CN104899837A - Correction method for OCT(optical coherence tomography) image - Google Patents

Correction method for OCT(optical coherence tomography) image Download PDF

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CN104899837A
CN104899837A CN 201510279893 CN201510279893A CN104899837A CN 104899837 A CN104899837 A CN 104899837A CN 201510279893 CN201510279893 CN 201510279893 CN 201510279893 A CN201510279893 A CN 201510279893A CN 104899837 A CN104899837 A CN 104899837A
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scanning
angle
oct
image
method
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CN 201510279893
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CN104899837B (en )
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冯东亮
郭曙光
朱晓湘
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深圳市莫廷影像技术有限公司
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Abstract

The present invention discloses a correction method for an OCT (optical coherence tomography)image. The correction method comprises: determining a plurality of parameters which at least comprise a first scanning radius R and a first vibrating angle (alpha 1) of a scanning line; regulating a scanning line angle of a scanning unit, respectively scanning a sample in the horizontal direction and in the vertical direction, obtaining a long axis 2a and a short axis 2b, structuring out a scanning distance c and a second vibrating angle alpha 2 when an ellipse formula (as shown in the description) is defined at a random scanning angle theta, and according to a formula (as shown in the description), obtaining the scanning distance c and alpha 2; and finally, according to a second scanning radius (2c) and alpha 2, restoring the OCT image obtained by scanning at the random scanning angle theta into a real shape of the scanned sample. Therefore, according to the method, the OCT image, which is obtained when the scanning line of the scanning unit of an OCT apparatus scans the sample at the random scanning angle theta, can be corrected and restored, and the defect of only correcting and restoring the OCT image obtained by scanning the sample at a single scanning angle is overcome.

Description

一种OCT图像的校正方法 A method of correcting OCT images

技术领域 FIELD

[0001] 本发明属于0CT图像的处理领域,具体涉及到一种0CT系统扫描单元的扫描线以任意扫描角度扫描样品得到的0CT图像的校正方法。 [0001] The present invention belongs to the field of 0CT image processing, and more particularly to an image correction method 0CT 0CT one kind of scanning lines of the scanning unit system at any scan angle scanning the sample obtained.

背景技术 Background technique

[0002] 光学相干层析成像技术(OpticalCoherenceTomography,0CT)是一种非侵入的探测技术。 [0002] Optical coherence tomography (OpticalCoherenceTomography, 0CT) is a non-invasive detection techniques. 0CT技术现已被广泛应用于生物组织的活体截面结构成像。 0CT technique has been widely used cross-sectional structure of the imaging living biological tissue. 通过测量与深度有关的散射光,0CT可以提供高分辨,高灵敏度的组织结构。 By measuring the scattered light is related to the depth, 0CT may provide high resolution, high sensitivity organization.

[0003] 参考图1,图1只是示例性的表示0CT系统扫描样品时,0CT系统的扫描单元1到样品4之间的光路图。 [0003] Referring to FIG 1, FIG 1 is merely exemplary representations when scanning the sample 0CT system, the optical scanning unit 0CT system of FIG path between the sample 1 to 4. 当利用扫描单元1扫描样品4时,从0CT系统的其他装置(未图示) 出射的光线入射至扫描单元1。 When scanning the sample by a scanning unit 4, 0CT systems from other devices (not shown) emitted light is incident to the scanning unit 1. 当扫描单元1处于其中某一振动角度时,经扫描单元1反射,光线传播的路线与透镜2和透镜3构成的主光轴重合,即沿着〇l〇2S,得到的被扫描样品4的0CT断层图经计算机显示为图2中的圆弧形状。 When the scanning unit 1 in which a vibration angle, reflected by the scanning unit 1, the main optical axis line of the light propagating lens 2 and the lens 3 coincides configuration, i.e. along 〇l〇2S, obtained by scanning the sample 4 0CT computer tomogram displayed by arc shape in FIG. 在图1中,〇1为透镜2的光心,〇2 为透镜3的光心,S点为入射光线沿〇1、〇2方向照射扫描样品4入射面的入射点,T点为与样品4上与S点相对的另一面的点,ST之间的距离即为样品4的厚度。 In FIG 1, the optical center of the lens 〇1 2, 〇2 optical center of the lens 3, S 〇1 points along the line of the incident light, the incident surface of the incident point 4 〇2 scanning direction irradiated sample, T is the sample point point 4 on the other side opposite to the point S, the thickness of the sample is the distance between the ST 4. 参考图2,图1中样品4的S点,在图2所示的0CT断层图上显示为上圆弧的最高点S1.当扫描单元1转动到另外的角度时,入射光线在扫描单元1到样品4之间的传播路线变为KLM或者EFU时,光线在样品4上的入射点M对应图2所示的0CT断层图上Ml,U点对应图2所示的0CT断层图上U1.由图1可以看出,样品4被扫描的断面MNVU是矩形状,但是经过计算机却显示成两同心的圆弧状U1S1M1、V1T1N1,即:线段USM在0CT断层图上显示为圆弧U1SIM1,线段VTN 在0CT断层图上显示为圆弧V1T1N1 ;因此,样品4的0CT断层图不能反应其真实的断面形状。 Referring to Figure 2, a sample point S in FIG. 4, showing the highest point of the arc fault 0CT S1 in FIG. 2 as shown in FIG. 1, when the scanning unit is rotated to another angle, the incident light scanning unit 1 when the propagation path between the sample becomes 4 or KLM EFU, the upper shown in FIG 0CT fault light is incident on the point M corresponds to sample 4 in FIG. 2 Ml, U 0CT fault point map shown in FIG. 2 U1. as can be seen from Figure 1, the sample being scanned MNVU section 4 is rectangular, but after the computer has a display into two concentric arcuate U1S1M1, V1T1N1, namely: an arc segment display USM U1SIM1, fault line in FIG 0CT VTN displayed on an arc V1T1N1 0CT tomographic; Thus, the sample 0CT tomographic 4 can reflect the true cross-sectional shape. 造成样品4断面本来的形状和计算机显示的0CT断层图形状不相同的原因在于:光程olo2S和光程KLM不同,光程olo2S和EFU也不同。 0CT tomogram shape is not the same cross section 4 causes sample computer displays original shape and in that: different optical path and the optical path olo2S KLM, and the optical path olo2S EFU different. 图1中光程olo2S和光程KLM的差值在图2中0CT断层图中反映为S1和U1在垂直方向的位置差;同样的,光程〇l〇2S和光程EFU 在图2中0CT断层图中反映为S1和U1在垂直方向的位置差;若光程KLM和光程EFU不同, 在图2中0CT断层图中反映为Ml和U1在垂直方向的位置差。 And the optical path difference between the optical path olo2S KLM reflected in FIG 1 FIG 2 FIG 0CT tomographic as U1 and S1 in a vertical position difference; the same optical path and the optical path 〇l〇2S EFU faults in FIG. 2 0CT FIG U1 and S1 is reflected in the difference in the vertical position; KLM if the optical path and the optical path EFU different, as reflected in the difference between Ml and U1 position in the vertical direction in FIG. 2 0CT tomogram. 因此,正是因为光程差造成了样品被扫描断面的形状和在0CT图像显示的形状不相同的原因。 Thus, it is because the optical path difference is caused by the shape of the scanned cross-section and the shape of the image display 0CT not sample the same reason. 需要说明的是,图1中只是示例性的例举了凸透镜为两个情形,事实上,凸透镜的数量不受限至,可以是1个或者多个,因为只要在扫描单元1和样品4之间有凸透镜,就会存在光程差,就会出现样品被扫描断面的形状和在0CT图像显示的形状不相同的现象。 Note that, in FIG 1 is merely exemplary of two convex exemplified case, in fact, is not limited to the number of the convex lens, may be one or more, as long as the scanning unit 1 and the sample 4 between the convex lens, the optical path difference will exist, the sample appears to be not the same shape and cross-sectional shape of the scan image display 0CT phenomenon.

