I stumbled upon the notion of “time varying channels” in signal processing and after reading a bit I realized that these are really interesting objects and appear in many different parts of mathematics. In this post I collect of few of their realizations and relations.

In signal processing a “time varying channel” is a mapping which maps a signal ${x:{\mathbb R}^d\rightarrow {\mathbb C}}$ to an output ${y:{\mathbb R}^d\rightarrow{\mathbb C}}$ via

$\displaystyle y(t) = \int_{{\mathbb R}^d}\int_{{\mathbb R}^d} s(\tau,\nu)x(t-\tau)\mathrm{e}^{\mathrm{i}\nu t}d\nu d\tau.$

Before we explain the name “time varying channel” we fix the notation for the Fourier transform I am going to use in this post:

Definition 1 For ${f:{\mathbb R}^d\rightarrow{\mathbb C}}$ we define the Fourier transform by

$\displaystyle \mathcal{F}(f)(\omega) = \hat f(\omega) = (2\pi)^{-d/2}\int f(t)\mathrm{e}^{-\mathrm{i}\omega t}dt$

and denote the inverse Fourier transform by ${\mathcal{F}^{-1}f}$ or ${\check f}$. For a function ${f:{\mathbb R}^d\times{\mathbb R}^d\rightarrow{\mathbb C}}$ we denote with ${\mathcal{F}_1}$ and ${\mathcal{F}_2}$ the Fourier transform with respect to the first and second ${{\mathbb R}^d}$-component, respectively.

Remark 1 In all what follows we do formal calculations with integral not caring about integrability. All calculation are justified in the case of Schwartz-functions and often hold in a much broader context of tempered distributions (this for example happens if the integrals represent Fourier transforms of functions).

The name “time varying channel” can be explained as follows: Consider a pure frequency as input: ${x(t) = \mathrm{e}^{-\mathrm{i}\omega t}}$. A usual linear channel gives as output a damped signal and the damping depends on the frequency ${\omega}$: ${y(t) = h(\omega) \mathrm{e}^{-\mathrm{i}\omega t}}$. If we send the pure frequency in our time varying channel we get

$\displaystyle \begin{array}{rcl} y(t) & = &\int\int s(\tau,\nu) \mathrm{e}^{-\mathrm{i}\omega(t-\tau)}e^{\mathrm{i}\nu t}d\nu dt\\ & =& \int\int s(\tau,\nu) \mathrm{e}^{\mathrm{i}(\omega\tau +\nu t)}d\nu dt\, \mathrm{e}^{-\mathrm{i}\omega t}\\ & =& (2\pi)^d \hat s(-\omega,-t) \mathrm{e}^{-\mathrm{i}\omega t}. \end{array}$

Hence, the time varying channel also damps the pure frequencies but with a time dependent factor.

Let’s start quite far away from signal processing:

1. Pseudo-differential operators

A general linear differential operator of order ${N}$ on functions on ${{\mathbb R}^d}$ is defined with multiindex notation as

$\displaystyle Af(t) = \sum_{|\alpha|\leq N} \sigma_\alpha(t) D^\alpha f(t)$

with coefficient functions ${\sigma_\alpha}$. Using Fourier inversion we get

$\displaystyle D^\alpha f(t) - (2\pi)^{-d/2} \int \hat f(\omega) (\mathrm{i} \omega)\mathrm{e}^{\mathrm{i}\omega t}d\omega$

and hence

$\displaystyle \begin{array}{rcl} Af(t) & =& \int \Big((2\pi)^{-d/2} \sum_{|\alpha|\leq N} \sigma_\alpha(t)(\mathrm{i} \omega)^\alpha) \hat f(\omega)\mathrm{e}^{\mathrm{i}\omega t}d\omega\\ &=& \int \sigma(\omega,t) \hat f(\omega)\mathrm{e}^{\mathrm{i}\omega t}d\omega \\ & = & K_\sigma f(t). \end{array}$

For a general function (usually obeying some restrictions) ${\sigma:{\mathbb R}^d\times{\mathbb R}^d\rightarrow{\mathbb C}}$ we call the corresponding ${K_\sigma}$ a pseudo-differential operator with symbol ${\sigma}$.

2. As integral operators

By integral operator I mean something like

$\displaystyle F_k f(t) = \int k(t,\tau)f(\tau)d\tau.$

We plug the definition of the Fourier transform into ${K_\sigma}$ and obtain

$\displaystyle \begin{array}{rcl} K_\sigma f(t) &=& \int \sigma(\omega,t) (2\pi)^{-d/2} \int f(\tau)\mathrm{e}^{-\mathrm{i} \omega \tau}d\tau \mathrm{e}^{\mathrm{i} \omega t}d\omega\\ & = & \int \underbrace{(2\pi)^{-d/2} \int \sigma(\omega,t)\mathrm{e}^{-\mathrm{i}\omega(\tau-t)}d\omega}_{=k(t,\tau)} f(\tau)d\tau\\ &=& \int k(t,\tau)f(\tau)d\tau. \end{array}$

Using the Fourier transform we can express the relation between ${\sigma}$ and ${k}$ as

$\displaystyle k(t,\tau) = (2\pi)^{-d/2}\int \sigma(\omega,\tau)\mathrm{e}^{-\mathrm{i}\omega(\tau-t)}d\omega = \mathcal{F}_1(\sigma(\cdot,t))(\tau-t). \ \ \ \ \ (1)$

3. As “time varying convolution”

The convolution of two functions ${f}$ and ${g}$ is defined as

$\displaystyle (f*g)(t) = \int f(\tau) g(t-\tau)d\tau$

and we write “the convolution with ${g}$” as an operator ${C_g f = f * g}$.

