2009Efficient test pattern compression techniques based on complementary Huffman coding
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March 2009High Quality DXT Compression using OpenCL for CUDAIgnacio Castaño*******************Document Change HistoryVersion Date Responsible Reason for Change0.1 02/01/2007 Ignacio Castaño First draft0.2 19/03/2009 Timo Stich OpenCL versionAbstractDXT is a fixed ratio compression format designed for real-time hardware decompression of textures. While it’s also possible to encode DXT textures in real-time, the quality of the resulting images is far from the optimal. In this white paper we will overview a more expensivecompression algorithm that produces high quality results and we will see how to implement it using CUDA to obtain much higher performance than the equivalent CPU implementation.MotivationWith the increasing number of assets and texture size in recent games, the time required to process those assets is growing dramatically. DXT texture compression takes a large portion of this time. High quality DXT compression algorithms are very expensive and while there are faster alternatives [1][9], the resulting quality of those simplified methods is not very high. The brute force nature of these compression algorithms makes them suitable to be parallelized and adapted to the GPU. Cheaper compression algorithms have already been implemented [2] on the GPU using traditional GPGPU approaches. However, with the traditional GPGPU programming model it’s not possible to implement more complex algorithms where threads need to share data and synchronize.How Does It Work?In this paper we will see how to use CUDA to implement a high quality DXT1 texturecompression algorithm in parallel. The algorithm that we have chosen is the cluster fit algorithm as described by Simon Brown [3]. We will first provide a brief overview of the algorithm and then we will describe how did we parallelize and implement it in CUDA.DXT1 FormatDXT1 is a fixed ratio compression scheme that partitions the image into 4x4 blocks. Each block is encoded with two 16 bit colors in RGB 5-6-5 format and a 4x4 bitmap with 2 bits per pixel. Figure 1 shows the layout of the block.Figure 1. DXT1 block layoutThe block colors are reconstructed by interpolating one or two additional colors between the given ones and indexing these and the original colors with the bitmap bits. The number of interpolated colors is chosen depending on whether the value of ‘Color 0’ is lower or greater than ‘Color 1’.BitmapColorsThe total size of the block is 64 bits. That means that this scheme achieves a 6:1 compression ratio. For more details on the DXT1 format see the specification of the OpenGL S3TCextension [4].Cluster FitIn general, finding the best two points that minimize the error of a DXT encoded block is a highly discontinuous optimization problem. However, if we assume that the indices of the block are known the problem becomes a linear optimization problem instead: minimize the distance from each color of the block to the corresponding color of the palette.Unfortunately, the indices are not known in advance. We would have to test them all to find the best solution. Simon Brown [3] suggested pruning the search space by considering only the indices that preserve the order of the points along the least squares line.Doing that allows us to reduce the number of indices for which we have to optimize theendpoints. Simon Brown provided a library [5] that implements this algorithm. We use thislibrary as a reference to compare the correctness and performance of our CUDAimplementation.The next section goes over the implementation details.OpenCL ImplementationPartitioning the ProblemWe have chosen to use a single work group to compress each 4x4 color block. Work items that process a single block need to cooperate with each other, but DXT blocks are independent and do not need synchronization or communication. For this reason the number of workgroups is equal to the number of blocks in the image.We also parameterize the problem so that we can change the number of work items per block to determine what configuration provides better performance. For now, we will just say that the number of work items is N and later we will discuss what the best configuration is.During the first part of the algorithm, only 16 work items out of N are active. These work items start reading the input colors and loading them to local memory.Finding the best fit lineTo find a line that best approximates a set of points is a well known regression problem. The colors of the block form a cloud of points in 3D space. This can be solved by computing the largest eigenvector of the covariance matrix. This vector gives us the direction of the line.Each element of the covariance matrix is just the sum of the products of different colorcomponents. We implement these sums using parallel reductions.Once we have the covariance matrix we just need to compute its first eigenvector. We haven’t found an efficient way of doing this step in parallel. Instead, we use a very cheap sequential method that doesn’t add much to the overall execution time of the group.Since we only need the dominant eigenvector, we can compute it directly using the Power Method [6]. This method is an iterative method that returns the largest eigenvector and only requires a single matrix vector product per iteration. Our tests indicate that in most cases 8iterations are more than enough to obtain an accurate result.Once we have the direction of the best fit line we project the colors onto it and sort them along the line using brute force parallel sort. This is achieved by comparing all the elements against each other as follows:cmp[tid] = (values[0] < values[tid]);cmp[tid] += (values[1] < values[tid]);cmp[tid] += (values[2] < values[tid]);cmp[tid] += (values[3] < values[tid]);cmp[tid] += (values[4] < values[tid]);cmp[tid] += (values[5] < values[tid]);cmp[tid] += (values[6] < values[tid]);cmp[tid] += (values[7] < values[tid]);cmp[tid] += (values[8] < values[tid]);cmp[tid] += (values[9] < values[tid]);cmp[tid] += (values[10] < values[tid]);cmp[tid] += (values[11] < values[tid]);cmp[tid] += (values[12] < values[tid]);cmp[tid] += (values[13] < values[tid]);cmp[tid] += (values[14] < values[tid]);cmp[tid] += (values[15] < values[tid]);The result of this search is an index array that references the sorted values. However, this algorithm has a flaw, if two colors are equal or are projected to the same location of the line, the indices of these two colors will end up with the same value. We solve this problem comparing all the indices against each other and incrementing one of them if they are equal:if (tid > 0 && cmp[tid] == cmp[0]) ++cmp[tid];if (tid > 1 && cmp[tid] == cmp[1]) ++cmp[tid];if (tid > 2 && cmp[tid] == cmp[2]) ++cmp[tid];if (tid > 3 && cmp[tid] == cmp[3]) ++cmp[tid];if (tid > 4 && cmp[tid] == cmp[4]) ++cmp[tid];if (tid > 5 && cmp[tid] == cmp[5]) ++cmp[tid];if (tid > 6 && cmp[tid] == cmp[6]) ++cmp[tid];if (tid > 7 && cmp[tid] == cmp[7]) ++cmp[tid];if (tid > 8 && cmp[tid] == cmp[8]) ++cmp[tid];if (tid > 9 && cmp[tid] == cmp[9]) ++cmp[tid];if (tid > 10 && cmp[tid] == cmp[10]) ++cmp[tid];if (tid > 11 && cmp[tid] == cmp[11]) ++cmp[tid];if (tid > 12 && cmp[tid] == cmp[12]) ++cmp[tid];if (tid > 13 && cmp[tid] == cmp[13]) ++cmp[tid];if (tid > 14 && cmp[tid] == cmp[14]) ++cmp[tid];During all these steps only 16 work items are being used. For this reason, it’s not necessary to synchronize them. All computations are done in parallel and at the same time step, because 16 is less than the warp size on NVIDIA GPUs.Index evaluationAll the possible ways in which colors can be clustered while preserving the order on the line are known in advance and for each clustering there’s a corresponding index. For 4 clusters there are 975 indices that need to be tested, while for 3 clusters there are only 151. We pre-compute these indices and store them in global memory.We have to test all these indices and determine which one produces the lowest error. In general there are indices than work items. So, we partition the total number of indices by the number of work items and each work item loops over the set of indices assigned to it. It’s tempting to store the indices in constant memory, but since indices are used only once for each work group, and since each work item accesses a different element, coalesced global memory loads perform better than constant loads.Solving the Least Squares ProblemFor each index we have to solve an optimization problem. We have to find the two end points that produce the lowest error. For each input color we know what index it’s assigned to it, so we have 16 equations like this:i i i x b a =+βαWhere {}i i βα, are {}0,1, {}32,31, {}21,21, {}31,32 or {}1,0 depending on the index and the interpolation mode. We look for the colors a and b that minimize the least square error of these equations. The solution of that least squares problem is the following:∑∑⋅∑∑∑∑= −i i i i i ii i i i x x b a βαββαβαα122 Note: The matrix inverse is constant for each index set, but it’s cheaper to compute it everytime on the kernel than to load it from global memory. That’s not the case of the CPU implementation.Computing the ErrorOnce we have a potential solution we have to compute its error. However, colors aren’t stored with full precision in the DXT block, so we have to quantize them to 5-6-5 to estimate the error accurately. In addition to that, we also have to take in mind that the hardware expands thequantized color components to 8 bits replicating the highest bits on the lower part of the byte as follows:R = (R << 3) | (R >> 2); G = (G << 2) | (G >> 4); B = (B << 3) | (B >> 2);Converting the floating point colors to integers, clamping, bit expanding and converting them back to float can be time consuming. Instead of that, we clamp the color components, round the floats to integers and approximate the bit expansion using a multiplication. We found the factors that produce the lowest error using an offline optimization that minimized the average error.r = round(clamp(r,0.0f,1.0f) * 31.0f); g = round(clamp(g,0.0f,1.0f) * 63.0f); b = round(clamp(b,0.0f,1.0f) * 31.0f); r *= 0.03227752766457f; g *= 0.01583151765563f; b *= 0.03227752766457f;Our experiment show that in most cases the approximation produces the same solution as the accurate solution.Selecting the Best SolutionFinally, each work item has evaluated the error of a few indices and has a candidate solution.To determine which work item has the solution that produces the lowest error, we store the errors in local memory and use a parallel reduction to find the minimum. The winning work item writes the endpoints and indices of the DXT block back to global memory.Implementation DetailsThe source code is divided into the following files:•DXTCompression.cl: This file contains OpenCL implementation of the algorithm described here.