## 2 Problem Formulation  
Unbalanced three-phase grid voltages comprise of positive (\({}^{+}\)), negative (\({}^{-}\)) and zero (\({}^{0}\)) sequence components are given by:\[U_{a} = \underbrace{A^{+}\sin\bigl{(}\omega t+\phi^{+}\bigr{)}}_{U_{a}^{+}} +\underbrace{A^{-}\sin\bigl{(}\omega t+\phi^{-}\bigr{)}}_{U_{a}^{-}}+ \underbrace{A^{0}\sin\Bigl{(}\omega t+\phi^{0}\Bigr{)}}_{U_{a}^{0}}, \tag{1}\] \[U_{b} = \underbrace{A^{+}\sin\biggl{(}\omega t-\frac{2\pi}{3}+\phi^{+} \biggr{)}}_{U_{b}^{+}}+\underbrace{A^{-}\sin\biggl{(}\omega t+\frac{2\pi}{3}+ \phi^{-}\biggr{)}}_{U_{b}^{-}}+\underbrace{A^{0}\sin\Bigl{(}\omega t+\phi^{0} \Bigr{)}}_{U_{a}^{0}},\] (2) \[U_{c} = \underbrace{A^{+}\sin\biggl{(}\omega t+\frac{2\pi}{3}+\phi^{+} \biggr{)}}_{U_{c}^{+}}+\underbrace{A^{-}\sin\biggl{(}\omega t-\frac{2\pi}{3}+ \phi^{-}\biggr{)}}_{U_{c}^{-}}+\underbrace{A^{0}\sin\Bigl{(}\omega t+\phi^{0} \Bigr{)}}_{U_{c}^{0}}, \tag{3}\]  
where the amplitude is denoted by \(A\), the initial phase angle is denoted by \(\phi\), and the angular frequency is denoted by \(\omega\) with the nominal value being given by \(\omega_{n}=100\pi\). In this paper, the problem is to estimate the unknown angular frequency, \(\omega\) and extract the positive \(U^{+}=\left[ \begin{array}{cc}U_{a}^{+}&U_{b}^{+}&U_{c}^{+}\end{array}  \right]^{T}\), negative \(U^{-}=\left[ \begin{array}{cc}U_{a}^{-}&U_{b}^{-}&U_{c}^{-}\end{array} \right] ^{T}\), and zero \(U^{0}=\left[ \begin{array}{cc}U_{a}^{0}&U_{b}^{0}&U_{c}^{0}\end{array} \right] ^{T}\) sequence components from the measured grid voltages (1)-(3). For this purpose, let us consider the 90\({}^{\circ}\) phase shifted version of the grid voltages (1)-(3):  
\[qU_{a} = \underbrace{A^{+}\cos\bigl{(}\omega t+\phi^{+}\bigr{)}}_{qU_{a}^ {+}}+\underbrace{A^{-}\cos\bigl{(}\omega t+\phi^{-}\bigr{)}}_{qU_{a}^{-}}+ \underbrace{A^{0}\cos\Bigl{(}\omega t+\phi^{0}\Bigr{)}}_{qU_{a}^{0}}, \tag{4}\] \[qU_{b} = \underbrace{A^{+}\cos\biggl{(}\omega t-\frac{2\pi}{3}+\phi^{+} \biggr{)}}_{qU_{b}^{+}}+\underbrace{A^{-}\cos\biggl{(}\omega t+\frac{2\pi}{3} +\phi^{-}\biggr{)}}_{qU_{b}^{-}}+\underbrace{A^{0}\cos\Bigl{(}\omega t+\phi^{ 0}\Bigr{)}}_{qU_{b}^{0}},\] (5) \[qU_{c} = \underbrace{A^{+}\cos\biggl{(}\omega t+\frac{2\pi}{3}+\phi^{+} \biggr{)}}_{U_{c}^{+}}+\underbrace{A^{-}\cos\biggl{(}\omega t-\frac{2\pi}{3}+ \phi^{-}\biggr{)}}_{U_{c}^{-}}+\underbrace{A^{0}\cos\Bigl{(}\omega t+\phi^{0} \Bigr{)}}_{U_{c}^{0}}. \tag{6}\]  
From the grid voltages (1)-(3) and their quadrature version (4)-(6), the positive, negative and zero sequence components can be extracted by the following equations [39]:  
\[\left[ \begin{array}{c}U_{a}^{+}\\ U_{b}^{+}\\ U_{c}^{+}\end{array} \right] = \frac{1}{3}\left[ \begin{array}{ccc}1&-\frac{1}{2}&-\frac{1}{2} \\ -\frac{1}{2}&1&-\frac{1}{2}\\ -\frac{1}{2}&-\frac{1}{2}&1\end{array} \right]\left[ \begin{array}{c}U_{a}\\ U_{b}\\ U_{c}\end{array} \right]+\frac{1}{2\sqrt{3}}\left[ \begin{array}{ccc}0&1&-1\\ -1&0&1\\ 1&-1&0\end{array} \right]\left[ \begin{array}{c}qU_{a}\\ qU_{b}\\ qU_{c}\end{array} \right], \tag{7}\] \[\left[ \begin{array}{c}U_{a}^{-}\\ U_{b}^{-}\\ U_{c}^{-}\end{array} \right] = \frac{1}{3}\left[ \begin{array}{ccc}1&-\frac{1}{2}&-\frac{1}{2} \\ -\frac{1}{2}&1&-\frac{1}{2}\\ -\frac{1}{2}&-\frac{1}{2}&1\end{array} \right]\left[ \begin{array}{c}U_{a}\\ U_{b}\\ U_{c}\end{array} \right]-\frac{1}{2\sqrt{3}}\left[ \begin{array}{ccc}0&1&-1\\ -1&0&1\\ 1&-1&0\end{array} \right]\left[ \begin{array}{c}qU_{a}\\ qU_{b}\\ qU_{c}\end{array} \right],\] (8) \[\left[ \begin{array}{c}U_{a}^{0}\\ U_{b}^{0}\\ U_{c}^{0}\end{array} \right] = \left\{ \begin{array}{ccc}1&0&0\\ 0&1&0\\ 0&0&1\end{array} \right]-\frac{1}{3}\left[ \begin{array}{ccc}1&-\frac{1}{2}&- \frac{1}{2}\\ -\frac{1}{2}&1&-\frac{1}{2}\\ -\frac{1}{2}&-\frac{1}{2}&1\end{array} \right]\right\}\left[ \begin{array}{c}U_{ a}\\ U_{b}\\ U_{c}\end{array} \right]. \tag{9}\]  
From Equations (7)-(9), it is clear that to extract the positive, negative, and zero sequence components, quadrature signals are required.