[0004] 目前用0CT系统对样品扫描时,采用单一的定标参数。 [0004] When a current 0CT sample scan system, a single scaling parameter. 也就是说,只能够对0CT系统的扫描单元(例如X振镜或Y振镜)用单一的扫描角度扫描样品得到的0CT图像进行校正还原。 That is, reduction can only be corrected by a single scan angle 0CT image scanning unit to scan the sample obtained 0CT system (e.g. X or Y galvanometer galvanometer). 当0CT系统的扫描单元的扫描线以多角度扫描样品时,也只能准确校正扫描线以其中的一个角度扫描得到的0CT图像,对于扫描线以其余任一扫描角度扫描样品得到的0CT图像,由于无法准确得到与其对应的相关参数,所以无法对相应的0CT图像进行校正还原,从而使得校正图像和真实图像会存在略微差异,无法反映被扫描样品真实断面形状。 When the scanning line of the scanning unit scanning the sample 0CT system to a multi-angle, accurate correction can only 0CT image scan line at an angle which scans obtained 0CT image scanning lines for any remaining samples obtained by scanning the scan angle, Unable to obtain accurate parameters corresponding thereto, it is not possible to carry out image correction corresponding 0CT reduction, so that the real image and the corrected image will be slight differences, can not reflect the true cross sectional shape scan the sample. 这里所说的定标,是指求得与OCT系统扫描单元以任一扫描角度扫描样品得到的一组参数值,包括但不限于:扫描半径、扫描范围、探测深度、扫描角度和扫描单元的中心位置。 Scaling here, refers to the OCT system is obtained with a scanning unit to any angular scan a set of parameter values ​​obtained sample scan, including but not limited to: the radius of scan, the scanning range, the depth of the probe, and the scanning angle of the scanning unit Central location. 也就是说,当扫描单元的其中一条扫描线的扫描角度一旦确定后,则与之对应的探测深度、扫描角度和扫描单元的中心位置均能确定。 That is, once the scan angle when the one scanning line wherein the scanning unit is determined, the depth of the center position corresponding thereto, and the scanning angle of the scanning unit can determine. 在本专利申请中,扫描线的扫描角度是指扫描线扫描方向和水平方向的夹角9,该角度通过扫描设备得知,为已知值。 In the present patent application, the scan angle refers to the angle between the scanning lines 9 and the scanning direction of the scanning lines in the horizontal direction, the angle through which scanning device that, to a known value.

发明内容 SUMMARY

[0005] 本发明公布了一种0CT图像的校正方法,其目的在于解决用0CT装置的扫描单元的扫描线以任意角度扫描样品得到的由计算机显示的0CT图像,和样品的被扫描断面的真实形状存在的细微差异。 [0005] The present invention discloses a method for correcting an image 0CT, and its object is to solve the real image of the scanned cross-section 0CT, and sample scanning line scanning unit 0CT apparatus at any angle scan samples obtained by a computer display subtle differences in shape.

[0006] 本发明的技术方案是这样的: [0006] aspect of the present invention is such that:

[0007] -种0CT图像的校正方法,包括如下步骤: [0007] - species 0CT image correcting method, comprising the steps of:

[0008] 0CT系统的扫描单元扫描样品,得到0CT图像,确定出至少包括有第一扫描半径R 和第一振动角度(a1)的若干参数; [0008] The scanning unit scanning the sample 0CT system, to obtain 0CT images, the number of parameters determining a first scan and a first radius R of the vibration angle (a1) comprises at least;

[0009] 调整所述扫描单元的扫描线角度,分别在水平方向和垂直方向扫描样品,求出长轴2a和短轴2b,构建出椭圆公式 [0009] adjusting the angle of the scanning unit scanning lines, respectively, in the horizontal direction and the vertical direction scanning the sample, the major axis is obtained and a minor axis 2a 2B, constructed ellipse formula

Figure CN104899837AD00051

[0010] 定义扫描单元的扫描线在任意扫描角度9时的第二扫描半径为2c,根据椭圆公式 [0010] The scanning line scanning unit defined in the second scanning radius of 0900 arbitrary scan angle 2c, in accordance with an elliptical equation

Figure CN104899837AD00052

,求得扫描距离c和第二振动角度(a2); , Obtained scanning distance c and the second vibration angle (A2);

[0011] 利用所述扫描距离C和所述第二振动角度(a2)将在任意扫描角度0扫描样品得到的0CT图像还原成被扫描样品真实的形状; [0011] C with the scanning distance and said second angular vibration (a2) will be reduced to the real shape of the sample in the scanned image of any scan angle of 0 0CT scanning the sample obtained;

[0012] 其中,0为扫描单元的扫描线和水平方向构成的角度,为已知值。 [0012] wherein the configuration of the angle 0 and the horizontal scanning line scanning unit, a known value.

