Hilbert transform as an infinite matrix

Posted on by John

The previous post linked to a post I wrote a few years ago about the Hilbert transform and Fourier series. That post says that if the Fourier series of a function is

f(t) = \sum_{n=1}^\infty \left\{ a_n \sin(nt) + b_n\cos(nt) \right\}

then the Fourier series of its Hilbert transform is

f_H(x) = \sum_{n=1}^\infty \left\{ -b_n \sin(nx) + a_n\cos(nx) \right\}

When I looked back at that post I thought about how if you thought of the Fourier coefficients as elements of an infinite vector then the Hilbert transform can be represented as multiplying by an infinite block matrix.

\left[ \begin{array}{cc|cc|cc|c} 0 & -1 & 0 & 0 & 0 & 0 & \cdots \\ 1 & 0 & 0 & 0 & 0 & 0 & \cdots \\ \hline 0 & 0 & 0 & -1 & 0 & 0 & \cdots \\ 0 & 0 & 1 & 0 & 0 & 0 & \cdots \\ \hline 0 & 0 & 0 & 0 & 0 & -1 & \cdots \\ 0 & 0 & 0 & 0 & 1 & 0 & \cdots \\ \hline \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \right] \left[ \begin{array}{c} a_1 \\ b_1 \\ \hline a_2 \\ b_2 \\ \hline a_3 \\ b_3 \\ \hline \vdots \end{array} \right]

I rarely see infinite matrices except in older math books. Apparently they were more fashionable a few decades ago than they are now. I suppose the notation falls between two stools, too concrete for some tastes and not concrete enough for others. The former folks would prefer something like H and the latter would prefer the sum above.

2 thoughts on “Hilbert transform as an infinite matrix”

  1. Joris

    This also makes it very clear that the Hilbert transform is a complex/symplectic structure (rotation by 90 degrees), where the as and the bs are conjugate coordinates.

  2. Brian Oxley

    Under what conditions do the higher-order weights on sin and cos fall away and contribute increasingly less? I’m wondering about good approximations with a finite matrix of just the lower-order weights.

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