A source trajectory differential image reconstruction method for linear CT scanning of a radiation source

Abstract

A source trajectory differential image reconstruction method for linear CT scanning of a ray source, including two-dimensional and three-dimensional methods, comprising the following steps: S1. initializing parameters i=1, the number of linear scanning segments T and the rotation angle interval Δθ, and a zero space of an image to be reconstructed; S2. obtaining a projection of the i-th linear scanning of the ray source; S3. pre-weighting the projection data; S4. differentiating the weighted projection along the source trajectory direction; S5. performing weighted back-projection on the differential projection to obtain the i-th Hilbert image; S6. performing finite Hilbert inverse transformation on the i-th Hilbert image along the i-th ray source trajectory to reconstruct the i-th limited angle image; S7. making i=i+1; S8. if i≤T, rotating the linear CT scanning system by (i-1)·Δθ, and jumping to S2, and sequentially circulating until i>T, and completing the reconstruction of the image. The image reconstruction method is oriented to the linear CT scanning trajectory, can directly avoid the artifacts caused by truncated projection, and reconstructs a high-quality image.

Description

Technical Field

[0001] This invention belongs to the field of CT image reconstruction, and more specifically, relates to a source trajectory differential image reconstruction method for linear CT scans oriented towards X-ray sources. Background Technology

[0002] In recent years, various CT scanning methods have emerged to meet the demand for higher resolution. Liu Fenglin and Yu Haijun et al. proposed a structure for expanding the imaging field of view in linear CT scanning with X-ray source. In this imaging geometry, the detector is fixed, the X-ray source is translated linearly along a direction parallel to the detector row, the object under test is placed at the center of the rotating stage, and an iterative image reconstruction algorithm based on TV minimization was designed [1,2]. This linear CT scanning structure with X-ray source can expand the imaging field of view by controlling the source translation and is easy to achieve high-precision control. It has great development potential in the field of CT scanning technology, and a series of image reconstruction algorithms applicable to this scanning model have been studied.

[0003] The first proposed imaging algorithm for linear CT scans using X-ray sources is an iterative image algorithm based on TV minimization. While it can reconstruct artifact-free images, it suffers from long reconstruction times and high computational costs, making it difficult to meet practical engineering needs [1,2]. Although traditional FBP (filtered backprojection) analytical reconstruction algorithms can perform high-efficiency reconstruction, in linear CT scans with expanded imaging fields of view, the acquired projection data is truncated, resulting in severe truncation artifacts in the reconstructed images obtained by filtered backprojection [2-5]. To avoid truncation artifacts when using FBP-type reconstruction algorithms, existing research employs virtual projection to achieve an equivalent globally truncation-free projection before image reconstruction [3-5]. While this method of first recombining the projection and then converting it into a virtual projection can avoid truncation artifacts, the reconstruction method is still not simple and direct enough.

[0004] References:

[0005] [1] Liu Fenglin, Yu Haijun, Li Lei, Tan Chuandong. A novel large field-of-view linear scanning CT system and image reconstruction method [P]. Chongqing: CN111839568A, 2020-10-30.

[0006] [2]H.Yu,L.Li,C.Tan,F.Liu,R.Zhou,X-ray source translation basedcomputed tomography(STCT),Optics Express,29(2021)19743-19758.R.Clackdoyle,F.Noo,A large class of inversion formulae for the 2D Radon transform of functions of compact support,Inverse Problems,20(2004)1281.

[0007] [3] Ge Wenjie, Yu Haijun, Chen Jie, et al. Analytical reconstruction of source line scanning computed tomography based on derivative-Hilbert transform-back projection [J]. Acta Optica Sinica, 2022, 42(11):292-303.

[0008] [4] Li Lei, Yu Haijun, Tan Chuandong, Duan Xiaojiao, Liu Fenglin. Research on analytical reconstruction algorithm of X-ray source translation scanning CT[J]. Journal of Instrumentation, 2022, 43(02):187-195.DOI:10.19650 / j.cnki.cjsi.J2108157.

[0009] [5]Yu H,Ni S,Chen J,et al.Analytical reconstruction algorithm for multiple source-translation computed tomography(mSTCT)[J].AppliedMathematical Modelling,2023,117:251-266. Summary of the Invention

[0010] In order to overcome the shortcomings of the prior art, the present invention aims to provide a source trajectory differential image reconstruction method for linear CT scanning oriented towards X-ray sources, which directly avoids truncation artifacts and reconstructs high-quality artifact-free images.

[0011] To achieve the above objectives, this invention provides two image reconstruction methods: two-dimensional and three-dimensional. The technical solution employed in the two-dimensional method is as follows:

[0012] S1. Initialize parameters i = 1, number of linear scan segments T and rotation angle interval Δθ, null space of the 2D image to be reconstructed.

[0013] S2. Obtain the projection of the i-th segment of the linear scan of the ray source.

