On the harmonic map equation
一种风电场HAPF后向线性谐波电流预测方法
一种风电场HAPF后向线性谐波电流预测方法李圣清;张彬;匡洪海;徐天俊;杨峻【摘要】风电场中谐波源的突发性、随机性给谐波电流治理带来较大的困难。
阐述风电场HAPF工作原理，提出基于后向最佳线性预测理论的风电场HAPF谐波电流预测方法。
通过求解最佳预测系数满足的正交、正则方程，进而推导出最小后向误差功率方程。
定义了前向预测误差和后向预测误差的相关系数，由Levinson-Durbin算法得出最小预测误差及系数的阶更新方程。
建立后向线性预测模型，根据现在时刻的谐波电流，通过显示预测控制方法得出前面时刻的谐波电流，为制定后续谐波电流控制方法提供依据。
仿真和实验验证了该方法的正确性和可行性。
%The burstiness and randomness of harmonic sources bring major difficulty to harmonic current suppression in wind farm. The principle of HAPF in wind farm was expounded, and a harmonic current forecasting method for hybrid active power filters based on the optimum backward linear prediction theory in wind farm was proposed. By solving the orthogonal and canonical equations which the optimal predic-tion coefficients satisfy, the smallest backward error power equation is deduced. Correlation coefficients of the forward prediction error and the backward prediction error was defined. The rank update equations of minimum prediction errors and coefficients were obtained through Levinson-Durbin algorithm. A backward linear prediction model was established. According to the present harmonic currents this method predicts the harmonic currents on the front time through display predictive control method, and can provide the basis for formulating subsequent harmonic currentcontrolling method. The simulation results and experi-ment show that the method is correct and feasible.【期刊名称】《电机与控制学报》【年(卷),期】2015(000)004【总页数】6页(P28-33)【关键词】后向线性预测;风电场;谐波;HAPF【作者】李圣清;张彬;匡洪海;徐天俊;杨峻【作者单位】湖南工业大学电气与信息工程学院，湖南株洲412007;湖南工业大学电气与信息工程学院，湖南株洲412007;湖南工业大学电气与信息工程学院，湖南株洲412007;湖南工业大学电气与信息工程学院，湖南株洲412007;湖南工业大学电气与信息工程学院，湖南株洲412007【正文语种】中文【中图分类】TM6140 引言风电场中许多类型的分布式发电电源受制于自然条件，运行不确定性强，具有间歇性、复杂性、多样性及不稳定性特点，其电能质量特征与传统电力系统有很大差异［1－3］，给谐波电流治理带来了较大的困难。
代数英语
(0,2) 插值||(0,2) interpolation0#||zero-sharp; 读作零井或零开。
0+||zero-dagger; 读作零正。
1-因子||1-factor3-流形||3-manifold; 又称“三维流形”。
AIC准则||AIC criterion, Akaike information criterionAp 权||Ap-weightA稳定性||A-stability, absolute stabilityA最优设计||A-optimal designBCH 码||BCH code, Bose-Chaudhuri-Hocquenghem codeBIC准则||BIC criterion, Bayesian modification of the AICBMOA函数||analytic function of bounded mean oscillation; 全称“有界平均振动解析函数”。
BMO鞅||BMO martingaleBSD猜想||Birch and Swinnerton-Dyer conjecture; 全称“伯奇与斯温纳顿－戴尔猜想”。
B样条||B-splineC*代数||C*-algebra; 读作“C星代数”。
C0 类函数||function of class C0; 又称“连续函数类”。
CAT准则||CAT criterion, criterion for autoregressiveCM域||CM fieldCN 群||CN-groupCW 复形的同调||homology of CW complexCW复形||CW complexCW复形的同伦群||homotopy group of CW complexesCW剖分||CW decompositionCn 类函数||function of class Cn; 又称“n次连续可微函数类”。
Cp统计量||Cp-statisticC。
非线性常微分方程高阶谐波平衡法傅里叶展开的简化
非线性常微分方程高阶谐波平衡法傅里叶展开的简化唐元璋;翁雪涛;楼京俊;林雄伟;张晖【摘要】简化了一种求取非线性常微分方程高阶谐波解的近似解析计算方法。
对平方和立方非线性项的傅里叶展开过程进行改进和简化，使计算过程变为两次矩阵运算即可完成展开过程，且两次矩阵运算过程一致，易于编程。
以Duffing方程为算例，计算结果与数值方法一致，运算效率有所提高。
%A simplified computation method of the high-order harmonic solution for nonlinear ordinary differential equations is discussed. Fourier expansion procedure of the equation with quadratic or cubic terms is improved and simplified. The procedure consists of two steps of matrix operation with the same computation process so that the algorithm is easier to program than that of the previous equation. Results of the solution for the Duffing equation using this method show that the high-order harmonic solution is in good agreement with its numerical solution, but more efficient than the latter one.【期刊名称】《噪声与振动控制》【年(卷),期】2014(000)002【总页数】6页(P28-33)【关键词】振动与波;非线性常微分方程;Duffing方程;傅立叶展开;谐波平衡法【作者】唐元璋;翁雪涛;楼京俊;林雄伟;张晖【作者单位】海军工程大学动力工程学院，武汉 430033; 船舶振动噪声重点实验室，武汉 430033;海军工程大学动力工程学院，武汉 430033; 船舶振动噪声重点实验室，武汉 430033;海军工程大学动力工程学院，武汉 430033; 船舶振动噪声重点实验室，武汉 430033;海军工程大学动力工程学院，武汉 430033; 船舶振动噪声重点实验室，武汉 430033;海军工程大学动力工程学院，武汉 430033; 船舶振动噪声重点实验室，武汉 430033【正文语种】中文【中图分类】O175.14线性隔振系统存在共振难避免等问题[1，2]。
华科奥本海姆讲义三
T
H(s) or H(z) is in general a function of the complex variable s or z .
Eigenfunction( of the system): an input signal for which the system output is a constant times the input.
1, xt 0, t T1 T1 t T 2
…
-T -T/2 -T1 T1 T/2 T
x(t) … t
T1 T1
Represent it in Fourier series.
