BrainFlux

Tomography

Scanner

X-ray source and detector turn round the object

Sinogram

the X-ray shadows, one row per angle

Reconstruction

More panels: Fourier space, the back-projection without the filter, and a profile through the pictures

Fourier space

Back-projection

shadows smeared back, no filter · this angle

Object

Scanner

More: the detector bins’ width, a beam of two energies

Reconstruction filter

Each shadow is sharpened by this filter before it is smeared back. Ram-Lak is the plain ramp, |k|: sharp, but it amplifies noise. The others roll the ramp off towards the cutoff: less noise, a little blur.

More: the cutoff, and the filter drawn

Show

More: the grey scale, the profile, Fourier space

Experiments

Start here

A CT scanner sends X-rays through an object from many directions and works out a picture of its inside from the shadows. The Scanner shows the source (orange) and the detector (blue) turning round three disks; the Sinogram records what the detector saw, one row per angle, bright where the rays were absorbed most; the Reconstruction is the picture worked out from that record, building up as the scan goes.

Point at a disk to see the path of its shadow (yellow) in the sinogram, or at a row of the sinogram to turn the scanner to that angle; press Step to measure one angle at a time.

Look for the straight band of the disk at the centre, the wavy bands of the other two, and the streaks in the reconstruction until enough angles are in.

Follow one angle through the panels by its colour. The angle just measured is cyan, like the detector: its shadow drawn on the detector, its row of the sinogram and, under More panels, its line of Fourier space and the smear it adds to each back-projection (orange where the filtered projection is negative). A row of the sinogram you point at is magenta: the scanner turns to that angle and shows the ray; so is the line of Fourier space you point at. A place you point at in the object or in a reconstruction is yellow, with the path of its shadow in the sinogram. Click to pin a ray or a place; Esc lets go.

1. Each row of the sinogram is a shadow

Three disks, as the page opens. At each angle the detector records how much every ray was absorbed: three bumps, drawn in cyan on the detector, which become one row of the sinogram. The disk at the centre casts its shadow at the middle of the detector whatever the angle, so its band in the sinogram is straight. The other two swing to and fro as the scanner turns: point at one of them and follow the yellow curve, which is where its shadow falls at every angle. Each band is as wide as its disk, and as bright as its disk is absorbing and thick. Point at a row of the sinogram: the scanner turns there, and the magenta ray shows which line through the object that one measurement sums. The reconstruction is streaky after a few angles and clean after all 180: press Scan again and watch it build.

2. One point, one sinusoid

A single small spot, off centre (a Gaussian one pixel wide: the ring shows where). As the scanner turns, its shadow on the detector moves to and fro: at angle $\theta$ it falls at $$s=x_0\cos\theta+y_0\sin\theta=r_0\cos(\theta-\phi_0).$$ So its record, row by row, is a sinusoid whose amplitude is the spot's distance $r_0$ from the centre and whose phase is its direction $\phi_0$: hence the name sinogram. Move the spot with its sliders. Point anywhere on the scanner: the yellow dot on the detector is where that place's shadow falls, and the dashed curve is its sinusoid. Every object is a sum of points, so every sinogram is a sum of sinusoids, which is what experiment 1's bands were.

3. One angle at a time: back-projection

Three disks, twelve angles, and the scan waits before the first. Press Step: one projection, smeared back along its rays (cyan, in the Back-projection panel, now open under More panels). The scanner cannot tell where along a ray the absorption happened, so back-projection spreads it evenly along the whole ray: stripes. Step again, and again: the stripes of each new angle cross the earlier ones, and they add up most where the disks are. After the twelfth, raise Angles to 180: the stripes merge into a blurred picture of the disks. Point at the back-projection: its value at the yellow point is the sum of the sinogram along that point's sinusoid.

4. Back-projection blurs a point into 1/r

One spot, 360 angles. Every smear passes through the spot, so the back-projection is a star of lines that merge into a halo. How bright is the halo at a distance $r$ from the spot? The smear of one angle covers that place only if the place lies within half a detector bin, $\Delta s/2$, of the ray through the spot, and that is so for a range $\Delta s/r$ of angles. Each of those smears has the value $m/\Delta s$ (the spot's mass $m$ falls in one bin), so $$b=\frac{\Delta s}{r}\cdot\frac{m}{\Delta s}=\frac{m}{r}.$$ The profile (the last panel) shows the back-projection along the line through the spot (orange, right-hand axis) and $m/r$ (dotted), with nothing adjusted: they lie on top of each other. Every point of an object is blurred this way, which is why the back-projection of the head looks foggy.

