CN107800149A - 一种三相电网不平衡下c‑mmc多变量保护控制方法 - Google Patents

一种三相电网不平衡下c‑mmc多变量保护控制方法 Download PDF

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CN107800149A
CN107800149A CN201711181453.9A CN201711181453A CN107800149A CN 107800149 A CN107800149 A CN 107800149A CN 201711181453 A CN201711181453 A CN 201711181453A CN 107800149 A CN107800149 A CN 107800149A
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杭丽君
邱键
何远彬
朱明琳
王文杰
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Hangzhou Dianzi University
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    • HELECTRICITY
    • H02GENERATION; CONVERSION OR DISTRIBUTION OF ELECTRIC POWER
    • H02JCIRCUIT ARRANGEMENTS OR SYSTEMS FOR SUPPLYING OR DISTRIBUTING ELECTRIC POWER; SYSTEMS FOR STORING ELECTRIC ENERGY
    • H02J3/00Circuit arrangements for ac mains or ac distribution networks
    • H02J3/36Arrangements for transfer of electric power between ac networks via a high-tension dc link
    • HELECTRICITY
    • H02GENERATION; CONVERSION OR DISTRIBUTION OF ELECTRIC POWER
    • H02JCIRCUIT ARRANGEMENTS OR SYSTEMS FOR SUPPLYING OR DISTRIBUTING ELECTRIC POWER; SYSTEMS FOR STORING ELECTRIC ENERGY
    • H02J3/00Circuit arrangements for ac mains or ac distribution networks
    • H02J3/24Arrangements for preventing or reducing oscillations of power in networks
    • HELECTRICITY
    • H02GENERATION; CONVERSION OR DISTRIBUTION OF ELECTRIC POWER
    • H02JCIRCUIT ARRANGEMENTS OR SYSTEMS FOR SUPPLYING OR DISTRIBUTING ELECTRIC POWER; SYSTEMS FOR STORING ELECTRIC ENERGY
    • H02J3/00Circuit arrangements for ac mains or ac distribution networks
    • H02J3/26Arrangements for eliminating or reducing asymmetry in polyphase networks
    • HELECTRICITY
    • H02GENERATION; CONVERSION OR DISTRIBUTION OF ELECTRIC POWER
    • H02JCIRCUIT ARRANGEMENTS OR SYSTEMS FOR SUPPLYING OR DISTRIBUTING ELECTRIC POWER; SYSTEMS FOR STORING ELECTRIC ENERGY
    • H02J3/00Circuit arrangements for ac mains or ac distribution networks
    • H02J3/36Arrangements for transfer of electric power between ac networks via a high-tension dc link
    • H02J2003/365Reducing harmonics or oscillations in HVDC
    • YGENERAL TAGGING OF NEW TECHNOLOGICAL DEVELOPMENTS; GENERAL TAGGING OF CROSS-SECTIONAL TECHNOLOGIES SPANNING OVER SEVERAL SECTIONS OF THE IPC; TECHNICAL SUBJECTS COVERED BY FORMER USPC CROSS-REFERENCE ART COLLECTIONS [XRACs] AND DIGESTS
    • Y02TECHNOLOGIES OR APPLICATIONS FOR MITIGATION OR ADAPTATION AGAINST CLIMATE CHANGE
    • Y02EREDUCTION OF GREENHOUSE GAS [GHG] EMISSIONS, RELATED TO ENERGY GENERATION, TRANSMISSION OR DISTRIBUTION
    • Y02E40/00Technologies for an efficient electrical power generation, transmission or distribution
    • Y02E40/50Arrangements for eliminating or reducing asymmetry in polyphase networks
    • YGENERAL TAGGING OF NEW TECHNOLOGICAL DEVELOPMENTS; GENERAL TAGGING OF CROSS-SECTIONAL TECHNOLOGIES SPANNING OVER SEVERAL SECTIONS OF THE IPC; TECHNICAL SUBJECTS COVERED BY FORMER USPC CROSS-REFERENCE ART COLLECTIONS [XRACs] AND DIGESTS
    • Y02TECHNOLOGIES OR APPLICATIONS FOR MITIGATION OR ADAPTATION AGAINST CLIMATE CHANGE
    • Y02EREDUCTION OF GREENHOUSE GAS [GHG] EMISSIONS, RELATED TO ENERGY GENERATION, TRANSMISSION OR DISTRIBUTION
