GATE EE 2024 Set 1 — Question 46

For a two-phase system the sequence (symmetrical-component) operator is $a = e^{j2\pi/n}$ with $n = 2$, i.e. $a = e^{j\pi} = -1$. The two possible sequence modes are therefore:

• the common (zero-sequence) mode, in which both phases are equal: $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$,
• the differential (positive-sequence) mode, in which the two phases are in antiphase: $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$.

So $\mathbf{S}$ is an admissible transformation only if it is non-singular and its two columns are these two mode vectors (in either order and to within a non-zero scale factor, since which component is called $V_\alpha$ and which $V_\beta$ is only a labelling choice).(A) $\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$: columns $\begin{bmatrix}1\\1\end{bmatrix}$ (common mode) and $\begin{bmatrix}1\\-1\end{bmatrix}$ (differential mode); $\det \mathbf{S} = -2 \neq 0$. This is the standard two-phase sequence transformation, so it is possible.(B) $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$: $\det \mathbf{S} = 0$. It is singular — it would force $V_p = V_q$ always and cannot be inverted to obtain the sequence voltages. Not possible.(C) $\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$: $\det \mathbf{S} = -1 \neq 0$, so it is invertible, but its second column $\begin{bmatrix}1\\0\end{bmatrix}$ excites phase $p$ alone and is not a sequence mode of a two-phase system. Not possible.(D) $\begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix}$: columns $\begin{bmatrix}-1\\1\end{bmatrix} = -\begin{bmatrix}1\\-1\end{bmatrix}$ (differential mode) and $\begin{bmatrix}1\\1\end{bmatrix}$ (common mode); $\det \mathbf{S} = -2 \neq 0$. This is (A) with the two sequence components interchanged and one of them sign-reversed, so it is equally valid.Answer: (A) and (D)