NMR Measurement of T1T2 Spectra with Partial Measurements using Compressive Sensing


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1 NMR Measurement of T1T2 Spectra with Partial Measurements using Compressive Sensing Alex Cloninger Norbert Wiener Center Department of Mathematics University of Maryland, College Park
2 Outline Basics of Nuclear Magnetic Resonance 1 Basics of Nuclear Magnetic Resonance 2 3 Using Compressive Sensing to Speed Up Data Collection 4
3 Outline Basics of Nuclear Magnetic Resonance 1 Basics of Nuclear Magnetic Resonance 2 3 Using Compressive Sensing to Speed Up Data Collection 4
4 Basics of Nuclear Magnetic Resonance We will consider multidimensional correlations as measured by NMR Specifically look at T1T2 relaxation properties Relaxation measures how fast magnetic spins forget orientation and return to equalibrium T1 is decay constant for zcomponent of nuclear spin T2 is decay constant for xycomponent perpendicular to magnetic field Analysis of multidimensional correlations requires multidimensional inverse Laplace transform Due to illconditioning of 2D ILT, require a large number of measurements Problem is getting T1T2 map is incredibly slow
5 SpinEcho Pulse Figure: Animation of SpinEcho Pulse
6 T1T2 Correlations
7 Math Behind NMR Echo measurements are related to T1T2 correlations via Laplace Transform M(τ 1, τ 2 ) = (1 2e τ 1/T 1 )e τ 2/T 2 F(T 1, T 2 )dt 1 dt 2 + E(τ 1, τ 2 ) We ll consider more general 2D Fredholm Integral M(τ 1, τ 2 ) = k 1 (τ 1, T 1 )k 2 (τ 2, T 2 )F(T 1, T 2 )dt 1 dt 2 + E(τ 1, τ 2 ) where E(τ 1, τ 2 ) N (0, ɛ) Discretize to M = K 1 FK 2 + E
8 Problems with Inverse Laplace Transform k 1 (τ1, T 1 ) and k 2 (τ 2, T 2 ) are smooth continuous Means K 1 and K 2 are illconditioned Makes inversion difficult
9 Outline Basics of Nuclear Magnetic Resonance 1 Basics of Nuclear Magnetic Resonance 2 3 Using Compressive Sensing to Speed Up Data Collection 4
10 Algorithm for Approximation of F(T 1, T 2 ) Definition (Objective Function for Recovery) Wish to find minimizer to min F 0 M K 1FK α F 2 Because inversion unstable, α F 2 term smooths solution Called Tikhonov regularization with parameter α Possible to determine choice of α that minimizes bias Computationally unwieldy
11 Outline of Preexisting Algorithm 1 Reduce the dimension of the problem in order to make the minimization manageable 2 Holding alpha fixed, solve the Tikhonov regularization problem 3 Having solved the minimization problem, determine a more optimal alpha to use in the regularization problem 4 Loop between Steps 2 & 3 until convergence upon the optimal alpha
12 Step 1: Dimension Reduction Assume M is N 1 N 2 data matrix To reduce the dimension of the problem, let K 1 = U 1 Σ 1 V1, K 2 = U 2 Σ 2 V2 where Σ 1 R s 1 s 1 and Σ 2 R s 2 s 2. Choose s 1, s 2 to account for 99% of the energy Because of fast decay of kernels, s 1 N 1 and s 2 N 2.
13 Step 1: Dimension Reduction Change objective function by min F 0 U 1 (M K 1 FK 2 )U α F 2 = min F 0 U 1 MU 2 Σ 1 V 1 FV 2 Σ α F 2 = min F 0 M K 1 F K α F 2. ek i = Σ i V i em = U 1 MU 2 R s 1 s 2 Note that now minimization only depends on M
14 Step 2: Tikhonov Regularization Now need to solve minimization with compressed data M min M K 1 F K α F 2. F 0 Reform problem into unconstraned optimization for f = max(0, ( K 1 K 2 ) c), c R s 1 s 2. Solve new unconstrained problem using Newton s method
15 Step 3: Choosing Optimal Alpha There are multiple ways to choose the optimal alpha: 1 Choose new alpha to be α new = s1 s 2 c 2 After determining F α for fixed α, calculate the fit error χ(α) = M K 1F αk2 F, N1 N 2 1 which is the standard deviation of the noise in reconstruction. Choose α such that d log χ d log α =.1
16 Outline Basics of Nuclear Magnetic Resonance 1 Basics of Nuclear Magnetic Resonance 2 3 Using Compressive Sensing to Speed Up Data Collection 4
17 Intuition Behind Compressive Sensing Collected N 1 N 2 data points in M, then compressed to s 1 s 2 matrix M. s In practice, 1 s 2 1% N 1 N 2 Key Question Why collect large amounts of data only to throw away over 99% of it? At end of day, only M matters in minimization problem If we had way of measuring e M directly, speedup would be massive
18 Basic ideas Since postsensing compression is inefficient, why not compress while sensing to boost efficiency? Key is that observations in traditional sensing are of form M i,j = M, δ i,j. Question: which functions should replace δ i,j in order to minimize the number of samples needed for reconstruction? Answer: they should not match image structures; they should mimic the behavior of random noise.
