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设三相交流系统各相电压为:u=Vcosωt u=Vcos(ωt−120°)u=Vcos(ωt+120°)(1.1)ua、ub、uc分别指ABC三相电压的瞬时值Vm指相电压基波幅值u=Vcosωt =Vu=Vcos(ωt−120°) =Vcos120°+Vsin120° =−V+√V(1.2)u=Vcos(ωt+120°) =Vcos120°−Vsin120° =−V−√Vuuu=10−12⁄√32⁄−12⁄−√32⁄VV(1.3) u+u+u=0(1.4)现在要求的是如何找到一个矩阵P使VV=Puuu(1.5)书上有两种表达式P=1−12⁄−12⁄0√32⁄−√32⁄与P=1−12⁄−12⁄0√32⁄−√32⁄(1.6)为什么有这两种表达式?当P=时,坐标变换前后电压空间矢量幅值不变;当P=时,坐标变换前后电机功率不变。推导过程:①恒功率变换iα=i−0.5i−0.5i幅值:1++=倍iβ=√32i−√32i=k∗=1=k=②等幅值变换在复平面上的矢量V⃗总能用互差120度的abc三轴系中的分量xa、xb、xc等效表示(a轴与复平面实轴重合),如下所示(x⃗和x⃗将合成矢量V⃗)。x⃗=k(x+ρx+ρx)``````````````(1)x⃗=k(x+x+x)```````````````(2)其中,ρ=e=−+j√、ρ=e=e=−−j√;x⃗的方向与复平面的实轴方向一致。所以有式(2)可以表示为x=k(x+x+x)````````````````(3)写出式(1)的实部与虚部如下:R{x⃗}=kx−x−x=kx−(x+x)```````(4)I{x⃗}=k√(x−x)``````````````````````````````````````````(5)由式(3)可得:x+x=−x```````````````````````````````````````````````(6)将(6)代入式(4)中可得:R{x⃗}=kx−−x=1.5kx−0.5`````````(7)等幅值变换时,规定x=R{x⃗}+x,所以有:R{x⃗}=x−x`````````````````````````````````````````````````(8)将(8)代入式(7)中可得:1.5kx−0.5=x−x````````````````````````````````````(9)对比式(9)两端的x和x的系数可解得:k=、k=将实轴用α轴代替、虚轴用β轴代替,将k、k代入式(3)(4)(5)得到Clark变换的等幅值变换形式:⎩⎪⎪⎨⎪⎪⎧x=23x−12(x+x)=23x−13x−13xx=23∗√32(x−x)=1√3(x−x) x=13(x+x+x) 写成矩阵形式为:xxx=23⎣⎢⎢⎢⎢⎡1−12−120√32−√32121212⎦⎥⎥⎥⎥⎤xxxClark变换的另一种求解方法:磁动势的等效也就代表着电流的等效,即iA/iB/iB、ia/ib和id/iq等效,他们三者能产生相同的磁动势。为方便起见,将A相与α相重合,当两者磁动势相等时,两套绕组瞬时磁动势在ab轴上的投影相等。即有以下关系式:Ni=Ni−Nicos60°−Nicos60° =Ni−12i−12iNi=Nisin60°−Nisin60° =√32N(i−i)由陈伯时书籍附录4所证明,变换前后功率不变,三相和两相的匝数比为:NN=23结合以上二式可得变换矩阵为:ii=23⎣⎢⎢⎡1−12−120√32−√32⎦⎥⎥⎤iii★反Clark变换uuu=P10−12⁄√32⁄−12⁄−√32⁄VV对应上面,当P=时,P=;当P=时,P=1。由Clark变换推出Park变换(2s/2r变换)由上图可得:icosφ−isinφ=iisinφ+icosφ=i(1.7)推出两相旋转变两相静止的变换(反Park变换)矩阵为:ii=cosφ−sinφsinφcosφii(1.8)由式(1.7)可得:icosφ−isinφcosφ=icosφisinφ+isinφcosφ=isinφ(1.9)两式相加,有:i=icosφ+isinφ(1.10)同理,由式(1.7)可得:icosφsinφ−isinφ=isinφicosφsinφ+icosφ=icosφ(1.11)两式相减,有:i=−isinφ+icosφ(1.12)可得两相静止变两相旋转坐标为:ii=cosφsinφ−sinφcosφii(1.13)
本文标题:Clark变换与Park变换
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