HARVESTING PERFORMANCE AND DYNAMIC RESPONSES OF THE BISTABLE HARVESTER WITH A NONLINEAR RESONANT CIRCUIT
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摘要: 双稳态俘能器可实现宽频和高效的俘能效果. 目前的研究主要在双稳态结构中接入单一电阻电路进行俘能. 本文将非线性RLC (电阻−电感−电容)谐振电路引入到三弹簧式双稳态结构中, 构建两自由度非线性系统, 以实现俘能特性的提升. 设计永磁体与线圈的构型, 获得了非线性机电耦合系数. 推导并得到了两自由度非线性俘能器的控制方程. 利用谐波平衡法推导得到了系统的电流与位移的频率响应关系. 基于雅可比矩阵对解的稳定性进行了判别. 将解析解与数值解进行了对比验证. 结果表明, 在双稳态俘能器中引入非线性二阶谐振电路不仅有利于低频俘能, 还可进一步提升俘能响应, 拓宽俘能带宽. 相同的电路参数下, 与线性电路相比非线性电路可通过电流的倍频现象实现结构更低频率的能量俘获. 减小谐振电路与双稳态结构共振频率之比, 增加基础激励幅值, 减小静平衡点之间的距离均可提升俘能器的俘能效果. 通过调控谐振电路与双稳态共振频率之比和基础激励幅值等参数, 可实现系统单倍周期响应、多倍周期响应及混沌响应之间的切换.Abstract: The bistable harvester can achieve wide band and high-efficiency energy harvesting performance under low frequency and low excitation levels. Previous studies mainly use a simple resistor circuit to capture the energy in the bistable structures. This paper proposes a two-degree-of-freedom (DOF) nonlinear system formed by coupling a three-spring bistable structure with a nonlinear RLC (resistance-inductance-capacitance) resonant circuit for energy harvesting enhancement. The nonlinear electromagnetic coupling coefficient between the circuit and structure is obtained by the special configuration between permanents and coils. The governing equation of the two DOF nonlinear systems is acquired. The analytical responses of the current and displacement are derived by the harmonic balance method, whose stability is judged by the Jacobin matrix. The analytical solution is compared with the numerical solution. Results demonstrate that introducing a nonlinear two-order resonant circuit into the bistable energy harvester can further improve the harvesting responses and broaden the energy bandwidth. With the same circuit parameters, the nonlinear resonant circuit can achieve lower frequency energy harvesting performance through frequency doubling of the current compared with the traditional linear circuit. One can enhance the energy harvester performance by decreasing the resonant ratio between the resonant circuit and bistable structure, increasing the excitation amplitude, and decreasing the distance between two static equilibrium points. The system can realize the switching of single-period response, multi-period response, and chaotic responses by adjusting the resonant ratio between circuit and bistable structure, and excitation amplitude.
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引 言
在物联网快速发展的时代背景下, 为数以亿计的网络设备提供源源不断的电力供应, 以保证它们在各种环境中正常的持续运行, 已经被公认为最大的挑战之一[1]. 俘能技术作为一种能使低功率物联网设备实现能量自供给的方式, 被广泛地认为是一种有前途的方法[2]. 在过去二三十年中, 学者们大量研究了如何从周围环境中俘获振动能量. 通常情况下, 可采用压电效应[3-4]、电磁感应[5-6]、静电效应[7]和摩擦发电[8]等多种方法将振动能量转化为所需的电能. 这就要求所设计的俘能装置需要有高效的机电耦合系数以实现良好的俘能效果.
线性谐振器可用于增强俘能器的响应. 因此, 学者们设计了线性压电式和电磁式等俘能结构来提升来自机械设备、风力和海浪等激励的俘能效果[9]. 但其缺点也很明显, 线性俘能器只能在共振频率附近处获得很强的俘能效果[10]. 而通常情况下, 周围环境的激励频率在一定的带宽内随机变化, 这使得共振频率外的激励能量无法被有效俘获. 为了拓宽俘能带宽、提高俘能效率, 学者们设计了多悬臂梁阵列[11]、L型梁[12]、多自由度梁[13]等结构, 提出了被动[14]、半主动[15]、自适应[16]等方法. 主要原理为在系统内增加多个不同的共振频率点以拓宽共振频带. 但这些方法增加了系统的质量和使用空间, 使结构系统变得过于复杂, 降低了单位质量和单位体积的俘能效率, 且系统单一共振频率下狭窄的俘能范围这一问题仍未得到解决[17].
为了克服线性俘能器的不足, 学者们将非线性概念引入到俘能器中. 非线性系统复杂的动力学行为使得幅频响应曲线发生了向右或向左的倾斜, 从而拓展了单一共振结构的俘能带宽、增强了俘能效果[18]. 此外, 增加静平衡点可使结构成为双稳态[19-20]、三稳态[21]和多稳态[22]系统, 从而利用多稳态系统中的软化或硬化频率响应特性[23], 亚谐波、超谐波和组合共振效应[24], 井内运动、井间运动、簇发振荡运动和混沌运动等实现高效的俘能响应和宽频的能量俘获[3]. 学者们还发现, 利用非线性系统中的内共振原理亦可提升俘能效果, 拓宽俘能带宽[25]. 理论分析和实验结果表明, 非线性俘能器无论是在定频激励、扫频激励还是随机激励条件下的俘能特性均优于线性俘能器[26].
将永磁体的强非线性磁力引入到悬臂梁式俘能器是学者们研究机械俘能特性最为经典的结构之一[27]. 通过改变永磁体间的距离和个数等参数, 可实现结构的强非线性以及单/双/三/多稳态. 通常在悬臂梁根部粘贴压电片以对振动能量进行俘获. 电磁式俘能器则通过在永磁体周围布置线圈, 通过二者的相对运动来进行振动能量俘获[28]. 之前对非线性多稳态俘能器的研究中, 压电式和电磁式结构主要通过在压电片或线圈两端接入外部纯电阻电路来俘获外界振动能量. 然而, 对外接电路为非线性共振型电路时的研究却没有深入地探讨.
本文研究了双稳态俘能器耦合非线性谐振电路的俘能效果及动力学特性. 分析了结构与电路之间的非线性机电耦合系数, 建立了两自由度非线性控制方程, 利用谐波平衡法求得了系统的电流及位移响应, 利用雅可比矩阵分析了解的稳定性, 将理论解与数值解进行了对比, 最后对俘能特性和动力学行为进行了参数化分析, 以期为低频俘能领域提供新的解决思路.
1. 非线性谐振式电磁俘能器模型
非线性谐振式电磁俘能器的构造如图1所示. 其由经典的“三弹簧”系统[29]和非线性RLC电路系统两部分构成.