•permutations.h: This file contains the code used to precompute the indices.dds.h: This file contains the DDS file header definition. PerformanceWe have measured the performance of the algorithm on different GPUs and CPUscompressing the standard Lena. The design of the algorithm makes it insensitive to the actual content of the image. So, the performance depends only on the size of the image.Figure 2. Standard picture used for our tests.As shown in Table 1, the GPU compressor is at least 10x faster than our best CPUimplementation. The version of the compressor that runs on the CPU uses a SSE2 optimized implementation of the cluster fit algorithm. This implementation pre-computes the factors that are necessary to solve the least squares problem, while the GPU implementation computes them on the fly. Without this CPU optimization the difference between the CPU and GPU version is even larger.Table 1. Performance ResultsImage TeslaC1060 Geforce8800 GTXIntel Core 2X6800AMD Athlon64 DualCore 4400Lena512x51283.35 ms 208.69 ms 563.0 ms 1,251.0 msWe also experimented with different number of work-items, and as indicated in Table 2 we found out that it performed better with the minimum number.Table 2. Number of Work Items64 128 25654.66 ms 86.39 ms 96.13 msThe reason why the algorithm runs faster with a low number of work items is because during the first and last sections of the code only a small subset of work items is active.A future improvement would be to reorganize the code to eliminate or minimize these stagesof the algorithm. This could be achieved by loading multiple color blocks and processing them in parallel inside of the same work group.ConclusionWe have shown how it is possible to use OpenCL to implement an existing CPU algorithm in parallel to run on the GPU, and obtain an order of magnitude performance improvement. We hope this will encourage developers to attempt to accelerate other computationally-intensive offline processing using the GPU.High Quality DXT Compression using CUDAMarch 2009 9References[1] “Real-Time DXT Compression”, J.M.P. van Waveren. /cd/ids/developer/asmo-na/eng/324337.htm[2] “Compressing Dynamically Generated Textures on the GPU”, Oskar Alexandersson, Christoffer Gurell, Tomas Akenine-Möller. http://graphics.cs.lth.se/research/papers/gputc2006/[3] “DXT Compression Techniques”, Simon Brown. /?article=dxt[4] “OpenGL S3TC extension spec”, Pat Brown. /registry/specs/EXT/texture_compression_s3tc.txt[5] “Squish – DXT Compression Library”, Simon Brown. /?code=squish[6] “Eigenvalues and Eigenvectors”, Dr. E. Garcia. /information-retrieval-tutorial/matrix-tutorial-3-eigenvalues-eigenvectors.html[7] “An Experimental Analysis of Parallel Sorting Algorithms”, Guy E. Blelloch, C. Greg Plaxton, Charles E. Leiserson, Stephen J. Smith /blelloch98experimental.html[8] “NVIDIA CUDA Compute Unified Device Architecture Programming Guide”.[9] NVIDIA OpenGL SDK 10 “Compress DXT” sample /SDK/10/opengl/samples.html#compress_DXTNVIDIA Corporation2701 San Tomas ExpresswaySanta Clara, CA 95050 NoticeALL NVIDIA DESIGN SPECIFICATIONS, REFERENCE BOARDS, FILES, DRAWINGS, DIAGNOSTICS, LISTS, AND OTHER DOCUMENTS (TOGETHER AND SEPARATELY, “MATERIALS”) ARE BEING PROVIDED “AS IS.” NVIDIA MAKES NO WARRANTIES, EXPRESSED, IMPLIED, STATUTORY, OR OTHERWISE WITH RESPECT TO THE MATERIALS, AND EXPRESSLY DISCLAIMS ALL IMPLIED WARRANTIES OF NONINFRINGEMENT,MERCHANTABILITY, AND FITNESS FOR A PARTICULAR PURPOSE.Information furnished is believed to be accurate and reliable. However, NVIDIA Corporation assumes no responsibility for the consequences of use of such information or for any infringement of patents or otherrights of third parties that may result from its use. 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This trial examination produced by Insight Publications is NOT an official VCAA paper for the 2009 Specialist Mathematics written examination 2.This examination paper is licensed to be printed, photocopied or placed on the school intranet and used only within theconfines of the purchasing school for examining their students. No trial examination or part thereof may be issued or passed on to any other party including other schools, practising or non-practising teachers, tutors, parents, websites or publishing agencies without the written consent of Insight Publications. INSIGHTTrial Exam Paper2009SPECIALIST MATHEMATICSWritten examination 2STUDENT NAME:QUESTION AND ANSWER BOOKReading time: 15 minutes Writing time: 2 hoursStructure of bookSectionNumber of questionsNumber of questions to beansweredNumber of marks1 22 22 22 2 5 5 58Total 80• Students are permitted to bring the following items into the examination: pens, pencils, highlighters,erasers, sharpeners, rulers, a protractor, set-squares, aids for curve sketching, once bound reference, one approved graphics calculator or CAS (memory DOES NOT need to be cleared) and, if desired, one scientific calculator.• Students are NOT permitted to bring sheets of paper or white out liquid/tape into the examination.Materials provided• The question and answer book of 29 pages with a separate sheet of miscellaneous formulas. • Answer sheet for multiple-choice questions.Instructions• Write your name in the box provided and on the multiple-choice answer sheet. • Remove the formula sheet during reading time. • You must answer the questions in English.At the end of the exam• Place the multiple-choice answer sheet inside the front cover of this book.Students are NOT permitted to bring mobile phones or any other electronic devices into the examinatio n.This page is blankSECTION 1 – continuedTURN OVERSECTION 1Instructions for Section 1Answer all questions in pencil on the multiple-choice answer sheet provided. Choose the response that is correct for the question.One mark will be awarded for a correct answer; no marks will be awarded for an incorrect answer.Marks are not deducted for incorrect answers.No marks will be awarded if more than one answer is completed for any question. Take the acceleration due to gravity to have magnitude g m/s 2, where g = 9.8Question 1The parametric equations x = 2 sec(t + 4) – 2 and y = 3 tan(t + 4) + 1 define a relation given by A. 22(2)(1)= 149x y ++−B. 22(2)(1)149x y +−+= C. 22(2)(1)132x y +−−= D. 22(2)(1)149x y +−−= E. 22(2)(1)123x y +−−=SECTION 1 – continuedThe region of the complex plane defined by ()⎭⎬⎫⎩⎨⎧<−≤−4)1(4:ππz i Arg z isA. B.C. D.E.z )z )Im(z )SECTION 1 – continuedTURN OVERThe maximal domain and range of the function f (x ) = 3artan (2x – π) are given by A. d f = (π, 3π) and r f = RB. d f = R and r f = [π2−, π2]C. d f = R and r f = (3π2−, 3π2)D. d f = R and r f = (π4, 3π4)E.d f = (π2−, π2) and r f = RQuestion 4If z 2 – z – 2 is a factor of 432()310620,P z z z z z z C =−+−−∈, then all of the factors must be A. z – 2, z + 1, z – 1 + 3i and z – 1 – 3i B. z – 2, z + 1, z – 1 + 3i and z + 1 – 3i C. z + 2, z – 1, z – 3 + i and z – 3 – i D. z – 2, z + 1, z + 3 + i and z + 3 – i E . z – 2, z + 1, z + 2 and z + 5The graph of ()= is shown below.y f x′xSECTION 1 – Question 5 – continuedSECTION 1 – continuedTURN OVERIf f (0) = 1, then the graph of y = f (x ) could be A.xB.xC. E.D.Question 6The graph of1()yf x= is shown below.SECTION 1 – Question 6 – continuedSECTION 1 – continuedTURN OVERThe graph of y = f (x ) could be A.xB.xC.D.E.xxxSECTION 1 – continuedQuestion 7The solutions to i a a z 32+=, where z C ∈and a R +∈, are A.)2i ±+ B.)2i ±+ C.(1)2±+ D.(1)2±+ E.(1)2i ±−Question 8For the vectors ~~~~~~~~~~~~32, 22 and 710a i j k b i j k c x i j k =−+=+−=−+ to be linearly dependent, the value of x must be A. 4 B. 7 C. 2 D. – 7 E. –2Question 9The graph of the relation (){}C z z z z z ∈=−,8Re 2: would be A.a circle with centre (0, 0) and radius . B. a circle with centre (–1, 0) and radius 3.C. a straight line with gradient 2 and y -intercept of 8.D. a straight line with gradient 1 and y -intercept of 8.E. a circle with centre (1, 0) and radius 3.SECTION 1 – continuedTURN OVERThe rule for the function graphed above, where a > 0, could be A. πcosec x y a ⎛⎞=⎜⎟⎝⎠B. πsec x y a ⎛⎞=⎜⎟⎝⎠C. πsec 2a y x a ⎛⎞⎛⎞=−⎜⎟⎜⎟⎝⎠⎝⎠D. πcosec x y a ⎛⎞=−⎜⎟⎝⎠E.πcosec 2a y x a ⎛⎞⎛⎞=+⎜⎟⎜⎟⎝⎠⎝⎠SECTION 1 – continued120234x dx x −⎛⎞⎜⎟−⎝⎠∫ is equal to A.9log 2e ⎛⎞⎜⎟⎝⎠B. log 18eC. 0D. log 72eE.9log 8e ⎛⎞⎜⎟⎝⎠Question 12The gradient of the tangent to the curve 2log ()e x y x y −= at the point where y = e is A. 3 B. –1 C. –3 D. 1E. 21Question 13Using a suitable substitution, 110cos x dx −⎛⎞⎜⎟is equal to A.22 u du ∫B.π2π3u du ∫C.212u du ∫ D.π402 u du ∫E.π2π21 du u ∫SECTION 1 – continuedTURN OVERThe direction (slope) field for a certain first-order differential equation is shown above. The differential equation could be A. 211dy dx x =+ B. 1tan dy x dx −= C. 221dy x y dx =++D. 1dy x dx =+E. 11dy dxx y =++yxSECTION 1 – continuedIf log ()e dyx dx= and y (1) = 2, then the value of y when x = 3 can be found by evaluating A. 321log () e t dt +∫B. 3112dt t +∫C. 312log () e t dt +∫D. 321log () e t dt −∫E.213log () e t dt +∫Question 16The position vectors of two moving particles, R and S , at any time t seconds are given by~~~4r at i j =− and 2~~2, 0, s t i t j t a R =+≥∈, respectively. The angle between the directions of the two particles at t = 1 is A. 69.3° B. 45° C. 35.3° D. 19.5° E. dependent on the value of a .Question 17The volume of a tank is given by 520.4πV h =, where h cm is the depth of water in the tank at time t minutes. Water leaks from the tank at a rate of 16 cm 3/minute. The depth of water in thetank when the height is decreasing at a rate of 2π cm/minute isA. 16 cmB. 8 cmC. 4π cmD. 4 cmE. 8π cmSECTION 1 – continuedTURN OVERA skier of mass 80 kilograms slides from rest down a straight slope inclined at 60° to the vertical. Assuming it is a smooth slope, the speed of the skier after moving 100 metres down the slope is nearest to A. 41.2 m/s B. 22.1 m/s C. 31.3 m/s D. 44.3 m/s E. 10 m/sQuestion 19A mass of 4 kilograms is at rest when two forces, 1~~~(3) F i j =−newtons and2~~~(2) F i j =−newtons, act on it. The time taken for the mass to travel 10 metres isA. 1 sB. 2 sC. 4 sD. 5 sE. 8 sQuestion 20The velocity of a particle moving in a straight line is given by 2()cos()v x x =, where x is the displacement from the origin O . The acceleration of the particle is A. 2()2sin ()a x x x =−B. ()cos (2)a x x =C. 2()2tan ()a x x x =−D. 2()sin (2)a x x x =−E.22()2tan ()sec()a x x x x =−The magnitude of the horizontal force, F newtons, required to hold a 30 kilogram child in equilibrium on a swinging rope, as shown in the diagram above, isA.30 tan70gDB.30sin70 sin20 g DDC.30sin20g°D.30 sin70gDE. 30tan20g°Question 22A lift travelling upwards accelerates at a m/s2 (a > 0) with a person of mass 100 kilograms standing on a set of weight scales in the lift. It then decelerates at twice the magnitude of the acceleration. The magnitude of the change in the reading on the scales will beA. 100a kgB. 200g kgC. 300a kgD. 100(g + a) kgE. –100a kgEND OF SECTION 1SECTION 2 – Question 1 – continuedTURN OVERSECTION 2Instructions for Section 2Answer all the questions in the spaces provided.A decimal approximation will not be accepted if an exact answer is required to a question. In questions where more than one mark is available, approximate working must be shown. Unless otherwise indicated, the diagrams in this book have not been drawn to scale. Take the acceleration due to gravity to have magnitude g m/s 2, where g = 9.8Question 1The diagram below shows the profile of a symmetrical small bowl ABCD . The bowl isgenerated by rotating the area between the curve AB and the y -axis about the y -axis. The top and base of the bowl have radii of 4 cm and 2 cm, respectively, and the height of the bowl is π cm.The curve AB can be modelled by the function 1sin (), [2,4].y a bx c x −=−∈ a.Show that a = 2, b = 21and c = 1.