[0013] 进一步地,所述若干参数还包括0CT图像顶端宽度w,根据公式w= R*sin((a1)/2)*2 求得。 [0013] Further, the plurality of parameters further includes a top 0CT image width w, according to the formula w = R * sin ((a1) / 2) * 2 is obtained.

[0014] 进一步地,所述R=H1+H3 ;其中,H1为所述0CT图像顶端到被扫描样品处于第一位置时的距离,H3为所述第一位置到所述扫描单元的中心位置〇的距离。 [0014] Further, the R = H1 + H3; wherein, Hl is the top of the image to be scanned 0CT the sample is from a first position, H3 to the first position to the center position of the scanning unit square of the distance.

[0015] 进一步地,所述HI=hl*ratio_h;其中,hi为所述0CT图像顶端到所述第一位置的像素距离;ratio_h为实际距离和像素的比例关系,ratio_h=dist/(h2_hl) ;dist为所述样品由所述第一位置沿着扫描单元的位置〇移动到第二位置的距离,h2为所述OCT图像顶端到所述第二位置的像素距离;hi和h2通过对所述0CT图像进行边界检测得到;所述0CT图像的横向像素数定义为W,纵向像素数定义为H。 [0015] Further, the HI = hl * ratio_h; wherein, Hi is the top of the image to the pixel 0CT distance from the first position; ratio_h proportional to the actual distance and the pixel, ratio_h = dist / (h2_hl) ; dist said sample from said first position along a movement position of the scanning unit to the square of the distance to the second position, h2 is the top OCT image pixel distance to the second position; and h2 by the Hi said edge detection image obtained 0CT; 0CT laterally defining the number of pixels of the image is W, the number of vertical pixels is defined as H.

[0016] 进一步地,(a1)/2 =arctan((up_w-low_w)/2/dist),H3 = 2*tan(a/2)/up_ w; [0016] Further, (a1) / 2 = arctan ((up_w-low_w) / 2 / dist), H3 = 2 * tan (a / 2) / up_ w;

[0017] 其中,up_w为所述样品处于所述第一位置时,所述扫描单元扫描所述样品的范围; low_w为所述样品处于所述第二位置时,所述扫描单元扫描所述样品的范围。 When [0017] wherein, up_w the sample is in the first position, the range of the scanning unit scans the sample; low_w when said sample is in said second position, said scanning means scans said sample range.

[0018] 进一步地,所述第二振动角度(a2)根据公式R*sin((a1)/2)= 2c*sin((a2)/2)求得。 [0018] Further, the second vibration angle (a2) the formula R * sin ((a1) / 2) = 2c * sin ((a2) / 2) determined in accordance with.

[0019] 进一步地,所述样品分别处于所述第一位置和所述第二位置时,所述扫描单元扫描所述样品,得到第一OCT图像和第二OCT图像;所述第一OCT图像和所述第二OCT图像的横向像素数为所述W,纵向像素数为所述H。 When [0019] Further, the samples were in the first position and the second position, the scanning unit scans said sample to obtain a first image and a second OCT OCT image; OCT image of the first and the number of horizontal pixels of the second image is the OCT W, the longitudinal number of pixels is H.

[0020] 进一步地,所述若干参数还包括扫描单元的探测深度h,h=ratio_h*H。 [0020] Further, the plurality of parameters further includes a scanning unit depths of h, h = ratio_h * H.

[0021] 进一步地,所述样品设置在一夹具上,通过移动夹具实现所述样品由所述第一位置移动到所述第二位置。 [0021] Further, the sample is set on a jig, moving from said first position to said second position said sample is achieved by moving the clamp.

[0022] 进一步地,所述up_w和所述low_w的值通过距离测量装置读出;所述距离测量装置暴露在所述0CT图像中。 [0022] Further, the values ​​of the up_w low_w and readout by the distance measuring means; said distance measuring device is exposed in the 0CT image.

[0023] 进一步地,所述样品为条块状玻璃片。 [0023] Further, the glass sample is a bar form.

[0024] 进一步地,所述扫描单元为XY振镜。 [0024] Further, the XY scanning unit is a galvanometer.

[0025] 本发明的有益技术效果:先通过将样品由第一位置移动到第二位置,求出振镜的第一扫描半径R、第一振动角度a1和样品的0CT图像顶端的扫描范围w;接着利用扫描单元的扫描线分别在水平方向和垂直方向分两次扫描样品,从得到的0CT断层图中求出长轴2a和短轴2b,构建出椭圆公式, [0025] Advantageous effects of the invention: first by moving the sample from the first position to the second position, obtains a first scanning galvanometer of radius R, the top of the image 0CT first vibration angle a1 and the sample scan range w ; then using a scanning unit in the sub-scanning line are respectively horizontal and vertical directions twice scanning the sample, determined from a major axis 2a and 2b 0CT short axis tomogram obtained to construct the ellipse equation,

Figure CN104899837AD00061

;当扫描线以任意扫描角度0扫描样品时,根据椭圆公式 ; When the scan lines scanning the sample at any scan angle of 0 The elliptic equation

Figure CN104899837AD00062

,求出第二扫描半径2c和振镜的第二振动角(a2); 最后,根据求出的第二扫描半径2c的像素值和第二振动角(a2),将以任意扫描角度0扫描样品得到的0CT图像校正,还原被扫描样品本来的图像形状。 , Obtains a second oscillation angle and a second scanning galvanometer 2c radius (A2); and finally, a second scan based on the determined pixel values ​​and the radius 2c of the second vibration angle (A2), will scan at any scan angle 0 0CT sample obtained image correction, reducing the scanned image of the original shape of the sample. 因此,本方法能对0CT装置的扫描单元的扫描线以任意扫描角度0扫描样品得到的0CT图像校正还原,克服了现有技术中只能够对0CT系统的扫描单元(例如X振镜或Y振镜)以单一的扫描角度扫描样品得到的0CT图像进行校正还原的缺陷。 Accordingly, the present method can scan line of the scanning unit 0CT means an arbitrary scan angle obtained by scanning the sample 0 0CT reducing image correction, overcomes the prior art can only e.g. X or Y galvanometer scanning transducer unit 0CT system ( mirror) 0CT images in a single scan angle of scanning the sample was subjected to reduction correction of defects.

