil consequently gives a triangular incoherent OTF: shifted pupils overlap less as their separation grows. Aberrations enter through pupil phase, making the overlap a complex weighted integral rather than merely an overlap area. The coherent amplitude transfer function remains the pupil itself under the same scaling.
A simple discrete blur is hand computable
A sampled teaching model makes these ideas concrete. Give neighboring sample offsets minus one, zero, and plus one the PSF weights one quarter, one half, and one quarter. The weights sum to one, so a constant intensity remains unchanged. Adding the two conjugate frequency factors gives an OTF equal to one half plus one half cosine of twice pi times frequency in cycles per sample. This is a specified discrete convolution model, not a universal diffraction PSF or a claim about detector sampling without aliasing.
The spectral instrument helps distinguish coarse from fine spatial content. Reset and set both controls to mode count one and coefficient decay one; predict a single sine. Reset again and set mode count four and decay one; predict added shorter spatial periods with smaller coefficients. It computes a finite sum of sin(nx−t) divided by n to the selected decay power. It does not convolve an optical PSF, measure an OTF, or model the physical temporal frequencies of those spatial modes.
Work it through
Contrast through a three-sample PSF
Use the discrete normalized PSF weights (1/4,1/2,1/4) at offsets (−1,0,1). An intensity object is 1+0.8 cos(2πfj), where j is the sample index and f=1/4 cycle per sample. Find the image contrast and its maximum and minimum intensities.
Transform the weights: H(f)=1/2+(1/4)e^{i2πf}+(1/4)e^{−i2πf}=1/2+(1/2)cos(2πf). Symmetry makes this response real.
At f=1/4, cosine is zero, so H=1/2. The mean remains one while the sinusoidal amplitude becomes 0.8×0.5=0.4.
The extrema are therefore 1.4 and 0.6. Their difference divided by their sum is 0.8/2=0.4, confirming the image contrast directly.
The image has contrast 0.4, maximum intensity 1.4, and minimum intensity 0.6 in the specified common units.
The living notebook
Explore the model
Interactive lab · spectral
Identify spatial frequencies before transfer
Reset and set BOTH controls to N=1 and p=1. Predict a single sine. Reset and set N=4 and p=1 for a new run; predict shorter spatial features. The instrument is a fixed finite sine sum, with no optical PSF filter.
- Sum at x = π/2
- 0.867
- Smallest wavelength
- 2π / 5
- Highest-mode coefficient
- 0.2
A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA frequency decomposition identifies what an OTF would act on; the lab does not supply that OTF.
Pause & reason
Check your understanding
Your turn
Try it yourself
In the same infinite discrete convolution model, put intensity eight at sample zero and zero elsewhere. Find the nonzero image samples and their total. Explain what normalization guarantees here.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
A single nonzero sample produces a scaled copy of the PSF.
Compare with the solution
The image intensities at samples −1,0,1 are 2,4,2. Their sum is eight, equal to the input total because the weights sum to one. The peak has fallen although total intensity is conserved. Cropping the image could lose some redistributed light, so the total statement assumes all affected samples are retained.
Keep the idea
Take it with you
- The incoherent intensity PSF convolves object intensity.
- The normalized OTF is its Fourier transform and a pupil autocorrelation.
- MTF describes contrast loss; OTF phase also carries spatial information.
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