EL-GY 6813/BE-GY 6203 Medical Imaging, Fall 2016 Final Exam

Transcription

1 EL-GY 6813/BE-GY 6203 Medical Imaging, Fall 2016 Final Exam (closed book, 1 sheets of notes double sided allowed, no calculator or other electronic devices allowed) 1. Ultrasound Physics (15 pt) A) (9 pts, 3 pts each) Answer the following questions related to ultrasound physics: i) How is acoustic impedance (Z) related to the speed of sound (c) and density (ρ) of the tissue? How does the acoustic impedance of the tissues on either side of an interface (Z1 and Z2) affect the reflection of ultrasound at that interface? Define the amplitude reflectance coefficient in terms of Z1 and Z2. Answer: Z = c x ρ. The amplitude of the reflected ultrasound pulse is proportional to the difference of the acoustic impedances Z1-Z2. (Amplitude) R = Z1-Z2 /(Z1+Z2); [ Optional: (Intensity) R = { Z1-Z2 /(Z1+Z2)} 2.] ii) How is the ultrasound Doppler shift frequency (fd) related to blood velocity (v) and the angle between the blood flow direction and the ultrasound beam? Answer: fd is directly proportional to v. fd = 2 f0 v cosθ / c, where f0 is the transmitted frequency, θ is the angle between the blood flow direction and the ultrasound beam, and c is the speed of sound in the blood iii) Briefly describe how absorption and scattering of the ultrasound beam give rise to the phenomenon known as attenuation. How does attenuation depend on the ultrasound frequency? Answer: Attenuation is the reduction of the amplitude of the received ultrasound signal (compared to the transmitted signal) due to absorption (tissue heating) and scattering (ultrasound directed away from the transducer). The attenuation of ultrasound depends (approximately) as a linear function of frequency. [Extra points if they mention that the attenuation can be modeled approximately as 1dB/cm/MHz, i.e., the signal is reduced one db per cm at 1 MHz, 20 db per cm at 20 MHz, etc ] B) (4pt, 1 pt each) Which of the following parameters limit the ultrasound image frame rate (acquisition rate). Answer yes (Y) or no (N) for each: i) Speed of sound of tissue ii) Number of lines per image iii) Attenuation coefficient iv) Maximum propagation length/depth in image Answer: i,ii,iv=y; iii=n C) (2 pt) For a 100-line image, what is the maximum image frame rate if the (maximum) propagation depth is 15-mm? Show your work in this calculation. (Assume that the speed of sound is 1500 m/s = 1.5 mm/µs) : The 2-way propagation is 2 x 15 = 30-mm Time per line = 30/1.5 = 20 µs Time per image (100 lines) = 100 x 20 = 2000 µs = 2 ms Frame rate = 1image/(2x10-3 s) = 500 images per s 2. Ultrasound Imaging Modes (5pt, 1 pt each) Match the image mode (i-v) with the best description of the voxel/pixel in the image (a-e). Assume that (x,y,z) are spatial dimensions, t is time, v is (tissue) velocity. Choose one answer per mode: i) A-mode a) (x,y) ii) B-mode b) (x,t) iii) M-mode c) (x,y,v) iv) 3-D d) (x)

2 v) Color Doppler e) (x,y,z) Answers: i-d; ii-a; iii-b; iv-e; v-c (give 1 pt for each exactly correct answer. 0 pt for incorrect answer) 3. Ultrasound Transducers (5pt, 1 pt each) Consider the single-element ultrasound transducer shown in the diagram. Match each of the transducer properties (i-v) to the most relevant description of the property (a-e). Choose one answer per transducer property (λ is the wavelength of ultrasound): f-number = f/2a FWHM = 1.41Af-number i) Piezoelectric element a) Increase transmission into tissue ii) Matching layer b) Damp reverberations in transducer iii) Backing material c) Transmitter / receiver iv) Axial resolution d) f-number x λ v) Lateral resolution e) Nλ/2 (N= number of cycles in pulse) Answers: i-c; ii-a; iii-b; iv-e; v-d 4. (12 pt, 3pt each) T 2 relaxation describes the decay of the transverse component of magnetization (M ) due to dephasing according to the equation: a) If M (0) is the initial value of the transverse magnetization, immediately after an RF excitation), what is the solution of the equation above? b) What is the difference between T 2 and T 2 * relaxation? c) Plot M (t) vs. t for both the cases of T 2 and T 2 * relaxation d) If you want an image with T 2 -weighted contrast, would you use a gradient echo (GRE) or spin-echo (SE) sequence? Why? a) b) T 2 relaxation is the irreversible dephasing among spins due to microscopic interactions with neighboring molecules and nuclei. When there are also macroscopic effects, like magnetic field differences due, for example, to B 0 inhomogeneity and susceptibility variations among tissues, which contribute to the dephasing of the spins, the relaxation is faster and it s called T 2 *. c) For both cases the plot should be an exponential decay starting at M (0) for t = 0. The decay is faster for the case of T 2 *. d) SE, as the refocusing pulse undo the effect of the magnetic field differences due to macroscopic effects.