19 Incoherent Measurements of M Remember that M = U 1 MU 2, so M i,j = u i Mv j = M, u i v j. where u i is the i th row of U 1 and v j is the j th row of U 2 By structure of U 1 and U 2, rows u i and v i are very incoherent Incoherent basically means energy is spread out over all elements rather than concentrated Because of this, may be possible to observe small number of entries and recover M Even if number of measurements greater than size of e M, still could be much less than N 1 N 2
20 Random Sensing Consider observing a small number of elements of M randomly We will call set of elements observed Ω Fewer measurements directly reduces time spent collecting Problem is using these measurements to reconstruct e M
21 Singular Values and Rank of M Measurements of e M are underdetermined and noisy. Need a way to bring in more a priori knowledge of e M Because of illconditioned K 1 and K 2, singular values of e M decay quickly Can be closely approximated by only first r min(s 1, s 2 ) singular values Possible to recover M as solution to arg min rank(x) X such that u i Xv j = M i,j, (i, j) Ω
22 Problems and Reformulation Two problems: Rank minimization is computationally infeasible Definition (Nuclear Norm) Let σ i (M) be the i th largest singular value of M. If rank(m) = r, then M := rx σ i (M) i=1 Measurements aren t perfect (there is noise) Definition (Sampling Operator) For a set of measurements {φ i } i J, then for Ω J the sampling operator is R Ω : C s 1 s 2 C m, (R Ω (X)) j = φ j, X, j Ω
23 Problems and Reformulation Recovery Algorithm Let y = R Ω ( e M) + e, where e 2 < ɛ. Wish to recover e M by solving arg min X R s 1 s 2 such that X R Ω (X) y 2 < ɛ (P*) Definition Need to establish that (P*) with these measurements will recover e M A Parseval tight frame for a d dimensional Hilbert space H is a collection of elements {φ j } j J H such that X f, φ j 2 = f 2, f H. (1) j J
24 Reconstruction and Noise Bounds Because u i and v j are rows of U 1 and U 2 (who s columns are orthogonal), these measurements form a tight frame Definition A Fouriertype Parseval tight frame with incoherence µ is a Parseval tight frame {φ j } j J on C d d with operator norm satisfying φ j 2 ν d, j J. (2) J This definition was introduced by David Gross for orthonormal basis Can be thought of as singular values bounded in L norm
25 Theorem (Cloninger) Let the measurements in (P*) be a Fouriertype Parseval tight frame of size J, and the true matrix of interest be denoted M 0. If the number of measurements m satisfies m Cνnr log 5 n log J, where ν measures the incoherence of the measurements and n = max(s 1, s 2 ), then the result of (P*), denoted M, will satisfy M 0 M F C 1 M 0 M 0,r r + C 2 J m ɛ, where M 0,r is the best rank r approximation of M 0.
26 Noise in Practice Keep in mind, ɛ is the measure of noise between the ideal matrix M 0 and the compressed data M M 0 M F ɛ C 2 is a very small constant, and in practice N 1 N 2 m 5
27 Proof of Theorem Definition Let R Ω : R s 1 s 2 R m be a linear operator of measurements. A satisfies the restricted isometry property (RIP) of order r if there exists some small δ r such that for all matrices X of rank r. (1 δ r ) X F R Ω (X) 2 (1 + δ r ) X F Theorem (Candès, Fazel) Let X 0 be an arbitrary matrix in C m n and assume δ 5r < 1/10. If the measurements satisfy RIP, then b X obtained from solving (P*) obeys b X X 0 F C 0 X 0 X 0,r r + C 1 ɛ, where C 0, C 1 are small constants depending only on the isometry constant.