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________3 marksyxb. If h cm is the height of water in the bowl at any time, express the volume of water,V cm3, in terms of h.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________4 marksc.Hence, find the exact volume of water in a full bowl.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________1 markSECTION 2 – Question 1 – continuedd. To the nearest millimetre, what would be the height of the water when the bowl is filled tohalf its capacity?______________________________________________________________________________________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________2 marksTotal 3 + 4 + 1 + 2 = 10 marksQuestion 2A miniature racing car of mass 6 kilograms is propelled from rest up a rough ramp19.6 metres long and inclined at an angle of 30° to the horizontal. The car is powered up the ramp by a constant force of 10g newtons. This causes the car to accelerate at 9.8 m/s2.bel the forces acting on the car as it moves up the ramp.1 markb. Show that at the top of the ramp the car is g metres above the ground and its speed is2g m/s when it leaves the ramp.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________2 marksSECTION 2 – Question 2 – continuedTURN OVERSECTION 2 – Question 2 – continuedc.Calculate the exact value of the coefficient of friction.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________2 marksWhen the car leaves the ramp it is only subject to the force of gravity.Take ~i as the unit vector in the horizontal direction and ~j as the unit vector in the verticaldirection from the point on the ground, directly below the top of the ramp. d.Determine the velocity vector ~v and the position vector ~r of the car at any time t seconds.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________2 markse. Find the exact Cartesian equation of the path of the car after it leaves the ramp.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________2 marksf. Find the exact magnitude of the momentum of the car when it hits the ground.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________3 marksTotal 1 + 2 + 2 + 2 + 2 + 3 = 12 marksSECTION 2 – continuedTURN OVERSECTION 2 – Question 3 – continuedQuestion 3a. Given w a bi =+, where R b a ∈, and 0>b . If2=+w w and 2=w w , show that i w +=1. ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________2 marksb. If i v 31+=,i. Find wv in simplest exact Cartesian form. ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________1 markii. Find wv in polar form. ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________2 marksSECTION 2 – Question 3 – continuedTURN OVERc. Hence , express πtan 12⎛⎞⎜⎟⎝⎠in the form a where a and b are positive integers. ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________3 marksd. S is a subset of the complex plane, which is defined by{}C z w z z S ∈=−=,1:Plot the points v and w and sketch the relation defined by S on the Argand diagrambelow.2 marks Re zRe(z )SECTION 2 – continuede.T is a subset of the complex plane defined by{}C z w z v z z T ∈−=−=,: i. Express the equation for the relation defined by T in Cartesian form.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________1 markii. Part of T is a chord to the relation {}C z w z z S ∈=−=,1: Find the exact length of this chord in the form c ba , where a ,b andc areintegers.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________3 marksTotal 2 + 3 + 3 + 2 + 4 = 14 marksSECTION 2 – Question 4 – continuedTURN OVERQuestion 4A tank contains 100 litres of sugar solution with a concentration of 0.05 kg/L. A sugarsolution of concentration 0.1 kg/L flows into the tank at a rate of 2 L/min. The mixture, which is kept uniform by stirring, flows out of the tank at a rate of 2 L/min. After t minutes the tank contains x kilograms of sugar.a. Show that the differential equation for x in terms of t is 1050dx x dt −= kg/min. ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________1 markb. Solve this differential equation to give x as a function of t .___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________3 marksc. Calculate the amount of sugar in the tank after 2 minutes. Give your answer correct tothree decimal places.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________1 markd.If this situation continued for a long period of time, how much sugar would be presentin the tank?___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________1 marke. If the outflow from the tank was 1 L/min instead of 2 L/min, set up the new differentialequation for x in terms of t.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________1 markf. For the differential equation from part e. use Euler’s method, with increments of1 minute, to predict the amount of sugar in the tank after2 minutes. Give your answercorrect to three decimal places.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________2 marksTotal 1 + 3 + 1 + 1 + 1 + 2 = 9 marksSECTION 2 – continuedSECTION 2 – Question 5 – continuedTURN OVERQuestion 5At 10 a.m. an aircraft is flying at an altitude of (e 2 – e ) km, 500 km north and 440 km east of a point T (0, 0, 0), which is its touchdown point on a horizontal runway.The position of the aircraft relative to the point T is given by the vector20.02~~~~2420() =+(500240.28)+(), where , 5c t r t a i t t j e e k a c R t −⎛⎞+−+−∈⎜⎟+⎝⎠. ~r is in kilometres and t is the time in minutes after 10 a.m. ~i is the unit vector in an easterly direction, ~j is the unit vector in a northerly direction and ~k is the unit vector representing the altitude of the aircraft. (Treat the aircraft as a point in this problem.)a. Show that a = –44 and c = 2.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________1 markb. Show that the aircraft touches down at point T at 10.50 a.m.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________1 markc. Show that the exact velocity of the aircraft at touchdown is ~~~~0.840.02. r'i j e k =−+− ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________2 marksd.Find the vertical angle to the runway at which the aircraft lands. Give your answer tothe nearest hundredth of a degree.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________3 markse.Relative to the point T, find the position vector~()p t of the aircraft on the runway when the aircraft stops.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________3 marksSECTION 2 – Question 5 – continuedf. A stationary fire engine is positioned 1.5 kilometres north and 200 metres west of T.Determine, to the nearest metre, the minimum distance between the fire engine and the aircraft on its path after touchdown.___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________3 marksTotal 1 + 1 + 2 + 3 + 3 + 3 = 13 marksEND OF QUESTION AND ANSWER BOOK。
液体芯片-飞行时间质谱技术在肿瘤早期诊断中的研究进展字体: 小中大| 打印发布: 2009-11-19 12:45:14 来源: 美国布鲁克道尔顿公司(Bruker Daltonic Inc.)相关产品ultrafleX III MALDI TOF & TOF/TOF质谱仪[点击打开]1.前言世界卫生组织最新资料显示,2005年,全球大约有760万人死于各种癌症,约1100万人患有癌症;中国死于恶性肿瘤的约160万人,约200万人患有癌症。
中国每年因癌症造成的损失高达数百亿,给家庭和社会带来沉重的经济负担。
据世界卫生组织估计,早期肿瘤治愈率可达83%。
美国从1960至1990的30年间,肿瘤患者的五年存活率由50%上升为63%,主要归功于早期诊断技术的发展,而中晚期的患者的五年存活率,几乎没有提高。
即临床早期诊断率每增加1%,全球就有约7.6万人免于死亡。
因此,大力推进早期诊断技术的发展,具有非常紧迫的经济和社会意义。
生理医学诺贝尔奖获得者Lee Hartwell博士在2004年第21届国际肿瘤标志学术会议上,就如何治疗癌症指出:早期诊断可以治疗癌症,早期诊断癌症的最新、最有效的方法是从血液中寻找肿瘤标志(TM),特别是其中的蛋白标志!目前临床常用的TM有20多种,如,前列腺特异性抗原(PSA)、癌胚抗原(CEA)或癌抗原125(CA125)等,但通常用于疗效观察、预后评价及预测复发,但能用于大规模人群的普查的TM寥寥无几。
临床上迫切需要研发出新的TM应用于肿瘤的早期诊断。
在现代蛋白质组学研究中出现的双向电泳技术、蛋白芯片,电喷雾质谱和飞行时间质谱、四级杆飞行时间质谱、傅立叶变换质谱技术,给临床肿瘤早期标志物的检测带来了划时代的革命,这些技术将传统的1次只能测定1个蛋白标志的方法革命性地发展到1次可测定数十个,甚至数百个蛋白的多项检测的水平,同时还极大地提高了诊断的灵敏度和特异性[1]。
2009年超级计算大会在美国举行,GPU运算成为热门话题佚名
【期刊名称】《《微型计算机》》
【年(卷),期】2009(000)034
【摘要】2009年11月16日,本年度的超级计算大会在美国俄勒冈州波特兰市举行。
本次大会期间.在NVIDIA(英伟达)公司的展台上演示了全新Testa20系列产品,它们全部基于代号为“Fermi”的全新GPU架构。
同时,包括田纳西大学的Jack Dongarra、东京工业大学的禅洲松冈(Satoshi Matsuoka),
【总页数】1页(P93)
【正文语种】中文
【中图分类】TP334.7
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因版权原因,仅展示原文概要,查看原文内容请购买。
基于多叶准直器日志文件的鼻咽癌调强放疗的累积剂量验证
作者:曹瑛, 杨振, 邱小平, 杨晓喻, 唐慧敏, 吕知平, 张子健, 雷明军, 刘归
作者单位:曹瑛,邱小平,唐慧敏(南华大学核科学技术学院,衡阳,421000), 杨振,杨晓喻,吕知平,张子健,雷明军,刘归(中南大学湘雅医院肿瘤科)
刊名:
中华放射医学与防护杂志
英文刊名:Chinese Journal of Radiological Medicine and Protection
年,卷(期):2015,35(9)
引用本文格式:曹瑛.杨振.邱小平.杨晓喻.唐慧敏.吕知平.张子健.雷明军.刘归基于多叶准直器日志文件的鼻咽癌调强放疗的累积剂量验证[期刊论文]-中华放射医学与防护杂志 2015(9)。