2T1 1 a0 dt T T1 T
T1
1 T1 jk0t 1 ak e dt e jk 0t T T1 jk 0T
x(t )
k
a e
k
jk0 t
0 2 / T
where the coefficients ak is generally a complex function of k0 . The signals in the set e jk0 t , k 0, 1, 2, are harmonically related complex exponentials.
synthesis equation: analysis equation:
1 ak T
T
大学物理 英文版 第一章
Teaching Plan of Fundamentals of PhysicsTeaching Plan of Fundamentals of Physics 考核办法： 1. 平时成绩（２０分） • 作业 • ２． 出勤 （７０％）（学委负责制） （３０％）（班长负责制）考试成绩（８０分）：考试内容：课堂讲的例题，课后作业（７０～８０％）Teaching Plan of Fundamentals of PhysicsTeaching Plan of Fundamentals of PhysicsReference Books:1.《 Fundamentals of Physics 》 6th Edition, Halliday 2 《 大学物理 》第四版 马文尉 主编 3.《 大学物理 》第二版 张三慧 主编 4.《 大学物理 》第一版 廖耀发 主编Teaching Plan of Fundamentals of PhysicsI The Contents of PhysicsClassical physicsIntroductionMotion, fluids, heat, sound, light, electricity, magnetism Relativity, atomic structure, condensed matter, nuclear physics, elementary particles, astrophysicsIt deals with the behavior and structure of the matter.Modern physicsII Why to Learn Physics1 It’s interesting 2 It’s usefulTeaching Plan of Fundamentals of PhysicsMufanba PhenomenarefrigeratoryTeaching Plan of Fundamentals of PhysicsTeaching Plan of Fundamentals of PhysicsTeaching Plan of Fundamentals of PhysicsTeaching Plan of Fundamentals of PhysicsTeaching Plan of Fundamentals of Physics 强磁场强磁场的作用是改变一个系统的物理状态，即改变角动量（自旋） 和带电粒子的轨道运动，并破坏时间反演对称性，使它们具有更独特的 性质, 因此，也就改变了物理系统的状态． １ “黑洞”原来是超强磁场.天体物理学家发现 “磁场吸积”计算机 模型预言的“磁风” 与“钱德拉”X射线观测 望远镜拍摄银河中心黑洞GRO J655-40 的X射 线光谱中特殊的峰值相对应.２韩国科学家称在强磁场下水20摄氏度就能 结冰． ３磁化处理可提高水玻璃的粘结强度，使之最大可达50%； ４强流脉冲离子束使金属表面发生细晶强化，加工硬化和相变强化，提 高金属表面硬度，改善奈磨性．Teaching Plan of Fundamentals of Physics 超低温１ 超导态是一些物质在超低温下出现的特殊物态。
黎曼流形上具有凸约束最优控制问题的必要条件
2.2. 平行移动
对于一些细节的定义可以参考文献([8] [9] [13] [14])。
对于任意的 x ∈ M 及 r, s ∈ N ，所谓 M 在 x 处的一个 (r, s) 型张量 Γ 是指： r + s 重线性映射：
F : Tx*M × × Tx*M × TxM × × TxM → R
r
s
其中 r 称为 Γ 的反变阶数，s 称为 Γ 的协变阶数。 Γrs ( x) 表示 M 在 x 点的所有 (r, s) 型张量构成的集合，
集，因此做的是凸变分，[9]中的二阶必要条件是针对使得庞特里亚金形最大值原理退化的控制所满足的 条件。本文所得到的二阶必要条件，是针对庞特里亚金型最大值原理在古典意义下退化的控制所满足的 条件。参考文献[9]所考虑的控制集合是开集的情况。本文结果对文献[9]和文献[10]有重要补充。对不同 控制给予不同的条件得到的结果的强弱也是不同的，每种方法各有优势。因此在弯曲空间中依然还有事 情要做。
Open Access
1. 引言
设 n ∈ N , M 是一个具有黎曼度量 g 的完全单连通的黎曼流形。设 ∇ 是 M 上与 g 相关的 Levi-Civita
联络，ρ (⋅,⋅) 是 M 上的距离函数，TxM 是 M 在 x ∈ M 处的切空间，Tx*M 为该余切空间。我们用 ⋅,⋅ 和 ⋅
分别表示与 g 相关的内积和 TxM 上的范数。另外，= 分别用 TM
Necessary Conditions for Optimal Control Problems with Convex Constraints on Riemannian Manifolds
Yanjun Zeng Department of Mathematics, Southwest Jiaotong University, Chengdu Sichuan
15 Transformer Design
15Transformer Designspeci-maximum flux density are In the design methods of the previous chapter, copper loss andspecifically addressed. This approach is appropriate for a number of appli-fied, while core loss isnotcations, such as the filter inductor in which the dominant design constraints are copper loss and saturation flux density. However, in a substantial class of applications, the operating flux density is lim-ited by core loss rather than saturation. For example, in a conventional high-frequency transformer, it isusually necessary to limit the core loss by operating at a reduced value of the peak ac flux density This chapter covers the general transformer design problem. It is desired to design a k-windingAsoperat-theloss aretransformer as illustrated in Fig. 15.1. Both copper loss andcoremodeled.ing flux density is increased (by decreasing the number of turns), the copper loss is decreased but the core loss is increased. We will determine the operating flux density that minimizes the total power loss It is possible to generalize the core geometrical constant design method, derived in the previ-ous chapter, to treat the design of magnetic devices when both copper loss and core loss are significant.measure of the effective magnetic size of core in a trans-This leads to the geometrical constant aformer design application. Several examples of transformer designs via the method are given in this chapter. A similar procedure is also derived, for design of single-winding inductors in which core loss is significant.15.1TRANSFORMER DESIGN: BASIC CONSTRAINTSAs in the case of the filter inductor design, we can write several basic constraining equations. These equations can then be combined into a single equation for selection of the core size. In the case of trans-former design, the basic constraints describe the core loss, flux density, copper loss, and total power loss566Transformer Designvs. flux density. The flux density is then chosen to optimize the total power loss.15.1.1Core Lossoperating As described in Chapter 13, the total core loss depends on the peak ac flux density the frequency f, and the volume of the core. At a given frequency, we can approximate the core loss by a function of the formmeancorethemagnetic path length, and hence istheAgain, is thecore cross-sectional area, isconstant of proportionality which depends on the operating frequency. The volume of the core. isaexponent is determined from the core manufacturer’s published data.Typically,the value of for fer-rite power materials is approximately 2.6; for other core materials, this exponent lies in the range 2 to 3. Equation (15.1) generally assumes that the applied waveforms are sinusoidal; effects of waveform har-monic content are ignored here.15.1.2Flux Dens i tyAn arbitrary periodic primary voltage waveform is illustrated in Fig. 15.2. The volt-seconds appliedduring the positive portion of the waveform is denotedThese volt-seconds, or flux-linkages, cause the flux density to change from its negative peak to its posi-tive peak value. Hence, from Faraday's law, the peak value of the ac component of the flux density is15.1Transformer Design: Basic Constraints567Note that, for a given applied voltage waveform and we can reduce byincreasing the primaryturns This has the effect of decreasing the core loss according to Eq. (15.1). However, it also causes the copper loss to increase, since the new windings will be comprised of more turns of smaller wire. As a result, there is an optimal choice for in which the total loss is minimized. In the next sections, wewill determine the optimal Having done so, we can then use Eq. (15.3) to determine the primary follows:turns asIt should also be noted that, in some converter topologies such as the forward converter with conven-tional reset winding, the flux density B(t) and the magnetizing current are not allowed to be nega-tive. In consequence, the instantaneous flux density B(t)contains a dc bias.Provided that the core does not approach saturation, this dc bias does not significantly affect the core loss: core loss is determined bypeakthe the ac component of B(t). Equations (15.2) to (15.4) continue to apply to this case,since is value of the ac component of B(t).15.1.3Copper Lossallo-As shown in Section 14.3.1, the total copper loss is minimized when the core window area is cated to the various windings according to their relative apparent powers. The total copper loss is then given by Eq. (14.34). This equation can be expressed in the formwhere568Transformer Designis the sum of the rms winding currents, referred to winding 1. Use of Eq. (15.4) to eliminate from Eq.