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图 1 非线性电磁俘能器的构造及对应物理简化模型Figure 1. Configurations of nonlinear electromagnetic harvesters and corresponding simplified physical models“三弹簧”构型通过调节竖直弹簧的刚度k0, 水平倾斜弹簧的刚度kh以及倾斜参数q, p等可实现非线性准零刚度和双稳态等特性. 该系统的弹性恢复力在忽略重力影响的情况下可表达为
$$ F = {k_0}x + 2{k_h}\left[ {1 - {{{l_1}} \mathord{\left/ {\vphantom {{{l_1}} {\sqrt {{{\left( {p + x} \right)}^2} + {q^2}} }}} \right. } {\sqrt {{{\left( {p + x} \right)}^2} + {q^2}} }}} \right]\left( {p + x} \right) $$ (1) 式中, x为质量m相对于基平面的相对位移, l1为水平倾斜弹簧的自然长度. 在小位移情况下, 该式可用泰勒公式进行展开, 即
$F\left( x \right) \approx \displaystyle\sum\limits_{i = 1}^n {\dfrac{{{F^{\left( i \right)}}\left( 0 \right)}}{{i!}}} {x^i}$ . 式中, 上标(i)代表第i阶微分. 本文的分析中取前三阶级数对弹性恢复力进行简化. 且为了实现对称的势能函数, 需使p = 0. 故非线性弹性恢复力的最终近似表达式为$$ F \approx {k_1}x + {k_3}{x^3} $$ (2) 式中,
${k_1} = {k_0} + 2{k_h}\left( {1 - {{{l_1}} \mathord{\left/ {\vphantom {{{l_1}} q}} \right. } q}} \right)$ ,${k_3} = {{{k_h}{l_1}} / {{q^3}}}$ . 对式(2)积分可得到系统的势能函数$$ U = {{{k_1}{x^2}} /2} + {{{k_3}{x^4}} / 4} $$ (3) 式(2)和式(3)在F = 0情况下的解可分为两类. 第一类为
${{{k_1}} \mathord{\left/ {\vphantom {{{k_1}} {{k_3} \geqslant 0}}} \right. } {{k_3} \geqslant 0}}$ 条件下, 系统有且仅有一个实解x = 0. 第二类为${{{k_1}} \mathord{\left/ {\vphantom {{{k_1}} {{k_3} < 0}}} \right. } {{k_3} < 0}}$ 条件下, 系统具有三个实解${x_1} = 0$ ,${x_2} = \sqrt { - {{{k_1}} \mathord{\left/ {\vphantom {{{k_1}} {{k_3}}}} \right. } {{k_3}}}} $ ,${x_3} = - \sqrt { - {{{k_1}} \mathord{\left/ {\vphantom {{{k_1}} {{k_3}}}} \right. } {{k_3}}}} $ . 其中, x1为非稳定解, x2和x3为稳定解. 为了更加直观地研究解与弹性恢复力和势能的关系, 引入如下表达式$$ F = {{kx\left( {x - {x_2}} \right)\left( {x - {x_3}} \right)} \mathord{\left/ {\vphantom {{kx\left( {x - {x_2}} \right)\left( {x - {x_3}} \right)} {\left| {{x_2}{x_3}} \right|}}} \right. } {\left| {{x_2}{x_3}} \right|}} $$ (4) 展开之后可得
$$ F = - kx + {{k{x^3}} \mathord{\left/ {\vphantom {{k{x^3}} {\left| {{x_2}{x_3}} \right|}}} \right. } {\left| {{x_2}{x_3}} \right|}} $$ (5) 与式(2)对比可得,
${k_1} = - k$ ,${k_3} = {k \mathord{\left/ {\vphantom {k {\left| {{x_2}{x_3}} \right|}}} \right. } {\left| {{x_2}{x_3}} \right|}}$ . 对式(5)积分亦可得势能函数$$ F = - {{k{x^2}}/ 2} + {{k{x^4}} / {\left( {4\left| {{x_2}{x_3}} \right|} \right)}} $$ (6) 图2对“三弹簧”构型非线性恢复力的精确解和近似解进行了对比. 精确解的参数为k0 = 1.4061 × 103 N·m, kh = 2 × 103 N·m, p = 0, q = 1.17 × 10−4 m, 求得的近似解参数为k1 = −100 N·m, k3 = 2 × 107 N·m, x1 = 0, x2 = 2.236 × 10−3 m, x3 = −2.236 × 10−3 m. 可以看到, 近似解可以良好地吻合精确解.
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图 2 “三弹簧”构型双稳态恢复力的精确解与近似解Figure 2. Exact and approximate bistable restoring force of the “three springs” configuration非线性RLC电路由一对永磁体, 一对线圈和外接电路构成, 如图1(a)所示. 参数相同的永磁体轴向均匀磁化, 并且同轴相斥分布, 两磁体间的距离为H. 二者与质量块固连在一起, 总质量为m. 在永磁体周向两侧分布有均匀缠绕的线圈1和线圈2. 两个线圈除绕制方向相反外, 其他条件如线圈匝数、电阻、电感等均相同. 绕制后的线圈内径为R1, 外径为R2. 此外, 串联连接的两个线圈与外接电路相连以构成闭合电路. 当永磁体与线圈产生相对位移时, 线圈便会切割磁感线, 从而产生感应电动势Ve及感应电流. 此时线圈可等效为电压Ve, 电阻Re和电感Le串联模型, 若外接电路中再引入电容C以及额外的可调电阻Ra和可调电感La, 电路便构成了二阶谐振电路. 其与“三弹簧”模型串联便形成了非线性两自由度系统. 通过适当地调节谐振电路和“三弹簧”模型的共振频率比, 便可实现高效的俘能效果.
线圈和永磁体的具体参数列于表1中.
表 1 线圈和永磁体的参数配置Table 1. Parameters of coils and permanent magnets (PM)Name Value residual magnetic flux density Br/T 1.19 inner radius of PM Rin/mm 9 outer radius of PM Rout/mm 14 height of PM Ht/mm 10 distance between PMs H/mm 3 average radius of coil Ra/mm 16.5 number of coil turn 350 + 350 resistance of coil Re/Ω 24.9 inductance of coil Le/mH 2.01 2. 动力学建模
2.1 非线性电磁耦合模型
根据图1和基尔霍夫电压定律可求得电路内的控制方程
$$ {V_e} = L\ddot q + R\dot q + {q \mathord{\left/ {\vphantom {q C}} \right. } C} $$ (7) $$ {V_e} = {C_e}\dot x $$ (8) 式中, L = Le + La为电路中的总电感, R = Re + Ra为总电阻, q为电荷, x为线圈与永磁体间的相对位移, Ve为电路中的总电压. Ce为机电耦合系数, 可根据线圈圈数、线圈半径和剩磁强度等参数进行计算[30]. 当切割线圈的磁场均匀分布时, Ce可视为常数.
相对运动产生的洛伦兹力为
$$ {F_e} = {C_m}\dot q $$ (9) 式中, Cm为电磁耦合系数. 由文献[31]可知, Ce = Cm. 由于线圈1和线圈2绕制方向相反, 而切割二者的磁感线较为均匀, 变化较小. 因此, 两个线圈会产生相反的电流, 从而导致相反的电压. 由文献[32]可知, 系统的等效机电耦合系数为
$$ {C_e} = {C_{e2}} - {C_{e1}} $$ (10) 式中, Ce1和Ce2分别为两个线圈的机电耦合系数. 计算了表1参数下的机电耦合系数, 结果如图3所示, 其为奇函数形式, 该结果亦可用多项式近似拟合
$$ {C_e} \approx {c_1}x + {c_3}{x^3} $$ (11) 式中, c1和c3为拟合系数. 在表1参数下计算得到的c1, c3分别为1.0674 × 103, 9.2731 × 106.
