Medical Imaging Systems: X-Ray, CT, SPECT, PET, MRI and Ultrasound
2. Computed tomography
CT measures, along each ray through the patient, the line integral of μ: ln(I₀/I) = ∫μ dl. A set of such integrals at one angle is a projection, and the projections over 180° or 360° form the sinogram — the Radon transform of the slice. Simple back-projection smears each projection back across the image and gives a blurred result (a 1/r blur); filtered back-projection first convolves each projection with a ramp filter (|ω| in frequency, usually apodised to limit noise) and is exact for ideal data. The Fourier (central) slice theorem underlies it: the 1-D Fourier transform of a projection is a line through the 2-D Fourier transform of the slice. Iterative reconstruction models the physics and noise and allows lower dose.
| Generation | Geometry | Remark |
|---|---|---|
| First | pencil beam, one detector; translate then rotate | minutes per slice |
| Second | narrow fan, several detectors; translate–rotate | fewer, larger rotation steps |
| Third | wide fan covering the patient; tube and detector arc rotate together | the basis of modern multislice scanners; ring artefacts if a detector fails |
| Fourth | stationary ring of detectors; only the tube rotates | many detectors needed |
| Fifth (electron-beam) | no moving parts; a steered electron beam sweeps stationary target rings | very fast, for cardiac imaging |
In helical (spiral) CT the table moves continuously while the gantry rotates; pitch = table feed per rotation ÷ total beam collimation, so pitch above 1 lowers dose and scan time at some cost in z-resolution. Each reconstructed voxel is expressed as a CT number in Hounsfield units, HU = 1000(μ − μ_water)/μ_water, which fixes water at 0 and air at −1000; fat is roughly −100, soft tissues a few tens, and bone several hundred to over a thousand. The display maps a chosen window onto the grey scale: with window level L and width W, values from L − W/2 to L + W/2 span black to white. Common artefacts are beam hardening (the beam’s mean energy rises through the patient, giving cupping and dark streaks between dense objects), partial volume, motion and metal streaks.
3. Nuclear imaging: the gamma camera, SPECT and PET
Nuclear imaging maps function rather than anatomy: a radiotracer follows a physiological pathway and its emissions are detected outside the body. Activity decays as A = A₀e−λt with λ = ln 2/T½, and the tracer also leaves the body biologically, so the effective half-life obeys 1/T_eff = 1/T_phys + 1/T_bio. The workhorse of single-photon imaging is technetium-99m, emitting 140 keV gamma rays with a physical half-life of about 6 hours.
- The Anger gamma camera: a lead collimator (usually parallel-hole) admits only photons travelling roughly perpendicular to it, fixing where each came from; a large NaI(Tl) scintillation crystal converts each photon into a flash of light; an array of photomultiplier tubes detects the flash, and position logic computes its centroid; a pulse-height analyser accepts only photons near the photopeak energy, rejecting most scatter. The collimator throws away almost all the photons, which is why the camera’s sensitivity is low.
- SPECT rotates one or more gamma-camera heads around the patient, acquires projections at many angles and reconstructs slices tomographically, as CT does, with corrections for attenuation and scatter.
- PET uses positron emitters such as fluorine-18 (half-life about 110 minutes) in fluorodeoxyglucose (FDG). The positron travels a short distance, annihilates with an electron, and produces two 511 keV photons emitted nearly 180° apart. A ring of detectors (BGO or LSO crystals) records coincidences within a nanosecond-scale window, and each coincidence defines a line of response — electronic collimation, so no lead collimator is needed and sensitivity is far higher than in SPECT. Time-of-flight PET also uses the difference in arrival times to locate the event along the line. Resolution is limited by the positron range and the slight non-collinearity of the photon pair; random and scattered coincidences and attenuation must be corrected.
4. Magnetic resonance imaging
Hydrogen nuclei (protons) have spin and a magnetic moment. In a static field B₀ they precess at the Larmor frequency f = γ̄B₀, with γ̄ = 42.58 MHz/T for ¹H — 63.87 MHz at 1.5 T and 127.74 MHz at 3 T — and a slight excess aligns with B₀ to form a net magnetisation M₀. An RF pulse at the Larmor frequency, applied by a coil as a field B₁, tips the magnetisation by a flip angle θ = γB₁τ (γ = 2πγ̄ in rad s⁻¹ T⁻¹, τ the pulse duration). The rotating transverse magnetisation induces the signal in a receive coil.
