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Problem 회로망의 대칭 성분

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  • 대칭좌표법
  • 대칭 좌표법

그림과 같은 회로의 영상, 정상 및 역상 임피던스 Z0\mathbf Z_0, Z1\mathbf Z_1, Z2\mathbf Z_2는?

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1

Z0=Z+3Zn1+jωC(Z+3Zn)Z1=Z2=Z1+jωCZ \\[1ex] \begin{align*} \mathbf Z_0 &= \frac{\mathbf Z+3\mathbf Z_n}{1+j\omega C(\mathbf Z+3\mathbf Z_n)} \\ \mathbf Z_1 &= \mathbf Z_2 = \frac{\mathbf Z}{1+j\omega C\mathbf Z } \\[1ex] \end{align*}

2

Z0=Z+Zn1+jωC(Z+Zn)Z1=Z2=Z1+j3ωCZn \\[1ex] \begin{align*} \mathbf Z_0 &= \frac{\mathbf Z+\mathbf Z_n}{1+j\omega C(\mathbf Z+\mathbf Z_n)} \\ \mathbf Z_1 &= \mathbf Z_2 = \frac{\mathbf Z}{1+j3\omega C\mathbf Z_n} \\[1ex] \end{align*}

3

Z0=3Zn1+jωC(3Z+Zn)Z1=Z2=3Zn1+jωCZ \\[1ex] \begin{align*} \mathbf Z_0 &= \frac{3\mathbf Z_n}{1+j\omega C(3\mathbf Z+\mathbf Z_n)} \\ \mathbf Z_1 &= \mathbf Z_2 = \frac{3\boldsymbol{Z_n}}{1+j\omega C\mathbf Z} \\[1ex] \end{align*}

4

Z0=3Z1+jωC(Z+Zn)Z1=Z2=3Zn1+j3ωCZ \\[1ex] \begin{align*} \mathbf Z_0 &= \frac{3\mathbf Z}{1+j\omega C(\mathbf Z+\mathbf Z_n)} \\ \mathbf Z_1 &= \mathbf Z_2 = \frac{3\mathbf Z_n}{1+j3\omega C\mathbf Z} \end{align*}

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