[0014] S3. Projection of the acquired data Perform pre-weighting to obtain the weighted projection.

[0015] S4. Weighted projection Perform differentiation along the translational trajectory λ of the X-ray source;

[0016] S5. Perform a weighted back projection operation on the differential projection to obtain the i-th segment of the Hilbert image.

[0017] S6. Perform a finite inverse Hilbert transform on the i-th segment of the Hilbert image along the direction of the i-th segment of the ray source's straight line translation trajectory to reconstruct the i-th segment of the finite angle image.

[0018] S7. Execute i = i + 1 and complete the image overlay:

[0019] S8. If i ≤ T, rotate the linear CT scanning system by an angle of (i-1)·Δθ and jump to S2. Repeat this process until i > T to complete the image. The reconstruction.

[0020] Preferably, for the source trajectory differential two-dimensional image reconstruction method of linear CT scan facing the X-ray source, in step S3, the projection Pre-weighting yields weighted projection Its calculation formula is:

[0021]

[0022] Where, θ i θ is the direction angle of the linear scanning trajectory of the X-ray source. i = (i-1)·Δθ; λ represents the coordinates of the ray source on the straight trajectory, λ∈[-s,s]; u represents the coordinates of the detector, and u∈[-d,d], d represents half the length of the detector; l represents the distance from the ray source to the center of rotation along the direction perpendicular to the detector; h represents the distance from the center of rotation to the center of the detector; This is a redundant weighting factor.

[0023] Preferably, for the source trajectory differential two-dimensional image reconstruction method of linear CT scan facing the X-ray source, in step S4, the weighted projection Differential projection is obtained by performing a differentiation operation along the translation trajectory λ of the ray source. Its calculation formula is:

[0024]

[0025] in, The differential operator along the λ direction is implemented using finite difference operations. Data within the projection is obtained using central difference, while data at the projection boundary is obtained using one-sided difference; λ * Indicates passing through the point to be rebuilt The coordinates of the ray on the translational trajectory of the ray source.

[0026]

[0027] Where L represents the point to be rebuilt The perpendicular distance to the trajectory of the ray source is L = -xsinθ i +ycosθ i +l; H represents the vertical distance from the point x to be reconstructed to the detector, H = xsinθ i -ycosθ i +h.

[0028] Preferably, for the source trajectory differential two-dimensional image reconstruction method of linear CT scan facing the X-ray source, in step S5, the differential projection... The i-th Hilbert image is obtained by performing a weighted back projection operation. Its calculation formula is,

[0029]

[0030] In the formula, Indicates the point to be rebuilt; η i This represents the i-th segment of the Hilbert image. The direction angle for performing the finite Hilbert inverse transform; 1 / H 2 This represents the back-projection weighting factor; the back-projection operation is performed in a matrix larger than the image to be reconstructed, that is, adding a number of zeros p0 = 0.5 * I outside each edge of the matrix space of the image to be reconstructed, where I is the number of rows or columns of the image to be reconstructed.

[0031] Preferably, for the source trajectory differential two-dimensional image reconstruction method of linear CT scan facing the X-ray source, in step S6, the i-th segment of the Hilbert image... By performing a finite Hilbert inverse transform along the direction of the i-th segment of the ray source's linear translation trajectory, the i-th segment of the finite angle image can be reconstructed. Its calculation formula is:

[0032]

[0033] Where y1 represents the direction of the one-dimensional Hilbert transform, y1∈[L y +ε y U y -ε y], where [L y U y ] represents a finite interval of the Hilbert transform, ε y It is a small positive number; This represents the reconstruction of the i-th segment of the finite angle image. The unknown constants that need to be calculated are obtained by finding the known ones. Some locations are used as the mean of their finite Hilbert transforms. The i-th Hilbert image The specific implementation of the finite Hilbert inverse transform includes the following steps:

[0034] 1) Transfer the i-th segment of the Hilbert image Divide by -2π and rotate around the center of the image by -η. i The angle is used to obtain a rotated Hilbert image. Where η i =θ i -π / 2 (i = 1, 2, ..., T), positive angle values ​​rotate the image counterclockwise; negative angle values ​​rotate the image clockwise.

[0035] 2) Calculate the weight matrix W mat ,Right now

[0036]

[0037] 3) Search for rotated Hilbert images Find the rows in the array where all values ​​are 0, and obtain an array R whose elements are row numbers. i ;

[0038] 4) Rotating Hilbert images Weighted W mat And perform a Hilbert transform along the column direction of the image matrix to obtain the image.

[0039] 5) Based on R obtained in step S3 i Numerical values, from images Extract the element by indexing it, calculate its mean along the column direction and invert it to obtain the result.

[0040] 6) Execution

[0041] 7) Rotate the image Rotate around the image center by an angle η i Returning to the original state, we obtain

[0042] 8) Extract the middle region of the image to be reconstructed.