Two relaxations follow. T1 (spin–lattice) recovery restores the longitudinal magnetisation, M_z = M₀(1 − e−t/T1) after a 90° pulse. T2 (spin–spin) decay dephases the transverse magnetisation, M_xy = M₀e−t/T2; field inhomogeneity adds further dephasing, so the observed free-induction decay follows T2* ≤ T2. A spin echo (90° then 180° at TE/2) refocuses the static inhomogeneity and recovers T2 decay. Tissue contrast is chosen by timing: T1-weighting uses short TR and short TE, T2-weighting long TR and long TE, and proton-density weighting long TR with short TE. Fat is bright and water dark on T1-weighted images; fluid is bright on T2-weighted images.
Spatial encoding uses three gradient coils that make the field, and so the Larmor frequency, vary linearly with position. A slice-select gradient G during a band-limited RF pulse excites only the slab whose frequencies fall in the pulse bandwidth, so slice thickness Δz = BW/(γ̄G). A phase-encoding gradient applied briefly between excitation and readout, stepped from one repetition to the next, and a frequency-encoding (readout) gradient during acquisition label the other two directions. The data fill k-space, and a 2-D inverse Fourier transform gives the image. The hardware is a superconducting magnet (niobium–titanium windings in liquid helium), shim coils, gradient coils and RF coils. MRI uses no ionising radiation; its hazards are ferromagnetic projectiles, implants, RF heating (limited by the specific absorption rate) and acoustic noise.
5. Ultrasound imaging and Doppler
A piezoelectric transducer (usually PZT) converts a voltage pulse into a pressure pulse of a few megahertz and converts returning echoes back into voltage. Sound travels at about c = 1540 m/s in soft tissue, so the wavelength λ = c/f is 0.308 mm at 5 MHz. An echo returning after time t comes from depth d = ct/2 — the factor 2 because the pulse goes there and back. Each boundary reflects according to the mismatch of acoustic impedance Z = ρc: the intensity reflection coefficient is R = [(Z₂ − Z₁)/(Z₂ + Z₁)]². Soft-tissue boundaries reflect about a percent, so most energy goes on; tissue–air and tissue–bone boundaries reflect almost everything, which is why gel is needed and why bone and gas shadow what lies beyond. A quarter-wave matching layer of impedance √(Z_transducer Z_tissue) improves coupling.
- Attenuation in soft tissue rises roughly in proportion to frequency (of the order of 0.5 dB per cm per MHz), so higher frequencies give finer resolution but shallower penetration — the central trade-off of transducer choice. Time-gain compensation amplifies later echoes more to offset it.
- Axial resolution is half the spatial pulse length, so short, well-damped pulses resolve better; lateral resolution is the beam width, best at the focus. An unfocused disc of radius a has a near-field (Fresnel) length of about a²/λ.
- Display modes: A-mode (amplitude against depth), B-mode (brightness-coded 2-D image from a swept or electronically steered array), M-mode (one line against time, for moving valves and walls), and Doppler.
- A pulse must return before the next is sent, so the maximum imaging depth is c/(2·PRF).
Blood moving at speed v at angle θ to the beam shifts the echo frequency by f_D = 2f₀v cos θ/c, again with a factor 2 for the two-way path. Continuous-wave Doppler uses separate transmit and receive crystals and measures high velocities without range information; pulsed-wave Doppler samples a chosen depth but, being sampled at the PRF, aliases when f_D exceeds PRF/2 (the Nyquist limit); colour Doppler maps mean velocity over a region. At θ = 90° the shift vanishes, so the angle must be kept well below 90° and corrected for.
Key takeaways
- I = I₀e−μx, HVL = 0.693/μ; photoelectric ∝ Z³/E³ gives bone contrast, Compton degrades soft-tissue contrast; kVp sets energy, mAs sets photon count.
- HU = 1000(μ − μ_water)/μ_water; window L, W displays L ± W/2; filtered back-projection uses a ramp filter; pitch = feed/collimation.