(15.5) leads toThe right-hand side of Eq. (15.7) is grouped into three terms. The first group contains specifications, while the second group is a function of the core geometry. The last term is a function of to be chosen to optimize the design. It can be seen that copper loss varies as the inverse square of increasingreducesThe increased copper loss due to the proximity effect is not explicitly accounted for in this design procedure. In practice, the proximity loss must be estimated after the core and winding geome-tries are known. However, the increased ac resistance due to proximity loss can be accounted for in the design procedure. The effective value of the wire resistivity is increased by a factor equal to the esti-mated ratio When the core geometry is known, the engineer can attempt to implement the wind-ings such that the estimated is obtained. Several design iterations may be needed.15.1.4Total power loss vs.found by adding Eqs. (15.1) and (15.7):The total power loss isinFig. 15.3.sketchedThe dependence of and on is15.1Transformer Design: Basic Constraints569 15.1.5Opt i mum Flux Dens i tyLet us now choose the value of that minimizes Eq.(15.8).At the optimum wecan writeNote that the optimum does not necessarily occur where Rather, it occurs whereThe derivatives of the core and copper losses with respect to are given bySubstitution of Eqs. (15.11) and (15.12) into Eq. (15.10), and solution for leads to the optimum fluxdensityThe resulting total power loss is found by substitution of Eq. (15.13)into (15.1), (15.8), and (15.9). Sim-plification of the resulting expression leads toThis expression can be regrouped, as follows:The terms on the left side of Eq. (15.15) depend on the core geometry, while the terms on the right sidethedesired core material depend on specifications regarding the application andThe left side of Eq. (15.15) can be defined as the core geometrical constant570Transformer DesignHence, to design a transformer, the right side of Eq. (15.15) is evaluated. A core is selected whoseexceeds this value:The quantity is similar to the geometrical constant used in the previous chapter to design magnet-ics when core loss is negligible. is a measure of the magnetic size of a core, for applications in which core loss is significant. Unfortunately, depends on and hence the choice of core material affects the value of However, the of most high-frequency ferrite materials lies in the narrow range 2.6 to 2.8, and varies by no more than ± 5% over this range. Appendix D lists the values of for variousstandard ferrite cores, for the valueOnce a core has been selected, then the values of and MLT are known.The peak acfound using Eq. flux density can then be evaluated using Eq. (15.13), and the primary turns can be(15.4). The number of turns for the remaining windings can be computed using the desired turns ratios. The various window area allocations are found using Eq. (14.35). The wire sizes for the various windings can then be computed as discussed in the previous chapter,where is the wire area for winding j.15.2 A STEP-B Y-STEP TRANSFORMER DESIGN PROCEDUREThe procedure developed in the previous sections is summarized below. As in the filter inductor design procedure of the previous chapter, this simple transformer design procedure should be regarded as a first-pass approach. Numerous issues have been neglected, including detailed insulation requirements, con-ductor eddy current losses,temperature rise, roundoff of number of turns, etc.The following quantities are specified, using the units noted:Wire effective resistivityTotal rms winding currents, referred to primaryDesired turns ratiosApplied primary volt-seconds15.2 A Step-By-Step Transformer Design Procedure 571Allowed total power dissipationWinding fill factorCore loss exponentCore loss coefficientThe core dimensions are expressed in cm:Core cross-sectional areaCore window areaMean length per turnMagnetic path lengthPeak ac flux densityWire areasThe use of centimeters rather than meters requires that appropriate factors be added to the design equa-tions.15.2.1Procedure1.Determine core size.Choose a core that is large enough to satisfy this inequality. If necessary, it may be possible to use a smaller core by choosing a core material having lower loss, i.e., smaller Evaluate peak ac flux density.Check whether is greater than the core material saturation flux density. If the core operates with a flux dc bias, then the dc bias plus should not exceed the saturation flux density. Proceed to the next step if adequate margins exist to prevent saturation. Otherwise, (1) repeat the procedure using a core material having greater core loss, or (2) use the design method, in which the maximum flux density is speci-fied.2.3.Evaluate primary turns.572Transformer Design4.Choose numbers of turns for other windingsAccording to the desired turns ratios:5.Evaluate fraction of window area allocated to each winding.6.Evaluate wire sizes.Choose wire gauges to satisfy these criteriaA winding geometry can now be determined, and copper losses due to the proximity effect can be evaluated. If these losses are significant, it may be desirable to further optimize the design by reiterat-ing the above steps,accounting for proximity losses by increasing the effective wire resistivity to theactual copper loss including proximity effects, and is thethevalue where iscopper loss obtained when the proximity effect is negligible.If desired, the power losses and transformer model parameters can now be checked. For the simple model of Fig. 15.4, the following parameters are estimated:Magnetizing inductance, referred to winding 1:Peak ac magnetizing current, referred to winding 1:15.3Examples573Winding resistances:The core loss, copper loss, and total power