ReviewCationic amphiphilic peptides with cancer-selective toxicityFrank Schweizer ⁎Department of Chemistry,University of Manitoba,Winnipeg,MB,Canada R3T 2N2Department of Medical Microbiology,University of Manitoba,Winnipeg,MB,Canada R3T 2N2a b s t r a c ta r t i c l e i n f o Article history:Received 24April 2009Received in revised form 23July 2009Accepted 3August 2009Available online 14October 2009Keywords:Cationic amphiphilic peptide Cell-penetrating peptide Anticancer peptide Drug delivery PeptidomimeticDuring the last two decades cationic amphiphilic peptides and peptide sequences (CAPs)with cancer-selective toxicity have appeared.Based on their spectrum of anticancer activity CAPs can be divided into two major classes.The first class includes peptides that are highly potent against both bacteria and cancer cells,but not against mammalian cells.The second class includes peptides that are toxic to bacteria,and both mammalian cancer and non-cancer cells.Most antimicrobial and anticancer CAPs share a common membranolytic mode of action that results either in the selective disruption of the cancer cell membrane or permeation and swelling of mitochondria.The electrostatic attraction between the negatively charged membrane components of bacterial and cancer cells and CAPs is believed to play a crucial role in the disruption of bacterial and cancer cell membranes.This mode of action appears to bypass established resistance mechanisms.However,it is currently unclear as to why some CAPs kill cancer cells when others do not.In addition,non-membranolytic mode of actions of CAPs is increasingly recognized to contribute signi ficantly to the anticancer activity of certain CAPs.The development of CAP-based chemotherapeutics is complicated due to the traditionally poor pharmacokinetic properties and high manufacturing costs of peptides and the low intrinsic selectivity for cancer cells.Peptidomimetic approaches combined with novel selective delivery devices show promise in overcoming some of these obstacles.Furthermore,the ability of CAPs to bypass established resistance mechanisms provides an attractive strategy to develop novel lead structures for cancer treatment.©2009Elsevier B.V.All rights reserved.Contents1.Introduction ..............................................................1902.Membrane differences of cancer cells implicated in contributing to the selective permeability and toxicity of CAPs ................1912.1.Membranolytic mode of actions of CAPs against cancer cells ....................................1912.2.Non-membranolytic mode of actions of CAPs ...........................................1923.Chemical modi fications on CAPs designed to improve pharmacokinetics and reduce toxicity ..........................1924.Synergistic effects of CAPs with other anticancer drugs and occurrence of multidrug resistance .......................1935.Conclusions and outlook ........................................................193Acknowledgement..............................................................193References .. (193)1.IntroductionMost of the anticancer agents currently used are based on alkylating agents,antimetabolites and natural products that display little or no selectivity against normal mammalian cells,which results in severe side effects (Smith et al.,2007).These agents interact with intracellular targets and need to penetrate through membranes.Oncethey have reached the cytosol many of these alkylating agents are deactivated by resistance mechanisms that transport these com-pounds out of the cell before interacting with intracellular targets (Pérez-Tomás,2006).Clearly,in order to battle drug resistance and reduce normal cell cytotoxicity,the development of novel drugs and delivery systems with novel mode of actions and high cancer cell selectivity are crucial.In the search for new cancer agents and delivery systems innate immunity peptides and synthetic innate immunity peptidomimetics have recently attracted attention due to their novel mode of actions,decreased likelihood of resistance development,andEuropean Journal of Pharmacology 625(2009)190–194⁎Tel.:+12044747012;fax:+12044747608.E-mail address:schweize@cc.umanitoba.ca.0014-2999/$–see front matter ©2009Elsevier B.V.All rights reserved.doi:10.1016/j.ejphar.2009.08.043Contents lists available at ScienceDirectEuropean Journal of Pharmacologyj o u r n a l h o me p a g e :w w w.e l sev i e r.c om /l oc a te /e j p h a rlow intrinsic cytotoxicity (Hoskin and Ramamoorthy 2008;Papo and Shai,2005;Leuschner and Hansel,2004).Innate immunity peptides are produced by many organisms as a first line of defense to fight antimicrobial invaders (Zasloff,2002).In addition,many of these peptides modulate the immune response in mammals through a variety of mechanisms (Bowdish et al.,2005).Innate immunity peptides show great structural diversity,but have some common features,including:their relatively small size (usually less than 50amino acid residues),their cationic nature (due to the presence of mostly multiple arginine and lysine but also histidine residues),and a substantial portion of hydrophobic amino acids (around 50%)(Zasloff 2002;Hancock and Sahl,2006).For the remainder of this paper we will refer to these peptides as cationic amphiphilic peptides (CAPs).The exact mode of action of CAPs is not completely understood.However there is a consensus that most of the CAPs selectively disrupt cell membranes,possibly by transient pore formation or disruption of lipid packing (Giuliani et al.,2008).In addition,intracellular targets have also been suggested for some CAPs (Park et al.,2000).Many of these peptides are unstructured in solution.However,upon binding to bacterial membranes most of the longer peptides will adopt a well-de fined amphipathic structure that is crucial to enhance membrane permeability (Zasloff,2002;Hancock and Sahl,2006;Giuliani et al.,2008).It should be noted that a well-de fined amphipathic structure is not required.For instance,ultrashort di-,tri-and tetrapeptide sequences (Haug et al.,2008;Svensen et al.,2008;Strom et al.,2003)and ultrashort lipopeptides that contain a polar peptide-based cationic head group (consisting of positively charged arginine and lysine residues and a hydrophobic tail composed of a lipophilic acid or hydro-phobic amino acid(s))are also able to induce a strong antibacterial in vitro activity (Makovitski et al.,2008;Makovitzki and Shai,2005;Malina and Shai 2005).Some of these ultrashort antimicrobial peptides display little hemolytic activity,while others show strong hemolytic activity.It is believed that the selective bacterial cytotoxicity of CAPs is caused by the af finity of the net negative charge found on bacterial cell membranes in contrast to eukaryotic lipid bilayers,which are typically made up of zwitterionic phospholipids (Hancock and Sahl,2006).The overall positive charge of CAPs ensures accumulation at polyanionic microbial cell surfaces that contain acidic polymers such as lipopoly-sacharides and cell wall associated teichoic acids,in Gram-negative and Gram-positive bacteria,respectively.Subsequently,CAPs transit the outer membrane of Gram-negative bacteria via self-promoted uptake (Hancock and Lehrer,1998),insert themselves into the cytoplasmic membrane,and act by either disrupting the physical integrity of the bilayer via membrane thinning,transient poration and/or disruption of the barrier function,or translocating across the membrane and acting on internal targets (Hancock and Sahl,2006).Interestingly,many CAPs also have the ability to kill cancer cells.Based on their spectrum of anticancer activity CAPs can be divided into two major classes (Papo and Shai,2005).The first class includes peptides that are highly potent against both bacteria and cancer cells,but not against mammalian cells.This class of peptides consists ofinsect cecropins and magainins isolated from the skin of frogs and others.The second class includes peptides that are toxic to bacteria,and both mammalian cancer and non-cancer cells.Members of this class include the bee venom melittin,tachyplesin II isolated from the horseshoe crab,human neutrophil defensins,insect defensins and the human LL-37.However,it must be pointed out that many innate immunity-derived CAPs do not possess antitumor activity (Papo et al.,2003;Cruciani et al.,1991;Papo and Shai,2003;Shin et al.,2000;Srisailam et al.,2000).This raises the question,what are the molecular and structural factors of CAPs responsible for inducing cancer toxicity (Table 1)?2.Membrane differences of cancer cells implicated in contributing to the selective permeability and toxicity of CAPsA variety of fundamental differences exists between the cell membranes of cancer cells and normal cells that may explain the cancer-selective toxicity of certain CAPs.Cancer cell membranes typically carry a net negative charge due to elevated expression of anionic molecules such as phosphatidylserine (Utsugi et al.,1991;Dobrzynska et al.,2005),O-glycosylated mucins (high molecular weight glycoproteins consisting of O-glycosidic linkages between serine and threonine that are rich in negatively charged saccharides comprised of sialic acids and sulfates)(Yoon et al.,1997;Burdick et al.,1997),sialilated gangliosides (Lee et al.,2008)and heparan sulfates (Kleeff et al.,1998).In contrast,the surface of normal mammalian cell membranes is mainly composed of neutral zwitterionic phospholipids and sterols (Hoskin and Ramamoorthy,2008).It has been suggested that cholesterol,which is a major component of eukaryotic cell membranes (Simons and Ikonen,2000)may protect eukaryotic cells from the cytolytic effects of antimicrobial CAPs by altering membrane fluidity and thereby interfering with the membrane insertion of lytic peptides.This hypothesis is supported by the observation that membrane insertion of the CAP cecropin,is reduced when the cholesterol content of the membrane is increased (Steiner et al.,1988;Silvestro et al.,1997).Moreover,some breast and prostate cancer cell lines have recently been shown to possess elevated levels of cholesterol-rich lipid rafts (Li et al.,2006),which make these cell lines less susceptible to lysis by CAPs.Another difference between the membrane of cancer and non-cancer cells is based on the relatively higher number of microvilli on tumorigenic cells compared to normal cells (Zwaal and Schroit,1997).This increases the surface area of the tumor cell which may allow the cancer cell to interact with an increased number of CAPs (Chan et al.,1998).2.1.Membranolytic mode of actions of CAPs against cancer cells Similar to the mode of action of antimicrobial peptides that disrupt the bacterial cell wall,the anticancer properties of CAPs may involve selective lysis of the cancer cell membrane.Initial evidence of such a membranolytic mechanism was discovered in a study of magaininTable 1Selected naturally occurring CAPs with anticancer activities.Peptide SequenceAnticancer activityMelittin GIGAVLKVLTTGLPALISWIKRKRQQ Membranolytic Hristova et al.(2001);phospholipase A 2activator Sharma (1992,1993);phospholipase D activator Saini et al.(1999)Tachyplesin KWC 1FRVC 2YRGIC2YRRC 1RActivation of complement C1q Chen et al.(2005);induction of cancer cell differentiation Ouyang et al.(2002)LL-37LLGDFFRKSKEKIGKEFKRIVQRIKDFLRNLVPRTES Membranolytic Henzler-Wildmann et al.(2003)Cecropin B KWKVFKKIEKMGRNIRNGIVKAGPAIAVLGEAKAL Membranolytic Hung et al.(1999)Magainin 2GIGKFLHSAKKFGKAFVGEIMNS Membranolytic Cruciani et al.(1991)Buforin IIb RAGLQFPVGRLLRRLLRRLLR Mitochondria-dependent apoptosis,caspase 9activation Lee et al.(2008))Alloferon-1HGVSGHGOHGVHG Immunomodulatory Chernysh et al.(2002)Alloferon-2GVSGHGQHGVHGImmunomodulatory Chernysh et al.