[0043] The technical solution adopted by the three-dimensional method provided by this invention is as follows:

[0044] S1. Initialize parameters i = 1, number of linear scan segments T and rotation angle interval Δθ, null space of the 3D image to be reconstructed.

[0045] S2. Obtain the cone-beam projection of the i-th segment of the linear scan of the X-ray source.

[0046] S3. Projection of the acquired cone beam Pre-weighting is performed to obtain the weighted cone-beam projection.

[0047] S4. Weighted cone-beam projection Perform differentiation along the λ direction of the ray source trajectory;

[0048] S5. Perform a weighted back projection operation on the differential cone-beam projection to obtain the i-th segment of the 3D Hilbert image.

[0049] S6. Perform a finite Hilbert inverse transform layer by layer along the axial direction of the linear translation trajectory of the ray source in the i-th segment of the 3D Hilbert image to reconstruct the finite angle 3D image of the i-th segment.

[0050] S7. Execute i = i + 1 and complete the overlay of the three-dimensional images:

[0051] S8. If i ≤ T, rotate the linear CT scanning system by an angle of (i-1)·Δθ and jump to S2. Repeat this process until i > T to complete the three-dimensional image. The reconstruction.

[0052] Preferably, for the source trajectory differential three-dimensional image reconstruction method of linear CT scanning facing the X-ray source, in step S3, the cone-beam projection... Pre-weighting yields weighted cone-beam projection Its calculation formula is:

[0053]

[0054] Where, θ i θ is the direction angle of the linear scanning trajectory of the X-ray source. i= (i-1)·Δθ; λ represents the coordinates of the ray source on the straight trajectory, λ∈[-s,s]; u represents the row coordinates of the panel detector, and u∈[-d,d], d represents half of the row length of the panel detector; l represents the distance from the ray source perpendicular to the direction of the panel detector to the rotation center; h represents the distance from the rotation center to the center of the panel detector. This is a redundant weighting factor.

[0055] Preferably, for the source trajectory differential three-dimensional image reconstruction method of linear CT scan facing the X-ray source, in step S4, the weighted cone-beam projection... Differential cone-beam projection is obtained by performing a differentiation operation along the λ direction of the ray source trajectory. Its calculation formula is:

[0056]

[0057] in, The differential operator along the λ direction is implemented using finite difference operations. Data within the projection is obtained using central difference, while data at the projection boundary is obtained using one-sided difference; λ * Indicates passing through the point to be rebuilt The ray's direction coordinates on the ray source trajectory λ, where v represents the point to be reconstructed. The column coordinates of the rays on the panel detector,

[0058]

[0059]

[0060] Where L represents the point to be rebuilt The perpendicular distance to the trajectory of the ray source is L = -xsinθ i +ycosθ i +l; H represents the point to be rebuilt. The vertical distance to the panel detector is H = xsinθ i -ycosθ i +h.

[0061] Preferably, for the source trajectory differential three-dimensional image reconstruction method of linear CT scan facing the X-ray source, in step S5, the differential cone-beam projection... The i-th segment of the 3D Hilbert image is obtained by performing a weighted back projection operation. Its calculation formula is:

[0062]

[0063] In the formula, Indicates the point to be rebuilt; η i This represents the i-th segment of the 3D Hilbert image. The direction angle for performing the finite Hilbert inverse transform; 1 / H 2 The back projection weighting factor is used; the back projection operation is performed in a matrix larger than the three-dimensional image to be reconstructed, that is, p0 = 0.5 * I zeros are added outside each edge of the matrix space of the three-dimensional image to be reconstructed, where I is the number of voxels in a certain dimension of the image to be reconstructed.

[0064] Preferably, for the source trajectory differential three-dimensional image reconstruction method of linear CT scan facing the X-ray source, in step S6, the three-dimensional Hilbert image of the i-th segment... A finite Hilbert inverse transform is performed layer by layer along the axial direction of the i-th segment of the linear translation trajectory of the ray source, i.e., layer by layer. k The two-dimensional Hilbert image is inversely transformed layer by layer (k = 1, 2, ..., I) to reconstruct the i-th segment of the finite-angle three-dimensional image. The inverse transform of a 2D Hilbert image is also along x j (j=1,2,...,I) is processed one-dimensionally line by line, and its calculation formula is:

[0065]

[0066] Where y1 represents the direction of the one-dimensional Hilbert transform, y1∈[L y +ε y U y -ε y ], where [L y U y ] represents a finite interval of the Hilbert transform, ε y It is a small positive number; Indicates reconstruction The unknown constants that need to be calculated are obtained by finding the known ones. Some locations are used as the mean of their finite Hilbert transforms as C. yi Regarding the i-th segment of the 3D Hilbert image The axial layer-by-layer finite Hilbert inverse transform is specifically implemented by the following steps:

[0067] 1) The i-th segment of the 3D Hilbert image Divide by -2π and rotate -η around the central axis of the 3D image. i Angle, to obtain a rotated 3D Hilbert image Where η i =θ i -π / 2 (i = 1, 2, ..., T), positive angle values ​​rotate the 3D image counterclockwise; negative angle values ​​rotate the 3D image clockwise.