    • Y02E60/00Enabling technologies; Technologies with a potential or indirect contribution to GHG emissions mitigation
    • Y02E60/60Arrangements for transfer of electric power between AC networks or generators via a high voltage DC link [HVCD]

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Abstract

本发明公开了一种三相电网不平衡下C‑MMC多变量保护控制方法,由于交流电网故障时***的三相电流会不平衡且在直流侧会存在较大电压纹波及功率波动,严重时危及变换器安全、缩短变换器寿命,因此必须要合理控制电网负序分量对***的影响。本发明利用改进的灵活正负序控制算法(FPNSC),采用d‑q坐标系下灵活正负和负序控制策略,通过参变量的调节,合理调整网侧电流参考的正负序含量,使得网侧电流波形质量和***瞬时有功功率、无功功率得到合理控制,从而实现对C‑MMC的保护控制。

Description

一种三相电网不平衡下C-MMC多变量保护控制方法
技术领域
本发明属于电力电子功率变换器领域,具体涉及一种三相电网不平衡下C-MMC多变量保护控制方法。
背景技术
柔性直流输电的应用,扩大了电力电子变流技术在电力***输电领域的应用,C-MMC以其较好的延展性被认为是较有应用前景的拓扑电路,广泛应用在柔性直流输电(VSC-HVDC)领域,尤其是在近年来兴起的多端柔性直流输电领域。
在交直流混合型输配电中,电网故障时由于电网负序分量的存在会导致***三相电流不平衡或非正弦、直流侧存在较大电压纹波、功率波动等问题,危及变流器安全、缩短变流器寿命。
目前电网不平衡时变流器的功率控制主要有如下几种:瞬时有功-无功控制,对称正序控制,正负序控制,平均有功-无功控制,灵活正负和负序控制。瞬时有功-无功控制可精确的控制***的有功和无功功率,但如果网侧电压出现不平衡跌落,该方法将导致网侧电流波形畸变严重;对称正序控制策略不考虑电网电压负序分量,可得到对称的网侧电流;正负序控制、平均有功-无功控制以及灵活正序和负序控制均考虑电网电压负序分量,灵活正序和负序控制可以更灵活的控制电压负序分量的比例。正负序控制以及平均有功-无功控制均为单一控制目标的方法,而灵活正序和负序控制可以灵活的调节电流参考中网侧电压负序分量的含量。但目前灵活正序和负序控制主要用在新能源发电中,而且缺少对***运行性能的分析说明。
发明内容
本发明针对现有技术的不足,提出一种三相电网不平衡下C-MMC多变量保护控制方法,以柔性直流输电(VSC-HVDC)为应用目标,从交流网侧电流质量、C-MMC桥臂电流、直流侧电压纹波以及中点电位平衡多个控制变量出发,提出了灵活的多变量保护控制方法。
本发明提出多变量保护控制方法,即选择一种较为灵活的控制方法,将恒有功功率控制的对称正序控制,正负序控制,平均有功-无功控制三种方法统一,通过调节控制变量,达到动态调节电流参考中的网侧电压负序分量部分的目的,在一定程度上消除***瞬时有功功率的二倍频振荡,相对减小电流参考的幅值,实现对网侧电流波形质量和***功率波动的灵活调节与控制;引入了多变量保护控制方法中的控制参数k,由于控制参数k对C-MMC运行状态具有较大影响,根据该参数能估算出电网故障时满足***具体需求时允许传输的功率最大值,保证***运行在安全稳定的状态。
本发明具体包括以下步骤:
步骤1.计算出电网不对称时MMC的瞬时有功功率和无功功率,公式如下:
其中:P0、Q0表示变换器中瞬时有功功率和无功功率的平均值;Pc2、Ps2分别为变换器有功功率的振荡分量的幅值;Qc2、Qs2分别表示变换器无功功率的振荡分量的幅值;ω表示电网频率,t表示时间;
步骤2.在同步旋转坐标系下求出步骤1公式中功率量的幅值,公式如下:
其中分别为电网电压的d轴正序、d轴负序、q轴正序、q轴负序分量幅值;分别为网侧电流的d轴正序、d轴负序、q轴正序、q轴负序分量幅值。
步骤3.假设要求***有功功率没有二倍频振荡,设定控制目标P0为常量,且Pc2=Ps2=Q0=0,引入调节参数k,d-q坐标系下电流控制,可得出网侧电流的参考值,公式如下:
步骤4.由步骤3的结果计算出Pc2、Ps2、Qc2和Qs2,公式如下:
本文提出多变量保护控制方法是一种较为灵活的控制方法,通过调节控制参数k,动态调节电流参考中的正序和负序分量,实现对网侧电流波形质量和***功率波动等的灵活调节和控制。
附图说明
图1为本发明所述C-MMC电路原理图;
图2为本发明所述C-MMC子模块拓扑结构图;
图3为本发明所述三相电网不平衡下C-MMC多变量保护控制方法***控制框图;
图4为本发明所述三相电网不平衡下C-MMC多变量保护控制方法流程图。
具体实施方式
以下结合说明书附图对本发明作进一步说明。
采用箝位双子模块的模块化多电平变换器(C-MMC)的拓扑结构图如图1所示,其子模块拓扑结构图如图2所示。当网侧交流电压出现不平衡故障时,三相电流会不平衡且在直流侧会存在较大电压纹波及功率波动,严重时危及变换器安全。采用本发明提出的三相电网不平衡下C-MMC多变量保护控制方法可以有效避免问题。本发明提出的方法其***控制框图如图3所示。方法主要步骤如下:
(1)***变量初始化,设定所有电流电压相关参数,包括正负序电流在dq坐标系下的给定值i+ dref、i+ qref、i- dref、i- qref,保护控制参数k(-1<k<1)。
(2)对电网电压和网侧输出电流分别进行正负序分离,得到电网电压d轴正序分量Vd +,d轴负序分量Vd -,q轴正序分量Vq +,q轴负序分量Vq -,网侧电流d轴正序分量Id +,d轴负序分量Id -,q轴正序分量Iq +,q轴负序分量Vq -
(3)在电网不平衡故障下进行锁相环计算以及直流母线电压环调节,锁相环对网侧电压锁相之后输出的相位角度θ以及C-MMC给定有功功率参考值P0
(4)根据步骤(1)(2)(3)所得的Vd +、Vd -、Vq +、Vq -、P0、k,计算出dq坐标轴下网侧电流参考值,其公式如下:
(5)分别进行正序d轴电流环调节、正序q轴电流环调节、负序d轴电流环调节、负序q轴电流环调节,输出相应的控制电压U+ d,U+ q,U- d,U- q
(6)根据步骤(5)所得出的U+ d,U+ q,U- d,U- q,获得C-MMC***三相调制信号,实现变换器的瞬时有功功率、无功功率和直流电压的控制。
本发明所提出的方法流程图如图4所示。由此可见,要使***稳定运行,需合理调整变量k,从而实现对电网以及C-MMC进行保护的目的。

Claims (1)

1.一种三相电网不平衡下C-MMC多变量保护控制方法,其特征在于,该方法具体包括以下步骤:
步骤1.计算出电网不对称时MMC的瞬时有功功率和无功功率,公式如下:
<mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mi>p</mi> <mo>=</mo> <msub> <mi>P</mi> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>P</mi> <mrow> <mi>c</mi> <mn>2</mn> </mrow> </msub> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mo>(</mo> <mn>2</mn> <mi>&amp;omega;</mi> <mi>t</mi> <mo>)</mo> <mo>+</mo> <msub> <mi>P</mi> <mrow> <mi>s</mi> <mn>2</mn> </mrow> </msub> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mo>(</mo> <mn>2</mn> <mi>&amp;omega;</mi> <mi>t</mi> <mo>)</mo> </mtd> </mtr> <mtr> <mtd> <mi>q</mi> <mo>=</mo> <msub> <mi>Q</mi> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>Q</mi> <mrow> <mi>c</mi> <mn>2</mn> </mrow> </msub> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mo>(</mo> <mn>2</mn> <mi>&amp;omega;</mi> <mi>t</mi> <mo>)</mo> <mo>+</mo> <msub> <mi>Q</mi> <mrow> <mi>s</mi> <mn>2</mn> </mrow> </msub> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mo>(</mo> <mn>2</mn> <mi>&amp;omega;</mi> <mi>t</mi> <mo>)</mo> </mtd> </mtr> </mtable> </mfenced>
其中:P0、Q0表示变换器中瞬时有功功率和无功功率的平均值;Pc2、Ps2分别为变换器有功功率的振荡分量的幅值;Qc2、Qs2分别表示变换器无功功率的振荡分量的幅值;ω表示电网频率,t表示时间;
步骤2.在同步旋转坐标系下求出步骤1公式中功率量的幅值,公式如下:
<mrow> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <msub> <mi>P</mi> <mn>0</mn> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mi>P</mi> <mrow> <mi>c</mi> <mn>2</mn> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mi>P</mi> <mrow> <mi>s</mi> <mn>2</mn> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mi>Q</mi> <mn>0</mn> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mi>Q</mi> <mrow> <mi>c</mi> <mn>2</mn> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mi>Q</mi> <mrow> <mi>s</mi> <mn>2</mn> </mrow> </msub> </mtd> </mtr> </mtable> </mfenced> <mo>=</mo> <mfrac> <mn>3</mn> <mn>2</mn> </mfrac> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <msubsup> <mi>v</mi> <mi>d</mi> <mo>+</mo> </msubsup> </mtd> <mtd> <msubsup> <mi>v</mi> <mi>q</mi> <mo>+</mo> </msubsup> </mtd> <mtd> <msubsup> <mi>v</mi> <mi>d</mi> <mo>-</mo> </msubsup> </mtd> <mtd> <msubsup> <mi>v</mi> <mi>q</mi> <mo>-</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <msubsup> <mi>v</mi> <mi>d</mi> <mo>-</mo> </msubsup> </mtd> <mtd> <msubsup> <mi>v</mi> <mi>q</mi> <mo>-</mo> </msubsup> </mtd> <mtd> <msubsup> <mi>v</mi> <mi>d</mi> <mo>+</mo> </msubsup> </mtd> <mtd> <msubsup> <mi>v</mi> <mi>q</mi> <mo>+</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <msubsup> <mi>v</mi> <mi>q</mi> <mo>-</mo> </msubsup> </mtd> <mtd> <mrow> <mo>-</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>-</mo> </msubsup> </mrow> </mtd> <mtd> <mrow> <mo>-</mo> <msubsup> <mi>v</mi> <mi>q</mi> <mo>+</mo> </msubsup> </mrow> </mtd> <mtd> <msubsup> <mi>v</mi> <mi>d</mi> <mo>+</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <msubsup> <mi>v</mi> <mi>q</mi> <mo>+</mo> </msubsup> </mtd> <mtd> <mrow> <mo>-</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>+</mo> </msubsup> </mrow> </mtd> <mtd> <msubsup> <mi>v</mi> <mi>q</mi> <mo>-</mo> </msubsup> </mtd> <mtd> <mrow> <mo>-</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>-</mo> </msubsup> </mrow> </mtd> </mtr> <mtr> <mtd> <msubsup> <mi>v</mi> <mi>q</mi> <mo>-</mo> </msubsup> </mtd> <mtd> <mrow> <mo>-</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>-</mo> </msubsup> </mrow> </mtd> <mtd> <msubsup> <mi>v</mi> <mi>q</mi> <mo>+</mo> </msubsup> </mtd> <mtd> <mrow> <mo>-</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>+</mo> </msubsup> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>-</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>-</mo> </msubsup> </mrow> </mtd> <mtd> <mrow> <mo>-</mo> <msubsup> <mi>v</mi> <mi>q</mi> <mo>-</mo> </msubsup> </mrow> </mtd> <mtd> <msubsup> <mi>v</mi> <mi>d</mi> <mo>+</mo> </msubsup> </mtd> <mtd> <msubsup> <mi>v</mi> <mi>q</mi> <mo>+</mo> </msubsup> </mtd> </mtr> </mtable> </mfenced> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <msubsup> <mi>i</mi> <mi>d</mi> <mo>+</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <msubsup> <mi>i</mi> <mi>q</mi> <mo>+</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <msubsup> <mi>i</mi> <mi>d</mi> <mo>-</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <msubsup> <mi>i</mi> <mi>q</mi> <mo>-</mo> </msubsup> </mtd> </mtr> </mtable> </mfenced> </mrow>
其中分别为电网电压的d轴正序、d轴负序、q轴正序、q轴负序分量幅值;分别为网侧电流的d轴正序、d轴负序、q轴正序、q轴负序分量幅值;
步骤3.假设要求***有功功率没有二倍频振荡,设定控制目标P0为常量,且Pc2=Ps2=Q0=0,引入调节参数k,d-q坐标系下电流控制,可得出网侧电流的参考值,公式如下:
<mrow> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <msubsup> <mi>i</mi> <mrow> <mi>d</mi> <mi>r</mi> <mi>e</mi> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <msubsup> <mi>i</mi> <mrow> <mi>q</mi> <mi>r</mi> <mi>e</mi> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <msubsup> <mi>i</mi> <mrow> <mi>d</mi> <mi>r</mi> <mi>e</mi> <mi>f</mi> </mrow> <mo>-</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <msubsup> <mi>i</mi> <mrow> <mi>q</mi> <mi>r</mi> <mi>e</mi> <mi>f</mi> </mrow> <mo>-</mo> </msubsup> </mtd> </mtr> </mtable> </mfenced> <mo>=</mo> <mfrac> <mn>2</mn> <mn>3</mn> </mfrac> <mfrac> <msub> <mi>P</mi> <mn>0</mn> </msub> <mrow> <mo>&amp;lsqb;</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>+</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>q</mi> <mo>+</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <mi>k</mi> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>-</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <mi>k</mi> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>q</mi> <mo>-</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>&amp;rsqb;</mo> </mrow> </mfrac> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <msubsup> <mi>v</mi> <mi>d</mi> <mo>+</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <msubsup> <mi>v</mi> <mi>q</mi> <mo>+</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <mrow> <msubsup> <mi>kv</mi> <mi>d</mi> <mo>-</mo> </msubsup> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msubsup> <mi>kv</mi> <mi>q</mi> <mo>-</mo> </msubsup> </mrow> </mtd> </mtr> </mtable> </mfenced> </mrow>
步骤4.由步骤3的结果计算出Pc2、Ps2、Qc2和Qs2,公式如下:
<mrow> <msub> <mi>P</mi> <mrow> <mi>c</mi> <mn>2</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>k</mi> <mo>)</mo> <msub> <mi>P</mi> <mn>0</mn> </msub> </mrow> <mrow> <mo>&amp;lsqb;</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>+</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>q</mi> <mo>+</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <mi>k</mi> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>-</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <mi>k</mi> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>q</mi> <mo>-</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>&amp;rsqb;</mo> </mrow> </mfrac> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>-</mo> </msubsup> <msubsup> <mi>v</mi> <mi>d</mi> <mo>+</mo> </msubsup> <mo>+</mo> <msubsup> <mi>v</mi> <mi>q</mi> <mo>-</mo> </msubsup> <msubsup> <mi>v</mi> <mi>q</mi> <mo>+</mo> </msubsup> <mo>)</mo> </mrow> </mrow>