loss can be determined using Eqs. (15.1), (15.7), and (15.8), respectively.15.3E X AMPLES15.3.1Example 1: S i ngle-Output Isolated ConverterAs an example, let us consider the design of a simple two-winding transformer for the converter of Fig. 15.5. This transformer is to be optimized at the operating point shown, corresponding to D = 0.5. The steady-state converter solution is The desired transformer turns ratio is574Transformer DesignThe switching frequency is corresponding to A ferrite pot coreconsisting of Magnetics, Inc. P-material is to be used; at 200 kHz, this material is described by the fol-lowing parameters: A fill factor of is assumed. Total power loss ofW is allowed. Copper wire, having a resistivity of is to be used.Transformer waveforms are illustrated in Fig. 15.6. The applied primary volt-seconds areThe primary rms current isIt is assumed that the rms magnetizing current is much smaller than the rms winding currents. Since the transformer contains only two windings, the secondary rms current is equal toThe total rms winding current, referred to the primary, isThe core size is evaluated using Eq. (15.19):15.3Examples575Eq.of The pot core data of Appendix D lists the 2213 pot core with forEvaluationany2213 is the smallest standardevent,core, when In(15.16) shows that forthisTheincreased value of should lead to lower total power loss.pot core size having ~The peak ac flux density is found by evaluation of Eq. (15.20), using the geometrical data for the 2213pot core:This flux density is considerably less than the saturation flux density of approximately 0.35 Tesla. Theprimary turns are determined by evaluation of Eq. (15.21):The secondary turns are found by evaluation of Eq. (15.22). It is desired that the transformer have a 5:1turns ratio, and henceslightly higherIn practice, we might select and This would lead to a slightly higher andloss.The fraction of the window area allocated to windings 1 and 2 are determined using Eq. (15.23):For this example, the window area is divided equally between the primary and secondary windings, sincethe ratio of their rms currents is equal to the turns ratio. We can now evaluate the primary and secondarywire areas, via Eq. (15.24):The wire gauge is selected using the wire table of Appendix D. AWG #16 has area andis suitable for the primary winding. AWG #9 is suitable for the secondary winding, with areaThese are very large conductors, and one turn of AWG #9 is not a practical solution! Wecan also expect significant proximity losses, and significant leakage inductance. In practice, interleavedfoil windings might be used. Alternatively, Litz wire or several parallel strands of smaller wire could be employed.576Transformer DesignIt is a worthwhile exercise to repeat the above design at several different switching frequencies, to determine how transformer size varies with switching frequency. As the switching frequency is increased, the core loss coefficient increases. Figure 15.7 illustrates the transformer pot core size, for various switching frequencies over the range 25 kHz to 1 MHz, for this converter example using PPeak flux densities in Tesla are also plotted. For switching frequencies below material with W.250 kHz, increasing the frequency causes the core size to decrease. This occurs because of the decreasedessentially independent of switching fre-applied volt-seconds Over this range, the optimal isquency; the variations shown occur owing to quantization of core sizes.For switching frequencies greater than 250 kHz, increasing frequency causes greatly increasedthen requires that be reduced, and hence the core size is core loss. Maintaining Wincreased. The minimum transformer size for this example is apparently obtained at 250 kHz.In practice, several matters complicate the dependence of transformer size on switching fre-quency. Figure 15.7 ignores the winding geometry and copper losses due to winding eddy currents. Greater power losses can be allowed in larger cores. Use of a different core material may allow higher or lower switching frequencies. The same core material, used in a different application with different speci-fications, may lead to a different optimal frequency. Nonetheless, examples have been reported in the lit-erature [1–4] in which ferrite transformer size is minimized at frequencies ranging from several hundred kilohertz to several megahertz. More detailed design optimizations can be performed using computer optimization programs [5, 6].15.3.2Example 2: Mult i ple-Output Full-Br i dge Buck Convertermultiple-output full-bridgetheAs a second example, let us consider the design of transformer forbuck converter of Fig.15.8. This converter has a 5 V and a 15 V output, with maximum loads as shown. The transformer is to be optimized at the full-load operating point shown, corresponding to D = 0.75. Waveforms are illustrated in Fig.15.9. The converter switching frequency is In the full-bridge configuration, the transformer waveforms have fundamental frequency equal to one-half of the switching frequency, so the effective transformer frequency is 75 kHz. Upon accounting for losses15.3Examples577caused by diode forward voltage drops, one finds that the desired transformer turns ratios are 110:5:15.A ferrite EE consisting of Magnetics,Inc.P-material is to be used in this example;at75kHz, this material is described by the following parameters: A fill factor of is assumed in this isolated multiple-output application.Total power loss of or approximately0.5% of the load power,is allowed.Copper wire,having a resistivity of is to be used.The applied primary volt-seconds are578Transformer DesignThe primary rms current isThe 5 V secondary windings carry rms currentThe 15 V secondary windings carry rms currentThe total rms winding current, referred to the primary, isThe core size is evaluated using Eq. (15.19):Eq.ofEvaluation The EE core data of Appendix D lists the EE40 core with forevent, EE40 is the smallest standardanythiscore,when In(15.16) shows that forpeak ac flux density is found by evaluation of Eq. (15.20), usingEE core size having Thethe geometrical data for the EE40 core:This flux density is less than the saturation flux density of approximately 0.35 Tesla. The primary turnsare determined by evaluation of Eq. (15.21):The secondary turns are found by evaluation of Eq. (15.22). It is desired that the transformer have a110:5:15 turns ratio, and hence15.3Examples579This would lead to a reduced with reduced In practice, we might select andsuboptimal, the total power loss will be core loss and increased copper loss. Since the resulting isincreased. According to Eq. (15.3), the peak ac flux density for the EE40 core will beThe resulting core and copper loss can be computed using Eqs.