[0068] 2) Calculate the one-dimensional weight sequence W mat ,Right now

[0069]

[0070] 3) for(k=1; k≤1; k++) (i.e., enter the for loop body):

[0071] 3.1) From rotated 3D Hilbert images Indexing the k-th layer 2D Hilbert image

[0072] 3.2) Searching for two-dimensional Hilbert images Find the rows in the array where all values ​​are 0, and obtain an array R whose elements are row numbers. i ;

[0073] 3.3) Two-dimensional Hilbert images Weighted W mat And perform a Hilbert transform along the column direction of the image matrix to obtain the image.

[0074] 3.4) Based on R obtained in step S3 i Numerical values, from images Extract the element by indexing it, calculate its mean along the column direction and invert it to obtain the result.

[0075] 3.5) Execution

[0076] 3.6) Rotating a two-dimensional image Rotate around the image center by an angle η i Returning to the original state, we obtain the reconstructed k-th layer 2D.

[0077] 4) Extract the middle region of the 3D image to be reconstructed, i.e.

[0078]

[0079] The beneficial effects of this invention are as follows:

[0080] Compared with existing reconstruction algorithms for linear CT scans oriented towards X-ray sources, this method can perform accurate reconstruction using truncated projection data, directly avoiding truncation artifacts and reconstructing nearly artifact-free two-dimensional or three-dimensional CT images. Attached Figure Description

[0081] Figure 1 This is a simplified flowchart of the present invention.

[0082] Figure 2This is a schematic diagram of the linear CT two-dimensional scanning trajectory of the X-ray source addressed in this invention.

[0083] Figure 3 This is a flowchart of the two-dimensional image reconstruction process of the present invention using the Shepp-Logan phantom as the test object.

[0084] Figure 4 These are two-dimensional reconstructed images of different projection numbers using the FORBILD phantom as the test object, according to the present invention.

[0085] Figure 5 This is a schematic diagram of the linear cone-beam CT three-dimensional scanning trajectory of the X-ray source addressed in this invention.

[0086] Figure 6 This is a schematic diagram of the finite inverse Hilbert transform of the three-dimensional Hilbert image along the axial direction of the ray source trajectory, according to the present invention.

[0087] Figure 7 This is a flowchart of the three-dimensional CT image reconstruction process of the present invention.

[0088] Figure 8 This invention presents the image reconstruction results using a three-dimensional Shepp-Logan phantom as the test object under different projection number conditions. Detailed Implementation

[0089] The embodiments of the present invention will now be described in detail with reference to the accompanying drawings and examples.

[0090] A method for source trajectory differential image reconstruction in linear CT scans oriented towards X-ray sources, such as... Figure 1 As shown, its two-dimensional process includes the following steps:

[0091] S1. Initialize parameters i = 1, number of linear scan segments T and rotation angle interval Δθ, null space of the 2D image to be reconstructed.

[0092] S2. Obtain the projection of the i-th segment of the linear scan of the ray source.

[0093] S3. Projection of the acquired data Perform pre-weighting to obtain the weighted projection.

[0094] S4. Weighted projection Perform differentiation along the translational trajectory λ of the X-ray source;

[0095] S5. Perform a backprojection operation on the differential projection to obtain the i-th segment of the Hilbert image.

[0096] S6. Perform a finite inverse Hilbert transform on the i-th segment of the Hilbert image along the direction of the i-th segment of the ray source's straight line translation trajectory to reconstruct the i-th segment of the finite angle image.

[0097] S7. Execute i = i + 1 and complete the image overlay:

[0098] S8. If i ≤ T, rotate the linear CT scanning system by an angle of (i-1)·Δθ and jump to S2. Repeat this process until i > T to complete the image. The reconstruction.

[0099] Preferably, for the source trajectory differential two-dimensional image reconstruction method of linear CT scan facing the X-ray source, in step S3, the projection Pre-weighting yields weighted projection Its calculation formula is:

[0100]

[0101] Where, θ i θ is the direction angle of the linear scanning trajectory of the X-ray source. i = (i-1)·Δθ; λ represents the coordinates of the ray source on the straight trajectory, λ∈[-s,s]; u represents the coordinates of the detector, and u∈[-d,d], d represents half the length of the detector; l represents the distance from the ray source to the center of rotation along the direction perpendicular to the detector; h represents the distance from the center of rotation to the center of the detector; This is a redundant weighting factor.