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图 3 非线性电磁耦合系数的理论解和多项式拟合结果Figure 3. Analytical and approximate nonlinear electromagnetic coupling coefficient2.2 俘能器的幅频响应关系
由牛顿第二定律和图1(b)可求得俘能器的振动控制方程
$$ m\ddot x + c\dot x + {k_1}x + {k_3}{x^3} + \left( {{c_1}x + {c_3}{x^3}} \right)\dot q = m{x_0}{\omega ^2}\cos \left( {\omega t} \right) $$ (12) $$ L\ddot q + R\dot q + {q \mathord{\left/ {\vphantom {q C}} \right. } C} - \left( {{c_1}x + {c_3}{x^3}} \right)\dot x = 0 $$ (13) 式中, m为运动系统的总质量, c为黏性阻尼, x0为基础激励幅值,
$ \omega $ 为基础激励频率, t为激励时间.式(12)和式(13)为非线性耦合系统, 因此经典的叠加原理已无法适应该系统的求解. 为了研究非线性系统的响应问题, 学者们提出了摄动法、平均法、多尺度法、渐进法和谐波平衡法等多种方法[33]. 其中, 谐波平衡法由于能够分析强非线性问题, 且特别适用于大范围参数变化的定量分析, 因此得到了广泛的应用. 本节采用该方法对该非线性系统的频域响应进行了求解分析. 可设俘能器中的质量系统m的相对位移和电路内的电荷为
$$ x\left( t \right) = {a_1}\left( t \right)\sin \left( {\omega t} \right) + {b_1}\left( t \right)\cos \left( {\omega t} \right) $$ (14) $$ q\left( t \right) = {a_2}\left( t \right)\sin \left( {2\omega t} \right) + {b_2}\left( t \right)\cos \left( {2\omega t} \right) $$ (15) 式中, a1(t), b1(t), a2(t), b2(t)为关于时间t缓慢变化的系数. 可以看到, 电荷的频率为位移频率的两倍, 这在文献[32]中已经得到了验证.
将式(14)和式(15)带入式(13)中并忽略
$\sin \left( {4\omega t} \right)$ 和$\cos \left( {4\omega t} \right)$ , 则$\sin \left( {2\omega t} \right)$ 和$\cos \left( {2\omega t} \right)$ 的常数项分别为$$ \begin{split} & \left( { - 4{\omega ^2}L + {1 \mathord{\left/ {\vphantom {1 C}} \right. } C}} \right){a_2} - 2R\omega {b_2} + \frac{1}{2}\omega {c_1}\left( { - {a_1}^2 + {b_1}^2} \right) +\\ &\qquad \frac{1}{4}\omega {c_3}\left( { - {a_1}^4 + {b_1}^4} \right) = \frac{1}{2}{c_1}{b_1}{a'_1} + \frac{3}{4}{c_3}{a_1}^2{b_1}{a'_1} + \\ & \qquad \left.\frac{1}{4}{c_3}{b_1}^3{a'_1} + \frac{1}{2}{c_1}{a_1}{b'_1} +\right. \frac{1}{4}{c_3}{a_1}^3{b'_1} + \\ &\qquad \frac{3}{4}{c_3}{a_1}{b^{2}_1}{b'_1} + 4\omega L{b'_2} - R{a'_2} - L{a''_2} \end{split} $$ (16) $$ \begin{split} & 2R\omega {a_2} + \left( { - 4{\omega ^2}L + {1 \mathord{\left/ {\vphantom {1 C}} \right. } C}} \right){b_2} - \left( {c_1} + \frac{1}{2}{c_3}{a_1}^2 + \right. \\ &\qquad\left.\frac{1}{2}{c_3}{b_1}^2 \right)\omega {a_1}{b_1} = - \frac{1}{2}{c_1}{a_1}{a'_1} - \frac{1}{2}{c_3}{a_1}^3{a'_1} + \\ &\qquad \frac{1}{2}{c_3}{b_1}^3{b'_1} + \frac{1}{2}{c_1}{b_1}{b'_1} - 4\omega L{a'_2} - R{b'_2} - L{b''_2} \end{split} $$ (17) 类似地, 将式(14)和式(15)带入式(12)中, 也忽略结果中的高次谐波项, 保留
$\sin \left( {\omega t} \right)$ 和$\cos \left( {\omega t} \right)$ 的系数$$ \begin{split} & - \left(- m{\omega ^2}{a_1} + {k_1}{a_1} + \frac{3}{4}{k_3}{a_1}^3 - \omega {c_1}{a_1}{a_2} - \omega {c_3}{a_1}^3{a_2} - \right.\\ &\qquad c\omega {b_1} + \frac{3}{4}{k_3}{a_1}{b_1}^2 - \omega {c_1}{b_1}{b_2} - \frac{3}{2}\omega {c_3}{a_1}^2{b_1}{b_2} - \\ &\qquad \left. \frac{1}{2}\omega {c_3}{b_1}^3{b_2} \right) = c{a'_1} - 2m\omega {b'_1} - \left( {\frac{1}{2}{c_1}{a_1} + \frac{1}{2}{c_3}{a_1}^3} \right)\cdot \\ &\qquad {b'_2} + \left( {\frac{1}{2}{c_1}{b_1} + \frac{3}{4}{c_3}{a_1}^2{b_1} + \frac{1}{4}{c_3}{b_1}^3} \right){a'_2} \\[-12pt]\end{split} $$ (18) $$ \begin{split} &- \left( - m{\omega ^2}{z_0} - m{\omega ^2}{b_1} + {k_1}{b_1} + \frac{3}{4}{k_3}{b_1}^3 + \omega {c_1}{a_2}{b_1} + \right.\\ &\qquad \omega {c_3}{a_2}{b_1}^3 + c\omega {a_1} + \frac{3}{4}{k_3}{a_1}^2{b_1} - \omega {c_1}{a_1}{b_2} - \\ &\qquad \left.\frac{3}{2}\omega {c_3}{a_1}{b_1}^2{b_2} - \frac{1}{2}\omega {c_3}{a_1}^3{b_2} \right)= \\ &\qquad c{b'_1} + 2m\omega {a'_1} + \left( {\frac{1}{2}{c_1}{b_1} + \frac{1}{2}{c_3}{b_1}^3} \right){b'_2} + \left( \frac{1}{2}{c_1}{a_1} + \right.\\ &\qquad \left.\frac{1}{4}{c_3}{a_1}^3 + \frac{3}{4}{c_3}{a_1}{b_1}^2 \right){a'_2} \\[-12pt] \end{split} $$ (19) 当系统的响应稳定时, 随时间变化的系数可以忽略. 故式(16) ~ 式(19)等号右边为零. 因此可用a1, b1表示a2, b2, 即
$$ {a_2} = - AV - UB $$ (20) $$ {b_2} = AU - BV $$ (21) 式中, r2 = a2 + b2,