(2002)191F.Schweizer /European Journal of Pharmacology 625(2009)190–194and its synthetic analogues against hematopoietic and solid tumors (Cruciani et al.,1991).A crucialfinding of this study was the discovery that the negative charge gradient across the plasma membrane was crucial for membrane disruption and cytotoxicity.Furthermore, replacement of the L-amino acids in magainin by D-amino acids retained anticancer activity,thereby ruling out a receptor-mediated pathway.Similar results were obtained with other antimicrobial CAPs including melittin,cecropin,roctonin and others(Wade et al.,1990; Bessalle et al.,1990;Merrifield et al.,1995;Hetru et al.,2000). Subsequent studies using antimicrobial peptides containing both D-and L-amino acids against various cancer cell types revealed that the cells died after perturbation of the plasma membrane.Two poten-tial mechanisms can account for the membranolytic properties of CAPs;these are referred to as‘carpet’and‘barrel-stave’mechanisms. According to the carpet mechanism(Oren and Shai,1998),CAPs align themselves parallel to the negatively charged membrane surface in a carpet-like fashion until a critical threshold concentration is reached, after which the CAP permeates the membrane.Alternatively,lytic CAPs self aggregate via hydrophobic interactions in the membrane and form transmembrane channels or pores via the‘barrel-stave’mechanism(Shai,1999).The membranolyic mode of action of CAPs is not limited to the cell membrane,and can also be extended to the permeation and swel-ling of mitochondria,which results in a release of cytochrome c and induces apoptosis(Mai et al.,2001;Li et al.,2000).The release of cytochrome c from damaged mitochondria also induces Apaf-1 oligomerization,caspase9activation and conversion of pro-caspase3 to caspase3(Cory and Adams,1998).This mitochondrial pathway of apoptosis takes place with the CAPs granulysin(Pardo et al.,2001)and (KLAKLAKKLAKLAK)fused to a CNGRC homing domain(Ellerby et al., 1999).Recently,it has been shown that the CAP RAGLQFPVG[RLLR]3 (buforin IIb),a histone H2A-derived peptide,displayed cancer-selective toxicity against60human tumor cell lines(Lee et al.,2008).Buforin IIb traversed cancer cell membranes without damaging them and induced mitochondria-dependent apoptosis,as confirmed by caspase9activa-tion and cytochrome c release to the cytosol.However,the exact mechanism for buforin IIb-mediated apoptosis is not clear(Lee et al., 2008).Moreover,the mitochondrial pathway of apoptosis can also be associated with the death receptor pathway(Kaufmann and Earnshaw, 2000).For instance,the CAP tachyplesin conjugated to the integrin RGD homing domain,induces apoptosis via both mechanisms.It rup-tures mitochondrial membranes while at the same time up-regulating members of the death receptor pathway,including Fas ligand,FADD and caspase8(Chen et al.,2001;Park et al.,1992).2.2.Non-membranolytic mode of actions of CAPsBesides the CAP-induced disruption of mitochondrial or plasma membranes in cancer cells there is growing evidence that alternative mechanisms exist.For instance,melittin(GIGAVLKVLTTGLPALIS-WIKRKRQQ)is a26amino acid channel-forming CAP,which specifically counter-selects cells with the overexpressed ras oncogene.Melittin acts through hyperactivation of phospholipase A2(PLA2)in the ras oncogene-transformed cells,resulting in their selective destruction (Sharma,1992).In another study it was shown that the CAPs melittin and cecropin reduce human immunodeficiency virus1(HIV-1)rep-lication and gene expression in acutely infected cells at subtoxic concentrations as either intact peptides,or when expressed in cells using retroviral expression plasmids.Analysis of the effect of melittin on cell-associated virus production revealed decreased levels of Gag antigen and HIV-1mRNAs.Transient transfection assays with HIV long terminal repeat(LTR)-driven reporter gene plasmids indicated that melittin has a direct suppressive effect on the activity of the HIV LTR. This study concluded that both melittin and cecropin are capable of inhibiting cell-associated production of HIV-1by suppressing HIV-1 gene expression(Wachinger et al.,1998).Moreover,a recent study has revealed the cytokine-like modulating properties of insect-derived, histidine-rich antitumor CAPs,termed alloferons.In vitro studies using synthetic versions of HGVSGHGOHGVHG(alloferon1)and GVSGH-GQHGVHG(alloferon-2)revealed that these peptide sequences have stimulatory activities on natural killer lymphocytes,whereas in vivo trials indicate induction of interferon(IFN)production in mice.These results suggest that alloferons have immunomodulatory properties which mayfind future use in the development of novel anticancer and antiviral chemotherapeutic approaches(Chernysh et al.,2002).3.Chemical modifications on CAPs designed to improve pharmacokinetics and reduce toxicityThe development of CAP-based antitumor drugs is hampered by the poor pharmacodynamic properties of peptide-based drugs(Hancock and Sahl,2006).CAPs based on naturally occurring L-amino acids are prone to proteolytic cleavage that limits the half time of these drugs in serum and reduces their oral bioavailability.Increased resistance to proteolytic cleavage can be achieved by partial or full replacement of L-amino acids by D-amino acids or incorporation of unnatural amino acids containing unusual side chains without affecting their antitumor effects. For instance,mice bearing several types of tumors were intraperitone-ally-injected with the CAP magainin2and its all-D amino acid ana-logue MSI-238;both were found to be active against P388leukemia, S180ascites and spontaneous ovarian tumor.While MSI-238showed a10-fold higher in vitro activity when compared to magainin,it was only2-fold more active in vivo against murine tumors(Baker et al., 1993).Serum components such as negatively charged albumins and high-and low-density lipoproteins can also affect the antitumor potency of CAPs.For example,the antitumor activity of human defensins is abolished by low levels of serum due to inactivation by low-density lipoprotein(Lichtenstein et al.,1986).Similar observations have been made with antibacterial CAPs.In this case non-specific binding of CAPs to albumin and lipoproteins significantly decreases their antibacterial activity in vitro(Peck et al.,1993).Another approach to overcome the poor pharmacokinetics and potential toxicity of CAPs involves the vector-mediated delivery of genes encoding membranolytic or apoptotic CAPs into tumor cells. For example,expression of cecropin in tumor cells resulted in either a complete loss or reduced tumorgenicity(Winder et al.,1998).Another strategy to reduce the toxicity of cell-penetrating CAPs can be achieved by conjugation to a homing peptide.Homing peptides are suitable carriers for therapeutic and diagnostic agents to tumors due to their selectivity for a specific site in the body(Ruoslahti et al.,2005).Previous studies have shown that several vasculature homing peptides conju-gated to anticancer drugs improve the anti-angiogenic efficacy by inhibiting tumor angiogenesis,while reducing drug effects on other organs(Arap et al.,1998;Sacchi et al.,2006;Sacchi et al.,2004). However,tumor homing peptides without cell-penetrating properties significantly reduce their applicability,as a result of their inability to carry drug molecules inside cancer cells.In order to overcome this drawback chimeric peptide conjugates containing a cell-penetrating CAP sequence linked to homing peptide sequences such as protein transduction domains(Wadia and Dowdy,2002),integrin receptor ligands(Chen et al.,2001)and the linearfive amino acid long peptide CREKA that selectively recognizes clotted plasma proteins on tumor blood vessels and stroma(Simberg et al.,2007;Mäe et al.,2009)have been studied.Many other tumor homing peptide sequences are known(Shadidi and Sioud,2003).A recent study describes the use of the cell-penetrating cationic amphiphilic peptide p Vec(LLIILRRRI-RKQAHAHSK)that is the conjugated homing peptide CREKA and the DNA alkylating agent Cbl(chorambucil).In vitro studies of the corre-sponding chimeric peptide-chlorambucil conjugate Cbl-CREKA-p Vec reduced the percentage of proliferating MDA-MB-231cells from 100%to approximately40%.At the same time Cbl,CREKA or pVec192 F.Schweizer/European Journal of Pharmacology625(2009)190–194alone did not show any toxicity,confirming the Cbl-mediated effect on cancer cells.This study demonstrated that CREKA-pVec is a suitable vehicle for targeted intracellular delivery of a DNA alkylating agent into tumor cells(Mäe et al.,2009).4.Synergistic effects of CAPs with other anticancer drugs and occurrence of multidrug resistanceThe membranolytic mode of action of antitumor CAPs can result in additive or synergistic effects when combined with conventional chemotherapeutics.For instance,the CAP cecropin which displays anticancer activity against ovarian,carcinoma,breast carcinoma and leukemia cells(Chen et al.,1997)synergizes the anticancer drugs S-fluorouracil and cytarabine against acute lymphoblastic leukemia cells(Hui et al.,2002).Similarly,the CAPs OLP-1(Ac-ILKKWPWW-PWRRK)and a diastereomeric version containing two D-amino acids such as D-isoleucine(i)and D-lysine(k)OLP-4(Ac-iLKKWP-WWPWRRk)enhance the in vitro cytotoxic activity of doxorubicin against tumor cells(Johnstone et al.,2000).Previously,magainin analogues in combination with chemotherapeutic agents such as cisplatin,doxorubicin and etoposide were found to have an additive effect against lung cancer cells(Ohsaki et al.,1992).Interestingly,the potency of these anticancer peptides was not affected by the presence of the multidrug resistance1(mdr1)gene.Similar results were obtained with the magainin2against human BRO melanoma cells transfected with the mdr1gene(Lincke et al.,1990).Based on these results,it appears that membranolytic CPAs may be able to bypass established resistance mechanisms thus providing new lead structures to delay or reduce the risk of resistance occurrence.5.Conclusions and outlookCurrent cancer treatments have many harmful side effects and give rise to multidrug resistance.Clearly,novel therapeutic ap-proaches with reduced risk of resistance development and higher cancer-selectivity are urgently required.Over the last two decades several CAPs with high cancer-selective toxicity have appeared.Most of these CAPs also kill microbial pathogens.However,many CAPs that kill pathogens do not have anticancer activity.In addition,antimi-crobial CAPs with cancer-selective toxicity may or may not be hemolytic.The reasons for this selectivity are poorly understood.It is anticipated that biophysical studies dealing with the structure, dynamics,topology,mechanisms of membrane disruption and the role of individual components of membranes will provide new insight and rationalizations to explain the observed selectivity differences.While a variety of CAPs displays cancer-selective toxicity it is currently difficult to develop peptide-based drugs due to poor pharmacokinetics,potential systemic toxicity and the high cost of manufacturing(Hancock and Sahl,2006).Ultrashort antimicrobial CAP sequences 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Cansmart 2009International WorkshopSMART MATERIALS AND STRUCTURES22 - 23 October 2009, Montreal, Quebec, Canada HIGH TEMPERATURE INTEGRATED ULTRASONIC TRANSDUCERS FOR ENGINE CONDITION MONITORINGM. Kobayashi1, K.-T. Wu2, C.