[0102] Preferably, for the source trajectory differential two-dimensional image reconstruction method of linear CT scan facing the X-ray source, in step S4, the weighted projection Differential projection is obtained by performing a differentiation operation along the translation trajectory λ of the ray source. Its calculation formula is:

[0103]

[0104] in, The differential operator along the λ direction is implemented using finite difference operations. Data within the projection is obtained using central difference, while data at the projection boundary is obtained using one-sided difference; λ * Indicates passing through the point to be rebuilt The coordinates of the ray on the translational trajectory of the ray source.

[0105]

[0106] Where L represents the point to be rebuilt The perpendicular distance to the trajectory of the ray source is L = -xsinθ i+ycosθ i +l; H represents the vertical distance from the point x to be reconstructed to the detector, H = xsinθ i -ycosθ i +h.

[0107] Preferably, for the source trajectory differential two-dimensional image reconstruction method of linear CT scan facing the X-ray source, in step S5, the differential projection... The i-th Hilbert image is obtained by performing a weighted back projection operation. Its calculation formula is,

[0108]

[0109] In the formula, Indicates the point to be rebuilt; η i This represents the i-th segment of the Hilbert image. The direction angle for performing the finite Hilbert inverse transform; 1 / H 2 This represents the back-projection weighting factor; the back-projection operation is performed in a matrix larger than the image to be reconstructed, that is, adding a number of zeros p0 = 0.5 * I outside each edge of the image matrix space to be reconstructed, where I is the number of rows or columns of the image matrix to be reconstructed.

[0110] Preferably, for the source trajectory differential two-dimensional image reconstruction method of linear CT scan facing the X-ray source, in step S6, the i-th segment of the Hilbert image... By performing a finite Hilbert inverse transform along the direction of the i-th segment of the ray source's linear translation trajectory, the i-th segment of the finite angle image can be reconstructed. Its calculation formula is:

[0111]

[0112] Where y1 represents the direction of the one-dimensional Hilbert transform, y1∈[L y +ε y U y -ε y ], where [L y U y ] represents a finite interval of the Hilbert transform, ε y It is a small positive number; This represents the reconstruction of the i-th segment of the finite angle image. The unknown constants that need to be calculated are obtained by finding the known ones. Some locations are used as the mean of their finite Hilbert transforms. The i-th Hilbert image The specific implementation of the finite Hilbert inverse transform includes the following steps:

[0113] 1) Transfer the i-th segment of the Hilbert image Divide by -2π and rotate around the center of the image by -η. i The angle is used to obtain a rotated Hilbert image. Where η i =θ i -π / 2 (i = 1, 2, ..., T), positive angle values ​​rotate the image counterclockwise; negative angle values ​​rotate the image clockwise.

[0114] 2) Calculate the weight matrix W mat ,Right now

[0115]

[0116] 3) Search for rotated Hilbert images Find the rows in the array where all values ​​are 0, and obtain an array R whose elements are row numbers. i ;

[0117] 4) Rotating Hilbert images Weighted W mat And perform a Hilbert transform along the column direction of the image matrix to obtain the image.

[0118] 5) Based on R obtained in step S3 i Numerical values, from images Extract the element by indexing it, calculate its mean along the column direction and invert it to obtain the result.

[0119] 6) Execution

[0120] 7) Rotate the image Rotate around the image center by an angle η i Returning to the original state, we obtain

[0121] 8) Extract the middle region of the image to be reconstructed.

[0122] Figure 2 The diagram shows the first two segments of the linear CT scan trajectory of the X-ray source, corresponding to the first and second segments of the linear X-ray source scan in this invention.

[0123] Figure 3 The image reconstruction process of the present invention is demonstrated when the Shepp-Logan phantom is used as the test object, confirming the effectiveness of the method and demonstrating its ability to reconstruct artifact-free images.

[0124] Furthermore, a FORBILD phantom with a pixel count of 512×512 and a size of 8.4mm×8.4mm was used. The image reconstruction effect of this invention was evaluated through numerical experiments. The projection number N of each segment of the linear CT scan from the X-ray source was set to 50, 100, 200, 400, 600, and 800. The orthographic projection of the FORBILD phantom was simulated and reconstructed using this invention. Figure 4 The image reconstruction results shown demonstrate that this method can reconstruct high-quality images without artifacts using the truncated projection data during linear CT scanning of the X-ray source, and the image reconstruction quality increases with the increase of the number of projections.

[0125] Table 1 Numerical simulation test parameters

[0126]

[0127] A method for source trajectory differential image reconstruction in linear CT scans oriented towards X-ray sources, such as... Figure 1 As shown, its three-dimensional method flow includes the following steps:

[0128] S1. Initialize parameters i = 1, number of linear scan segments T and rotation angle interval Δθ, null space of the 3D image to be reconstructed.

[0129] S2. Obtain the cone-beam projection of the i-th segment of the linear scan of the X-ray source.

[0130] S3. Projection of the acquired cone beam Pre-weighting is performed to obtain the weighted cone-beam projection.