$U = {{\omega \left( {{c_1} + {{{c_3}{r^2}} \mathord{\left/ {\vphantom {{{c_3}{r^2}} 2}} \right. } 2}} \right)\left( {b_1^2 - a_1^2} \right)} \mathord{\left/ {\vphantom {{\omega \left( {{c_1} + {{{c_3}{r^2}} \mathord{\left/ {\vphantom {{{c_3}{r^2}} 2}} \right. } 2}} \right)\left( {b_1^2 - a_1^2} \right)} 2}} \right. } 2}$ ,$V = - \left( {{c_1}{\text{ + }}{{{c_3}{r^2}} \mathord{\left/ {\vphantom {{{c_3}{r^2}} 2}} \right. } 2}} \right)\omega {a_1}{b_1}$ ,$A = \dfrac{{2 R\omega }}{{{{\left( {{1 \mathord{\left/ {\vphantom {1 C}} \right. } C} - 4{\omega ^2}L} \right)}^2} + {{\left( {2 R\omega } \right)}^2}}}$ ,$B = \dfrac{{{1 \mathord{\left/ {\vphantom {1 C}} \right. } C} - 4{\omega ^2}L}}{{{{\left( {{1 \mathord{\left/ {\vphantom {1 C}} \right. } C} - 4{\omega ^2}L} \right)}^2} + {{\left( {2 R\omega } \right)}^2}}}$ .将式(20)和式(21)带入式(12)中, 之后将结果中
$\sin \left( {\omega t} \right)$ 和$\cos \left( {\omega t} \right)$ 的系数平方相加可得到$$ \begin{split} &{r^2}\left( - m{\omega ^2} + {k_1} + \frac{3}{4}{k_3}{r^2} - \frac{1}{2}B{\omega ^2}c_1^2{r^2} - \frac{3}{4}B{\omega ^2}{c_1}{c_3}{r^4} -\right. \\ &\qquad \left. \frac{1}{4}B{\omega ^2}c_3^2{r^6} \right)^2 + {r^2}\left( c\omega + \frac{1}{2}A{\omega ^2}c_1^2{r^2} + \frac{1}{2}A{\omega ^2}{c_1}{c_3}{r^4} + \right.\\ &\qquad \left.\frac{1}{8}A{\omega ^2}c_3^2{r^6} \right)^2 = {\left( {m{\omega ^2}{x_0}} \right)^2}\\[-12pt] \end{split} $$ (22) 由式(22)可知, 给定激励频率和激励幅值, 便可求得系统的相对位移r. 之后, 据此可求得电流大小与相对位移之间的关系为
$$ I = \sqrt {{{\left( {2\omega {a_2}} \right)}^2} + {{\left( {2\omega {b_2}} \right)}^2}} = {\omega ^2}{r^2}\left| {{c_1} + {{{c_3}{r^2}} \mathord{\left/ {\vphantom {{{c_3}{r^2}} 2}} \right. } 2}} \right|\sqrt {{A^2} + {B^2}} $$ (23) 2.3 稳定性分析
式(22)求得的响应位移r存在部分不稳定解, 本部分基于雅可比矩阵对其进行了稳定性分析. 雅可比矩阵可表达为
$$ {{\boldsymbol{J}}}=\left[\begin{array}{llll} \dfrac{\partial \dot{a}_1}{\partial a_1} & \dfrac{\partial \dot{a}_1}{\partial b_1} & \dfrac{\partial \dot{a}_1}{\partial a_2} & \dfrac{\partial \dot{a}_1}{\partial b_2} \\ \dfrac{\partial \dot{b}_1}{\partial a_1} & \dfrac{\partial \dot{b}_1}{\partial b_1} & \dfrac{\partial \dot{b}_1}{\partial a_2} & \dfrac{\partial \dot{b}_1}{\partial b_2} \\ \dfrac{\partial \dot{a}_2}{\partial a_1} & \dfrac{\partial \dot{a}_2}{\partial b_1} & \dfrac{\partial \dot{a}_2}{\partial a_2} & \dfrac{\partial \dot{a}_2}{\partial b_2} \\ \dfrac{\partial \dot{b}_2}{\partial a_1} & \dfrac{\partial \dot{b}_2}{\partial b_1} & \dfrac{\partial \dot{b}_2}{\partial a_2} & \dfrac{\partial \dot{b}_2}{\partial b_2} \end{array}\right] $$ (24) 根据式(16) ~ 式(19)求解了
$ {\dot a_1} $ ,$ {\dot b_1} $ ,$ {\dot a_2} $ ,$ {\dot b_2} $ 的表达式, 结果见附录. 当矩阵的特征值所有解的实数值均为负数时, 解稳定.2.4 数值解
上一节利用谐波平衡法对俘能器的频域响应进行了求解. 为了验证该方法的准确性, 利用四五阶龙格库塔法(OED45)对式(12)和式(13)进行了数值分析, 从而求得其振动位移及电流的数值响应. 数值分析中采用线性扫频方式和定频方式进行激励. 一些固定不变的参数列于表2中.
表 2 仿真分析中用到的固定参数数值Table 2. Parameters values used in simulationParameters name Value m/kg 0.5 k0/(N·m) 1.4061 × 103 kh/(N·m) 2 × 103 k1/(N·m) −2100 k3/(N·m) 2 × 107 c/(N·s·m−1) 2.6515 c1/(N·A−1·m−1) 1.0674 × 103 c3/ (N·A−1·m−3) 9.2731 × 106 C/F 1/(8π2) 图4对理论解和数值解进行了比较分析. 参数设置如下, x0 = 3 mm, R = 24.9 Ω,
${{{\omega _e}} \mathord{\left/ {\vphantom {{{\omega _e}} {{\omega _k}}}} \right. } {{\omega _k}}} = 0.649\;0$ . 其中,${\omega _e} = {1 \mathord{\left/ {\vphantom {1 ({2\sqrt {LC} })}} \right. } {(2\sqrt {LC}) }}$ 为电路的谐振频率,${\omega _k} = \sqrt {{{{k_0}} \mathord{\left/ {\vphantom {{{k_0}} m}} \right. } m}} $ 为“三弹簧”系统线性情况下的固有频率. 采用定位移幅值线性正扫频进行激励, 扫频速率为0.25 Hz/s. 可以看到, 理论解良好地预测了电流和相对位移的数值响应, 系统的非线性突跳现象也得到了良好的体现. 需要注意的是, 图中蓝色方框内的理论解并不合理. 这主要是因为外界激励输入的能量过小, 导致系统只能产生井内振荡, 而无法出现井间振荡响应. 而本文中采用谐波平衡法的前提条件为系统为井间振荡. 因此, 在此之后的分析中, 该部分解不予考虑分析.![]()
图 4 理论解与数值解对比Figure 4. Comparison between theoretical solution and numerical solution此外, 利用以上参数也分析了相对位移与电流频率之间的关系. 在5 Hz定频激励的条件下, 由图5(b)的相轨迹和庞加莱截面可知系统为周期响应. 对应的位移和电流结果如图5(a)所示. 可知, 电流频率为位移频率的两倍, 这与文献[32]中的结果相同. 需要注意的是, 该结果使得电路的实际谐振频率由原来的
${1 / {\sqrt {LC} }}$ 降为${1 / ({2\sqrt {LC} })}$ . 因此, 可用于低频俘能.![]()
图 5 5 Hz定频激励下相对位移与电流之间的频率关系Figure 5. Relationship between relative displacement and current under 5 Hz constant frequency excitation3. 俘能特性分析
本部分验证了共振式俘能器的俘能效果, 分析了电路谐振频率、激励幅值以及静平衡位置对俘能特性的影响, 同时也分析了俘能器的动力学响应.