Defining

$\displaystyle h_t(\tau) = (2\pi)^{-d/2}\int \sigma(\omega,t)\mathrm{e}^{\mathrm{i}\omega \tau}d\omega$

we deduce from (1)

$\displaystyle K_\sigma f(t) = \int f(\tau) h_t(t-\tau)d\tau= (f * h_t)(t) = C_{h_t}f(t).$

4. As superposition of time-frequency shifts

Using that iterated Fourier transforms with respect to components give the Fourier transform, i.e. ${\mathcal{F}_2\mathcal{F}_1 = \mathcal{F}}$, we obtain

$\displaystyle \mathcal{F}_1\sigma = \mathcal{F}_2^{-1}\mathcal{F}_2\mathcal{F}_1\sigma = \mathcal{F}_2^{-1}\hat\sigma.$

From (1) we get

$\displaystyle \begin{array}{rcl} k(t,\tau) &=& \mathcal{F}_2^{-1}\hat\sigma(\tau-t,t)\\ &=& (2\pi)^{-d/2} \int\hat\sigma(\tau-t,\nu)\mathrm{e}^{\mathrm{i} t\nu}d\nu. \end{array}$

Before plugging this into ${K_\sigma}$ we define time shifts ${T_u}$ and frequency shifts (or modulations) ${M_\nu}$ as

$\displaystyle T_u f(t) = f(t-u),\quad M_\nu f(t) = \mathrm{e}^{\mathrm{i} t\nu}.$

With this we get

$\displaystyle \begin{array}{rcl} K_\sigma f(t) &=& (2\pi)^{-d/2}\int\int \hat\sigma(\tau-t,\nu)\mathrm{e}^{\mathrm{i} t\nu}f(\tau)d\nu d\tau\\ &=& (2\pi)^{-d/2}\int\int \hat\sigma(u,\nu)\mathrm{e}^{\mathrm{i} t\nu}f(t+u)d\nu du\\ &=& (2\pi)^{-d/2}\int\int\hat\sigma(u,\nu) M_\nu T_{-u} f(t) d\nu du\\ &=& \int\int w(u,\nu) M_\nu T_{-u} f(t)d\nu du = S_wf(t). \end{array}$

Hence, ${K_\sigma}$ is also a weighted superposition of ${M_\nu T_{-u} f}$ (time-frequency shifts) with weight ${(2\pi)^{-d/2} \hat\sigma(u,\nu)}$.

5. Back to time varying channels

Simple substitution brings us back the situation of a time varying channel

$\displaystyle \begin{array}{rcl} K_\sigma f(t) &=& (2\pi)^{-d/2} \int\int \hat\sigma(u,\nu)\mathrm{e}^{\mathrm{i} t\nu}f(t+u)d\nu du\\ &=& \int \int s(\tau,\nu) f(t-\tau) \mathrm{e}^{\mathrm{i} t\nu} d\nu d\tau\\ &=& H_s f(t) \end{array}$

with ${s(\tau,\nu) = (2\pi)^{-d/2} \hat\sigma(-\tau,\nu)}$.

6. As superposition of product-convolution operators

Finally, I’d like to illustrate that this kind of operators can be seen as superposition of product convolution operators.

Introducing product operators ${P_g f(t) = g(t)f(t)}$ we define a product-convolution operator as a convolution with a function ${h}$ followed by a multiplication with another function ${g}$: ${P_g C_h f}$.

To express ${K_\sigma}$ with product-convolution operators we choose an orthonormal basis which consists of tensor-products of functions ${(\phi_n\otimes\psi_k)}$ and develop ${\sigma}$ into this basis as

$\displaystyle \sigma(\omega,t) = \sum_{n,k} a_{n,k} \phi_n(\omega)\psi_k(t).$

Then

$\displaystyle \begin{array}{rcl} K_\sigma f(t) &=& \sum_{n,k} a_{n,k}\psi_k(t) \int \phi_n(\omega) \hat f(\omega) \mathrm{e}^{\mathrm{i} \omega t}d\omega\\ & = & \sum_{n,k} a_{n,k}\psi_k(t) (2\pi)^{d/2}\mathcal{F}^{-1}(\phi_n\, \hat f)(t)\\ &=& \sum_{n,k} a_{n,k}\psi_k(t) (\check \phi_n * f)(t) \end{array}$

and we obtain

$\displaystyle K_\sigma f(t) = \sum_{n,k} a_{n,k}P_{\psi_k} C_{\check\phi_n} f (t).$

Remark 2 The integral operators of the form ${F_k}$ are indeed general objects as can be seen from the Schwartz kernel theorem. Every reasonable operator mapping Schwartz functions linearly onto the tempered distributions is itself a generalized integral operator.

I tried to capture the various relation between the first five representations time varying channels in a diagram (where I went out of motivation before filling in all fields…):