附图说明 BRIEF DESCRIPTION

[0026] 图1为0CT系统扫描样品时,扫描单元1到被扫描样品4之间的光路示意图; [0026] FIG. 1 is a scanning system 0CT sample, to the scanning unit 1 is a schematic view of an optical path between the sample 4 scan;

[0027] 图2为计算机系统显示的被扫描样品4的0CT断层图,该图不能反映被扫描样品4的真实断面形状。 [0027] FIG. 2 is a scanned sample 0CT tomographic 4 shows a computer system, which FIG not reflect the true cross sectional shape of the scanned sample 4.

[0028] 图3为本发明的流程图; [0028] FIG. 3 is a flowchart of the present invention;

[0029] 图4为扫描单元的真实扫描区域并延伸至扫描单元位置的示意图; [0029] FIG. 4 is a real scanning region and the scanning unit extends to the schematic diagram of the scanning unit location;

[0030]图5为被扫描的板状样品处于第一位置时,扫描单元扫描样品的实际扫描位置图; When [0030] FIG. 5 is in the first position of the plate-like sample to be scanned, the scanning unit scanning the actual scanning position of FIG sample;

[0031] 图6为0CT系统的计算机显示的被扫描的板状样品处于第一位置时的第一0CT图像; Plate-like sample to be scanned [0031] FIG. 6 is a computer display system 0CT first 0CT image in a first position;

[0032] 图7为被扫描的板状样品处于第二位置时,扫描单元扫描样品的实际扫描位置图; When [0032] FIG. 7 is a plate-like sample in a second position to be scanned, the scanning unit scanning the actual scanning position of FIG sample;

[0033] 图8为0CT系统的计算机显示的被扫描的板状样品处于第二位置时的第二0CT图像; Plate-like sample to be scanned [0033] Figure 8 is a computer display 0CT system 0CT in a second image in a second position;

[0034] 图9为被扫描的板状样品分别处于第一位置和第二位置时,将0CT系统扫描被扫描的板状样品得到的0CT图构建的数学模型; When [0034] FIG. 9 is a plate-like sample to be scanned in the first and second positions, respectively, the mathematical model of the system scans the scanned 0CT plate-like sample obtained 0CT FIG constructed;

[0035] 图10为将0CT图像扫描校正回正常的扫描范围所得到的0CT图像; [0035] FIG. 10 is the image 0CT 0CT image scanning correction back to the normal scanning range obtained;

[0036] 图11为OCT系统的扫描单元以任一扫描范围和扫描角度扫描得到的椭圆轨迹。 [0036] FIG. 11 is a scanning unit in any OCT system elliptical locus a scanning range and the scanning angle scanned.

具体实施方式 detailed description

[0037] 为了使本发明所要解决的技术问题、技术方案及有益效果更加清楚明白,以下结合附图及实施例,对本发明进行进一步详细说明。 [0037] In order to make the technical problem to be solved by the present invention, technical solutions and beneficial effects clearer, the accompanying drawings and the following embodiments, the present invention will be further described in detail. 应当理解,此处所描述的具体实施例仅用以解释本发明,并不用于限定本发明。 It should be understood that the specific embodiments described herein are merely used to explain the present invention and are not intended to limit the present invention.

[0038] 参考图3,图3为本发明所示的流程图,包括如下步骤: [0038] Referring to FIG 3, a flowchart shown in FIG. 3, the invention comprises the steps of:

[0039] S101 :0CT系统的扫描单元扫描样品,得到0CT断层图,确定出至少包括第一扫描半径R和第一振动角度(a1)的若干参数; [0039] S101: a scanning unit scanning the sample 0CT system, to obtain tomographic 0CT, determining a plurality of scanning parameters including at least a first and a first radius R vibration angle of (a1);

[0040] S102 :调整所述扫描单元的扫描线角度,分别在水平方向和垂直方向扫描样品,求出长轴2a和短轴2b,构建出椭圆公式 [0040] S102: adjusting the angle of the scanning unit scanning lines, respectively, in the horizontal direction and the vertical direction scanning the sample, the major axis is obtained and a minor axis 2a 2B, constructed ellipse formula

Figure CN104899837AD00071

[0041] S103 :定义扫描单元的扫描线在任意扫描角度0时的第二扫描半径为2c,根据椭圆公式 [0041] S103: the scanning unit scanning lines defined in the second scanning radius 0:00 arbitrary scan angle 2c, in accordance with an elliptical equation

Figure CN104899837AD00072

> 求得扫描距离c和第二振动角度(a2); > C and the second scanning distance obtained by the vibration angle (A2);

[0042] S104 :利用所述扫描距离c和所述第二振动角度(a2)将在以任意扫描角度0扫描样品得到的0CT图像还原成被扫描样品真实的形状。 [0042] S104: using said second scanning distance c and the vibration angle (a2) to be reduced to the real shape of the sample in the scanned image 0CT scanning angle 0 at any scanning the sample obtained.

[0043] 下面对这4个步骤展开具体分析。 [0043] The following describes the four steps to expand specific analysis.