28 Proof of Theorem Lemma Let R Ω be defined as before. Fix some 0 < δ < 1. Let m satisfy m Cνrn log 5 n log J, (3) where C only depends on δ like C = O ( 1/δ 2). Then with high J probability, m R Ω satisfies the RIP of rank r with isometry constant δ. Furthermore, the probability of failure is exponentially small in δ 2 C. J Since, m R Ω satisfies the RIP, can reform (P*) to say arg min such that J m R Ω(X) X J m y 2 < J m ɛ
29 Outline of Proof Can restate RIP as ɛ r (A) = sup X U 2 X, (A A I)X 2δ δ 2 where U 2 = {X C s 1 s 2 : X F 1, X r X F } Notation: Norm: M (r) = sup X U 2 X, MX ; Simplify Terms: A = q J m R Ω Eɛ r (A) = E A A I (r) X E Ω E ɛ ɛi (φ i φ i (φ i ) φ i ) J m (r) 2 n r r m E X J J ΩE ɛ ɛi n φ i φ i n (r)
30 Outline of Proof Lemma Let {V i } m i=1 Cs 1 s 2 have uniformly bounded norm, V i K. Let n = max(s 1, s 2 ) and let {ɛ i } m i=1 be iid uniform ±1 random variables. Then m X E ɛ ɛ i Vi m X 1/2 V i C 1 Vi V i i=1 (r) i=1 (r) where C 1 = C 0 rk log 5/2 n log 1/2 m and C 0 is a universal constant. q J For our purposes, V i = n φ i. Then n Eɛ r (A) 2C 1 m E X r r J J 1/2 Ω n φ i φ i n (r) = 2C 1 r n m (E A A ) 1/2 2C 1 r n m (Eɛr (A) + 1)1/2.
31 Outline of Proof Fix some λ 1 and choose This makes Eɛ r (A) 1 λ + 1 λ. We can now use result by Talagrand. Theorem m Cλµrn log 5 n log J λn(2c 1 ) 2 Let {Y i } m i=1 be independent symmetric random variables on some Banach space such P that Y i R. Let Y = m Y i. Then for any integers l q and any t > 0 i=1 Pr( Y 8qE Y + 2Rl + te Y ) (K /q) l + 2e t2 /256q, (4) where K is a universal constant. Use theorem to establish Pr( ɛ r (A) ɛ) e Cɛ2 λ
32 Revised Algorithm 1 Randomly subsample data matrix M at indices Ω 2 Reconstruct M by solving arg min such that X R Ω (X) y 2 < ɛ 3 Holding alpha fixed, solve the Tikhonov regularization problem 4 Having solved the minimization problem, determine a more optimal alpha to use in the regularization problem 5 Loop between Steps 2 & 3 until convergence upon the optimal alpha
33 Outline Basics of Nuclear Magnetic Resonance 1 Basics of Nuclear Magnetic Resonance 2 3 Using Compressive Sensing to Speed Up Data Collection 4
34 Recovery from Small Number of Entries
35 Error Analysis
36 Simple T1T2 Map with 30dB SNR Figure: (TL) T1T2 Map, (TR) Original Algorithm, (BL) 30% Measurements, (BR) 10% Measurements
37 Simple T1T2 Map with 15dB SNR Figure: (TL) T1T2 Map, (TR) Original Algorithm, (BL) 30% Measurements, (BR) 10% Measurements
38 Correlated T1T2 Map with 30dB SNR Figure: (TL) T1T2 Map, (TR) Original Algorithm, (BL) 30% Measurements, (BR) 10% Measurements
39 Two Peak T1T2 Map with 35dB SNR Figure: (TL) Original Algorithm, (TR) 20% Measurements, (BL) 1D ILT of T1, (BR) 1D ILT of T2
40 Two Peak T1T2 Map with 35dB SNR Figure: (TL) T1T2 Map, (TR) Original Algorithm, (BL) 30% Measurements, (BR) 20% Measurements
41 Two Peak Skewing Only keeping largest singular values causes fattening of peaks Necessary for computational efficiency of any 2D ILT algorithm Can be shown by taking 1D ILT of M, U 1 U 1 MU 2U 2, and reconstructions of M using CS
42 Preliminary Real Data Figure: (L) Original Algorithm, (R) 20% Measurements
43 Conclusion recover measurement matrix M almost perfectly While small noise is added in compressive sensing reconstruction, it is dwarfed by algorithmic errors Next step is to implement on machine and run thorough testing
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