3.1 共振式俘能器性能
图6对电阻式俘能器和共振式俘能器的俘能效果进行了对比. 参数设置如下, x0 = 3 mm, R = 24.9 Ω,
${{{\omega _e}} \mathord{\left/ {\vphantom {{{\omega _e}} {{\omega _k}}}} \right. } {{\omega _k}}} = 0.245\;3$ , 电阻式俘能器中, L = 0.![]()
图 6 电阻式俘能器与共振式俘能器性能对比Figure 6. Energy harvesting comparison between the energy harvesters with a pure resistor and with a resonant circuit由图可知, 一阶电路的电阻式俘能器由于未能与“三弹簧”式双稳态系统发生共振耦合, 响应的峰值位移(x = 12.9 mm)和峰值电流处于较低值(I = 0.41 A), 对应的俘能带宽(0 ~ 11 Hz)也较窄. 当电路系统变为二阶谐振电路时, 由于系统的共振耦合效应, 系统的相对位移增加到15 mm, 电流增加到0.72 A, 俘能带宽也拓宽至14.8 Hz. 此外, 响应曲线在引入非线性共振电路之后向高频方向上发生了较大的偏移, 即系统的硬刚度特性更加明显, 这表明系统更强的非线性更有利于俘能特性的提升. 总的来说, 引入二阶谐振电路有利于提升俘能器的俘能性能.
3.2 共振频率对俘能特性的影响
上节验证了俘能器中引入非线性谐振电路可提升俘能性能. 本节则具体分析电路谐振频率对俘能特性的影响趋势, 如图7所示. 参数设置如下, x0 = 3 mm, R = 24.9 Ω.
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图 7 电路谐振频率对俘能特性的影响Figure 7. Effect of circuit resoance frequency on energy harvesting performance由图可知, 随着系统频率比由0.2902减小到0.2052, 响应位移由13.9 mm增加到18.9 mm, 响应电流由0.56 A增加到1.30 A, 俘能带宽由[0 13.1] Hz拓宽到[0 20.2] Hz. 因此, 减小电路的谐振频率可有效提升俘能特性, 拓宽俘能带宽. 需要注意的是, 传统线性电路的谐振频率为
${\omega _{et}} = {1 / {\sqrt {LC} }}$ , 系统的频率比为${{{\omega _{et}}} / {{\omega _k}}}$ . 而本文所设计的非线性电路的谐振频率仅为线性电路的一半, 即${1 / ({2\sqrt {LC} })} = {{{\omega _{et}}}/ 2}$ , 这使得系统的频率比也降至${{{\omega _{et}}} / ({2{\omega _k}}})$ . 因此, 在电容电感不变的条件下, 与传统的线性电路相比, 本文所设计的非线性俘能器可进一步提升俘能器的俘能特性, 并用于低频俘能.为了更好地分析频率比对俘能特性的影响, 本节也讨论了系统的动力学特性, 如图8 ~ 图11所示. 参数设置如下, x0 = 3 mm, R = 3 Ω, f = 15 Hz. 需要注意的是, 通常情况下, 分叉图内所选取的点通过间隔一个外界激励周期进行确定, 如图8所示的位移分叉图. 但由于本文所设计的非线性电路中电流变化的频率为激励频率的两倍, 因此在绘制电流分叉图时点的选取周期应为激励周期的一半, 即每隔1/30 s选取一个点. 之后电流的相轨迹图和庞加莱截面图也采用该原则.
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图 8 响应位移和电流随频率比的分叉图Figure 8. Birfication diagram of displacement and current with respect to frequency ratio![]()
图 9 频率比0.25情况下位移和电流的时域响应、傅里叶频谱、相轨迹及庞加莱截面Figure 9. Output responses, Fourier frequency responses, phase orbit, and Poincare map of displacement and current when the frequency ratio is 0.25由图8可知, 系统在频率比0 ~ 0.65范围内的运动状态总体上处于单倍周期运动和多倍周期运动交替出现的情况, 少部分区域如0.2427 ~ 0.2621等会出现混沌运动情况. 由于电路与“三弹簧”构型的耦合作用, 位移响应和电流响应在相同频率比下具有同样的运动类型, 即均为单倍/多倍周期运动或混沌运动.
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图 10 频率比0.5情况下位移和电流的时域响应、傅里叶频谱、相轨迹及庞加莱截面Figure 10. Output responses, Fourier frequency responses, phase orbit, and Poincare map of displacement and current when the frequency ratio is 0.5![]()
10 频率比0.5情况下位移和电流的时域响应、傅里叶频谱、相轨迹及庞加莱截面 (续)10. Output responses, Fourier frequency responses, phase orbit, and Poincare map of displacement and current when the frequency ratio is 0.5 (continued)也分析了图8中特定频率比情况下的动力学响应. 当频率比为0.25时, 响应位移和电流的时域结果、傅里叶频谱图、相轨迹图及庞加莱截面如图9所示. 从位移频谱图可得, 除了15 Hz基频响应外, 还有1/3倍的亚谐波响应以及0 ~ 15 Hz内连续的幅值较低的频带. 结合相应的相轨迹及庞加莱截面可知, 位移响应处于3倍周期振动和混沌振动之间的准周期振动. 相比于位移响应, 电流响应呈现出更加杂乱的趋势. 电流响应频谱中除了30 Hz的基频和1/3倍(10 Hz)的亚谐波以及0 ~ 30 Hz幅值较低的连续频带外, 还出现了2/3倍(20 Hz)的亚谐波, 且其在所有频率中响应幅值最大(0.08 A). 此外, 电流的相轨迹及庞加莱截面也表明系统处于周期振动和混沌振动之间的准周期振动. 当频率比增至0.5时, 响应结果如图10所示. 可以看到, 位移和电流响应在一个周期内的变化基本一致, 其可以从频率响应中消失的连续频带、相轨迹中重合的轨迹以及庞加莱截面重合的点得到验证. 位移频谱图中出现了15 Hz的基频响应以及1/3倍的亚谐波响应, 而电流频谱图中除了出现30 Hz的基频响应, 1/3倍和2/3倍的亚谐波响应外, 还出现了幅值较低的4/3倍和5/3倍超谐波响应. 此外, 由庞加莱截面结果可知, 位移出现了3倍周期运动, 而电流则出现了1.5倍周期运动. 当频率比增至0.75时, 位移及电流响应结果如图11所示. 图中结果均表明两种响应为规律的周期振动, 位移为单倍周期振动, 电流为0.5倍周期振动.