[0044] 对于步骤S101,确定若干参数的具体过程如下: [0044] For the step S101, the determined number of parameters of the specific process is as follows:

[0045] 第一步,将被扫描的样品(未图示)设置在0CT系统的扫描范围内,也就是置于0CT系统的扫描单元的扫描范围内。 [0045] The first step, the sample to be scanned (not shown) disposed in the scanning range 0CT system, which is placed within the scanning range of the scanning unit 0CT system. 在0CT系统处于工作状态下,让样品处于如图5所示的第一位置,扫描单元扫描样品,得到如图6所示的计算机显示的呈两同心圆弧的第一0CT 图像。 In 0CT system is in working condition, so that the sample is in the first position as shown in FIG. 5, the scanning unit scanning the sample, to obtain a first image 0CT form two concentric arcs of a computer display as shown in Fig. 但是,第一0CT图像并不能反映样品被扫描断面的本来形状,具体原因见背景技术的相关解释。 However, the image does not reflect the first 0CT sample was scanned original shape of the cross section, the specific reasons explained in the relevant background art. 当样品经过移动,由图5所示的第一位置移动到图7所述的第二位置时,记录其移动距离dist。 When the sample is moved to said second position moves through 7, a first position shown in FIG. 5, which records a moving distance dist. 样品在第二位置时,扫描单元扫描样品,得到如图8所示的计算机显示的第二0CT图像,同样的,第二0CT图像也不能反映样品本来的形状,也显示为两条同心的圆弧形状。 0CT second image of the sample in the second position, the scanning unit scanning the sample, to give a computer display shown in FIG. 8, the same, the second 0CT image does not reflect the shape of the original sample, it is also shown as two concentric circular arc shape. 接着用距离测量装置测出样品在图5所示的第一位置时被扫描单元扫描的范围,标注为up_w;测出样品在图7所示的第二位置时被扫描单元扫描的范围,标注为low_w。 Followed by the distance measuring device measuring the sample at the first position as shown in FIG. 5 is a scanning range of the scanning unit, denoted up_w; measured range of the scanning unit is scanning the sample in the second position shown in FIG. 7 are denoted by as low_w.

[0046] 第二步,参考图6和图8,定义图6中的第一OCT图像的横向像素数为W,纵向像素数为H;同样的,图8中的第二0CT图像的横向像素数也定义为W,纵向像素数也定义为H。 [0046] The second step, with reference to FIGS. 6 and 8, the number of horizontal pixels of the first OCT image in FIG. 6 is defined as W, the number of vertical pixels is H; Likewise, the second lateral 0CT pixel image in FIG. 8 number is also defined as W, also defined as the number of vertical pixels H. 此处所述的横向像素数,即扫描线数。 Number of horizontal pixels as described herein, i.e., the number of scanning lines. 通过图像处理办法,进行边界检测,分别求出图6 中显示的样品在第一位置时经扫描得到的0CT断层图的圆弧层中心位置到第一0CT图像的高度为hi(见图9),图8显示的样品在第二位置时经扫描得到的0CT断层图的圆弧层中心位置距离第二0CT图像的高度为h2(见图9)。 By way of image processing, for boundary detection, respectively, to obtain the center position of the circular arc 0CT tomographic layer of the sample shown in Figure 6 is obtained by scanning in the first position to the height of the first image is 0CT hi (see Fig. 9) arc center position 0CT tomographic layer of the sample shown in FIG. 8 in the second position obtained from the scanned image is 0CT second height H2 (see FIG. 9). 需要说明的是,hi和h2的单位为像素。 It should be noted, hi and h2 pixels. 当样品在第二位置时,扫描得到的第二0CT断层图中的圆弧层到第二0CT断层图顶端的位置, 和样品在第一位置时,扫描得到的第一0CT断层图中的圆弧层到第一0CT断层图顶端的位置相比,有所下移。 When the sample is in the second position, the second arcuate 0CT layer tomogram obtained by scanning in the first round to the position of the second tomographic 0CT 0CT tomogram tip, and the sample in a first position, scanned in position of the first layer to the arc 0CT tomogram as compared to the top, down somewhat. 也就是说,随着样品的下移,样品的0CT图相对其0CT图像顶端下移了。 That is, as the down sample, the sample relative to its FIG 0CT 0CT down the top of the image. 由于样品由第一位置移动到第二位置的距离为dist,其像素距离差为(h2-hl),据此可以求出毫米和像素的比例关系ratio_h=dist/(h2 -hi)。 Since the sample is moved from the first position to the second position as distance dist, which pixel distance difference (h2-hl), whereby the pixel can be determined mm and the ratio between ratio_h = dist / (h2 -hi).

[0047] 第三步,构建数学模型,求出OCT图像顶端的扫描范围w。 [0047] The third step, construct a mathematical model, to obtain an OCT scan image of the top range w. 参考图9,由于已经求出ratio_h的值为dist/ (h2 -hi),则AB线段对应的实际毫米数HI=ratio_h*hl可以求出, 0CT系统的探测深度h=AD=ratio_h*H可以求出。 Referring to Figure 9, since the value has been determined ratio_h dist / (h2 -hi), the segment AB corresponding to the actual number of millimeters HI = ratio_h * hl can be obtained, the system 0CT depths h = AD = ratio_h * H can obtained. 其中,H为0CT图像的纵向像素数,这里的0CT图像既可以是第一0CT图像,也可以是第二0CT图像。 Where, H is the number of pixels in the longitudinal 0CT image, where the image 0CT may 0CT either a first image, the image may be the second 0CT. 因为第一0CT图像和第二0CT图像均是由同一个0CT系统扫描样品得到的,横向像素数和纵向像素数均不会发生改变,即:横向像素数均为W,纵向像素数均为H(见图6和图8)。 Because the first image and the second 0CT 0CT images are obtained by scanning the sample with a 0CT system, lateral and vertical pixels will not change the number of pixels, namely: number of horizontal pixels are W, the number of vertical pixels are both H (see Figures 6 and 8).