- SPECT: single photons (Tc-99m, 140 keV) through a collimator; PET: two 511 keV photons in coincidence, no collimator, higher sensitivity.
- f = 42.58 MHz/T × B₀; T1: short TR and TE, T2: long TR and TE; slice thickness = BW/(γ̄G).
- Ultrasound: λ = c/f, d = ct/2, R = [(Z₂ − Z₁)/(Z₂ + Z₁)]², f_D = 2f₀v cos θ/c, depth limit c/(2·PRF), aliasing above PRF/2.
Practice questions (18)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A material has a linear attenuation coefficient of 0.2 cm⁻¹ for a monoenergetic X-ray beam. Its half-value layer (in cm), to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 3.47
HVL = ln 2/μ = 0.693/0.2 = 3.47 cm. Taking 1/μ = 5 cm gives the mean free path, the thickness that reduces intensity to 1/e (37%), not to one half.A monoenergetic beam passes through 2 cm of bone (μ = 0.5 cm⁻¹) and then 4 cm of soft tissue (μ = 0.2 cm⁻¹). The transmitted fraction (in %), to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 16.5
Exponents add along the path: μ₁x₁ + μ₂x₂ = 1.0 + 0.8 = 1.8, so I/I₀ = e−1.8 = 0.165 = 16.5%. Adding the transmitted fractions of each layer (36.8% + 44.9%) or averaging μ over the two layers without weighting by thickness gives wrong answers.A tissue has μ = 0.21 cm⁻¹ where water has μ = 0.20 cm⁻¹. Its CT number (in HU) is ____.
Numerical answer — type the value.
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Answer: 50
HU = 1000(μ − μ_w)/μ_w = 1000 × 0.01/0.20 = 50 HU, a typical soft-tissue value. Dividing by the tissue’s μ gives 47.6, and leaving out the 1000 gives 0.05.A CT image is displayed with window level 40 HU and window width 400 HU. The CT number (in HU) at and above which pixels appear white is ____.
Numerical answer — type the value.
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Answer: 240
The window spans L ± W/2 = 40 ± 200, i.e. −160 to +240 HU, so everything at +240 HU and above is white and everything at −160 HU and below is black. Adding the whole width, 40 + 400 = 440, is the usual slip.Using γ̄ = 42.58 MHz/T for ¹H, the Larmor frequency (in MHz) in a 1.5 T scanner, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 63.87
f = γ̄B₀ = 42.58 × 1.5 = 63.87 MHz. Using γ = 2π × 42.58 = 267.5 × 10⁶ rad s⁻¹ T⁻¹ and quoting the result as hertz gives 401 MHz, which is an angular frequency, not f.A slice-selection gradient of 10 mT/m is applied with an RF pulse of bandwidth 1 kHz. Using γ̄ = 42.58 MHz/T, the slice thickness (in mm), to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 2.35
The frequency changes by γ̄G = 42.58 × 10⁶ × 0.01 = 425 800 Hz per metre, so a 1000 Hz band covers Δz = 1000/425 800 m = 2.35 mm. A stronger gradient or a narrower bandwidth gives a thinner slice; using γ = 2πγ̄ gives 0.37 mm.The wavelength (in mm) of 5 MHz ultrasound in soft tissue (c = 1540 m/s), to three decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.308
λ = c/f = 1540/(5 × 10⁶) = 3.08 × 10⁻⁴ m = 0.308 mm, which sets the scale of the achievable resolution. Using the speed of sound in air (about 343 m/s) gives 0.069 mm, which does not apply in tissue.An echo returns 100 µs after the pulse is transmitted. Taking c = 1540 m/s, the depth of the reflector (in cm) is ____.
Numerical answer — type the value.
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Answer: 7.7
d = ct/2 = 1540 × 100 × 10⁻⁶/2 = 0.077 m = 7.7 cm. Forgetting that the pulse travels there and back gives 15.4 cm.A 5 MHz Doppler beam insonates blood flowing at 0.5 m/s at 60° to the beam (c = 1540 m/s). The Doppler shift (in Hz), rounded to the nearest whole number, is ____.
Numerical answer — type the value.
