# How to find the impulse response from the following input/output relation

Suppose we have a system defined with input $$x(t)$$, and output $$y(t)$$, related by

$$y(t) = \sum_i \alpha_i(t)x(t-\tau_i(t)) \tag 1$$

This is a linear system, (according to the textbook), so we can write it as a convolution with an impulse response $$h(\tau,t)$$

$$y(t) = \int_{-\infty}^{\infty} h(\tau,t)x(t-\tau)d\tau \tag 2$$

According to the text this tell us that by comparing the two previous equations,

$$h(\tau,t) = \sum_i \alpha_i(t)\delta(\tau-\tau_i(t)) \tag 3$$

but I cant understand how they came to this relationship? I thought id sub it back in to verify it, and then use the shifting theorem but I get as far as:

\begin{align} y(t) & = \int_{-\infty}^{\infty} \sum_i \alpha_i(t)\delta(\tau-\tau_i(t))x(t-\tau)d\tau \\ & = \sum_i \alpha_i(t)\int_{-\infty}^{\infty} \delta(\tau-\tau_i(t))x(t-\tau)d\tau \end{align} \tag 4

Then the $$\int_{-\infty}^{\infty} \delta(\tau-\tau_i(t))x(t-\tau)d\tau$$ term would be some sort of convolution with the dirac delta which equals $$x(t-\tau_i(t))$$, but I can't find a convolution in which the input signal is shifted by a function of $$t$$, as opposed to a constant value $$t_0$$.

Thanks to anyone who can offer help on this problem.

Your trouble lies with $$\int_{-\infty}^{\infty} \delta(\tau-\tau_i(t))x(t-\tau)d\tau. \tag a$$
Because $$t$$ is an independent variable in (a), then so is $$\tau_i(t)$$. So if you just look at the math, $$\int_{-\infty}^{\infty} \delta(\tau-\tau_i(t))x(t-\tau)d\tau = x(t-\tau_i(t))$$.
• be careful Jonah, playing fast and loose with the Dirac delta when you're around pure mathematicians. we engineers and them math guys have a slightly different spin on it. We say that $$\int\limits_{-\epsilon}^{+\epsilon} \delta(t) \, \mathrm{d}t = 1$$ The math guys say that any function that is zero "almost everywhere" integrates to zero. Jan 4 at 1:56
• I'm not sure how the math folks treat it, but you can take just about any function $f(t)$ that integrates to one and falls to zero at infinity, then call the Dirac delta $\delta(t) = \lim_{a \to 0} f(a \cdot t)/a$. This is actually handy if you want to reconcile the four flavors of the Fourier transform with one another without just waving your hands and scattering $\delta(t)$ all over. Jan 4 at 2:27