[0048] 请继续参考图9,根据几何知识,定义扫描单元以扇形扫描区域扫描样品时的第一振动角度为a1,可以求出(a1)/2 =arctan((up_w-low_w)/2/dist),从而得到第一振动角度为al。 [0048] Please continue to refer to FIG. 9, according to the knowledge of geometry, the scanning unit is defined in the first vibration angle sector scanning is scanning the sample area a1, can be determined (a1) / 2 = arctan ((up_w-low_w) / 2 / dist), whereby the first vibration angle al. 参考图9,又由于tan((al)/2) =up_w/2/B0,所以,可以计算出B0的线段长度H3,S卩BO=up_w/2/tan((a1)/2);扫描单元第一扫描半径R=AB+B0 =H1+H3,即可求出扫描单元的扫描半径R的毫米数。 Referring to Figure 9, and since tan ((al) / 2) = up_w / 2 / B0, it is possible to calculate the length of the line B0, H3, S Jie BO = up_w / 2 / tan ((a1) / 2); Scan a first scanning unit radius R = AB + B0 = H1 + H3, a few millimeters to obtaining the scan radius R of the scanning unit. 根据公式R/ratio_h,求得到扫描单元像素级别的扫描半径。 According to the formula R / ratio_h, seek to obtain the radius of the scanning unit scans the pixel level. 最后,用w=R*sin((a1)/2)*2求得0CT图像顶端的扫描范围w。 Finally, w = R * sin ((a1) / 2) * 2 0CT top of the image obtained by the scanning range w. 需要说明的是,在图9中,〇代表扫描单元的位置,A点表示扫描单元所做的扇形扫描区域的最高点,AD表示0CT系统的探测深度H。 Incidentally, in FIG. 9, the position of the representative square of the scanning unit, A represents the highest point of the sector scanning region of the scanning unit made, showing depth of the AD 0CT system H.

[0049] 在本发明中,被扫描样品由第一位置移动到第二位置,是通过夹具的移动来实现的。 [0049] In the present invention, the sample is moved from a first scanned position to a second position, it is achieved by moving the jig. 也就是说,被扫描样品装夹在夹具上,通过夹具的移动来带动被扫描样品由第一位置移动到第二位置。 That is, the sample is scanned clamped in the fixture, by moving the jig to drive movement of the sample is scanned from a first position to a second position. 需要说明的是,样品在处于第一位置和第二位置时,均需要位于0CT系统的扫描范围内。 Incidentally, the sample when in the first position and the second position, is located within the scanning range are required 0CT system. 被扫描样品在图5所示的第一位置时,被扫描单元扫描的范围标记为up_w,通过距离测量装置测出;样品在图7所示的第二位置时,被扫描单元扫描的范围标记为low_ w,也是通过距离测量装置测出。 The sample is scanned in a first position shown in Figure 5, the range of the scanning unit scans labeled up_w, measured by the distance measuring means; samples in the second position shown in Figure 7, the scanning unit is scanning range marker is low_ w, is measured by the distance measuring means. 由于up_w、low_w均是由距离测量装置测量得到,因此距离测量装置需要暴露在0CT图像中。 Since up_w, low_w are obtained by the distance measuring means, and therefore needs to be exposed in the image 0CT distance measuring device.

[0050] 作为其中的一个实施例,距离测量装置优先选择为卡尺;当然,up_w、low_w也可以通过OCT系统中的距离测量软件求出。 [0050] As an embodiment wherein the embodiment, the distance measuring device is preferentially selected caliper; of course, up_w, low_w software may be determined by measuring the distance of the OCT system.

[0051] 综上所述,在0CT图像的横向像素数W,纵向像素数H为已知情况下,通过图像的处理办法,对0CT图像进行边界检测,记录被扫描的样品在第一0CT图中得到像素距离为hl,经过移动后样品在第二0CT图中的像素距离为h2 ;记录样品由第一位置移动到第二位置时的位移dist,利用公式disV(h2 -hi),求得ratio_h,S卩毫米和像素的比值;在求出ratio_h后,最后求出扫描单元的第一扫描半径R、扫描范围w和探测深度h参数,完成定标过程;在第一扫描半径R求出后,通过乘以ratio_h的倒数,得到扫描单元的第一扫描半径R的像素距离。 [0051] As described above, the number of pixels in the lateral direction W is 0CT image, H is the vertical pixel number, if known, by the approach of the image, the image boundary detecting 0CT of the records are scanned in a first sample 0CT FIG. HL pixel distance is obtained, after moving the pixel in the second sample distance h2 is 0CT figure; the sample is moved from the first recorded position to a second position when the displacement dist, using the formula disV (h2 -hi), obtained ratio ratio_h, S and pixels Jie mm; after obtaining ratio_h, finally obtaining a first scan of the radius R of the scanning unit, the scanning range and depth of h w parameters, complete calibration process; determined in a first scan radius R after multiplying by the inverse of ratio_h obtain a first pixel from the scanning radius R of the scanning unit.

[0052] 在完成上述参数定标后,将0CT图像扫描校正回如图10所示的正常的范围。 [0052] After the above calibration parameters, the image scanning correction 0CT normal range 10 back as shown in FIG. 在完成步骤S101后,执行步骤S102。 After completion of step S101, the step S102.

[0053] 在步骤S102中,调整扫描单元的扫描线的扫描角度,即分别取水平方向(0度)和垂直方向(90度方向),再次用步骤S101的方法重新确定扫描单元在水平方向一组数据: 扫描范围w',扫描半径R'和探测深度h';在垂直方向取得的另一组数据:扫描范围w",扫描半径R"和探测深度h"。可以发现,除R'、R"、R不同外,三组扫描范围值:w'、w"和w相同;三组探测深度值:h'、h"和h相同。 [0053] In step S102, adjust the scan angle of the scan line of the scanning unit, i.e. taken in the horizontal direction (0 degrees) and a vertical direction (90 ° direction) to redefine the method step S101 again the scanning unit in a horizontal direction a group data: scanning range w ', the scan radius R' and a depth of h '; the other set of data acquired in the vertical direction: scanning range w ", the scan radius R" and a depth of h "can be found, in addition to R', R "different from, R, the three groups of scan range values: w ', w" and the same as W; depth of three values: h', the same h "and h. 也就是说,扫描单元的扫描线的第一扫描半径R和扫描角度0的变化并不会影响扫描单元扫描样品的扫描范围w和探测深度h的变化。 That is, the radius R of the first scanning and the scanning angle of the scanning line of the scanning unit 0 and changes will not affect the scanning range of the scanning unit scans the sample w and the depth of h.