[0131] S4. Weighted cone-beam projection Perform differentiation along the λ direction of the ray source trajectory;

[0132] S5. Perform a weighted back projection operation on the differential cone-beam projection to obtain the i-th segment of the 3D Hilbert image.

[0133] S6. Perform a finite Hilbert inverse transform layer by layer along the axial direction of the linear translation trajectory of the ray source in the i-th segment of the 3D Hilbert image to reconstruct the finite angle 3D image of the i-th segment.

[0134] S7. Execute i = i + 1 and complete the overlay of the three-dimensional images:

[0135] S8. If i ≤ T, rotate the linear CT scanning system by an angle of (i-1)·Δθ and jump to S2. Repeat this process until i > T to complete the three-dimensional image. The reconstruction.

[0136] Preferably, for the source trajectory differential three-dimensional image reconstruction method of linear CT scanning facing the X-ray source, in step S3, the cone-beam projection... Pre-weighting yields weighted cone-beam projection Its calculation formula is:

[0137]

[0138] Where, θ i θ is the direction angle of the linear scanning trajectory of the X-ray source. i = (i-1)·Δθ; λ represents the coordinates of the ray source on the straight trajectory, λ∈[-s,s]; u represents the row coordinates of the panel detector, and u∈[-d,d], d represents half of the row length of the panel detector; l represents the distance from the ray source perpendicular to the direction of the panel detector to the rotation center; h represents the distance from the rotation center to the center of the panel detector. This is a redundant weighting factor.

[0139] Preferably, for the source trajectory differential three-dimensional image reconstruction method of linear CT scan facing the X-ray source, in step S4, the weighted cone-beam projection... Differential cone-beam projection is obtained by performing a differentiation operation along the λ direction of the ray source trajectory. Its calculation formula is:

[0140]

[0141] in, The differential operator along the λ direction is implemented using finite difference operations. Data within the projection is obtained using central difference, while data at the projection boundary is obtained using one-sided difference; λ * Indicates passing through the point to be rebuilt Let v represent the row coordinate of the ray on the ray source trajectory λ, and v represent the column coordinate of the ray passing through the point x to be reconstructed on the panel detector.

[0142]

[0143]

[0144] Where L represents the point to be rebuilt The perpendicular distance to the trajectory of the ray source is L = -xsinθ i +ycosθ i +l; H represents the point to be rebuilt. The vertical distance to the panel detector is H = xsinθ i -ycosθ i +h.

[0145] Preferably, for the source trajectory differential three-dimensional image reconstruction method of linear CT scan facing the X-ray source, in step S5, the differential cone-beam projection... The i-th segment of the 3D Hilbert image is obtained by performing a weighted back projection operation. Its calculation formula is:

[0146]

[0147] In the formula, Indicates the point to be rebuilt; η i This represents the i-th segment of the 3D Hilbert image. The direction angle for performing the finite Hilbert inverse transform; 1 / H 2 The back projection weighting factor is used; the back projection operation is performed in a matrix larger than the three-dimensional image to be reconstructed, that is, p0 = 0.5 * I zeros are added outside each edge of the matrix space of the three-dimensional image to be reconstructed, where I is the number of voxels in a certain dimension of the image to be reconstructed.

[0148] Preferably, for the source trajectory differential three-dimensional image reconstruction method of linear CT scan facing the X-ray source, in step S6, the three-dimensional Hilbert image of the i-th segment... A finite Hilbert inverse transform is performed layer by layer along the axial direction of the i-th segment of the linear translation trajectory of the ray source, i.e., layer by layer. k The two-dimensional Hilbert image is inversely transformed layer by layer (k = 1, 2, ..., I) to reconstruct the i-th segment of the finite-angle three-dimensional image. The inverse transform of a 2D Hilbert image is also along x j (j=1,2,...,I) is processed one-dimensionally line by line, and its calculation formula is:

[0149]

[0150] Where y1 represents the direction of the one-dimensional Hilbert transform, y1∈[L y +ε y U y -ε y ], where [L y U y ] represents a finite interval of the Hilbert transform, ε y It is a small positive number; Indicates reconstruction The unknown constants that need to be calculated are obtained by finding the known ones. Some locations are used as the mean of their finite Hilbert transforms. Regarding the i-th segment of the 3D Hilbert image The axial layer-by-layer finite Hilbert inverse transform is specifically implemented by the following steps:

[0151] 1) The i-th segment of the 3D Hilbert image Divide by -2π and rotate -η around the central axis of the 3D image. i Angle, to obtain a rotated 3D Hilbert image Where η i =θ i -π / 2 (i = 1, 2, ..., T), positive angle values ​​rotate the 3D image counterclockwise; negative angle values ​​rotate the 3D image clockwise.

