已知永磁同步电机的电机d、q轴电压方程为

在d、q轴分别注入不同频率，不同幅值的高频正弦电压信号

由于在高频下， 此项在电机运行的稳态过程中可以看做是直流项，因此此项可以忽略，则电压方程的高频模型变为

同时，注入电压信号幅值一般为5V~15V，得到的电流响应幅值为0.01pu（额定电流为21A的电机，的高频响应电流幅值仅为0.2A），并且该信号是由多个频率的正弦信号合成的复杂信号，因此电阻压降可以忽略掉，上式简化为

因此该问题成为求解此方程组！

clear
syms Udh Uqh w1 w2 t
syms Ld Lq we  idh iqh


[x,y] = dsolve('Udh*sin(w1*t) = Ld*Didh+we*Lq*iqh', ...
    'Uqh*sin(w2*t) = Lq*Diqh-we*Ld*idh','t');
    
    %转为LaTeX格式便于Markdown文件展示
 latex(x)
latex(y)

得到的公式为：
− L q   e − t   w e   1 i   ( C 1 + ( − U d h   w 1   e t   w e   1 i   cos ⁡ ( t   w 1 ) w 1 2 − w e 2 + U q h   w 2   e t   w e   1 i   cos ⁡ ( t   w 2 )   1 i w 2 2 − w e 2 + U d h   w e   e t   w e   1 i   sin ⁡ ( t   w 1 )   1 i w 1 2 − w e 2 + U q h   w e   e t   w e   1 i   sin ⁡ ( t   w 2 ) w 2 2 − w e 2 )   1 i 2   L q )   1 i L d + L q   e t   w e   1 i   ( C 2 + ( U d h   w 1   e − t   w e   1 i   cos ⁡ ( t   w 1 ) w 1 2 − w e 2 + U q h   w 2   e − t   w e   1 i   cos ⁡ ( t   w 2 )   1 i w 2 2 − w e 2 + U d h   w e   e − t   w e   1 i   sin ⁡ ( t   w 1 )   1 i w 1 2 − w e 2 − U q h   w e   e − t   w e   1 i   sin ⁡ ( t   w 2 ) w 2 2 − w e 2 )   1 i 2   L q )   1 i L d -\frac{\mathrm{Lq}\,{\mathrm{e}}^{-t\,\mathrm{we}\,1{}\mathrm{i}}\,\left(C_{1}+\frac{\left(-\frac{\mathrm{Udh}\,w_{1}\,{\mathrm{e}}^{t\,\mathrm{we}\,1{}\mathrm{i}}\,\cos\left(t\,w_{1}\right)}{{w_{1}}^2-{\mathrm{we}}^2}+\frac{\mathrm{Uqh}\,w_{2}\,{\mathrm{e}}^{t\,\mathrm{we}\,1{}\mathrm{i}}\,\cos\left(t\,w_{2}\right)\,1{}\mathrm{i}}{{w_{2}}^2-{\mathrm{we}}^2}+\frac{\mathrm{Udh}\,\mathrm{we}\,{\mathrm{e}}^{t\,\mathrm{we}\,1{}\mathrm{i}}\,\sin\left(t\,w_{1}\right)\,1{}\mathrm{i}}{{w_{1}}^2-{\mathrm{we}}^2}+\frac{\mathrm{Uqh}\,\mathrm{we}\,{\mathrm{e}}^{t\,\mathrm{we}\,1{}\mathrm{i}}\,\sin\left(t\,w_{2}\right)}{{w_{2}}^2-{\mathrm{we}}^2}\right)\,1{}\mathrm{i}}{2\,\mathrm{Lq}}\right)\,1{}\mathrm{i}}{\mathrm{Ld}}+\frac{\mathrm{Lq}\,{\mathrm{e}}^{t\,\mathrm{we}\,1{}\mathrm{i}}\,\left(C_{2}+\frac{\left(\frac{\mathrm{Udh}\,w_{1}\,{\mathrm{e}}^{-t\,\mathrm{we}\,1{}\mathrm{i}}\,\cos\left(t\,w_{1}\right)}{{w_{1}}^2-{\mathrm{we}}^2}+\frac{\mathrm{Uqh}\,w_{2}\,{\mathrm{e}}^{-t\,\mathrm{we}\,1{}\mathrm{i}}\,\cos\left(t\,w_{2}\right)\,1{}\mathrm{i}}{{w_{2}}^2-{\mathrm{we}}^2}+\frac{\mathrm{Udh}\,\mathrm{we}\,{\mathrm{e}}^{-t\,\mathrm{we}\,1{}\mathrm{i}}\,\sin\left(t\,w_{1}\right)\,1{}\mathrm{i}}{{w_{1}}^2-{\mathrm{we}}^2}-\frac{\mathrm{Uqh}\,\mathrm{we}\,{\mathrm{e}}^{-t\,\mathrm{we}\,1{}\mathrm{i}}\,\sin\left(t\,w_{2}\right)}{{w_{2}}^2-{\mathrm{we}}^2}\right)\,1{}\mathrm{i}}{2\,\mathrm{Lq}}\right)\,1{}\mathrm{i}}{\mathrm{Ld}} −LdLqe−twe1i ​C1​+2Lq(−w1​2−we2Udhw1​etwe1icos(tw1​)​+w2​2−we2Uqhw2​etwe1icos(tw2​)1i​+w1​2−we2Udhweetwe1isin(tw1​)1i​+w2​2−we2Uqhweetwe1isin(tw2​)​)1i​ ​1i​+LdLqetwe1i ​C2​+2Lq(w1​2−we2Udhw1​e−twe1icos(tw1​)​+w2​2−we2Uqhw2​e−twe1icos(tw2​)1i​+w1​2−we2Udhwee−twe1isin(tw1​)1i​−w2​2−we2Uqhwee−twe1isin(tw2​)​)1i​ ​1i​