[0054] 另外,在步骤S102中,扫描单元的扫描线的扫描方向与水平方向的构成的角度为0°,与垂直方向构成的角度为90° ;当角度为0°时,扫描线扫描半径R= 2a,其中2a为长轴;当9为90°时,扫描线扫描半径R= 2b,其中2b为短轴。 [0054] Further, in step S102, the angle formed by the scanning direction of the horizontal scanning line of the scanning unit is 0 °, an angle formed with the vertical direction of 90 °; when the angle is 0 °, scan line radius R = 2a, where 2a is the major axis; when 9 is 90 °, the scanning line scan radius R = 2b, where the minor axis 2b. 由此,可以构建出椭圆方 Thereby, it is possible to build an elliptic

Figure CN104899837AD00091

[0055] 在完成步骤S102后,进入步骤S103,求出扫描单元的扫描线的任一第二扫描半径2c和扫描角度0的数学关系。 [0055] After completion of step S102, enters the step S103, the determined one of the scanning lines of the scanning unit scans a second radius 2c and scan angle of 0 mathematical relationships.

[0056] 假定扫描线在任意角度0的扫描距离为c,扫描单元的振动角度为第二振动角a2,第二扫描半径为2c。 [0056] If the scan lines in the scan angle of 0 is an arbitrary distance vibration angle c, the scanning unit is a second vibration angle a2, the radius of the second scanning 2c. 在这些参数中,0为扫描单元的扫描线的扫描方向与水平方向的夹角,可以通过扫描单元得知,为已知值,其余参数为未知值。 Within these parameters, 0 is the angle to the horizontal scanning direction of the scanning lines of the scanning unit, the scanning unit that can, to a known value, the unknown values ​​of the remaining parameters. 参考图11,利用椭圆公式: 11, the elliptic equation:

Figure CN104899837AD00092

可以得到 You can get

Figure CN104899837AD00093

,由于a和b在步骤S102的时候已经求出,0为已知值,因此可以通过解方程得到扫描距离c值,进而求得第二扫描半径2c的值。 Since a and b have been obtained in step S102 when, as a known value 0, c can be obtained from the scan values ​​by solving equations, then obtain the values ​​of the second scanning radius 2c.

[0057] 参考图4,由于前面说过,OCT图像顶端的扫描范围w不会随着扫描半径R和扫描角度9的变化而变化,前面已经求出w=R*sin((al)/2)*2;则在扫描线的扫描角度为9、第二扫描半径为2c、扫描单元的第二振动角度为a2的情况下,w= 2c*sin((a2)/2) *2 =R*sin((a1)/2)*2,又因为a1和R在步骤S101中的第三步已经求出,因此可以求出扫描单元的第二振动角度a2。 [0057] Referring to FIG 4, since the said earlier, the OCT scan image of the top of the range does not vary with w scans the scan angle of the radius R and 9, it has already been determined w = R * sin ((al) / 2 ) * 2; then the scan angle of scanning lines 9, scanning the second radius of 2C, the vibration angle of the second scanning unit is in the case of a2, w = 2c * sin ((a2) / 2) * 2 = R * sin ((a1) / 2) * 2, and R a1 and because step S101 in the third step has been determined, it is possible to obtain the vibration angle of the second scanning unit a2.

[0058] 最后,根据第二扫描半径2c和第二振动角度a2,对扫描线在以任意扫描角度9 和扫描距离c扫描样品得到的OCT断层图像,校正还原成反应样品被扫描断面真实形状的图像。 [0058] Finally, according to a second scan angle and the second vibration 2c radius a2, the scanning lines 9 and the scanning angle at any scanning distance c OCT scan samples obtained tomographic image, the correction reduction reaction sample is scanned into the true shape of the cross section image.

[0059] 需要再次强调的的是,正如前面所说,即使扫描单元的扫描线的扫描半径R和扫描角度0不断变化,但是扫描单元的位置〇、0CT系统探测深度h和样品OCT图像顶端的扫描范围w(即OCT系统的扫描范围)均不会变化。 [0059] It bears repeating that, as previously mentioned, even if the scan lines of the scan radius R of the scanning unit and changing the scanning angle 0, but the position of the scanning unit square, and the depth h 0CT system detects an OCT image at the top of the sample scanning range w (i.e., the scanning range of an OCT system) will not change.

[0060] 在本专利申请中,被扫描样品为小块的长条状,其断面呈如图5所示的两条平行线形状。 [0060] In the present patent application, it is scanned into small elongated samples, which were cross-sectional shape shown in two parallel lines in FIG. 5. 进一步地,被扫描样品优先选择为小块的长条玻璃片。 Further, the preferred sample is scanned into small glass strip.

[0061] 在本专利申请中,扫描单元优先选择为XY振镜。 [0061] In the present patent application, the scanning unit preferentially selects the XY galvanometer.

[0062] 以上所述仅为本发明的较佳实施例而已,并不用以限制本发明,凡在本发明的精神和原则之内所作的任何修改、等同替换和改进等,均应包含在本发明的保护范围之内。 [0062] The foregoing is only preferred embodiments of the present invention but are not intended to limit the present invention, any modifications within the spirit and principle of the present invention, equivalent substitutions and improvements should be included in the present within the scope of the invention.