[0152] 2) Calculate the one-dimensional weight sequence W mat ,Right now

[0153]

[0154] 3) for(k=1; k≤1; k++) (i.e., enter the for loop body):

[0155] 3.1) From rotated 3D Hilbert images Indexing the k-th layer 2D Hilbert image

[0156] 3.2) Searching for two-dimensional Hilbert images Find the rows in the array where all values ​​are 0, and obtain an array R whose elements are row numbers. i ;

[0157] 3.3) Two-dimensional Hilbert images Weighted W mat And perform a Hilbert transform along the column direction of the image matrix to obtain the image.

[0158] 3.4) Based on R obtained in step S3 i Numerical values, from images Extract the element by indexing it, calculate its mean along the column direction and invert it to obtain the result.

[0159] 3.5) Execution

[0160] 3.6) Rotating a two-dimensional image Rotate around the image center by an angle η i Returning to the original state, we obtain the reconstructed k-th layer 2D.

[0161] 4) Extract the middle region of the 3D image to be reconstructed, i.e.

[0162]

[0163] Figure 5 A schematic diagram of the linear cone-beam CT scan trajectory of the X-ray source is shown. Figure 6 Showing the three-dimensional Hilbert image of the i-th segment A schematic diagram of performing a finite Hilbert inverse transform layer by layer along the axial direction of the linear translation trajectory of the i-th segment of the ray source.

[0164] Figure 7 The basic process of three-dimensional CT image reconstruction of the present invention is demonstrated, which can prove the effectiveness of the method and reconstruct high-quality, nearly artifact-free three-dimensional CT images.

[0165] Furthermore, a three-dimensional Shepp-Logan phantom was used, with a pixel count of 512×512×512 and a size of 8.4mm×8.4mm×8.4mm. The image reconstruction effect of this invention was evaluated through quantitative numerical experiments. The reconstructed image results are as follows: Figure 8 As shown, different projection numbers N were set for each segment of the X-ray source under a straight-line CT scan. The cone-beam orthographic projection of the three-dimensional Shepp-Logan phantom was simulated and reconstructed using this invention. The quality of the reconstructed image was observed. N was set to 200 and 1600 respectively. The parameter settings for the numerical experiment are shown in Table 2. Figure 8 As can be seen from the reconstruction results, the present invention can reconstruct high-quality, nearly artifact-free three-dimensional CT images when the number of projections is large.

[0166] Table 2 Numerical simulation test parameters

[0167]

[0168] This invention is not limited to the specific embodiments described above. The specific embodiments described above are merely illustrative and not restrictive. Those skilled in the art can make many other modifications based on the teachings of this invention without departing from the spirit and scope of the claims. All of these modifications are within the protection scope of this invention.

Claims

1. A method for reconstructing the source trajectory differential image in linear CT scanning oriented towards a radiation source, characterized in that, Its two-dimensional or three-dimensional method includes the following steps: S1. Initialize parameters i=1, number of linear scan segments T, and rotation angle interval. The null space of the two-dimensional image to be reconstructed ; S2. Obtain the projection of the i-th segment of the linear scan of the ray source. or ; S3. Projection of the acquired data or Perform pre-weighting to obtain the weighted projection. or ; S4. Weighted projection or Translational trajectory along the X-ray source Differentiate the direction to obtain the differential projection, and the differential calculation formula is as follows: ,in, Indicates along Differential operators of direction, Indicates passing through the point to be rebuilt The coordinates of the ray on the translational trajectory of the ray source; ,in, Indicates the point to be rebuilt The vertical distance to the trajectory of the radiation source. ; Indicates the point to be rebuilt Vertical distance to the detector ; S5. Perform a weighted back projection operation on the differential projection to obtain the i-th segment of the Hilbert image. ; S6. Perform a finite inverse Hilbert transform on the i-th segment of the Hilbert image along the direction of the i-th segment of the ray source's straight line translation trajectory to reconstruct the i-th segment of the finite angle image. The formula for calculating the finite Hilbert inverse transform is: ,in, Indicates the direction of the one-dimensional Hilbert transform. ,in Describes a finite interval of the Hilbert transform. It is a small positive number; This represents the reconstruction of the i-th segment of the finite angle image. The unknown constants that need to be calculated; S7. Execute i = i + 1 and complete the image overlay: ; S8. If i ≤ T, then rotate the linear CT scanning system by (i-1). The angle is determined, and the process jumps to S2, repeating this loop until i > T, thus completing the image. The reconstruction.

2. The source trajectory differential image reconstruction method for linear CT scanning oriented towards a radiation source according to claim 1, characterized in that, In step S3, the projection or Pre-weighting yields weighted projection or Its calculation formula is or , in, The direction angle of the linear scanning trajectory of the X-ray source. ; This represents the coordinates of the ray source on the straight line trajectory. ; Represents the coordinates of the detector, and , This indicates half the length of the detector; This represents the distance from the ray source perpendicular to the detector direction to the center of rotation; This represents the distance from the center of rotation to the center of the detector; This is a redundant weighting factor.