e − t   w e   1 i   ( C 1 + ( − U d h   w 1   e t   w e   1 i   cos ⁡ ( t   w 1 ) w 1 2 − w e 2 + U q h   w 2   e t   w e   1 i   cos ⁡ ( t   w 2 )   1 i w 2 2 − w e 2 + U d h   w e   e t   w e   1 i   sin ⁡ ( t   w 1 )   1 i w 1 2 − w e 2 + U q h   w e   e t   w e   1 i   sin ⁡ ( t   w 2 ) w 2 2 − w e 2 )   1 i 2   L q ) + e t   w e   1 i   ( C 2 + ( U d h   w 1   e − t   w e   1 i   cos ⁡ ( t   w 1 ) w 1 2 − w e 2 + U q h   w 2   e − t   w e   1 i   cos ⁡ ( t   w 2 )   1 i w 2 2 − w e 2 + U d h   w e   e − t   w e   1 i   sin ⁡ ( t   w 1 )   1 i w 1 2 − w e 2 − U q h   w e   e − t   w e   1 i   sin ⁡ ( t   w 2 ) w 2 2 − w e 2 )   1 i 2   L q ) {\mathrm{e}}^{-t\,\mathrm{we}\,1{}\mathrm{i}}\,\left(C_{1}+\frac{\left(-\frac{\mathrm{Udh}\,w_{1}\,{\mathrm{e}}^{t\,\mathrm{we}\,1{}\mathrm{i}}\,\cos\left(t\,w_{1}\right)}{{w_{1}}^2-{\mathrm{we}}^2}+\frac{\mathrm{Uqh}\,w_{2}\,{\mathrm{e}}^{t\,\mathrm{we}\,1{}\mathrm{i}}\,\cos\left(t\,w_{2}\right)\,1{}\mathrm{i}}{{w_{2}}^2-{\mathrm{we}}^2}+\frac{\mathrm{Udh}\,\mathrm{we}\,{\mathrm{e}}^{t\,\mathrm{we}\,1{}\mathrm{i}}\,\sin\left(t\,w_{1}\right)\,1{}\mathrm{i}}{{w_{1}}^2-{\mathrm{we}}^2}+\frac{\mathrm{Uqh}\,\mathrm{we}\,{\mathrm{e}}^{t\,\mathrm{we}\,1{}\mathrm{i}}\,\sin\left(t\,w_{2}\right)}{{w_{2}}^2-{\mathrm{we}}^2}\right)\,1{}\mathrm{i}}{2\,\mathrm{Lq}}\right)+{\mathrm{e}}^{t\,\mathrm{we}\,1{}\mathrm{i}}\,\left(C_{2}+\frac{\left(\frac{\mathrm{Udh}\,w_{1}\,{\mathrm{e}}^{-t\,\mathrm{we}\,1{}\mathrm{i}}\,\cos\left(t\,w_{1}\right)}{{w_{1}}^2-{\mathrm{we}}^2}+\frac{\mathrm{Uqh}\,w_{2}\,{\mathrm{e}}^{-t\,\mathrm{we}\,1{}\mathrm{i}}\,\cos\left(t\,w_{2}\right)\,1{}\mathrm{i}}{{w_{2}}^2-{\mathrm{we}}^2}+\frac{\mathrm{Udh}\,\mathrm{we}\,{\mathrm{e}}^{-t\,\mathrm{we}\,1{}\mathrm{i}}\,\sin\left(t\,w_{1}\right)\,1{}\mathrm{i}}{{w_{1}}^2-{\mathrm{we}}^2}-\frac{\mathrm{Uqh}\,\mathrm{we}\,{\mathrm{e}}^{-t\,\mathrm{we}\,1{}\mathrm{i}}\,\sin\left(t\,w_{2}\right)}{{w_{2}}^2-{\mathrm{we}}^2}\right)\,1{}\mathrm{i}}{2\,\mathrm{Lq}}\right) e−twe1i ​C1​+2Lq(−w1​2−we2Udhw1​etwe1icos(tw1​)​+w2​2−we2Uqhw2​etwe1icos(tw2​)1i​+w1​2−we2Udhweetwe1isin(tw1​)1i​+w2​2−we2Uqhweetwe1isin(tw2​)​)1i​ ​+etwe1i ​C2​+2Lq(w1​2−we2Udhw1​e−twe1icos(tw1​)​+w2​2−we2Uqhw2​e−twe1icos(tw2​)1i​+w1​2−we2Udhwee−twe1isin(tw1​)1i​−w2​2−we2Uqhwee−twe1isin(tw2​)​)1i​ ​
可以看出非常复杂
需要采用matlab内置函数进行简化

simplify(x)
simplify(y)
latex(x)
latex(y)

dq轴的两个初值项在同步旋转轴系下是直流项，旋转方向不同，特解项包含高频响应分量以及dq轴之间的耦合响应分量。因此，与初值有关的通解可以忽略，可以得到