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图 11 频率比0.75情况下位移和电流的时域响应、傅里叶频谱、相轨迹及庞加莱截面Figure 11. Output responses, Fourier frequency responses, phase orbit, and Poincare map of displacement and current when frequency ratio is 0.75总的来说, 电路谐振频率与主结构的共振频率之比可显著影响俘能器的俘能特性、系统位移和电流的动力学行为.
3.3 激励幅值对俘能特性的影响
本节分析了外界激励幅值对俘能特性的影响, 如图12所示. 参数设置如下, R = 24.9 Ω,
${{\omega _e}} \mathord{\left/ {\vphantom {{{\omega _e}} {{\omega _k} = 0.3245}}} \right. } {\omega _k} = 0.324\;5$ .![]()
图 12 激励幅值对俘能特性的影响Figure 12. Effect of excitation amplitude on energy harvesting performance由图可知, 随着激励幅值从2.75 mm增加到3.75 mm, 响应位移由13 mm增至15 mm, 对应的电流由0.4296 A增至0.772 A, 俘能带宽由[0 11.79] Hz拓宽至[0 14.26] Hz. 因此, 增加输入激励幅值可有效提升俘能效果, 拓宽俘能带宽.
同样讨论了激励幅值对系统动力学响应特性的影响, 如图13 ~ 图16所示. 参数设置如下, R = 3 Ω, f = 5 Hz,
${{{\omega _e}} \mathord{\left/ {\vphantom {{{\omega _e}} {{\omega _k} = 0.175}}} \right. } {{\omega _k} = 0.175}}$ .![]()
图 13 响应位移和电流随激励幅值的分叉图Figure 13. Birfication diagram of displacement and current with respect to excitation amplitude由图可知, 在激励幅值不断增加的条件下, 响应位移和电流同样出现了单倍周期运动、多倍周期运动和混沌运动交替出现的现象. 与图8结果不同的是, 混沌运动出现的范围明显增多, 例如[0.54 0.64] mm, [0.94 1.19] mm, [1.27 1.52] mm. 需要注意的是, 系统一般不会从单倍周期运动突变为混沌运动, 而是先从单倍周期运动变为多倍周期运动, 最后过渡到混沌运动.
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图 14 激励幅值x0 = 0.75 mm情况下位移和电流的时域响应、傅里叶频谱、相轨迹及庞加莱截面Figure 14. Output responses, Fourier frequency responses, phase orbit, and Poincare map of displacement and current when x0 = 0.75 mm![]()
14 激励幅值x0 = 0.75 mm情况下位移和电流的时域响应、傅里叶频谱、相轨迹及庞加莱截面 (续)14. Output responses, Fourier frequency responses, phase orbit, and Poincare map of displacement and current when x0 = 0.75 mm (continued)![]()
图 15 激励幅值x0 = 1 mm情况下位移和电流的时域响应、傅里叶频谱、相轨迹及庞加莱截面Figure 15. Output responses, Fourier frequency responses, phase orbit, and Poincare map of displacement and current when x0 = 1 mm![]()
图 16 激励幅值x0 = 2 mm情况下位移和电流的时域响应、傅里叶频谱、相轨迹及庞加莱截面Figure 16. Output responses, Fourier frequency responses, phase orbit, and Poincare map of displacement and current when x0 = 2 mm接下来分析了图13特定激励幅值下的动力学响应. 激励幅值为0.75 mm下位移和电流的响应结果如图14所示. 由时域响应可知, 位移和电流具有明显的周期性. 由频域响应可知, 位移和电流除了具有5 Hz和10 Hz的基频响应外, 还分别具有3倍和2倍的超谐波响应. 由相轨迹和庞加莱截面可知, 位移和电流具有确定的运动轨迹以及分别为3倍周期响应和1.5倍周期响应. 当幅值增加到1 mm时, 响应结果如图15所示. 位移和电流的时域响应变得杂乱无章, 周期无法判别, 频域响应出现了连续的无规律的频带. 对应的相轨迹及庞加莱截面上出现了有规律的奇怪吸引子, 这表明系统进入了混沌运动. 当幅值进一步增加时, x0 = 2 mm, 响应结果如图16所示. 有规律的时域振荡, 单一的频域响应, 重合的相轨迹以及庞加莱截面点均表明系统为单倍周期运动. 总的来说, 可通过调节外界激励幅值来控制系统的响应类型.
3.4 静平衡位置对俘能特性的影响
式(2) ~ 式(9)表明, 系统静平衡位置会影响势能特性, 从而影响俘能效果. 图17研究了静平衡位置和势能井深度对俘能特性的影响. 参数设置如下, R = 24.9 Ω,
${{{\omega _e}} \mathord{\left/ {\vphantom {{{\omega _e}} {{\omega _k} = 0.2052}}} \right. } {{\omega _k} = 0.205\;2}}$ , x0 = 2.75 mm. 由图可知, 随着静平衡位置的减小, 势能井的深度逐渐变浅, 系统的响应位移和电流均有所增大, 俘能带宽也有所增加. 但总的变化趋势不明显.![]()
图 17 静平衡位置和势能井深度对俘能特性的影响Figure 17. Effect of static equilibrium position and depth of potential energy on energy harvesting performance![]()
17 静平衡位置和势能井深度对俘能特性的影响 (续)17. Effect of static equilibrium position and depth of potential energy on energy harvesting performance (continued)4. 结 论
提出了一种非线性RLC电路耦合双稳态俘能器, 用于宽频带能量回收及俘能特性提升. 线圈中接入串联的电容、电阻和电感, 使电路形成二阶谐振电路. 从而使整体系统形成了包含非线性俘能结构和非线性谐振电路的两自由度系统. 分析了非线性机电耦合系数. 得到了系统的振动控制方程. 利用谐波平衡法推导得到了响应位移和响应电流. 借助雅可比矩阵对理论解的稳定性进行了判别. 将解析解与数值解进行了对比. 对俘能器的俘能特性及动力学行为进行了参数化分析. 结果表明, 相比于纯电阻电路, 二阶非线性谐振电路可实现结构更低频率的能量俘获, 有效提升俘能器的俘能响应, 拓宽俘能带宽. 与传统线性谐振电路相比, 非线性谐振电路可通过电流的倍频现象将结构的有效俘能频率降低一半. 降低谐振电路的共振频率
${\omega _e}$ , 增加输入激励的幅值, 减小静平衡位置的数值均可提升俘能性能, 拓宽俘能带宽. 调节共振频率${\omega _e}$ 和外界激励幅值可控制系统的动力学行为, 使响应位移和电流变为单倍周期运动, 多倍周期运动及混沌运动. 理论解和数值解吻合良好. 后续研究将从实验验证和电路谐振频率的优化两方面来提升俘能器的性能.附录
${\dot a_1}$ ,$ {\dot b_1} $ ,$ {\dot a_2} $ ,$ {\dot b_2} $ 的表达式如下$\; $ $$ {\dot a_1} = {{\left( {{X_4}{Y_1} + {X_2}{Y_2}} \right)} \mathord{\left/ {\vphantom {{\left( {{X_4}{Y_1} + {X_2}{Y_2}} \right)} {\left( {{X_1}{X_4} + {X_2}{X_3}} \right)}}} \right. } {\left( {{X_1}{X_4} + {X_2}{X_3}} \right)}}\tag{A1} $$ $$ {\dot b_1} = {{\left( {{X_1}{Y_2} - {X_3}{Y_1}} \right)} \mathord{\left/ {\vphantom {{\left( {{X_1}{Y_2} - {X_3}{Y_1}} \right)} {\left( {{X_1}{X_4} + {X_2}{X_3}} \right)}}} \right. } {\left( {{X_1}{X_4} + {X_2}{X_3}} \right)}}\tag{A2} $$ $$ {\dot a_2} = {{\left[ {\left( {R{W_1} - 4\omega L{W_2}} \right){{\dot a}_1} + \left( {R{W_3} + 4\omega L{W_4}} \right){{\dot b}_1} + 4\omega L{U_4} + R{U_3}} \right]} \mathord{\left/ {\vphantom {{\left[ {\left( {R{W_1} - 4\omega L{W_2}} \right){{\dot a}_1} + \left( {R{W_3} + 4\omega L{W_4}} \right){{\dot b}_1} + 4\omega L{U_4} + R{U_3}} \right]} {\left[ {{R^2} + {{\left( {4\omega L} \right)}^2}} \right]}}} \right. } {\left[ {{R^2} + {{\left( {4\omega L} \right)}^2}} \right]}} \tag{A3}$$ $$ {\dot b_2} = {{\left[ {\left( { - R{W_2} - 4\omega L{W_1}} \right){{\dot a}_1} + \left( {R{W_4} - 4\omega L{W_3}} \right){{\dot b}_1} - 4\omega L{U_4} + R{U_3}} \right]} \mathord{\left/ {\vphantom {{\left[ {\left( { - R{W_2} - 4\omega L{W_1}} \right){{\dot a}_1} + \left( {R{W_4} - 4\omega L{W_3}} \right){{\dot b}_1} - 4\omega L{U_4} + R{U_3}} \right]} {\left[ {{R^2} + {{\left( {4\omega L} \right)}^2}} \right]}}} \right. } {\left[ {{R^2} + {{\left( {4\omega L} \right)}^2}} \right]}} \tag{A4}$$ 式中