3 5. (10 pt, 2 pt each) a) Complete the table below to show what choice of TE/TR correspond to T1-weighted, T 2 -weighted, or Proton-Density-weighted contrast (one of the choice is not useful), in the case of a spin-echo pulse sequence. Explain your answers. Long TE Short TE Long TR Short TR b) Given that gray matter has T2 ~ 77ms and T1 ~ 760 ms, whereas cerebral spinal fluid (CSF) has T2 ~ 280 ms and T1 ~ 2650 ms, indicate which of the following two images is T1-weighted and which is T2-weighted. Explain your answers. # CSF# # gray# matter# # # a) Long TR Short TR Long TE 1) T2-weighted 2) Not useful Short TE 3) Proton-Density 4) T1-weighted 1) Short TE à minimize dephasing; long TR à no signal saturation. The combination of the two results in maximum signal from all tissues 2) Short TE à minimize dephasing; short TR à signal of tissues with long enough T 1 will be saturated. The combination of the two allows creating a contrast that depends on the T 1 of tissues. 3) Long TE à spins of tissues with long enough T 2 will diphase (i.e. no signal); long TR à no signal saturation. The combination of the two allows creating a contrast that depends on the T 2 of tissues (or T 2 * for gradient-echo sequences) 4) Long TE à spins of tissues with long enough T 2 will diphase (i.e. no signal); short TR à signal of tissues with long enough T 1 will be saturated. The combination of the two is not useful because the contrast will be a mix of T 1 and T 2 effects.

4 b) The image on the left is T 2 -weighted, the one on the right is T 1 -weighted. As CSF has long T 1, its signal is saturated by the use of a short TR and therefore it appears dark in T 1 -weighted image. As CSF has a long T 2, it will appear brighter than other tissues with smaller T 2 (therefore faster dephasing) on T 2 -weighted images, which use long TE values. 6. (12 pt) Match the essential components of an MRI system listed in the left column with one or more descriptions on the right column: A) excite the spins B) polarize the spins i) Gradient and shim coils C) encode spatial information ii) RF surface coils iii) Magnet D) incorporated in the MR system E) detect emitted signal F) compensate for B 0 inhomogeneities (give 2 pts for each exactly correct answer. 0 pt for incorrect answer) ii) A) excite the spins i) Gradient and shim coils iii) B) polarize the spins ii) RF surface coils i) ii) C) encode spatial information iii) Magnet iii) i) D) incorporated in the MR system ii) E) detect emitted signal i) F) compensate for B 0 inhomogeneities 7. (15 pt, 3pt each) a) Explain how frequency encoding works in MRI. b) What parameter of frequency encoding controls the extension of the corresponding imaging field of view? c) How is aliasing normally avoided in the frequency encoding direction? d) Why avoiding aliasing in the phase encoding direction can be more challenging? e) What imaging parameters can be adjusted to control the thickness of the selected slice? a) A magnetic field gradient is applied during the data acquisition. The gradient alters the Larmor frequency ω L of the spins in a spatially-dependent manner. The frequency of the signal emitted by each spin will therefore depend on its location along the direction of the gradient. The frequency thus provides a label to identify the spins location. b) It s ΔK x, which is the step size between samples. c) The scanner automatically adds a 2-fold oversampling (i.e., sampling with half the step size) along the frequency encoding direction and uses an antialiasing band-pass filter on the data. The oversampling does not lengthen scan time, since it can be achieved in the same time by sampling faster. d) Increasing the imaging FOV y to avoid aliasing requires a smaller ΔK y, therefore more k-space lines to achieve the same spatial resolution. However, every phase encoding step (K y line) requires an extra repetition of the pulse sequence between two RF pulses, i.e., an extra TR of scan duration. Before adding such phase oversampling and lengthen scan time, one should try first to swap frequency and phase encoding directions to make sure phase encoding occurs along the shortest dimension of the anatomy of interest. e) The bandwidth of the RF pulse and the strength of the slice selection gradient. 8. (16 pt, 4 pt each) a) What is the source of the contrast in fmri? Is it measured based on T2, T2* or T1 relaxation? b) What is a typical pulse sequence used in fmri and why? c) Describe a representative functional MRI (fmri) experiment, including as many steps as possible (from patient set up to image acquisition) and how fmri maps are generated. d) What factors limits the spatial resolution of fmri maps? What source of error is associated with that?

