I have recently read a number of derivations of the formula for convolution based on the idea of representing $x(t)$ as a series of scaled impulse functions $\delta(t)$.

Given that we know $\int_{-\infty}^{\infty}x(\tau)\delta(t-\tau)d\tau = x(t)$,

if we know the impulse response function $h(t)$ characterizing the linear time-invariant system in question we can "easily" get the output $y(t)$ for the input $x(t)$ as

$y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)h(t - \tau) d\tau$.

Great. This makes sense and jives well with graphical illustrations of how convolution works and is computed.

But given that $\delta(t - \tau) = \delta(\tau - t)$ isn't it also true that we could have just as easily flipped the lag of the impulse function

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

to get a convolution defined as

$x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)h(\tau - t) d\tau$ ?

$h$ isn't necessarily symmetric though, so this is a different function altogether. What am I missing here? Is there a reason why this parallel "alternate" derivation is wrong?


1 Answer 1


One of the fundamental properties of the impulse function is the following. For a function $h:\mathbb{R}\rightarrow \mathbb{R}$ $$ h(t) * \delta(t-\tau)=h(t-\tau). \text{ $(1)$} $$ If what you suggest was true, it would mean that $h(t) * \delta(t-\tau)=h(\tau - t)$, which, as you correctly observe is not the same quantity as $h(t-\tau)$.

What you are doing wrong is that you treat the arguments of the impulse function, $t$ and $\tau$, as constants when they are really not. When you apply the convolution operator, it is important to realize which of the two variables, $t$ or $\tau$, represents "time" (allowed to take every possible value), and which variable represents a constant. In other words, if you were to draw $\delta(t-\tau)$ on paper, would the horizontal axis correspond to $t$ with an impulse appearing at $t=\tau$ or would the horizontal axis correspond to $\tau$ with an impulse appearing at $\tau = t$?

The notation $\delta(t-\tau)$ alone does not provide this information, however, the signal function $h(t)$ that convolves with it does. In this case $t$ is the "time" variable and $\tau$ is the constant, which results in (1).

You can now see that it also holds

$$ h(\tau) * \delta(t-\tau) = h(\tau-t) $$ where now the notation of the signal function indicates that $\tau$ is to be treated as the time variable and $t$ as the shift.


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