Another look at recursive quadrature oscillators

In a recent post, we looked at which \(2\times 2\) Toeplitz real matrices \(T\) gave useful quadrature oscillators by the recurrence \(x_{n+1}=T x_n\). There, we computed their eigenvalues and solved the recurrence in terms of them. Of course, there are many other ways to approach this problem. Here we look at another approach that gives a good geometric picture of what happens, can be applied to general \(2\times 2\) matrices, and may be used as a starting point for the \(n\times n\) case.

We are interested in characterizing the \(2\times 2\) matrices \(T\) such that if we consider the sequence defined by \(x_{n+1}=T x_n\), \(x_0 \in \mathbb{R}^2\) fixed, then its two components are sinusoids of constant phase difference 90º (or perhaps just approximately 90º).

We will denote by \(R(\omega)\) the matrix of rotation by \(\omega\) in \(\mathbb{R}^2\):\[R(\omega)=\begin{pmatrix}\cos\omega &-\sin\omega\\\sin\omega&\cos\omega\end{pmatrix}.\] Reasoning as in the previous post, we see that the eigenvalues of \(T\) need to have modulus \(1\) and be complex conjugates. It follows that \(T\) is similar to a rotation matrix:\[T = PR(\omega)P^{-1}.\] Therefore\[x_{n} = T^n x_0 = PR(\omega n)P^{-1}x_0.\] In fact it makes sense and is worthy to consider the “continuous version” of the recurrence, which has the solution\[x(t) = PR(\omega t)P^{-1}x_0,\quad t\in\mathbb{R}.\]

We denote by \(P=U\Sigma V^*\) the singular value decomposition of \(P\). \(\Sigma\) is a diagonal matrix with strictly positive entries on the diagonal, and \(U\) and \(V\) are real orthogonal matrices. We note that \(V^*R(\omega t)V=R(\pm\omega t)\) depending on whether \(V\) preserves or reverses the orientation. Hence,\[x(t) = U\Sigma R(\pm\omega t) \Sigma^{-1}U^*x_0.\] We put \(y_0 = \Sigma^{-1}U^*x_0\) and make the change of variables \(s = \pm\omega t\). We have\[x(s)=U\Sigma R(s)y_0.\]

Now we can clearly see geometrically what happens. As \(s\) varies along \(\mathbb{R}\), \(R(s)y_0\) traces the circumference of centre \(0\) and radius \(\|y_0\|\). Then \(\Sigma\) stretches this circumference to produce an ellipse (except in the trivial case when \(\Sigma\) is a multiple of the identity, which implies that the \(T\) we started with was just a rotation matrix). The axes of this ellipse are the coordinate axes. Then \(U\), which is a rotation and perhaps an orthogonal symmetry, just rotates this ellipse, so that its axes may no longer be the coordinate axes.

Therefore, for any \(x_0 \in \mathbb{R}^2\setminus\{0\}\), the orbit \(x(s)\) traces an ellipse. Moreover, if we take two nonzero initial data \(x_0\) and \(\widetilde{x}_0\), their corresponding orbits \(X=\{x(s)\}\) and \(\widetilde{X}=\{\widetilde{x}(s)\}\), seen as sets, are proportional: \(\widetilde{X} = (\|\widetilde{x}\|/\|x\|)X\). So the initial data \(x_0\) that we choose is not important.

Of course, the components of \(x(s)\) are sinusoids of angular frequency \(1\), because they are linear combinations of the the components of \(R(s)y_0\), which are sinusoids of angular frequency \(1\) themselves. Hence,\[x(s) = \begin{pmatrix}a\cos(s+\theta_0)\\ b\sin(s+\theta_0+\varphi)\end{pmatrix},\] for some \(a> 0\), \(b\neq 0\), \(\theta_0 \in \mathbb{R}\) and \(\varphi\in (-\pi/2,\pi/2]\). The curve traced by \(x(s)\) is an ellipse whose axes are rotated \(\varphi/2\) with respect to the coordinate axes.

It follows that if we want the components of \(x(s)\) to be in quadrature, which amounts to \(\varphi=0\), then the axes of the ellipse traced by \(x(s)\) should be the coordinate axes. This implies that \(U\) performs no rotation at all. It can only permute and change the sign of the vectors in the standard basis. Therefore, we may as well assume that \(U\) is the identity, because these simple transformations of the vectors in the standard basis don’t make much difference.

We recall that \(T = U\Sigma R(\omega) \Sigma^{-1} U^*\). Putting \(U = I\), denoting the diagonal elements of \(\Sigma\) by \(\alpha,\beta > 0\), and computing the matrix product, we see that\[T=\begin{pmatrix}\cos\omega & -\frac{1}{\tau}\sin\omega\\ \tau\sin\omega & \cos\omega\\\end{pmatrix},\]where \(\tau=\beta/\alpha\). We see that the only matrices that produce a quadrature oscillator are Toeplitz: precisely those which we studied in the previous post.

Perhaps one is interested in matrices \(T\) that give oscillators which are almost in quadrature, meaning that \(\varphi\) is very small. Of course, this just means that \(U\) is essentially a rotation of very small angle, so the matrices \(T\) that satisfy this condition may be computed in a similar fashion.

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