-K. Jen1, J. Bird3, B. Galeote3 and N. Mrad41Industrial Materials Institute, National Research Council, BouchervilleQuebec, Canada J4B 6Y4****************************.ca; **************************.ca2Department of Electrical and Computer Engineering, McGill UniversityMontreal, Quebec, Canada H3A 2A7***********************.ca3Institute for Aerospace Research, National Research CouncilOttawa, Ontario, Canada K1A 0R6************************.ca; *************************.ca4Department of National Defence, Defence R&D Canada, Air Vehicles Research Section,National Defence Headquarters, Ottawa, Ontario, Canada K1A 0K2***********************.caABSTRACTEmploying a portable on-site fabrication kit, high temperature integrated ultrasonic transducers (IUTs) made of bismuth titanate piezoelectric film, of thickness greater than 50 µm, have been coated directly onto a modified CF700 turbojet engine outer casing, oil sump and supply lines and gaskets using sol-gel spray technology. Transducers top electrodes, electrical wires, conducting adhesive bond, connectors and cables have all been successfully tested for temperatures up to 500°C. The excellent ultrasonic performance of these IUTs is demonstrated in this paper and the potential applications for the non-intrusive real-time temperature and lubricant oil quality and metal debris monitoring have been identified and discussed.Keywords: Structural health monitoring, Non-destructive evaluation, Integrated ultrasonic transducers, Turbojet engine, Piezoelectric thick films, Sol-gel process.INTRODUCTIONThe increasing demand to improve the performance, reduce downtime, increase reliability and extend the life of engines requires the use of sensors for continuous engine conditions monitoring during development and service. It is established that methods employing piezoelectric ultrasonic transducers (UTs) are widely used for real-time, in-situ or off-line non-destructive evaluation (NDE) of large metallic structures including airplanes, automobiles, ships, pressure vessels, pipelines, etc. because of their subsurface inspection capability, fast inspection speed, simplicity and cost-effectiveness [1-3]. In this investigation the objective is to develop and evaluate effective integrated UT (IUT) technology to perform non-intrusive engine NDE and structural health monitoring (SHM). The intended applications include non-intrusive real-time temperature, lubricant oil quality, and metal debris monitoring within the turbojet engine environment. For this purpose a portable on-site IUT fabrication technique is most desirable. Additionally, because engines operate at elevated temperatures, the developed IUT technology including piezoelectric UTs together with electrical wire, conductive bonding agent, connectors and cables must be assessed. In this study the assessment is aimed and limited to temperatures up to 500°C.PORTABLE ON-SITE INTEGRATED ULTRASONIC TRANSDUCER FABRICATION The engine to be used is a modified (fan module removed) CF700 turbojet engine as shown in Figure 1 at the Institute for Aerospace Research (IAR), of the National Research Council (NRC) of Canada. In order to coat high temperature IUT directly onto the engine components and on-site, a portable IUT fabrication kit has been developed. The fabrication of IUTs involves a sol-gel based sensor fabrication process [4-6]. Such process consists of six main steps: (1) preparing high dielectric constant lead-zirconate-titanate (PZT) solution, (2) ball milling of piezoelectric bismuth titanate (BIT) powders to submicron size, (3) sensor spraying using slurries from steps (1) and (2) to produce the thin film UT, (4) heat treating to produce a solid BIT composite (BIT-c) thin film UT, (5) Corona poling to obtain piezoelectricity, and (6) electrode painting for electrical connectivity. Steps (3) and (4) are used multiple times to produce optimal film thickness for specified ultrasonic operating frequency and performance. Silver or platinum paste was used to fabricate top electrodes. BIT-c was used because of its high temperature (500°C) endurance [5, 6]. Figure 2 shows the lower and upper levels of the developed portable IUT fabrication kit. The kit with dimensions of 0.8×0.53×0.3 m3, consists of the following item:a.BIT powders and PZT solution (used in step 1).b.One ball milling device, sand papers, detergent, acetone and methanol for samplecleaning (used in step 2).c.One air brush, one compressor air device, and two glass beakers for cleaning of airbrush (used in step 3).d.One heat gun, one propane torch, two high temperature gloves, and one thermo-couple(used in step 4).e.One Corona poling device (used in step 5).f.One silver and one platinum paste pen, and one multi-meter to measure resistance ofthe electrode and temperature together with a thermocouple. Special high temperatureelectrical connection accessories are included (used in step 6).g.One suite case for housing all the components of the portable kit.Fig. 1:Modified CF700turbojet engine.(a) (b)Fig. 2:A portable IUT fabrication kit; (a) interior lower level and (b) interior upper level.Portable on-site Fabrication KitFig. 3:A portable fabrication kit.Such a developed portable IUT fabrication kit (Figure 3) can be conveniently carried for on-site application of the technology onto target components. Figure 4 illustrates the instrumentation of two IUTs, consisting of BIT-c films with 500°C platinum paste top electrodes, onto the outer casing of the modified CF700 turbojet engine. This Figure furthershows the bonding conductive material that was applied onto the platinum paste, and the 500°C signal transmitting wires (Figure. 4(b)). In general one wire would be sufficient for one IUT; however, for redundancy two wires were installed. The principle of the use of the IUT for the non-intrusive real-time and continuous internal engine temperature measurements is based on using ultrasonic pulses and determining the time of flight as reported in [7, 8].500\C IUT s(a) 500o C Conducting Bond500o C IUT sPlatinum PasteT op Electrode500o C Conducting Wire(b)Fig. 4: Two IUTs installed on-site at the top of the engine outer casing and (b) 500°C wiresbonded to the platinum electrodes .Additionally, for ultrasonic temperature monitoring of 500°C, IUTs were also coated onto two different engine access covers as shown in Figure 5. These access covers have also been instrumented with conventional thermocouples. In such a configuration, temperature measured by ultrasound can be compared with those measured by thermocouples. These sensors have been installed onto the modified CF700 turbojet engine as shown in Figure 6. Figure 7 shows the typical measured ultrasonic signals traversed back and forth within the gasket thickness in pulse/echo mode at room temperature for the IUT coated onto the gaskets as shown in Fig. 6. The center frequency of these IUTs is around 11 MHz. L n is the n th round trip echo through the thickness of the gasket.500o C Conducting Bond500o C IUTsSilver Paste Top Electrode500o C ConductingWire 500o C Conducting Bond500o C IUTsSilver Paste Top Electrode500o C ConductingWireThermocouple Installation LocationThermocouple Installation LocationFig. 5:500o C IUTs + Wire Connection(a)500o C IUTs + Wire Connection(b) Fig. 6: Installed access covers on the modified CF700 turbojet engine.Time Delay (µs)A m p l i t u d e (A r b . U n i t )L 2L 1L 3L 4L 5(a) L 2L 1L 3L 4L 5Time Delay (µs)A m p l i t u d e (A r b . U n i t )(b)Fig. 7: Typical ultrasonic signals of the IUT coated onto the access covers shown in Fig. 6.HIGH TEMPERATURE TESTS ON THE ELECTRICAL IUT CONNECTORSIn order to test the fabricated IUTs made of BIT-c film and platinum top electrodes, electrical connectors, and conductive adhesive, a 25.4 mm diameter 150 mm long steel rod, simulating engine component, has been used for the deposition of IUTs (Figure 8(a)). The center frequency of these IUTs is about 10 MHz at room temperature. Such an assembly was placed into a furnace (Figure 8(b)) and thermally cycled from room temperature to 500°C. The results of two thermal cycles are shown in Figure 9. The upper two pictures on the left and two on the right indicate the ultrasonic signals with high signal to noise ratio (SNR) which traveled a distance of 300 mm and 600 mm, respectively. The lowest and right part in Figure 9 indicates the measured ultrasonic velocity profile in the steel rod during the thermal cycles. In the middle of Figure 9, the attenuation number (Attn) indicates the total loss of IUT, electrical connections and ultrasonic attenuation propagating via a round trip of the 150 mm long steel rod. At 500°C this attenuation was only 6 dB.500o C 500o C Conducting BondWire(a)500o C Thermal Cycle Test(b)Fig. 8: (a) One IUT made of BIT-c films on top of a steel rod (delay line) together with topelectrode, wire connection, conducting adhesive bond, electrical connector and cable; (b) 500°C thermal cycle test chamber .Room temperature to 500°C Thermal Cycle TestUltrasonic Velocity ProfileUltrasonic signal at 500°CUltrasonic signal at 500°CZoomed SignalZoomed SignalAfter a distance of 300 mm After a distance of 600 mmFig. 9: Thermal cycle tests of 500°C BIT-c film IUT directly coated on top of a 150 mmlong steel rod including top electrode, conducting bond, wire, connector and cable .IUTS FOR LUBRICANT OIL QUALITY AND METAL DEBRIS MONITORINGIt is also the intention of this work to perform non-intrusive real-time continuous monitoring of the lubricant oil quality and metal debris in the engine oil system. For such purpose one oil sump line and one oil supply line of the turbojet engine, shown in Figure 10, are instrumented. Figure 11 illustrates the instrumented IUTs together with the necessary prototype electrical connections for the oil sump and supply line, respectively. The cables used can only sustain temperatures of up to 200°C; hence they were properly installed as shown in figure 12. Figure13 illustrates typical measured ultrasonic signals traversed back and forth within the pipe thickness in pulse/echo mode at room temperature for the IUT coated onto the oil sump and supply lines, respectively. The center frequencies of the L echo, shown in Fig. 13, are at 10 and 17 MHz, respectively. L is the n round trip echo through the wall thickness of the pipe. During actual operations, the ultrasonic measurements will be carried out in the transmission mode and ultrasonic signals with higher SNR than those shown in Fig. 13 are expected. Such an expectation comes from previous experiments which simulated the monitoring of lubricant quality and metal debris [9].1n thOil Sump Line Pipe #1Oil Supply Line Pipe #2(a)O.D: 7.92mm Wall: 1.09mmO.D: 15.60mm Wall: 0.80mmSupply Line Pipe #2o Sump Line Pipe #1500o C IUTs(b) Fig. 10: (a) Standard engine oil sump and supply lines, (b) IUTs coated onto special oilsump and supply lines .Oil Sump Line Pipe #1O.D: 15.6mm; Wall: 0.80mm(a)Oil Supply Line Pipe #2(b)Fig. 11: (a) Oil sump line (b) Oil supply line equipped with IUTs and prototype electricalconnections .Oil Sump Line Pipe #1Oil Supply Line Pipe #2Fig. 12: Oil sump and supply lines installed onto the modified CF700 turbojet engine.Time Delay (µs)A m p l i t u d e (A r b . U n i t )L 2L 1L 3L 4L 5(a)Time Delay (µs)L 2L 1L 3L 4L 5A m p l i t u d e (A r b . U n i t )(b)Fig. 13: Typical ultrasonic signals of the IUT coated onto the (a) Oil sump and (b) supplylines and operated in pulse/echo mode at room temperature .CONCLUSIONS AND DISCUSSIONSEmploying a developed portable on-site fabrication kit, high temperature integrated ultrasonic transducers (IUTs) made of thick BIT composite piezoelectric film have been coated directly onto a modified CF700 turbojet engine outer casing, oil sump and supply lines and gaskets using a sol-gel spray technology that consists of six main steps. The top electrodes, electrical wires, conducting adhesive bond, connectors and cables have been also tested successfully at temperatures of up to 500°C. The center frequencies of these IUTs were around 10 to 17 MHz. Ultrasonic signals obtained in pulse/echo measurements of these IUTs are excellent and it is expected that high temperature ultrasonic performance will be obtained in the transmission mode as well. The potential applications of the developed IUTs include non-intrusive real-time temperature, lubricant oil quality and metal debris monitoring which will be carried out in pulse/echo and transmission mode, respectively.ACKNOWLEDGMENTFinancial support from the Natural Sciences and Engineering Research Council of Canada is acknowledged.REFERENCES1.Gandhi, M.V. and Thompson, B.S., “Smart Materials and Structures”, Chapman & Hall,NY, 1992.2.Ihn, J.-B. and Chang, F.-K., “Ultrasonic non-destructive evaluation for structure healthmonitoring: built-in diagnostics for hot-spot monitoring in metallic and composite structures”, Chapter 9 in Ultrasonic Nondestructive Evaluation Engineering and Biological Material Characterization, edited by Kundu T., CRC Press, NY, 2004.3.Birks, A.S., Green, R.E. Jr. and McIntire, P. ed., “Nondestructive Testing Handbook”, 2ndEdition, vol.7, Ultrasonic Testing, ASNT, 1991.4.Barrow, D.A., Petroff, T.E., Tandon, R.P. and Sayer, M., “Characterization of thick leadzirconate titanate films fabricated using a new sol gel based process”, J. Appl. Phys., vol.81, 1997, pp. 876-881.5.Kobayashi, M. and Jen, C.-K., “Piezoelectric thick bismuth titanate/PZT composite filmtransducers for smart NDE of metals”, Smart Materials and Structures, vol.13, 2004, pp.951-956.6.Kobayashi, M., Jen, C.-K., Bussiere, J.F. and Wu, K.-T., “High temperature integratedand flexible ultrasonic transducers for non-destructive testing”, NDT&E Int., vol.42, 2009, pp.157-161.7.Walker, D.G., Myers, M.R., Yuhas, D.E. and Mutton, M.J., “Heat flux determinationsfrom ultrasonic pulse measurements”, Proc. IMECE, ASME, Boston, MA, 2008.8.Yuhas, D.E., Mutton, M.J., Remiasz, J.R. and Vorres, C.L., “Ultrasonic measurements ofbore temperature in large caliber guns”, Proc. Review of QNDE, vol.28, 2008, pp.1759-1766.9.Kobayashi, M., Jen, C.-K., Moisan, J.-F., Mrad, N. and Nguyen, S.B., “Integratedultrasonic transducers made by sol-gel spray technique for structural health monitoring,”Smart Materials and Structures, vol.16, 2007, pp.317-322.。
Efficient Test Pattern Compression Techniques Based on Complementary Huffman CodingShyue-Kung Lu, Hei-Ming Chuang, Guan-Ying Lai, Bi-Ting Lai, and Ya-Chen HuangDepartment of Electronic Engineering, Fu-Jen Catholic UniversityTaipei, Taiwan, 242, R.O.Csklu@.twAbstract⎯In this paper, complementary Huffman encodingtechniques are proposed for test data compression of complex SOC designs during manufacturing testing. The correlations of blocks of bits in a test data set are exploited such that moreblocks can share the same codeword. Therefore, beside the compatible blocks used in previous works, the complementary property betweens blocks can also be used. Based on thisproperty, two methods are proposed for Huffman encoding. According to this technique, more blocks will share the same codeword and the size of the Huffman tree can be reduced. This will not only reduce the area overhead of the decoding circuitry but also substantially increase the compression ratio. In order to facilitate the proposed complementary encoding techniques, a don’t-care assignment algorithm is also proposed. According to experimental results, the area overhead of the decompression circuit is lower than that of the full Huffman coding technique. Moreover, the compression ratio is higher that that of the selective and optimal selective Huffman coding techniques.I. I NTRODUCTIONDue to the ever-increasing complexity of SoC designs, a large amount of test patterns should be sent from the ATE’s to test embedded cores through the A TE channels. However, since the bandwidth of ATE channels and the capacity of ATE memory are limited, the cost of testing and test application time will increase considerably. To alleviate these problems, test compression techniques are usually adopted in today’s SOC era. There are three kinds of test compression techniques⎯linear-decompression-based scheme, broadcast-scan-based scheme [1], and code-based scheme [2-10].The code-based scheme is appropriate to compress the pre-computed test set without requiring any structural information of IP cores. In [3, 4], new dictionary-based compression schemes are proposed. Due to the regularity of the schemes, the advantages of a multiple-scan architecture can be preserved, and very low test time can be achieved. In [2], the RLE (run-length encoding) techniques used in JPEG still image compression systems are used to reduce test data and to make the area overhead almost negligible. Test compression techniques based on full Huffman encoding are proposed in [5, 6, 7, 8]. However, the size of the decoder will increase exponentially as the encoded symbols increase [5]. To cure this problem, the selective Huffman encoding technique was proposed in [6]. It only encodes the symbols with higher occurrence frequencies. The size of the decompression circuit can be reduced substantially. The technique was further improved in [7] by the optimal selective Huffman coding technique, which regards all non-encoded symbols as one symbol.In this paper, complementary Huffman encoding techniques are proposed. The correlations between blocks of bits in the test set are exploited to further reduce the size of Huffman tree. That is, more blockss that are compatible or complementary compatible can be merged and encoded with the same codeword. Two methods are proposed to encode compatible and complementary compatible blocks⎯Method 1 and Method 2. In order to maximum the merging process, a don’t-care assignment algorithm is also proposed. The size of the Huffman tree can be reduced significantly. Moreover, according to experimental results, the compression ratio will achieve 45% to 87% with Method 1, and achieve66% to 93% with Method 2 by using ISCAS89 benchmark circuits. The area overhead for implementing our decompression circuit is lower than that of the full Huffman technique.II. O VERVIEW O F T HE C OMPLEMENTARY H UFFMANC ODING T ECHNIQUESHuffman coding technique is basically a statistical coding scheme, which is a fix-to-variable coding scheme. It represents data blocks of fixed length by variable length codewords. Symbols occur more frequently have shorter codewords than symbols that occur less frequently. This will reduce the average number of bits for a codeword. Therefore, compression is achieved. An example of deriving the Huffman code is shown in Fig. 1 [6]. In this example, the test set is divided into 4-bit blocks. Table 1 [6] shows the frequency of occurrence of each of the possible blocks. The compression ratio is determined by how skewed the frequency of occurrence is.978-1-4244-2587-7/09/$25.00 ©2009 IEEE0010 0100 0010 0110 0000 0010 1011 0100 0010 0100 0110 0010 0010 0100 0010 0110 0000 0110 0010 0100 0110 0010 0010 0000 0010 0110 0010 0010 0010 0100 0100 0110 0010 0010 1000 0101 0001 0100 0010 0111 0010 0010 0111 0111 0100 0100 1000 0101 1100 0100 0100 0111 0010 0010 0111 1101 0010 0100 1111 0011Fig. 1: The example test set divided into 4-bit blocks.Table 1: Huffman Coding of the example [6]In this example, 13 symbols are used to construct the Huffman tree. The correlations between symbols are not exploited. We say that two symbols S i and S j are complementary compatible if the corresponding bits of these two symbols are bit-wise complementary and can be merged as a new symbol S i, j. For example, in Table 1, S1 (0100) and S7 (1011) are complementary compatible. Therefore, they can be merged as a new symbol S1,7. Similarly, S0 (0010) and S10 (1101) are complementary compatible and can be merged as S0, 10. The merged results of the symbols in Table 1 are shown in Table 2. The “Joint Frequency” column denotes the accumulated probability of symbols S i and S j. From this table we can see that instead of the original 16 symbols, we only need to encode the 8 merged symbols. Therefore, the size of the Huffman tree can be reduced significantly. Moreover, the skewing of the occurrence probabilities is helpful for the compression of test data. The average length of codewords can also be reduced significantly. This will in turn increase the compression ratio.III. D ON’T C ARE A SSIGNMENT T ECHNIQUESIn general, test patterns generated by commercial ATPG tools contain massive unspecified bits (don’t-care bits) that can be assigned with 1’s and 0’s in a way to skew the frequency distribution. This is helpful to maximize the compression efficiency. In [6], two don’t care assignment techniques are proposed. However, only compatible characteristics are used in their techniques. Two blocks areTable 2: Merged Symbols of Table 1 compatible if there is no conflict in any bit position of the two blocks. For our complementary coding techniques, beside the compatible blocks are exploited, complementary blocks are also searched. Therefore, the Alg_comp algorithm is proposed in this paper to perform don’t care assignment.In Alg_comp., we first search for the most frequently occurring unspecified block F1. Thereafter, it is compared with the next most frequently occurring unspecified block F2. If there is no conflict in any bit position, these two blocks are merged by specifying every bits position if there is specified bit in either block. All F2 blocks in test set are appended with 0 to indicate that it is merged with F1 block. For example, if block F1 00XX is merged with block F2 0X1X, the result is 001X. All F2 blocks in the test set will be changed to 0001X.Alternatively, if there is a conflict, F1 is compared with the complement of F2. If there is no conflict in any bit position, then they are merged. All