<mrow> <msub> <mi>P</mi> <mrow> <mi>s</mi> <mn>2</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>k</mi> <mo>)</mo> <msub> <mi>P</mi> <mn>0</mn> </msub> </mrow> <mrow> <mo>&amp;lsqb;</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>+</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>q</mi> <mo>+</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <mi>k</mi> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>-</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <mi>k</mi> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>q</mi> <mo>-</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>&amp;rsqb;</mo> </mrow> </mfrac> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>+</mo> </msubsup> <msubsup> <mi>v</mi> <mi>q</mi> <mo>-</mo> </msubsup> <mo>-</mo> <msubsup> <mi>v</mi> <mi>q</mi> <mo>+</mo> </msubsup> <msubsup> <mi>v</mi> <mi>d</mi> <mo>-</mo> </msubsup> <mo>)</mo> </mrow> </mrow>
<mrow> <msub> <mi>Q</mi> <mrow> <mi>c</mi> <mn>2</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>k</mi> <mo>)</mo> <msub> <mi>P</mi> <mn>0</mn> </msub> </mrow> <mrow> <mo>&amp;lsqb;</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>+</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>q</mi> <mo>+</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <mi>k</mi> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>-</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <mi>k</mi> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>q</mi> <mo>-</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>&amp;rsqb;</mo> </mrow> </mfrac> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>q</mi> <mo>-</mo> </msubsup> <msubsup> <mi>v</mi> <mi>d</mi> <mo>+</mo> </msubsup> <mo>-</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>-</mo> </msubsup> <msubsup> <mi>v</mi> <mi>q</mi> <mo>+</mo> </msubsup> <mo>)</mo> </mrow> </mrow>
<mrow> <msub> <mi>Q</mi> <mrow> <mi>s</mi> <mn>2</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mo>-</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>k</mi> <mo>)</mo> </mrow> <msub> <mi>P</mi> <mn>0</mn> </msub> </mrow> <mrow> <mo>&amp;lsqb;</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>+</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>q</mi> <mo>+</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <mi>k</mi> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>-</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <mi>k</mi> <msup> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>q</mi> <mo>-</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>&amp;rsqb;</mo> </mrow> </mfrac> <mrow> <mo>(</mo> <msubsup> <mi>v</mi> <mi>d</mi> <mo>+</mo> </msubsup> <msubsup> <mi>v</mi> <mi>d</mi> <mo>-</mo> </msubsup> <mo>+</mo> <msubsup> <mi>v</mi> <mi>q</mi> <mo>+</mo> </msubsup> <msubsup> <mi>v</mi> <mi>q</mi> <mo>-</mo> </msubsup> <mo>)</mo> </mrow> <mo>.</mo> </mrow>
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