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Rectangle

This page illustrates the Rect shape in the Julia package ImagePhantoms.

This page comes from a single Julia file: 03-rect.jl.

You can access the source code for such Julia documentation using the 'Edit on GitHub' link in the top right. You can view the corresponding notebook in nbviewer here: 03-rect.ipynb, or open it in binder here: 03-rect.ipynb.

Setup

Packages needed here.

using ImagePhantoms: rect
using ImagePhantoms: phantom, radon, spectrum
import ImagePhantoms as IP
using ImageGeoms: ImageGeom, axesf
using MIRTjim: jim, prompt
using FFTW: fft, fftshift, ifftshift
using Unitful: mm, unit, °
using Plots: plot, plot!, scatter!, default
default(markerstrokecolor=:auto)

The following line is helpful when running this file as a script; this way it will prompt user to hit a key after each figure is displayed.

isinteractive() ? jim(:prompt, true) : prompt(:draw);

Overview

A useful shape for constructing 2D digital image phantoms is the rectangle, specified by its center, widths, angle and value. All of the methods in ImagePhantoms support physical units, so we use such units throughout this example. (Using units is recommended but not required.)

Define a rectangle object, using physical units.

width = (20mm, 80mm)
ob = rect((40mm, 30mm), width, π/6, 1.0f0)
Object2d{Rect, Float32, Unitful.Quantity{Int64, 𝐋, Unitful.FreeUnits{(mm,), 𝐋, nothing}}, Float64, 1, Float64} (S, D, V, ...)
 center::NTuple{2,Unitful.Quantity{Int64, 𝐋, Unitful.FreeUnits{(mm,), 𝐋, nothing}}} (40 mm, 30 mm)
 width::NTuple{2,Unitful.Quantity{Int64, 𝐋, Unitful.FreeUnits{(mm,), 𝐋, nothing}}} (20 mm, 80 mm)
 angle::Tuple{Float64} (0.5235987755982988,)
 value::Float32 1.0

Phantom image using phantom

Make a digital image of it using phantom and display it.

dx, dy = 0.8mm, 1.0mm
M, N = (2^8, 2^8+2)
x = (-M÷2:M÷2-1) * dx
y = (-N÷2:N÷2-1) * dy
oversample = 2
img = phantom(x, y, [ob], oversample)
jim(x, y, img, "Rect image")
Example block output

Hereafter we use ImageGeoms to simplify the indexing.

M, N = (2^8, 2^8+17) # odd
ig = ImageGeom(dims=(M,N), deltas=(dx,dy), offsets=:dsp)
@assert axes(ig)[1] ≈ x
oversample = 2
img = phantom(axes(ig)..., [ob], oversample)
p1 = jim(axes(ig), img, "Rect phantom", xlabel="x", ylabel="y")
Example block output

The image integral should approximate the object area

area = IP.area(ob)
(sum(img) * prod(ig.deltas), area)
(1600.0 mm^2, 1600 mm^2)

Spectrum using spectrum

There are two ways to examine the spectrum of this image:

  • using the analytical Fourier transform of the object via spectrum
  • applying the DFT via FFT to the digital image.

Because the shape has units mm, the spectra axes have units cycles/mm.

vscale = 1 / area # normalize spectra by area
spectrum_exact = spectrum(axesf(ig)..., [ob]) * vscale
sp = z -> max(log10(abs(z)/oneunit(abs(z))), -6) # log-scale for display
clim = (-6, 0) # colorbar limit for display
(xlabel, ylabel) = ("ν₁", "ν₂")
p2 = jim(axesf(ig), sp.(spectrum_exact), "log10|Spectrum|"; clim, xlabel, ylabel)
Example block output

Sadly fft cannot handle units currently, so this function is a work-around:

function myfft(x)
    u = unit(eltype(x))
    return fftshift(fft(ifftshift(x) / u)) * u
end

spectrum_fft = myfft(img) * (prod(ig.deltas) * vscale)
p3 = jim(axesf(ig), sp.(spectrum_fft), "log10|DFT|"; clim, xlabel, ylabel)
Example block output

Compare the DFT and analytical spectra to validate the code

errf = maximum(abs, spectrum_exact - spectrum_fft) / maximum(abs, spectrum_exact)
@assert errf < 2e-2
p4 = jim(axesf(ig), 1e3*abs.(spectrum_fft - spectrum_exact),
    "|Difference| × 10³"; xlabel, ylabel)
jim(p1, p4, p2, p3)
Example block output

Radon transform using radon

Examine the Radon transform of the object using radon, and validate it using the projection-slice theorem aka Fourier-slice theorem.

dr = 0.2mm # radial sample spacing
nr = 2^10 # radial sinogram bins
r = (-nr÷2:nr÷2-1) * dr # radial samples
fr = (-nr÷2:nr÷2-1) / nr / dr # corresponding spectral axis
ϕ = deg2rad.(0:180)
sino = radon(ob).(r, ϕ') # sample Radon transform of a single shape object
smax = ob.value * sqrt(sum(abs2, maximum(width)))
p5 = jim(r, rad2deg.(ϕ), sino; title="sinogram",
    xlabel="r", ylabel="ϕ", clim = (0,1) .* smax)
Example block output

The maximum sinogram value is about sqrt(80^2+20^2) = 82.mm, which makes sense for a 80mm × 20mm rect.

The above sampling generated a parallel-beam sinogram, but one could make a fan-beam sinogram by sampling (r, ϕ) appropriately.

Fourier-slice theorem illustration

Pick one particular view angle (55°) and look at its slice and spectra.

ia = argmin(abs.(ϕ .- 55°))
slice = sino[:,ia]
slice_fft = myfft(slice) * dr
ϕd = round(rad2deg(ϕ[ia]), digits=1)

kx, ky = (fr * cos(ϕ[ia]), fr * sin(ϕ[ia])) # Fourier-slice theorem
slice_ft = spectrum(ob).(kx, ky)
errs = maximum(abs, slice_ft - slice_fft) / maximum(abs, slice_ft)
@assert errs < 3e-5

p3 = plot(r, slice, title="profile at ϕ = $ϕd", label="")
p4 = plot(title="1D spectra")
scatter!(fr, abs.(slice_fft), label="abs fft", color=:blue)
scatter!(fr, real(slice_fft), label="real fft", color=:green)
scatter!(fr, imag(slice_fft), label="imag fft", color=:red,
    xlims=(-1,1) .* (0.1/mm))

plot!(fr, abs.(slice_ft), label="abs", color=:blue)
plot!(fr, real(slice_ft), label="real", color=:green)
plot!(fr, imag(slice_ft), label="imag", color=:red)
plot(p1, p5, p3, p4)
Example block output

The good agreement between the analytical spectra (solid lines) and the DFT samples (disks) validates that phantom, radon, and spectrum are all self consistent for this shape.

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