3. The source trajectory differential image reconstruction method for linear CT scanning oriented towards a radiation source according to claim 1, characterized in that, In step S5, the differential projection or The i-th Hilbert image is obtained by performing a weighted back projection operation. Its calculation formula is, ， In the formula, or Indicates the point to be rebuilt; This represents the i-th segment of the Hilbert image. The direction angle for performing a finite Hilbert inverse transform; This represents the backprojection weighting factor; the backprojection operation is performed in a matrix larger than the image to be reconstructed, and an additional weight is added outside each edge of the image matrix space. =0.5 * I, the number of zeros, where I is the number of rows or columns of the image to be reconstructed.

4. The source trajectory differential image reconstruction method for linear CT scanning oriented towards a radiation source according to claim 1, characterized in that, In step S6, the i-th Hilbert image segment... A finite Hilbert inverse transform is performed along the direction of the linear translation trajectory of the i-th ray source segment to reconstruct the finite angle image of the i-th segment. The calculation formula is obtained by finding the known... Some locations are used as the mean of their finite Hilbert transforms. The i-th Hilbert image The specific implementation of the finite Hilbert inverse transform includes the following steps: 1) Transfer the i-th Hilbert image segment Dot removal and rotate around the center of the image. The angle is used to obtain a rotated Hilbert image. ,in , When the angle value is positive, the image rotates counterclockwise; when the angle is negative, the image rotates clockwise. 2) Calculate the weight matrix , ； 3) Search for rotated Hilbert images Find the rows in the array where all values ​​are 0, and obtain an array whose elements are row numbers. ; 4) Rotating Hilbert images Weighted And perform a Hilbert transform along the column direction of the image matrix to obtain the image. ; 5) Based on the results obtained in step S3 Numerical values, from images Extract the element by indexing it, calculate its mean along the column direction and invert it to obtain the result. ; 6) Execution ; 7) Rotate the image Rotation angle around the image center Returning to the original state, we obtain ; 8) Extract the middle region of the image to be reconstructed. .

5. The source trajectory differential image reconstruction method for linear CT scanning oriented towards a radiation source according to claim 1, characterized in that, In step S4, the weighted projection Along the trajectory of the radiation source Differential cone-beam projection is obtained by performing a differential operation on the direction. Its calculation formula is ， in, Indicates along The differential operator for the direction is implemented by finite difference operations. The data inside the projection uses central difference, and the data at the projection boundary uses one-sided difference. Indicates passing through the point to be rebuilt The trajectory of the ray source The row coordinates on, Indicates passing through the point to be rebuilt The column coordinates of the rays on the panel detector, ， ， in, Indicates the point to be rebuilt The vertical distance to the trajectory of the radiation source. ; Indicates the point to be rebuilt Vertical distance to the panel detector .

6. The source trajectory differential image reconstruction method for linear CT scanning oriented towards a radiation source according to claim 1, characterized in that, In step S6, the i-th Hilbert image segment... A finite Hilbert inverse transform is performed layer by layer along the axial direction of the linear translation trajectory of the i-th segment of the ray source, including layer by layer. The two-dimensional Hilbert image is inversely transformed layer by layer to reconstruct the i-th segment of the finite-angle three-dimensional image. The inverse transform of a two-dimensional Hilbert image is also along... The calculation formula is as follows: (The formula is missing from the original text.) in, Indicates the direction of the one-dimensional Hilbert transform. ,in Describes a finite interval of the Hilbert transform. It is a small positive number; Indicates reconstruction The unknown constants that need to be calculated are obtained by finding the known ones. Some locations are used as the mean of their finite Hilbert transforms. Regarding the i-th segment of the 3D Hilbert image The axial layer-by-layer finite Hilbert inverse transform is specifically implemented by the following steps: 1) Transform the i-th segment of the 3D Hilbert image Dot removal And rotate around the central axis of the three-dimensional image. Angle, to obtain a rotated 3D Hilbert image ,in , When the angle value is positive, the 3D image rotates counterclockwise; when the angle is negative, the 3D image rotates clockwise. 2) Calculate the one-dimensional weight sequence , ； 3) for (k=1; k≤1; k++): 3.1) From rotated 3D Hilbert images Indexing the k-th layer 2D Hilbert image ; 3.2) Searching for two-dimensional Hilbert images Find the rows in the array where all values ​​are 0, and obtain an array whose elements are row numbers. ; 3.3) Two-dimensional Hilbert images Weighted And perform a Hilbert transform along the column direction of the image matrix to obtain the image. ; 3.4) Based on the results obtained in step S3 Numerical values, from images Extract the element by indexing it, calculate its mean along the column direction and invert it to obtain the result. ; 3.5) Execution ; 3.6) Rotating a two-dimensional image Rotation angle around the image center Returning to the original state, we obtain the reconstructed k-th layer 2D. ; 4) Extract the central region of the 3D image to be reconstructed. 。

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