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Answer: 1623
f_D = 2f₀v cos θ/c = 2 × 5 × 10⁶ × 0.5 × 0.5/1540 = 2.5 × 10⁶/1540 = 1623 Hz, conveniently in the audible range. Dropping the 2 gives 812 Hz; using sin 60° instead of cos 60° gives 2812 Hz.An ultrasound scanner fires pulses at a repetition frequency of 5 kHz. Taking c = 1540 m/s, the maximum unambiguous imaging depth (in cm) is ____.
Numerical answer — type the value.
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Answer: 15.4
Each echo must return within the pulse interval 1/PRF = 200 µs, so d_max = c/(2·PRF) = 1540/10 000 = 0.154 m = 15.4 cm. Omitting the 2 gives 30.8 cm, which would let deep echoes arrive after the next pulse and appear falsely shallow.At a boundary between fat (Z = 1.38 MRayl) and muscle (Z = 1.70 MRayl), the fraction of incident ultrasound intensity reflected (in %), to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 1.08
R = [(Z₂ − Z₁)/(Z₂ + Z₁)]² = (0.32/3.08)² = (0.1039)² = 0.0108 = 1.08%, so about 99% of the energy continues deeper. Leaving out the square gives 10.4%, which is the amplitude, not intensity, reflection coefficient.In positron emission tomography, what is detected?
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Answer: A — two 511 keV photons travelling in nearly opposite directions, detected in coincidence
A positron travels only a millimetre or so before annihilating with an electron; the pair of 511 keV photons flies off about 180° apart and the coincidence defines a line of response. A single 140 keV photon through a collimator is SPECT with technetium-99m, and transmitted X-rays are CT.A T2-weighted spin-echo image is obtained with
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Answer: A — long TR and long TE
A long TR lets every tissue recover its longitudinal magnetisation, removing T1 differences, and a long TE lets tissues with different T2 dephase by different amounts, so contrast reflects T2. Short TR with short TE is T1-weighting, long TR with short TE is proton-density weighting, and short TR with long TE gives little signal of either kind.Which statements about diagnostic X-ray imaging are correct?
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Answer: A — raising the tube kVp raises the maximum photon energy; B — the mAs sets the number of photons and so the image noise; C — Compton scatter dominates in soft tissue over most of the diagnostic range and lowers contrast
The bremsstrahlung endpoint in keV equals the kVp; the tube current-time product sets how many photons reach the detector, and noise falls as their square root; Compton scatter dominates in soft tissue and fogs the image. Photoelectric absorption falls steeply with energy, roughly as 1/E³, which is why low kVp gives more contrast.Which statements comparing PET and SPECT are correct?
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Answer: A — PET locates events by coincidence detection and needs no lead collimator; B — SPECT images single-photon emitters such as technetium-99m through a collimator; C — PET resolution is limited partly by positron range and photon non-collinearity
Coincidence of the two annihilation photons is PET’s electronic collimation, while SPECT relies on a physical collimator for single photons. The positron’s travel before annihilation and the small deviation from 180° blur PET images. A collimator does not focus anything — it absorbs almost every photon not travelling along its holes, which is why SPECT sensitivity is lower.A dose of 400 MBq of an ¹⁸F tracer (physical half-life 110 min) is prepared. Ignoring biological clearance, its activity (in MBq) 220 minutes later is ____.
Numerical answer — type the value.
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Answer: 100
220 min is two half-lives, so A = 400 × (½)² = 100 MBq. Subtracting half the original activity twice, linearly, would reach zero; decay is exponential, halving the current amount each half-life.In helical CT the table advances 15 mm per gantry rotation with a total beam collimation of 10 mm. The pitch is
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Answer: A — 1.5
Pitch = table feed per rotation/beam collimation = 15/10 = 1.5, meaning gaps between successive turns of the helix that interpolation fills, with lower dose. Inverting the ratio gives 0.67, which would be an overlapping helix.An X-ray tube operates at 100 kVp. The minimum wavelength in its spectrum (in pm), to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 12.4
The most energetic photon carries the full 100 keV, and λ_min = hc/E = 1.24 nm·keV/100 keV = 0.0124 nm = 12.4 pm (the Duane–Hunt limit). Using the mean spectral energy, roughly a third to a half of the peak, would give a longer wavelength that is not the minimum.