$$\begin{split} & {X_1} = \left[ {{R^2} + {{\left( {4\omega L} \right)}^2}} \right]c + {W_1}\left( {R{W_1} - 4\omega L{W_2}} \right) - {W_2}\left( { - R{W_2} - 4\omega L{W_1}} \right) \\ & {X_2} = 2\left[ {{R^2} + {{\left( {4\omega L} \right)}^2}} \right]m\omega + {W_1}\left( {R{W_3} + 4\omega L{W_4}} \right) - {W_2}\left( {R{W_4} - 4\omega L{W_3}} \right) \\ & {X_3} = 2\left[ {{R^2} + {{\left( {4\omega L} \right)}^2}} \right]m\omega + {W_3}\left( {R{W_1} - 4\omega L{W_2}} \right) + {W_4}\left( { - R{W_2} - 4\omega L{W_1}} \right) \\ & {X_4} = \left[ {{R^2} + {{\left( {4\omega L} \right)}^2}} \right]c + {W_3}\left( {R{W_3} + 4\omega L{W_4}} \right) + {W_4}\left( {R{W_4} - 4\omega L{W_3}} \right) \\ & {Y_1} = \left[ {{R^2} + {{\left( {4\omega L} \right)}^2}} \right]{U_1} - {W_1}\left( {4\omega L{U_4} + R{U_3}} \right) + {W_2}\left( {R{U_4} - 4\omega L{U_3}} \right) \\ & {Y_2} = \left[ {{R^2} + {{\left( {4\omega L} \right)}^2}} \right]{U_2} - {W_3}\left( {4\omega L{U_4} + R{U_3}} \right) - {W_4}\left( {R{U_4} - 4\omega L{U_3}} \right) \end{split} $$ $$ \begin{split} &{U_1} = - \left( { - m{\omega ^2}{a_1} + {k_1}{a_1} + {{3{k_3}{a_1}^3} \mathord{\left/ {\vphantom {{3{k_3}{a_1}^3} 4}} \right. } 4} - \omega {c_1}{a_1}{a_2} - \omega {c_3}{a_1}^3{a_2} - c\omega {b_1} + {{3{k_3}{a_1}{b_1}^2} \mathord{\left/ {\vphantom {{3{k_3}{a_1}{b_1}^2} 4}} \right. } 4} - \omega {c_1}{b_1}{b_2} - {{3\omega {c_3}{a_1}^2{b_1}{b_2}} \mathord{\left/ {\vphantom {{3\omega {c_3}{a_1}^2{b_1}{b_2}} 2}} \right. } 2} - {{\omega {c_3}{b_1}^3{b_2}} \mathord{\left/ {\vphantom {{\omega {c_3}{b_1}^3{b_2}} 2}} \right. } 2}} \right) \\ &{U_2} = - \left( { - m{\omega ^2}{x_0} + c\omega {a_1} - m{\omega ^2}{b_1} + {k_1}{b_1} + {{3{k_3}{a_1}^2{b_1}} \mathord{\left/ {\vphantom {{3{k_3}{a_1}^2{b_1}} 4}} \right. } 4} + \omega {c_1}{a_2}{b_1} + {{3{k_3}{b_1}^3} \mathord{\left/ {\vphantom {{3{k_3}{b_1}^3} 4}} \right. } 4} + \omega {c_3}{a_2}{b_1}^3 - \omega {c_1}{a_1}{b_2} - {{\omega {c_3}{a_1}^3{b_2}} \mathord{\left/ {\vphantom {{\omega {c_3}{a_1}^3{b_2}} 2}} \right. } 2} - {{3\omega {c_3}{a_1}{b_1}^2{b_2}} \mathord{\left/ {\vphantom {{3\omega {c_3}{a_1}{b_1}^2{b_2}} 2}} \right. } 2}} \right)\\ &{U_3} = - \left( { - {{\omega {c_1}{a_1}^2} \mathord{\left/ {\vphantom {{\omega {c_1}{a_1}^2} 2}} \right. } 2} - {{\omega {c_3}{a_1}^4} \mathord{\left/ {\vphantom {{\omega {c_3}{a_1}^4} 4}} \right. } 4} + {{{a_2}} \mathord{\left/ {\vphantom {{{a_2}} C}} \right. } C} - 4{\omega ^2}L{a_2} + {{\omega {c_1}{b_1}^2} \mathord{\left/ {\vphantom {{\omega {c_1}{b_1}^2} 2}} \right. } 2} + {{\omega {c_3}{b_1}^4} \mathord{\left/ {\vphantom {{\omega {c_3}{b_1}^4} 4}} \right. } 4} - 2R\omega {b_2}} \right)\\ &{U_4} = - \left( {2R\omega {a_2} - \omega {c_1}{a_1}{b_1} - {{\omega {c_3}{a_1}^3{b_1}} \mathord{\left/ {\vphantom {{\omega {c_3}{a_1}^3{b_1}} 2}} \right. } 2} - {{\omega {c_3}{a_1}{b_1}^3} \mathord{\left/ {\vphantom {{\omega {c_3}{a_1}{b_1}^3} 2}} \right. } 2} + {{{b_2}} \mathord{\left/ {\vphantom {{{b_2}} C}} \right. } C} - 4{\omega ^2}L{b_2}} \right)\\ &{W_1} = {{{c_1}{b_1}} \mathord{\left/ {\vphantom {{{c_1}{b_1}} 2}} \right. } 2} + {{3{c_3}{a_1}^2{b_1}} \mathord{\left/ {\vphantom {{3{c_3}{a_1}^2{b_1}} 4}} \right. } 4} + {{{c_3}{b_1}^3} \mathord{\left/ {\vphantom {{{c_3}{b_1}^3} 4}} \right. } 4}\\ &{W_2} = {{{c_1}{a_1}} \mathord{\left/ {\vphantom {{{c_1}{a_1}} 2}} \right. } 2} + {{{c_3}{a_1}^3} \mathord{\left/ {\vphantom {{{c_3}{a_1}^3} 2}} \right. } 2}\\ &{W_3} = {{{c_1}{a_1}} \mathord{\left/ {\vphantom {{{c_1}{a_1}} 2}} \right. } 2} + {{{c_3}{a_1}^3} \mathord{\left/ {\vphantom {{{c_3}{a_1}^3} 4}} \right. } 4} + {{3{c_3}{a_1}{b_1}^2} \mathord{\left/ {\vphantom {{3{c_3}{a_1}{b_1}^2} 4}} \right. } 4}\\ &{W_4} = {{{c_1}{b_1}} \mathord{\left/ {\vphantom {{{c_1}{b_1}} 2}} \right. } 2} + {{{c_3}{b_1}^3} \mathord{\left/ {\vphantom {{{c_3}{b_1}^3} 2}} \right. } 2}\end{split}$$ -
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图 9 频率比0.25情况下位移和电流的时域响应、傅里叶频谱、相轨迹及庞加莱截面