(15.1) and (15.7):Hence, the total power loss would beSince this is 50% greater than the design goal of 4 W, it is necessary to increase the core size. The next larger EE core is the EE50 core, having of0.0284. The optimum ac flux density for this core, given by Eq. (15.3), is operation at this flux density would require and would lead to asimilar to Eqs. (15.45) to (15.48) lead to a peak flux total power loss of 2.3 W. With calculationsdensity of T. The resulting power losses would then beWith the EE50 core and the fraction of the available window area allocated to the pri-mary winding is given by Eq. (15.23) asThe fraction of the available window area allocated to each half of the 5 V secondary winding should beThe fraction of the available window area allocated to each half of the 15 V secondary winding should begiventhen The primary wire area 5Vsecondary wire area and 15 V secondary wire area are580Transformer Designby Eq.(15.24) asIt may be preferable to wind the15V outputs using two#19wires in parallel;this would lead to the same area but would be easier to wind.The5V windings could be wound using many turns of smaller par-alleled wires,but it would probably be easier to use a flat copper foil winding. If insulation requirements allow,proximity losses could be minimized by interleaving several thin layers of foil with the primary winding.15.4AC INDUCTOR DESIGNThe transformer design procedure of the previous sections can be adapted to handle the design of other magnetic devices in which both core loss and copper loss are significant. A procedure is outlined here for design of single-winding inductors whose waveforms contain significant high-frequency ac components (Fig.15.10). An optimal value of is found,which leads to minimum total core-plus-copper loss. Themajor difference is that we must design to obtain a given inductance, using a core with an air gap.The constraints and a step-by-step procedure are briefly outlined below.15.4.1Outl i ne of Der i vat i onAs in the filter inductor design procedure of the previous chapter, the desired inductance L must be obtained, given byThe applied voltage waveform and the peak ac component of the flux density are related according toThe copper loss is given bywhere I is the rms value of i(t).The core loss is given by Eq.(15.1).15.4AC Inductor Design581The value of that minimizesthetotal power loss is found in a manner similarto the transformer design derivation. Equation(15.54) is used to eliminate from the expression forThe optimal is then computed by setting the derivative of to zero. The result isevaluated at this value of and which is essentially the same as Eq. (15.13). The total power loss isthe resulting expression is manipulated to find The result iswhere is defined as in Eq. (15.16). A core that satisfies this inequality is selected.582Transformer Design15.4.2Step-by-step AC Inductor Des i gn ProcedureThe units of Section 15.2 are employed here.1.Determine core size.Choose a core that is large enough to satisfy this inequality. If necessary, it may be possible to use a smaller core by choosing a core material having lower loss, that is, smaller2.Evaluate peak ac flux density.3.Number of turns.4.Air gap length.with specified in and expressed in meters. Alternatively, the air gap can be indirectly expressed via (mH/1000 turns):5.Check for saturation.If the inductor current contains a dc component then the maximum total flux density isgreatermaximum total flux density, in Tesla, is given bythan the peak ac flux density TheIf is close to or greater than the saturation flux density then the core may saturate. The filter inductor design procedure of the previous chapter should then be used, to operate at a lower flux density.6.Evaluate wire size.15.5Summary 583A winding geometry can now be determined, and copper losses due to the proximity effect can be evalu-ated. If these losses are significant, it may be desirable to further optimize the design by reiterating the above steps, accounting for proximity losses by increasing the effective wire resistivity to the valuewhere is the actual copper loss including proximity effects, and is the copperloss predicted when the proximity effect is ignored.7.Check power loss.15.5SUMMAR Y 1.2.3.In a multiple-winding transformer, the low-frequency copper losses are minimized when the available window area is allocated to the windings according to their apparent powers, or ampere-turns.As peak ac flux density is increased, core loss increases while copper losses decrease. There is an optimumflux density that leads to minimum total power loss. Provided that the core material is operated near its intended frequency, then the optimum flux density is less than the saturation flux density. Minimization oftotal loss then determines the choice of peak ac flux density.The core geometrical constant is a measure of the magnetic size of a core, for applications in which core loss is significant. In the design method, the peak flux density is optimized to yield minimum total loss, as opposed to the design method where peak flux density is a given specification.R EFERENCES[1][2][3][4][5][6]W. J. G U and R. L IU , “A Study of Volume and Weight vs. Frequency for High-Frequency Transformers,”IEEE Power Electronics Specialists Conference, 1993 Record, pp. 1123–1129.K. D. T. N GO , R. P. A LLEY , A. J. Y ERMAN , R. J. C HARLES , and M. H. K UO , “Evaluation of Trade-Offs in Transformer Design for Very-Low-Voltage Power Supply with Very High Efficiency and Power Density,”IEEE Applied Power Electronics Conference, 1990 Record, pp. 344–353.A. F. G OLDBERG and M. F. S CHLECHT , “The Relationship Between Size and Power Dissipation in a 1-10MHz Transformer,” IEEE Power Electronics Specialists Conference, 1989 Record, pp. 625-634.K. D. T. N GO and R. S. L AI , “Effect of Height on Power Density in High-Frequency Transformers,” IEEE Power Electronics Specialists Conference, 1991 Record, pp. 667-672.R. B. R IDLEY and F. C. L EE , “Practical Nonlinear Design Optimization Tool for Power Converter Compo-nents,” IEEE Power Electronics Specialists Conference, 1987 Record, pp. 314-323.R. C. W ONG , H. A. O WEN , and T. G. W ILSON , “Parametric Study of Minimum Converter Loss in an Energy-Storage Dc-to-Dc Converter,” IEEE Power Electronics Specialists Conference, 1982 Record, pp.411-425.584Transformer DesignP ROBLEMS15.1Forward converter inductor and transformer design. The objective of this problem set is to design themagnetics (two inductors and one transformer) of the two-transistor, two-output forward converter shown in Fig.15.11. The ferrite core material to be used for all three devices has a saturation flux density of approximately 0.3 T at 120°C. To provide a safety margin for your designs, you should use a maxi-mum flux density that is no greater than 75% of this value. The core loss at 100 kHz is described by Eq. (15.1), with the parameter values and Calculate copper loss at 100°C.Steady-state converter analysis and design. You may assume 100% efficiency and ideal lossless compo-nents for this section.