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.


1 Answer 1


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, $\int_{-\infty}^{\infty} \delta(\tau-\tau_i(t))x(t-\tau)d\tau = x(t-\tau_i(t))$.

(Note that this gave me considerable pause, too -- but I'm pretty sure it's right. Keep in mind that the Dirac delta is something that a lot of mathematicians draw their skirts away from, at the same time that it is being embraced by engineers and physicists. To the point where you'll see it referred to as a "functional" by mathematicians, even as it is known as a "function" by engineers. Just keep in mind that you're hanging with trash from the wrong side of the tracks, and it's easier to keep things straight.)

  • $\begingroup$ Thank you Tim - it does make intuitive sense, I was just wondering if there was a particular equation I was missing, guess not! I love playing fast and loose with the dirac delta, brings a bit of fire to your life. $\endgroup$
    – Jonah F
    Dec 5, 2021 at 1:19
  • $\begingroup$ 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. $\endgroup$ Jan 4, 2022 at 1:56
  • $\begingroup$ 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. $\endgroup$
    – TimWescott
    Jan 4, 2022 at 2:27

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