Time-varying "impulse response"

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

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.