5. The ramp filter sharpens

The head phantom. Before smearing, each projection is convolved with the ramp filter. The last panel shows a projection (grey) and its filtered version $Q$ (blue, right-hand axis): the filter is a sharpener; it overshoots at every edge and goes negative beside it. Press Pause, then Step a few times: in the reconstruction each new smear is cyan, with orange stripes where $Q$ is negative. Added up, those negative stripes cancel the $1/r$ halos of the other angles' smears, and the picture comes out sharp, with the right values. Point at the sinogram to see other projections, and choose Profile in the last panel to compare the reconstruction with the object along a line.

6. Each projection is one line of Fourier space

The projection-slice theorem: the 1D Fourier transform of a projection, $P(\theta,k)=\int p(\theta,s)\,e^{-iks}\,ds$, is a slice through the 2D Fourier transform of the object, $F(k_x,k_y)=\iint\mu\,e^{-i(k_xx+k_yy)}\,dx\,dy$: $$P(\theta,k)=F(k\cos\theta,\,k\sin\theta),$$ along the line through $k=0$ in the direction $\hat n$, across the rays. (Why: on that line $k_xx+k_yy=k\,s$, the same for every point of a ray, so the 2D integral can integrate along each ray first, which gives $p$.) The Fourier space panel shows each slice with its magnitude as brightness and its phase as hue. Here the object is a disk with stripes, whose wavevector puts two bright spots in $F$, at $|k|=40$ in the direction 60°. Fourier space is magnified four times, with the object's own transform drawn faintly underneath. Watch the cyan slice turn, 5° per angle: the spots light up when it reaches 60°, when the rays run along the stripes. Point at a spot: the slice through it and its row of the sinogram turn magenta, and the scanner shows that projection, with the stripes' ripple on it. Collect slices at enough angles and you have $F$, and so the object.

7. Why the filter is |k|

Twelve angles, so the slices are easy to count. Each slice has the same number of samples, so near the centre of Fourier space they crowd together and far out they spread apart: at radius $k$ neighbouring slices are $k\,\Delta\theta$ apart, and the density of samples falls as $1/|k|$. Back-projection simply adds the slices up, so the low frequencies, where they crowd, count too much: the back-projection's transform is $2\pi F(\vec k)/|k|$. In real space, dividing by $|k|$ is the $1/r$ blur of experiment 4 (in 2D, $1/r$ and $2\pi/|k|$ are a Fourier pair). Weighting each slice by $|k|/2\pi$ undoes it: that is the ramp filter.

The last panel uses the theorem directly: it interpolates the slices onto a square grid of $(k_x,k_y)$ and takes one inverse 2D FFT. With these twelve slices its RMS error is 0.12, against 0.26 for filtered back-projection, whose streaks cost more. With 180 angles it does worse, 0.034 against 0.023: interpolating far from the centre, where the slices are sparse, is hard, and an error at one $\vec k$ spreads over the whole picture.

8. Too few angles: streaks

The head with 24 angles: see the gaps between the rows of the sinogram. Close to each sharp edge the smears still cancel, but further away they separate into streaks. Fourier space shows why. An object of radius $R$ (here the scan circle, $R=1$) is described fully by samples of $F$ at most $\pi/R$ apart: the sampling theorem, with $x$ and $k$ swapped. The slices are $k\,\Delta\theta=k\pi/n_\theta$ apart at radius $k$, so they sample fully only inside the dotted circle, $|k|<\pi/(R\,\Delta\theta)=n_\theta/R$, although the detector measures out to the dashed circle, $|k|=\pi/\Delta s$. With $M$ detector bins across the circle, $\Delta s=2R/M$, and the two circles meet when $$n_\theta=\frac{\pi}{2}M:$$ the number of angles that samples the whole scan circle fully. Raise the angles and watch the streaks retreat: the RMS error away from edges (under the picture) falls from 0.15 at 24 angles to 0.067 at 48, 0.023 at 96 and 0.009 at 180; beyond that the streaks are faint and little more is gained (0.008 at $\tfrac{\pi}{2}\times256\approx402$).