Figure 9. Output responses, Fourier frequency responses, phase orbit, and Poincare map of displacement and current when the frequency ratio is 0.25
![]()
图 1 非线性电磁俘能器的构造及对应物理简化模型
Figure 1. Configurations of nonlinear electromagnetic harvesters and corresponding simplified physical models
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图 2 “三弹簧”构型双稳态恢复力的精确解与近似解
Figure 2. Exact and approximate bistable restoring force of the “three springs” configuration
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图 3 非线性电磁耦合系数的理论解和多项式拟合结果
Figure 3. Analytical and approximate nonlinear electromagnetic coupling coefficient
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图 4 理论解与数值解对比
Figure 4. Comparison between theoretical solution and numerical solution
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图 5 5 Hz定频激励下相对位移与电流之间的频率关系
Figure 5. Relationship between relative displacement and current under 5 Hz constant frequency excitation
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图 6 电阻式俘能器与共振式俘能器性能对比
Figure 6. Energy harvesting comparison between the energy harvesters with a pure resistor and with a resonant circuit
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图 7 电路谐振频率对俘能特性的影响
Figure 7. Effect of circuit resoance frequency on energy harvesting performance
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图 8 响应位移和电流随频率比的分叉图
Figure 8. Birfication diagram of displacement and current with respect to frequency ratio
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图 10 频率比0.5情况下位移和电流的时域响应、傅里叶频谱、相轨迹及庞加莱截面
Figure 10. Output responses, Fourier frequency responses, phase orbit, and Poincare map of displacement and current when the frequency ratio is 0.5
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10 频率比0.5情况下位移和电流的时域响应、傅里叶频谱、相轨迹及庞加莱截面 (续)
10. Output responses, Fourier frequency responses, phase orbit, and Poincare map of displacement and current when the frequency ratio is 0.5 (continued)
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图 11 频率比0.75情况下位移和电流的时域响应、傅里叶频谱、相轨迹及庞加莱截面
Figure 11. Output responses, Fourier frequency responses, phase orbit, and Poincare map of displacement and current when frequency ratio is 0.75
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图 12 激励幅值对俘能特性的影响
Figure 12. Effect of excitation amplitude on energy harvesting performance
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图 13 响应位移和电流随激励幅值的分叉图
Figure 13. Birfication diagram of displacement and current with respect to excitation amplitude
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图 14 激励幅值x0 = 0.75 mm情况下位移和电流的时域响应、傅里叶频谱、相轨迹及庞加莱截面
Figure 14. Output responses, Fourier frequency responses, phase orbit, and Poincare map of displacement and current when x0 = 0.75 mm
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14 激励幅值x0 = 0.75 mm情况下位移和电流的时域响应、傅里叶频谱、相轨迹及庞加莱截面 (续)
14. Output responses, Fourier frequency responses, phase orbit, and Poincare map of displacement and current when x0 = 0.75 mm (continued)
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图 15 激励幅值x0 = 1 mm情况下位移和电流的时域响应、傅里叶频谱、相轨迹及庞加莱截面
Figure 15. Output responses, Fourier frequency responses, phase orbit, and Poincare map of displacement and current when x0 = 1 mm
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图 16 激励幅值x0 = 2 mm情况下位移和电流的时域响应、傅里叶频谱、相轨迹及庞加莱截面
Figure 16. Output responses, Fourier frequency responses, phase orbit, and Poincare map of displacement and current when x0 = 2 mm
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图 17 静平衡位置和势能井深度对俘能特性的影响
Figure 17. Effect of static equilibrium position and depth of potential energy on energy harvesting performance
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17 静平衡位置和势能井深度对俘能特性的影响 (续)
17. Effect of static equilibrium position and depth of potential energy on energy harvesting performance (continued)
表 1 线圈和永磁体的参数配置
Table 1 Parameters of coils and permanent magnets (PM)
Name Value residual magnetic flux density Br/T 1.19 inner radius of PM Rin/mm 9 outer radius of PM Rout/mm 14 height of PM Ht/mm 10 distance between PMs H/mm 3 average radius of coil Ra/mm 16.5 number of coil turn 350 + 350 resistance of coil Re/Ω 24.9 inductance of coil Le/mH 2.01 
下载: 导出CSV
表 2 仿真分析中用到的固定参数数值
Table 2 Parameters values used in simulation
Parameters name Value m/kg 0.5 k0/(N·m) 1.4061 × 103 kh/(N·m) 2 × 103 k1/(N·m) −2100 k3/(N·m) 2 × 107 c/(N·s·m−1) 2.6515 c1/(N·A−1·m−1) 1.0674 × 103 c3/ (N·A−1·m−3) 9.2731 × 106 C/F 1/(8π2)
下载: 导出CSV
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期刊类型引用(13)
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