F2 blocks in the test set are appended with 1 to indicate that it is merged with F1 block in its complementary form. For example, if F1 is 00XX and F2 is 1X0X, block 00XX is merged with block 0X1X, which is the complement of block 1X0X. The resulted block is 001X and all F2 blocks in the test set are changed to 1001X. Thereafter, F1 is compared with all the other unspecified blocks in decreasing order of frequency and merged if possible. This process is finished if there is no more block that can be merged with F1. This process is then repeated until there is no more blocks that can be merged in the test set. Any remaining X’s can be randomly assigned 0’s or 1’s since it has no effect on the compression ratio.Consider the example test set shown in Fig. 2(a), the test set contains five unique unspecified blocks⎯1X01, 10X1, 01XX, 01X1, and 101X. Let B uniq denote the set of unique blocks. In the test set, the frequency of occurrence of block 10X1 is 3, that of blocks 1X01 and 01XX is 2, and that of other blocks is 1. After applying the first step of Alg_comp., the most frequently occurring unspecified block 10X1 is found. It is compared with 1X01, which is the next most frequently occurring unspecified block. SinceSymbol Freq. Block Huff. Code S022 0010 10S113 0100 00S27 0110 110S3 5 0111 010S4 3 0000 0110 S5 2 1000 0111 S6 2 0101 11100 S7 1 1011 111010 S8 1 1100 111011 S9 1 001 111100 S10 1 1101 111101 S11 1 1111 111110 S12 1 0011 111111 S130 1110 -S140 1010 -S150 1001 - Symbol Joint Freq. Pattern S0, 1023 0010 S1, 714 0100 S2,157 0110 S3, 57 0111 S4, 11 4 0000 S6, 14 2 0101 S8, 12 2 1100 S9, 13 1 0001there are no conflicts, these two blocks can be merged and the result is 1001. All 1X01 in the test set are appended with 0. The merged block 1001 is compared with other unspecified blocks. It is merged with 10XX, which is the complement of the unspecified block 01XX. All 01XX blocks in the test set are appended with 1. At this time, B uniq = {1001, 01X1, 101X} and blocks 10X1, 1X01, and 01XX are transferred to 01001, 01001, and 11001, respectively.This process is repeated and 01X1 is merged with 010X, which is the complement of the unspecified block 101X. Since there are no more unspecified blocks that can be merged in the test set, therefore, the process is finished. The blocks 01X1 and 101X are changed to 00101 and 10101, respectively. Finally, the set B uniq is {1001, 0101}. The final test set is shown in Fig. 2(b).1X01 10X1 1X01 01001 01001 0100110X1 01XX 01X1 01001 11001 0010110X1 101X 01XX 01001 10101 11001(a) (b)Fig. 2: (a) Test set before don’t care assignment, and (b)after applying Alg_comp.IV. C OMPLEMENTARY H UFFMAN E NCODING T ECHNIQUES When all don’t-care bits of unspecified blocks are assigned by Alg_comp algorithm, the first bit of each merged block denotes whether the block is complementary merged or not. In this section, two methods⎯Method 1 and Method 2 are proposed for encoding the specified blocks. Test volume and area overhead of the decoding circuitry can be reduced significantly by using these two methods. Method 1: One Control Bit AddedAfter using the alg_comp algorithm for filling the don’t care bits, the blocks with complementary relationship are regard as one symbol to reduce the number of symbols such that it can reduce the size of Huffman tree. However, in order to distinguish the complementary pair, one control bit C is appended to the codeword of the complementary pair. This control bit is used to indicate the decoder whether the decoded blocks should be complemented or not.An example is shown in Fig. 3 and Table 3. The blocks 0110 and 1001 are a complementary pair. Therefore, they share the same symbol S0. Similarly, blocks 1010 and 0101 share the symbol S1. The occurrence frequencies of all symbols are shown in the third column. All symbols are encoded based on the conventional Huffman coding techniques. However, one control bit is appended to each codeword. For example, the original Huffman code is 0 for blocks 0110 and 1001. However, a “0” is appended to form the final codeword for block 0110 and an “1” is appended to form the final codeword for block 1001. In the decoding circuitry, this control bit is used to determine whether the decoded blocks should be complemented or not.00110 00110 10110 00110 00110 00110 10110 10110 00110 00110 00110 00110 00110 00110 10110 00110 00110 00110 00110 00110 00110 00110 01010 01010 00110 00110 00110 00110 00110 00110 10110 00110 00110 00110 00110 00110 01110 00110 00110 00111 11010 00111 10110 00110 00110 00110 00011 10110 10111 00110 11010 00110 00110 00110 00110 11010 01010 10110 00110 10110 Fig. 3: The test set after don’t-care assignment (block size= 4)Table 3: Complementary EncodingSymbol Freq. BlockMethod 1 Method 2HuffmanCodeFinalCodeHuffmanCodeFinalCode S0 49400110009100110 100S group 13------------------10 ------S1 6310101001011011030101110 10110 S2 320011110011011101110111001110 101110 S3 111111111001110111101111000000------------S4 110111111101111111111111101000------------Method 2: Regarding all complementary blocks as a symbolFor Method 1 encoding technique, one control bit is required for each codeword. This will impact the compression ratio. Therefore, in order to substantially improve the compression ratio and reduce the area overhead for Huffman decoding. Method 2 treats all complementary blocks as a symbol (S group). For example, in Table 3, the complementary blocks 1001, 0101, and 1100 are treated as symbol S group. The occurrence frequency of S group is 13. This symbol is then encoded along with other symbols by using the conventional Huffman coding technique. The Huffman code for S group is 10 as shown in this table. This control code is then appended to the Huffman code of other symbols to form the final code of complementary blocks. For example, The Huffman code for block 0110 is 0. When the control code “10” is appended, the final code for block 1001 is 100.V. EXPERIMENTAL RESULTS Experimental results based on ISCAS89 benchmark circuits are shown in Table 4. The block size can be 4, 8, 12, or 16 bits. The results of selective Huffman coding, optimal selective Huffman coding, and the methods proposed in this paper are compared as shown in this table. We can see thatTable 4: Comparisons of compression ratiothe compression ratio of Method 2 is better than that of Method 1. Moreover, the proposed two methods usually have higher compression ratio than the selective and optimal selective Huffman coding techniques.The results of hardware overhead of the decoder are shown in Table 5. Since Method 1 and Method 2 consider the complementary relationship between blocks, the number of leaf nodes is reduced. Therefore, the number of states will be reduced in the decoder. In this table, we can see that the area overhead of Method 1 and Method 2 is smaller than the full Huffman encoding technique. If the block size increases, the compression efficiency is better. However, the hardware overhead will increase. Therefore, the block size provides an easy way to tradeoff between area overhead and compression efficiency. Although the area overhead of the proposed complementary Huffman coding technique is larger than that of the selective Huffman coding technique, it has 10%~30% improvement in compression ratio.Table 5: Comparisons of area overhead.VI. C ONCLUSIONSIn this paper, complementary Huffman coding techniques are proposed for test compression of SOC designs. Instead of exploiting compatible blocks as the conventional techniques, complementary compatible blocks are also searched to further increase the compression ratio. Two methods are proposed for the encoding of blocks. For don’t-care assignment, the Alg_comp algorithm is proposed. According to experimental results, the area overhead of the decoding circuitry is lower than that of the full Huffman coding technique. Moreover, the compression ratio is higher that that of the selective and optimal selective Huffman coding techniques. Although the area overhead is larger than [6], however, the compression ratio is much better than it.R EFERENCE[1] K. J. Lee, J. J. Chen, C. H. Huang, ”Broadcasting test patternsto multiple circuits,” IEEE Trans. Computer-Aided Des. Integr.Circuits and Syst., vol. 18, no. 12, pp. 1793-1802, Dec. 1999. [2] H. Ichihara, Y. Setohara, Y. Nakashima, and T. Inoue, “Testcompression/decompression based on JPEG VLC algorithm,”in Proc. Asian Test Symposium., Oct. 2007,pp. 87-90.[3] A. Wurtenberger, C. S. Tautermann, and S. Hellebrand, Datacompression for multiple scan chains using dictionaries withcorrections,” in Proc. Int’l Test Conf. (ITC 2004), pp. 926-935, Oct. 2004.[4] F. G. Wolff, C. Papachristou, “Multiscan-based testcompression and hardware decompression using LZ77,” in Proc.Int’l Test Conf., Oct. 2002, pp. 331-339.[5] V. Iyengar, K. Chakrabarty, and B. T. Murray, ”Huffmanencoding of test sets for sequential circuits,” IEEE Trans.Instrumentation and Measurement, vol. 47, no. 1, pp. 21-25,Feb. 1998.[6] A. Jas, J. Ghosh-Dastidar, M. E. Ng, and N. A. Touba, “AnEfficient Test Vector Compression Scheme Using Selective Huffman Coding,” IEEE put.-Aided Des.Integr.Circuits Syst., vol. 22, no. 6, pp. 797-806, Jun. 2003.[7] X. Kavousianos, E. Kalligeros, D. Nikolos, “Optimal SelectiveHuffman Coding for Test-Data Compression,” IEEE Trans.Computers, vol. 56, no. 8, pp. 1146-1152, Aug. 2007.[8] X. Kavousianos, E. Kalligeros, D. Nikolos, “Test DataCompression Based on Variable-to-Variable Huffman Encoding With Codeword Reusability,” IEEE Trans.Comput.-Aided Des.Integr. Circuits Syst., vol. 27, no. 7, pp.1333-1338, Jul. 2008.[9] L. Lingappan, S. Ravi, A. 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IEEESymp.Circuits and System. vol. 2, pp. 681-684, May 2004.CircuitSelective Huffman Optimal Selective Huffman Method 1 Method 28 enc. dist. blocks 16 enc. dist. blocks 8 enc. dist. blocks 16 enc. dist. blocks--- ---4 8 12 16 8 12 16 4 8 12 16 8 12 16 4 8 12 16 4 8 12 16S5378 28.9 50.1 53.0 53.0 50.2 55.1 53.8 47.1 54.754.951.155.857.5 55.046.6967.6766.4057.81 66.15 75.54 70.8760.46 S9234 30.0 50.4 50.7 46.1 50.2 54.2 51.0 51.4 55.351.846.157.756.8 51.949.2773.4880.0782.11 70.41 82.65 85.4485.36 S13207 45.6 69.2 76.6 78.6 69.2 77.1 79.7 69.9 80.283.483.180.684.3 84.549.9374.8483.1587.12 74.53 86.92 91.1293.12 S15850 38.8 60.0 63.6 61.6 59.9 65.6 64.8 62.1 67.967.363.669.370.4 67.249.8874.4082.3485.79 73.49 85.00 88.9990.36 S38417 34.9 55.3 56.9 54.4 55.5 58.9 57.3 55.9 60.959.355.262.861.8 58.650.0074.8478.9780.08 74.65 86.99 87.3186.29 S38584 37.8 58.5 62.2 61.7 58.5 63.9 64.1 60.4 65.465.263.366.968.0 66.149.9474.8982.7186.15 72.31 84.82 88.7390.3Circuit Full Huff. SelectiveHuff. Method 1Method 2S5378 79.2 10.4 38.57 41.83S9234 51.1 8.6 16.8 17.78S13207 23.6 3.7 6.96 7.14S15850 26.1 4.5 8.22 8.90S38417 21.3 1.3 1.3 1.52S38584 19.7 1.3 1.75 1.88。