Time-varying “impulse response”

Question

According to many references [1,2], the time-varying "impulse response" can compute wireless channel output $y(t)$ at time $t$ using the following expression: $$ y(t) = \int h(\tau, t) x(t - \tau) d\tau $$

In both references, they state that this represents the response of the channel at time $t$ to an impulse applied at time $t-\tau$.

It seems reasonable to assume that there is some version of x(t) that involves a delta function that we can apply as an input that returns $h(\tau,t)$ as the output.

Trying: $$x(t) = \delta(t)$$ $$\implies y(t) = \int h(\tau, t) \delta(t - \tau) d\tau = h(t,t) \qquad \text{nope} $$

Trying: $$x(t) = \delta(t-\tau')$$ $$\implies y(t) = \int h(\tau, t) \delta(t - \tau - \tau') d\tau = h(t-\tau',t) \qquad \text{nope, but closer} $$

Is there a way to generate something resembling $h(\tau,t)$ as the output?

$$ $$

References:

[1] Proakis, Digital Communications, 5th ed, p.832

[2] Goldsmith, Wireless Communications, 1st ed, p.67

Answer 1

changing $\tau$ to $\tau'$ for ease of notation, we have $$y(t) = \int h(\tau', t) x(t - \tau') d\tau' \tag{1}$$

Now to get the output $h(\tau, t)$ it suffices to select

$$\color{red}{x(\tau')=\delta(\tau'-t+\tau)}$$

which is an impulse at $\tau'=t-\tau$.

To substitute it in the convolution equation: $$\begin{align} x(t-\tau')&= x(-\tau'+t)\\ &=\delta(-\tau'-t+\tau+t)\\ &=\delta(\tau-\tau') \end{align}$$

Hence the result of convolution in $(1)$ is $$y= \int h(\tau', t) \delta(\tau-\tau') d\tau'=h(\tau,t)$$.

which is resulted from the sifting property of Dirac delta.

Answer 2

In writing $h(\tau,t)$ with $t$ is time and $\tau$ is delay, we are in the model that $\tau$ varies "differently" from $t$. Or in other words, they are different notions in spite of the fact that they are both time unit.

Similarly, in this equation $$y(t) = \int h(\tau, t) x(t - \tau) d\tau$$

The time index of $y(t)$ varies "differently" from the time index of $h(\tau,t)$.

It is easier to write $h(\tau, t')$ and fix the measuring moment $t'=t_0$ for $h(\tau,t_0)$ and applying an input $\delta(t)$

$$y(t) = \int h(\tau, t_0) \delta(t - \tau) d\tau = h(t,t_0)$$

The index $t$ of $h(t,t_0)$ plays the role of delay $\tau$ that is exactly the same expression $h(\tau,t_0)$.

Redo the experiment at $t'=t_1, t_2, ..., $ we have the two dimensional time varying impulse reponse $h(\tau,t)$.

Answer 3

With $\tau$ being the integration variable in the convolution integral, your output (i.e., the result of the integration) will never be $h(\tau,t)$. Your second approach leads to the correct interpretation: since $h(t-\tau',t)$ is the response to an impulse at time $\tau'$, the function $h(\tau,t)$ is the response at time $t$ to an impulse at time $\tau'=t-\tau$, as already stated in your question.

Note that the function $h(\tau,t)$ as (implicitly) defined by the input-output relation in your question is often referred to as input delay-spread function. More than you probably ever want to know about the characterization of time-varying channels can be found in Fundamentals of Time-Varying Communication Channels by G.Matz and F.Hlawatsch.

Answer 4

The notation here is somewhat sloppy. The whole concept of a time variant impulse response only works if the time scale on which time variation happens is significantly slower than the length of the impulse response. That allows treating the impulse response as piece wise "time invariant" at least during the interval of the convolution.

The time invariant impulse responses has two arguments: one is really fast and one is really slow. Think about cell phone communication in a car: the first argument is nano seconds or thereabouts (GHz) bandwidth), the second is more in milliseconds or seconds (channel varies by the car driving).

Both have units of time, but the second should not be denoted by t, but rather some other symbol since it elapses at a much much smaller rate than the t in x(t) and y(t), which are again in the GHz range

Answer 5

so, for reference (and to remind myself), the LTI convolution is:

$$\begin{align} y(t) &= h(t) \circledast x(t) \\ &= \int_{-\infty}^{\infty} h(t-u) \, x(u) \, du \\ &= \int_{-\infty}^{\infty} x(t-u) \, h(u) \, du \\ \end{align}$$

if $x(t) = \delta(t)$ then the output is $y(t)=h(t)$.

if $x(t) = \delta(t-\tau)$ then the output is $y(t)=h(t-\tau)$.

i'm gonna generalize the first integral to linear time-varying, extending $h(t)$ to two arguments. so

$$\begin{align} h(t) &\leftarrow h(0,t) \\ h(t-\tau) &\leftarrow h(0,t-\tau) = h(\tau,t) \qquad \text{(LTI)} \\ \end{align}$$

$h(\tau,t)$ is the system response to an impulse applied at time $\tau$. so, the more general convolution integral is

$$\begin{align} y(t) &= \int_{-\infty}^{\infty} h(u,t) \, x(u) \, du \\ \end{align}$$

so, i think your first equation is wrong.