(a )(b )(c )Select the transformer turns ratios so that the desired output voltages are obtained when the duty cycle is D = 0.4.Specify values of and such that their current ripples and are 10%of their respective full-load current dc components and Determine the peak and rms currents in each inductor and transformer winding.Inductor design. Allow copper loss of 1 W in and 0.4 W in Assume a fill factor of Use ferrite EE cores—tables of geometrical data for standard EE core sizes are given in Appendix D. Design the output filter inductors and For each inductor, specify:(i )(ii )(iii )(iv )EE core size Air gap length Number of turns AWG wire sizeTransformer design. Allow a total power loss of 1 W. Assume a fill factor of (lower than for the filter inductors, to allow space for insulation between the windings). Use a ferrite EE core. You may neglect losses due to the skin and proximity effects, but you should include core and copper losses.Design the transformer, and specify the following:(i )(ii )EE core sizeTurnsandProblems585(iii)AWG wire size for the three windingsCheck your transformer design:(iv)Compute the maximum flux density.Will the core saturate?(v)Compute the core loss,the copper loss of each winding,and the total power loss15.2 15.3A single-transistor forward converter operates with an input voltage V,and supplies two out-puts:24V at2A,and15V at6A.The duty cycle is D =0.4.The turns ratio between the primary wind-ing and the reset winding is1:1.The switching frequency is100kHz.The core material loss equation parameters are You may assume a fill factor of0.25.Do not allow the core maximum flux density to exceed0.3T.Design a transformer for this application,having a total power loss no greater than 1.5W at 100°C. Neglect proximity losses. You may neglect the reset winding. Use a ferrite PQ core. Specify: core size,peak ac flux density,wire sizes,and number of turns for each pute the core and cop-per losses for your design.Flyback/SEPIC transformer design.The“transformer”of the flyback and SEPIC converters is an energy storage device,which might be more accurately described as a multiple-winding inductor.The magne-tizing inductance functions as an energy-transferring inductor of the converter,and therefore the “transformer” normally contains an air gap. The converter may be designed to operate in either the con-tinuous or discontinuous conduction mode.Core loss may be significant.It is also important to ensure that the peak current in the magnetizing inductance does not cause saturation.A flyback transformer is to be designed for the following two-output flyback converter application:Input:160VdcOutput1: 5 Vdc at 10 AOutput2:15 Vdc at 1 ASwitching frequency:100kHzMagnetizing inductance 1.33mH,referred to primaryTurns ratio:160:5:15Transformer power loss:Allow1W total(a)(b)(c)(d)(e)(f)Does the converter operate in CCM or DCM?Referred to the primary winding,how large are(i) the magnetizing current ripple(i i)the magnetizing current dc component I,and(iii)thepeak magnetizingcurrentDetermine (i) the rms winding currents, and (ii) the applied primary volt-seconds Ispro-portional toModify the transformer and ac inductor design procedures of this chapter,to derive a general procedure for designing flyback transformers that explicitly accounts for both core and copper loss,and that employs the optimum ac flux density that minimizes the total loss.Give a general step-by-step design procedure,with all specifications and units clearly stated.Design the flyback transformer for the converter of part(a),using your step-by-step procedureof part(d).Use a ferrite EE core,with and Specify:core size,air gap length,turns,and wire sizes for all windings.For your final design of part (e), what are (i) the core loss,(ii)the total copper loss, and (iii) the peak flux density?。
第4章 ANSYS求解设置
ANSYS-Solution
Solution-Sol’n Controls SPARSE -Sparse direct equation solver. Applicable to real symmetric and unsymmetric matrices. Available only for STATIC, HARMIC (full method only), TRANS (full method only), SUBSTR, and PSD spectrum analysis types [ANTYPE]. Can be used for nonlinear and linear analyses, especially nonlinear analysis where indefinite matrices are frequently encountered. Well suited for contact analysis where contact status alters the mesh topology. Other typical well-suited applications are: (a) models consisting of shell/beam or shell/beam and solid elements (b) models with a multi-branch structure, such as an automobile exhaust or a turbine fan. This is an alternative to iterative solvers since it combines both speed and robustness. Generally, it requires more memory than the frontal solver, but it is comparable in memory requirement to the PCG solver. When memory is limited, the solver works partly in and out-of-core memory without much increase in CPU time.
分数阶微分方程的高精度算法
博士学位论文
分数阶微分方程的高精度算法
学位申请人 ： 马晓华 学 科 专 业 ： 计算机软件与理论 指 导 教 师 ： 黄乘明 教授 答 辩 日 期 ： 2013 年 5 月 14 日
A Dissertation Submitted in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy in Science
In Chapter 3, we propose and analyze a spectral Jacobi-collocation method for the numerical solution of general linear fractional integro-differential equations. First, we use some function and variable transformations to change the equation into a Volterra integral equation defined on the standard interval [−1, 1]. Then the Jacobi-Gauss points are used as collocation nodes and the Jacobi-Gauss quadrature formula is used to approximate the integral equation. Later, the convergence order of the proposed method is investigated in the L∞ norm.
