Consider this derivation of the zero-state response \$y_{zs} \$ for an LTI system caused by an input \$x(t) \$: - $$\begin{smallmatrix}\begin{array}{r|cc} \text{Input} & \text{Output} & \text{comments}\\\\ \hline \delta(t) & h(t) & \text{Impulse in -> impulse response out}\\\\ \delta(t-\tau) & h(t-\tau) & \text{Because of time invariance}\\\\ x(\tau)\delta(t-\tau) & x(\tau)h(t-\tau) & \text{Scaled input -> scaled output}\\\\ \int^{\infty}_{-\infty}{x(\tau)\delta(t-\tau)} d\tau & \int^{\infty}_{-\infty}{x(\tau)h(t-\tau)} d\tau& \text{Superposition (linearity)}\\\\ x(t) & y_{zs}(t) & \end{array}\end{smallmatrix}$$ Which shows that the zero state response can be found by convolving the input with the impulse response of the system.

So the idea is that the input signal \$x(t) \$ can be split into infinitely many scaled and time-shifted impulses \$x(\tau)\delta(t-\tau) \$, where \$\tau \$ is the time-shift and \$x(\tau) \$ is the scaling factor and \$\delta(t) \$ is the Dirac delta function.

The thing I don't understand is that \$\delta(t) \$ has infinite height and zero width, an area of 1 and is undefined at \$t=0 \$.

If the height is infinite, then when we scale it, the height becomes \$x(\tau) \cdot \infty \$ which doesn't make much sense.

Am I missing something here, or is this something I just have to accept and move on? Is the derivation wrong? Hopefully you understand my confusion.


1 Answer 1


The Dirac impulse is a concept, and that concept says it has an infinite value at one and only one point, and zero everywhere else. That concept relies on another concept, the mathematical point, which has no dimensions1. As they are it's impossible to be used in any mathematical formula. So, in order to be able to use it, people have appealed to limits, where the Dirac function is expressed as a distribution:


Only now it can be used, by saying the \$a\rightarrow 0\$. In theory, the concept that an infinite value over a zero width area is chosen because that would give a completely flat Fourier transform, from DC to light (and beyond/before, if it's ever possible). In practice, evaluating a system's response is done with a band limited impulse, which is modelled as a distribution (or something close) with a bandwidth that's sufficiently large. Otherwise, as you noted, it simply doesn't make sense to involve infinity in a limited world.


In case the above is not clear enough, you can treat the \$\delta\$ function just like any regular function, with its awkward property. It all works because now it can be proven to be a distribution, so it makes sense. Therefore the convolution integral of \$\delta(t)\$ with \$h(t)\$ becomes (using integration by parts and from 0 to t):

$$\begin{align} \int_0^t{h(\tau)\delta(\tau-t)\text{d}\tau}&=h(t)\int_0^t{\delta(\tau-t)\text{d}\tau}-\int_0^t{h'(\tau-t)\left(\int_0^t{\delta(\tau)\text{d}\tau}\right)\text{d}\tau} \tag{1} \end{align}$$

The second term uses the derivative of the Dirac function, which is a positive impulse at \$0^-\$ and a negative impulse at \$0^+\$. That means it's an odd function and any function multiplied by it becomes odd, thus its integral will be zero. The first term is \$h(t)\$ multiplied by the integral of the Dirac function which, if it were indefinite, it would have been the Heaviside function, but here it's \$2\theta(t)-1\$. Since the time starts at 0 the output now becomes simply \$h(t)\$.

1As it is, it doesn't make sense to say a line is made of an infinity of points, because a zero width line is also made of an infinity of points. But if you think of it in terms of a limit, then it makes sense.

  • 1
    \$\begingroup\$ You should know that you're not the first one to ask this question and, at the very least on dsp.ee, there's a thread (which I can't find) which deals with this. I'm sure on math.ee, too. \$\endgroup\$ Nov 8, 2021 at 16:42
  • \$\begingroup\$ I've updated the answer, maybe it needed clarifications. \$\endgroup\$ Nov 9, 2021 at 17:10

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