9. A limited range of angles: the missing wedge

The scanner turns through only 110° instead of 180°. The slices leave a wedge of Fourier space empty, 70° wide on each side of the centre, and with it every detail that varies in those directions. An edge is seen by the rays that run along it; edges that no ray ran along are lost, so the picture smears in one direction. (Electron tomography of a thin slab, which cannot be tilted much beyond 60° to 70° either way, has exactly this problem.) Widen the range and the wedge closes. A scan building up shows the same thing as it goes: until it has turned through 180°, every reconstruction is a limited-angle one.

10. Noise: the ramp amplifies it

Each ray now counts photons: $n$ arrive out of $I_0=10^5$ sent, Poisson distributed, so $n$ is uncertain by $\sqrt n$, and the projection $-\ln(n/I_0)$ by $1/\sqrt n$. Along the longest paths through the 1974 head (realistic attenuations) only 2% of the photons get through, so those rays are the noisiest. The noise is spread evenly over all the frequencies of a projection, and the ramp multiplies the high frequencies by $|k|$: the picture is grainy, and the ventricles (2% darker than the brain) are lost. Choose Hann: the window rolls the filter smoothly to zero at the cutoff, the grain drops to about a third (the RMS error away from edges, under the picture, goes from 0.043 to 0.015), the ventricles reappear, and edges blur a little. Lower the cutoff (under More, below the filter) to trade more. With Ram-Lak, ten times the photons (ten times the dose) cut the noise by $\sqrt{10}$; the reading falls from 0.043 to 0.020, not 0.014, because 0.016 of it is streaks and ringing that are there without noise (turn the noise off to see).

11. Detector sampling and aliasing

A zone plate, $\cos(kr^2)$, whose rings get finer outwards (the local wavenumber is $d(kr^2)/dr=2kr$), scanned by only 64 detector bins, each a fifth of the spacing wide. 64 bins sample up to $|k|=\pi/\Delta s=100$, which the rings pass at $r=0.5$. Beyond that, the reconstruction shows rings that are not there, coarser and in the wrong places: detail finer than the sampling comes back at the wrong frequency (aliasing). Under More in the Scanner panel, widen the bins to 1: each now averages over its width, which removes much of that detail before it is sampled, and the false rings fade to grey. Then raise the bins to 256: every ring is there.

12. Metal makes streaks

Two metal pins, twenty times as absorbing as water, in the 1974 head, reconstructed with the Hann filter as soft tissue is. Turn off the noise and choose one energy: an ideal scanner gets the brain between the pins right, with only faint ripples round the pins (their edges are too sharp and bright for 180 angles and 256 bins to capture fully). A real scanner adds two worse problems. Few photons get through the metal (photon starvation), so the rays through it are very noisy, and filtered back-projection spreads their errors along the rays: streaks fanning out from the pins. And the beam has many energies. The scanner corrects for the hardening of the beam as water causes it, but metal absorbs the soft photons far more strongly than the hard ones, so rays through metal look less absorbing than they are: a dark band joins the pins, where rays pass through both, and the brain there comes out about 0.87 instead of 1.02. Turn each effect on and off.

13. The faint features of a real head

Shepp and Logan's head phantom of 1974 has a skull of attenuation 2 and a brain of 1.02, and its small ellipses differ from the brain by 0.01, 1%. In the Hounsfield units of CT, $1000\,(\mu/\mu_{\mathrm{water}}-1)$, that is a brain of +20, ventricles (fluid) of 0 and a skull of +1000, much like a real head. The grey scale runs from 1.00 to 1.05, as CT viewers use narrow "windows" for soft tissue; the skull is far off the scale. Turn on Photon noise, $10^6$ photons per ray: the grain from pixel to pixel, 0.014, is larger than the spots' 0.01, and the three small spots at the bottom disappear. Choose Hann: the grain falls to 0.005 and they come back, a little blurred. Or keep Ram-Lak and send $10^7$ photons (grain 0.004).