5 a) BOLD effect: deoxygenated blood is diamagnetic. The local change in B 0 induced by this effect changes the T 2 *, therefore T 2 * relaxation time can be used to measured it. fmri relies on the fact that oxygenated blood has different magnetic properties from de-oxygenated blood (optional: the latter is paramagnetic therefore distorts locally the magnetic field causing a signal loss). Therefore, blood oxygenation can be imaged with MRI as a local change in T 2 * and used to map areas of neuronal activation in the brain. b) Typically an Echo Planar Imaging (EPI) pulse sequence is used because it allows to acquire an entire image (all k-space) in a single shot, providing the necessary temporal resolution. c) IF a student included steps 3,4,5, give full credits. 1. Prepare subject Consent form Safety screening Instructions about the tasks to be completed during the scan (if any) 2. Shimming putting a body in the magnetic field makes the field non-uniform adjust 3 orthogonal weak magnets to make the magnetic field as homogenous as possible (Shimming is important because any inhomogeneity would cause a change in T 2 * indistinguishable from the BOLD signal) 3. Anatomical images high-resolution images covering the whole brain Take about 5 minutes 4. Functional T 2 * images Lower resolution images Fast acquisition (e.g., all volume in few seconds) The acquisition is repeated multiple times (4D dataset). If a task is involved, a part of the time frames is acquired before the task, the remaining during the task. 5. Activation statistics The time course of the signal for a voxel (or a region of interested) is analyzed to assess whether it was different before and during the task. The statistical significance of the difference (p-value) is mapped in color and superimposed to the anatomical images. d) Noise smaller voxels have lower SNR Head motion the smaller the voxels, the more contamination head motion induces Temporal resolution the smaller the voxels, the longer it takes to acquire the same volume (Optional: vasculature) A typical source of error is partial voluming, i.e. the combination, within a single voxel, of signal contributions from two or more distinct tissue types or functional regions. 9. (2pt) Name two features why PET imaging is considered non-invasive. Any two from the following: a. Most positron emitters such as C-11, F-18, N-13 or O-l8 are elements of life. Thus we can prepare structurally identical drugs or endogenous molecules and inject them into the living humans or animals without disturbing the biological systems. b. PET tracers are injected in minute amounts; e.g., microgram and nanomole level, but able to provide detailed pictures of what is happening inside the body at the molecular and cellular level. Thus, the procedures are safe and will not interfere with the biological function or metabolism inside the body. c. Most positron emitters have very short half-life. They will decay to stable isotope and background level after a short period

6 of time (several half-lives). 10. (5pt) Why can PET be used to study both drug PK and drug PD, but SPECT can only used for PD study? Positron emitters such as C-11, F-18, N-13, or O-18 are elements of life. They can be used to prepare structurally identical drugs. The labeled drug itself can be used to measure the absolute uptake, regional distribution, kinetics at its site of action in the body, and the metabolism of the drug (drug PK). In the case that a specific labeled tracer and a drug bind to the same molecular targets or relevant targets, the labeled tracer can be used to assess the effects of the drug on particular physiological or functional processes (drug PD). SPECT tracers, such as I-123 or Tc-99m, are not elements of life, and most of drugs do not contain I or Tc, thus, SPECT can only be used to study PD, not PK. 11. (3pt) For the most commonly used short-lived positron emitter, Fluorine-18, provide (a) its half-life, (b) the nuclear equation describing its production via cyclotron from oxygen-18 water, and (c) the stable nuclide it decays to.!"!!" Answer: (a) 110min, (b)! O +! P! F +!!"!"!! n, (c)! F! O +! e