# 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.

This site uses Akismet to reduce spam. Learn how your comment data is processed.