2016 AMC 12B 试题及解答
2016 AMC 12BProblem 1What is the value of when ?SolutionBy: DragonflyWe find that is the same as , since a number to the power of is just the reciprocal of that number. We then get the equation to beWe can then simplify the equation to getProblem 2The harmonic mean of two numbers can be calculated as twice their product divided by their sum. The harmonic mean of and is closest to which integer?SolutionBy: dragonflySince the harmonic mean is times their product divided by their sum, we get the equationwhich is thenwhich is finally closest to .Problem 3Let . What is the value of ?SolutionBy: dragonflyFirst of all, lets plug in all of the 's into the equation.Then we simplify to getwhich simplifies intoand finally we getProblem 4The ratio of the measures of two acute angles is , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?SolutionBy: dragonflyWe set up equations to find each angle. The larger angle will be represented as and the larger angle will we represented as , in degrees. This implies thatandsince the larger the original angle, the smaller the complement.We then find that and , and their sum isProblem 5The War of started with a declaration of war on Thursday, June , . The peace treaty to end the war was signed days later, on December , . On what day of the week was the treaty signed?SolutionBy: dragonflyTo find what day of the week it is in days, we have to divide by to see the remainder, and then add the remainder to the current day. We get that has a remainder of 2, so we increase the current day by to getProblem 6All three vertices of lie on the parabola defined by , with at the originand parallel to the -axis. The area of the triangle is . What is the length of ?SolutionBy: Albert471Plotting points and on the graph shows that they are at and , which is isosceles. By setting up the triangle area formula you get: Makingx=4, and the length of is , so the answer is .Problem 7Josh writes the numbers . He marks out , skips the next number , marksout , and continues skipping and marking out the next number to the end of the list. Then he goes back to the start of his list, marks out the first remaining number , skips the next number ,marks out , skips , marks out , and so on to the end. Josh continues in this manner until only one number remains. What is that number?SolutionBy Albert471Following the pattern, you are crossing out...Time 1: Every non-multiple ofTime 2: Every non-multiple ofTime 3: Every non-multiple ofFollowing this pattern, you are left with every multiple of which is only .Problem 8A thin piece of wood of uniform density in the shape of an equilateral triangle with sidelength inches weighs ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of inches. Which of the following is closest to the weight, in ounces, of the second piece?Solution 1By: dragonflyWe can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use.We can then solve the equation to get which is closest toSolution 2Another approach to this problem, very similar to the previous one but perhaps explained more thoroughly, is to use proportions. First, since the thickness and density are the same, we can set up aproportion based on the principle that , thus .However, since density and thickness are the same and (recognizing that the area of an equilateral triangle is ), we can say that .Then, by increasing s by a factor of , is increased by a factor of ,thus or .Problem 9Carl decided to fence in his rectangular garden. He bought fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?SolutionBy Albert471To start, use algebra to determine the number of posts on each side. You have (the long sides count for because there are twice as many) (each corner is double counted so you must add ) Making the shorter end have , and the longer endhave . . Therefore, the answer isProblem 10A quadrilateral has vertices , , , and , where and are integers with . The area of is . What is ?Solution 1By distance formula we have . SImplifying weget . Thus and have to be a factor of 8. The only way for them to be factors of and remain integers is if and . So the answer isSolution by I_Dont_Do_MathSolution 2Solution by e_power_pi_times_iBy the Shoelace Theorem, the area of the quadrilateral is , so .Since and are integers, and, so .Problem 11How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line , the line and the lineSolutionSolution by e_power_pi_times_i(Note: diagram is needed)If we draw a picture showing the triangle, we see that it would be easier to count the squaresvertically and not horizontally. The upper bound is squares , and the limit forthe -value is squares. First we count the squares. In the back row, there are squares withlength because generates squares from to , and continuing on wehave , , and for -values for , , and in the equation . So thereare squares with length in the figure. For squares, each square takes up un left and un up. Squares can also overlap. For squares, the back row stretchesfrom to , so there are squares with length in a by box. Repeating the process, the next row stretches from to , so there are squares. Continuing and adding up in the end, there are squares with length in the figure. Squares with length in the back row start at and end at , so there are such squares in the back row. As the front row starts at and ends at there are squares with length . As squares with length would not fit in the triangle, the answer is which is .Problem 12All the numbers are written in a array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to . What is the number in the center?SolutionSolution by Mlux: Draw a matrix. Notice that no adjacent numbers could be in the corners since two consecutive numbers must share an edge. Now find 4 nonconsecutive numbers that add up to . Trying works. Place each odd number in the corner in a clockwise order. Then fill in the spaces. There has to be a in between the and . There isa between and . The final grid should similar to this.is in the middle.Solution 2If we color the square like a chessboard, since the numbers altrenate between even and odd, and there are five odd numbers and four even numbers, the odd numbers must be in the corners/center, while the even numbers must be on the edges. Since the odd numbers add up to 25, and the numbers in the corners add up to 18, the number in the center must be 25-18=7Problem 13Alice and Bob live miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is from Alice's position and from Bob's position. Which of the following is closest to the airplane's altitude, in miles?SolutionLet's set the altitude = z, distance from Alice to airplane's ground position (point right below airplane)=y and distance from Bob to airplane's ground position=xFrom Alice's point of view, . . So,From Bob's point of view, . . So,We know that + =Solving the equation (by plugging in x and y), we get z= = about 5.5.So, answer issolution by sudeepnaralaSolution 2Non-trig solution by e_power_pi_times_iSet the distance from Alice's and Bob's position to the point directly below the airplane to be and , respectively. From the Pythagorean Theorem, . As both are triangles,the altitude of the airplane can be expressed as or . Solving the equation , we get . Plugging this into the equation , we get ,or ( cannot be negative), so the altitude is , which is closest to Problem 14The sum of an infinite geometric series is a positive number , and the second term in the series is . What is the smallest possible value ofSolutionThe second term in a geometric series is , where is the common ratio for the seriesand is the first term of the series. So we know that and we wish to find the minimum value of the infinite sum of the series. We know that: and substituting in , we getthat . From here, you can either use calculus or AM-GM.Calculus: Let , then .Since and are undefined . This means that we only need to find where thederivative equals , meaning . So , meaningthatAM-GM For 2 positive real numbers and , . Let and .Then: . This implies that . or .Rearranging : . Thus, the smallest value is . Solution 2A simple approach is to initially recognize that and . We know that , since the series must converge. We can start by observing the greatest answer choice, 4.Therefore, , because that would make , which would make the series exceed 4. In orderto minimize both the initial term and the rest of the series, we can recognize that is the opitimal ratio, thus the answer is .Problem15All the numbers are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?SolutionFirst assign each face the letters . The sum of the product of the facesis . We can factor thisinto which is the product of the sum of each pair of opposite faces. In order to maximize we use the numbers or . Problem 16In how many ways can be written as the sum of an increasing sequence of two or more consecutive positive integers?SolutionWe proceed with this problem by considering two