Claims (12)

  1. 1. 一种OCT图像的校正方法,其特征在于,包括如下步骤: OCT系统的扫描单元扫描样品,得到OCT图像,确定出至少包括有第一扫描半径R和第一振动角度(α 1)的若干参数; 调整所述扫描单元的扫描线角度,分别在水平方向和垂直方向扫描样品,求出长轴2a 和短轴2b,构建出椭圆公式, 1. A calibration method OCT images, wherein, comprising the steps of: a scanning unit scanning the sample OCT system, to obtain an OCT image, comprising determining at least a first scan and a first radius R vibration angle (α 1) of several parameters; adjusting the angle of the scanning unit scanning line, samples were scanned in the horizontal and vertical directions, to obtain a major axis and a minor axis 2a 2B, constructed ellipse equation,
    Figure CN104899837AC00021
    定义扫描单元的扫描线在任意扫描角度Θ时的第二扫描半径为2c,根据椭圆公式 A second radial scan line scanning of the scanning unit defined in an arbitrary scan angle Θ to 2c, according to the elliptical equation
    Figure CN104899837AC00022
    求得扫描距离c和第二振动角度(α 2); 利用所述扫描距离c和所述第二振动角度(α 2)将在以任意扫描角度Θ扫描样品得到的OCT图像还原成被扫描样品真实的形状; 其中,Θ为扫描单元的扫描线和水平方向构成的夹角,为已知值。 Obtained scanning distance c and the second vibration angle (α 2); c with the scanning distance and the second vibration angle (α 2) the sample is reduced to be scanned OCT images at any scan angle Θ scanning the sample obtained true shape; wherein, [Theta] is the angle between the horizontal scanning lines and the scanning unit is configured as a known value.
  2. 2. 如权利要求1所述的OCT图像的校正方法,其特征在于,所述若干参数还包括OCT 图像顶端宽度w,根据公式w = R*sin (( α 1)/2) *2求得。 2. The calibration method according to claim OCT image. 1, wherein the plurality of parameters further includes an OCT image tip width w, ((α 1) / 2) * 2 calculated according to the formula w = R * sin .
  3. 3. 如权利要求2所述的OCT图像的校正方法,其特征在于:所述R = Η1+Η3 ;其中,Hl 为所述OCT图像顶端到被扫描样品处于第一位置时的距离,Η3为所述第一位置到所述扫描单元的中心位置〇的距离。 3. The OCT image correction method according to claim 2, wherein: said R = Η1 + Η3; wherein, the OCT image on Hl is the top sample to be scanned at a distance when the first position, is not larger from the center position of the square of the scanning unit to the first position.
  4. 4. 如权利要求3所述的OCT图像的校正方法,其特征在于:所述Hl = hl*ratio_h ;其中,hi为所述OCT图像顶端到所述第一位置的像素距离;ratioj!为实际距离和像素的比例关系,:ratio_h = dist/(h2_hl) ;dist为所述样品由所述第一位置沿着扫描单元的位置〇移动到第二位置的距离,h2为所述OCT图像顶端到所述第二位置的像素距离;hi和h2通过对所述OCT图像进行边界检测得到;所述OCT图像的横向像素数定义为W,纵向像素数定义为H。 4. The calibration method according to OCT images claimed in claim 3, wherein: said Hl = hl * ratio_h; wherein, Hi is the top to the OCT image from the first pixel position; ratioj actual! and the ratio between the pixel distance,: ratio_h = dist / (h2_hl); dist said sample from said first position along a movement position of the scanning unit to the square of the distance to the second position, h2 is the OCT image to top the distance of the second pixel position; Hi and h2 by the boundary detection obtained OCT image; defining the number of horizontal pixels OCT image is W, the number of vertical pixels is defined as H.
  5. 5. 如权利要求4所述的OCT图像的校正方法,其特征在于:(a I) /2 = arctan ((up_ w - low_w)/2/dist), H3 = 2*tan ( α /2)/up_w ; 其中,up_w为所述样品处于所述第一位置时,所述扫描单元扫描所述样品的范围; low_w为所述样品处于所述第二位置时,所述扫描单元扫描所述样品的范围。 5. The OCT image correction method according to claim 4, wherein: (a I) / 2 = arctan ((up_ w - low_w) / 2 / dist), H3 = 2 * tan (α / 2) / up_w; wherein, up_w the sample is in the first position, the range of the scanning unit scans the sample; low_w when said sample is in said second position, said scanning means scans said sample range.
  6. 6. 如权利要求5所述的OCT图像的校正方法,其特征在于:所述第二振动角度(α 2) 根据公式R*sin(( a 1)/2) = 2c*sin(( a 2)/2)求得。 6. The OCT image correction method according to claim 5, wherein: said second vibration angle (α 2) according to the formula R * sin ((a 1) / 2) = 2c * sin ((a 2 ) / 2) is obtained.
  7. 7. 如权利要求5所述的OCT图像的校正方法,其特征在于:所述样品分别处于所述第一位置和所述第二位置时,所述扫描单元扫描所述样品,得到第一OCT图像和第二OCT图像;所述第一OCT图像和所述第二OCT图像的横向像素数为所述W,纵向像素数为所述H。 7. OCT image correction method according to claim 5, wherein: each said sample in said first position and said second position, the scanning unit scans the sample, to give a first OCT OCT image and the second image; the number of horizontal pixels of the first image and the second OCT OCT image of the W, the longitudinal number of pixels is H.
  8. 8. 如权利要求4所述的OCT图像的校正方法,其特征在于:所述若干参数还包括扫描单元的探测深度h,h = ratio_h*H。 8. OCT image correction method according to claim 4, wherein: said plurality of parameters further includes a scanning unit depths of h, h = ratio_h * H.
  9. 9. 如权利要求4-5中任一项所述的OCT图像的校正方法,其特征在于:所述样品设置在一夹具上,通过移动夹具实现所述样品由所述第一位置移动到所述第二位置。 9. The method of correcting an OCT images according to any of claims 4-5, characterized in that: said sample is set on a jig, the sample is achieved by moving the clamp from the first position to the said second position.
  10. 10. 如权利要求4-5中任一项所述的OCT图像的校正方法,其特征在于:所述up_w和所述low_w的值通过距离测量装置读出;所述距离测量装置暴露在所述OCT图像中。 10. The OCT image correction method according to any one of claims 4-5, wherein: said value of said low_w up_w and readout by the distance measuring means; said distance measuring device is exposed in the OCT image.
  11. 11. 如权利要求1-8中任一项所述的OCT图像的校正方法,其特征在于:所述样品为条块状玻璃片。 11 OCT image correction method according to any one of the 1-8 claims, wherein: the sample is a bulk glass sheet strip.
  12. 12. 如权利要求1-8中任一项所述的OCT图像的校正方法,其特征在于:所述扫描单元为XY振镜。 12. The OCT image correction method according to any one of the 1-8 claims, wherein: said XY scanning unit is a galvanometer.
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