14. Your own picture, or the camera

Fourier's portrait, scanned at 360 angles. Paint on it with the brush (the reconstruction follows), or choose Your picture… or Camera. With the camera the scanner keeps turning and each angle sees the picture of its moment: move during the scan and see what motion does to a CT image. (Pictures are projected numerically, by summing along each ray with linear interpolation, accurate to a fraction of a percent on average; the phantom, the disks and the spot exactly.)

How it works

A CT scanner sends a sheet of parallel X-rays through one slice of the body. Along a ray, tissue of attenuation coefficient $\mu(x,y)$ removes photons at the rate $dI/dl=-\mu I$, so the ray keeps the fraction $I/I_0=e^{-\int\mu\,dl}$ of its photons (the Beer–Lambert law). Counting the photons that reach each detector bin, the scanner gets the projection $$p(\theta,s)=-\ln\frac{I}{I_0}=\int_{\mathrm{ray}}\mu\,dl,$$ the integral of $\mu$ along the ray $x\cos\theta+y\sin\theta=s$: the ray at distance $s$ from the centre, across the direction $\hat n=(\cos\theta,\sin\theta)$. Turning the source and the detector through half a turn fills the sinogram, one row per angle. The task is to get $\mu(x,y)$ back from it.

Back-projection smears every projection back along its rays and adds them up, $$b(\vec r)=\int_0^\pi p(\theta,\vec r\cdot\hat n)\,d\theta:$$ a blurred picture. The point $\vec r$ lies on the ray $s=\vec r\cdot\hat n$ at every angle, which is a sinusoid in the sinogram (experiments 1 and 2), so $b(\vec r)$ is the sum of the sinogram along $\vec r$'s sinusoid. Filtered back-projection first convolves each projection with the ramp filter, whose Fourier transform is $|k|/2\pi$, and smears the result $Q$ back instead: $\int_0^\pi Q(\theta,\vec r\cdot\hat n)\,d\theta$ is $\mu(\vec r)$ itself.

Things to know

The pictures show $\mu$ in units of water's, $\mu_{\mathrm{water}}\approx0.2$ per cm (at about 70 keV), and lengths in units of 10 cm, the radius of the scan circle. In these units a projection is $p=-\tfrac12\ln(I/I_0)$.
Parallel beams: half a turn measures every line once, since $p(\theta+\pi,s)=p(\theta,-s)$. Real scanners use a fan of rays from one point and turn all the way round.
Filtered back-projection here is $\mu(\vec r)=\sum_j Q(\theta_j,\vec r\cdot\hat n_j)\,\Delta\theta$, with $Q$ the projection convolved with the band-limited ramp kernel (Kak and Slaney, 1988), interpolated linearly between detector bins. Each detector bin averages the projection over its width. The filters multiply the ramp by a window: Ram-Lak by 1, Shepp-Logan by a sinc, Cosine by $\cos(\pi k/2k_c)$, Hann by $\tfrac12(1+\cos(\pi k/k_c))$, all zero beyond the cutoff $k_c$.
The RMS error compares the reconstruction with the object pixel by pixel, inside 95% of the scan circle; "away from edges" counts only pixels whose surroundings (four pixels each way) are uniform in the object, so that it measures noise and streaks rather than blurred edges.
Photon counting is exact (Poisson). The two-energy beam is a simple model of beam hardening: water-like matter absorbs the soft photons 1.35 times as strongly as the hard ones, metal 4 times; the scanner corrects exactly for water, not for metal.
The scanner measures the object as it is at each angle, so changing the object during a scan (painting, sliders, the camera) gives the motion artefacts of a real scan. Once a scan is over, any change measures every angle again at once.
Radon showed in 1917 that a function is fixed by its integrals along lines. Cormack (1963) and Hounsfield, whose scanner imaged its first patient in 1971, shared the 1979 Nobel Prize in medicine. Filtered back-projection with the ramp is due to Ramachandran and Lakshminarayanan (1971); the phantom to L. A. Shepp and B. F. Logan, "The Fourier reconstruction of a head section", IEEE Trans. Nucl. Sci. 21, 21–43 (1974); its high contrast version to P. Toft's thesis (1996). The book to read is A. C. Kak and M. Slaney, Principles of Computerized Tomographic Imaging (IEEE Press, 1988).

The tomography idea from the old BrainFlux FFT2D to-do list, built in 2026. Source: github.com/gallicchio/brainflux, MIT licence.