cases, when: 1) There are an odd number of consecutive numbers, 2) There are an even number of consecutive numbers.For the first case, we can cleverly choose the convenient form of our sequence tobebecause then our sum will just be . We now have and will have asolution when is an integer, namely when is a divisor of 345. We checkthat work, and no more, because does not satisfy the requirements of two or more consecutive integers, and when equals the next biggest factor, , there must be negative integers in the sequence. Our solutionsare .For the even cases, we choose our sequence to be of theform:so the sum is . In this case, we find our solutions to be .We have found all 7 solutions and our answer is .Solution 2The sum from to where and are integers and isLet andIf we factor into all of its factor groups we will have several orderedpairs whereThe number of possible values for is half the number of factors of which isHowever, we have one extraneous case of because here, and we have the sum of one consecutive number which is not allowed by the question.Thus the answer is.Problem 17In shown in the figure, , , , and is an altitude.Points and lie on sides and , respectively, so that and are angle bisectors, intersecting at and , respectively. What is ?SolutionGet the area of the triangle by heron'sformula:Use the area to find the height AHwith known baseBC:Apply angle bisector theorem on triangle and triangle , weget and , respectively. From now, you can simply use the answer choices because only choice has in it and we know that the segments on it all have integral lengths so that will remain there. However, by scaling up the length ratio: and . weget .Problem 18What is the area of the region enclosed by the graph of the equationSolutionConsider the case when , .Notice the circle intersect the axes at points and . Find the area of this circle in the first quadrant.The area is made of a semi-circle with radius of and atriangle:Because of symmetry, the area is the same in all four quadrants. The answer isProblem 19Tom, Dick, and Harry are playing a game. Starting at the same time, each of them flips a fair coin repeatedly until he gets his first head, at which point he stops. What is the probability that all three flip their coins the same number of times?Solution 1By: dragonflyWe can solve this problem by listing it as an infinite geometric equation. We get that to have thesame amount of tosses, they have a chance of getting all heads. Then the next probability is all of them getting tails and then on the second try, they all get heads. The probability of that happeningis .We then get the geometric equationAnd then we find that equals to because of the formula of the sum for an infinite series, .Solution 2Call it a "win" if the boys all flip their coins the same number of times, and the probability that they win is . The probability that they win on their first flip is . If they don't win on their first flip, that meansthey all flipped tails (which also happens with probability ) and that their chances of winning have returned to what they were at the beginning. This covers all possible sequences of winning flips. So we haveSolving for gives .Problem 20A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won games and lost games; there were no ties. How many sets of threeteams were there in which beat , beat , and beatSolutionWe use complementary counting. Firstly, because each team played other teams, thereare teams total. All sets that do not have beat , beat , and beat have one team that beats both the other teams. Thus we must count the number of sets of three teams such that one team beats the two other teams and subtract that number from the total number of ways to choose three teams.There are ways to choose the team that beat the two other teams, and to choose two teams that the first team both beat. This is sets. There are sets of three teams total. Subtracting, we obtain as our answer.Problem 21Let be a unit square. Let be the midpoint of . For let be the intersection of and , and let be the foot of the perpendicular from to . WhatisSolution(By Qwertazertl)We are tasked with finding the sum of the areas of every where is a positive integer. We can start by finding the area of the first triangle, . This is equal to ⋅⋅. Notice that since triangle is similar to triangle in a 1 : 2 ratio, must equal (since we are dealing with a unit square whose side lengths are 1). is of course equal to as it is the mid-point of CD. Thus, the area of the first triangle is ⋅⋅.The second triangle has a base equal to that of (see that ~ ) andusing the same similar triangle logic as with the first triangle, we find the area to be ⋅⋅. If we continue and test the next few triangles, we will find that the sum of all is equalto orThis is known as a telescoping series because we can see that every term after the first is going to cancel out. Thus, the the summation is equal to and after multiplying by the half out in front, we find that the answer is .Problem 22For a certain positive integer less than , the decimal equivalent of is , a repeatingdecimal of period of , and the decimal equivalent of is , a repeating decimal of period . In which interval does lie?SolutionSolution by e_power_pi_times_iIf , must be a factor of . Also, by the same procedure, must be a factor of . Checking through all the factors of and that are less than , we seethat is a solution, so the answer is .Note: is also a solution, which invalidates this method. However, we need to examine all factors of that are not factors of , , or , or . Additionally, we need to be a factor of but not , , or . Indeed, satisfies these requirements.For anyone who wants more information about repeating decimals,visit: https:///wiki/Repeating_decimalProblem 23What is the volume of the region in three-dimensional space defined by theinequalities andSolution 1 (Non Calculus)The first inequality refers to the interior of a regular octahedron with top and bottomvertices . Its volume is . The second inequality describes an identicalshape, shifted upwards along the axis. The intersection will be a similar octahedron, linearly scaled down by half. Thus the volume of the intersection is one-eighth of the volume of the firstoctahedron, giving an answer of .Solution 2 (Calculus)Let , then we can transform the two inequalitiesto and . Then it's clearthat , consider , , then since the areaof is , the volume is . By symmetry, the casewhen is the same. Thus the answer is .Problem 24There are exactly ordered quadruplets suchthat and . What is the smallest possible value for ?SolutionLet , etc., so that . Then for each primepower in the prime factorization of , at least one of the prime factorizationsof has , at least one has , and all must have with .Let be the number of ordered quadruplets of integers suchthat for all , the largest is, and the smallest is . Then for the primefactorization we must have So let's take a look at the function by counting the quadruplets we just mentioned..There are quadruplets which consist only of and . Then there are quadruplets which include three different values, and with four.Thus and the first few valuesfrom onwards are Straight away we noticethat , so the prime factorization of can use the exponents . To make it as small as possible, assign the larger exponents to smaller primes. The resultis , so which is answer .Problem 25The sequence is defined recursively by , , and for . What is the smallest positive integer such that the product is an integer?Solution 1Let . Then and for all . The characteristic polynomial of this linear recurrence is , which has roots and .Therefore, for constants to be determined . Using the factthat we can solve a pair of linear equations for :.Thus , , and .Now, , so we are looking for the least value of so that.Note that we can multiply all by three for convenience, as the are always integers, and it does not affect divisibility by .Now, for all even the sum (adjusted by a factor of three) is . The smallest for which this is a multiple of is by Fermat's Little Theorem, as it is seen withfurther testing that is a primitive root .Now, assume is odd. Then the sum (again adjusted by a factor of three)is . The smallest for which this is a multiple of is , by the same reasons. Thus, the minimal value of is .Solution 2Since the product is an integer, the sum of the logarithms must be an integer. Multiply all of these logarithms by , so that the sum must be a multiple of . We take these vales modulo to save calculation time. Using therecursion :Listing the numbers out is expedited if you notice . Noticethat . The cycle repeats every terms.Since and , we only need the first terms to sum up to a multiple of : . (NOTE: This solution proves 17 is the upper bound, but since 17 is the lowest answer choice, it is correct. To rigorously prove it, you will have to add up the mods listed until you get .青山埋白骨，绿水吊忠魂。