# How is $\delta(at+b)=\frac{1}{|a|}\delta(t+b/a)$? This result has been used in the second last line of the pic. I don't know why it's true. Both functions are zero for $$t$$ not equal to $$-b/a$$. But at $$t=-b/a$$, a scaling factor $$1/|a|$$ has been introduced in the second function.

[While commenting on Matt's answer, I tried to find a different path. I somehow failed to do so, but it is written, so]

A folk (and false) interpretation of the Dirac $$\delta(t)$$ is that

• this is would be a function (false, in the classic sense, it cannot be evaluated; it should be understood as an application or operator on other functions, and called generalized function or distribution)
• which would be infinite at $$t=0$$ and zero elsewhere (nonsense: this is no "single" sense of infinity that can cope with that, as far as I know).

Assuming this false interpretation for a quick moment, one might think that with $$at$$, with $$a> 0$$, $$at = 0$$ only when $$t=0$$, and $$at \neq 0$$ only when $$t\neq0$$, so $$at$$ and $$t$$ are essentially similar for $$\delta(t)$$ evaluation, and thus, the $$0$$ and $$\infty$$ "values" of $$\delta(t)$$, at $$t=0$$ and $$t\neq 0$$ respectively, could be the same. However, this interpretation is wrong.

What makes a little more sense is to treat $$\delta(t)$$ as undefined at $$t=0$$, yet considering its area property. In other (yet mundane) words, its surface is unity: $$\int_{-\infty}^{\infty}\delta(t)dt=1$$

Then, $$at$$ is seen as a time stretch. Even at the vicinity of $$0$$, the time axis can be stretched. One interpretation of the Dirac is to consider it as a limit of standard function sequences of unit area. Examples are rectangle (left) or triangle (right) functions. If the triangles $$T_\epsilon(t)$$ of support $$[-\epsilon,\epsilon]$$ and height $$\epsilon$$ are dilated by $$a$$, they now have support $$[-a\epsilon,a\epsilon]$$, hence have area $$a$$ instead of $$1$$. So to preserve the unit area, we shall compensate by dividing the height by $$a$$: $$T_\epsilon(at)/a$$. What happens with $$a<0$$? Well (huge simplification) the triangle is symmetric, so the time reversion does not change its shape and area.

So, if (big if) $$\delta(t)$$ is a limit of unit area triangles (but this works for other functions) of support $$[-\epsilon,\epsilon]$$ and height $$\epsilon$$, $$\delta(at)$$ is a limit of unit area triangles (but this works for other functions) of support $$[-a\epsilon,a\epsilon]$$ and height $$\frac{1}{|a|}\epsilon$$, or $$\frac{1}{|a|}T_\epsilon(t)$$, hence with limit $$\frac{1}{|a|}\delta(t)$$.

Finally, the question of the shift is simpler than the stretch: $$\delta(at+b) = \delta(a(t+b/a))$$

and the result follows.

• honestly I didn't read your answer before posting mine :-)) – Fat32 Nov 17 '19 at 1:06
• Good to see some uniform convergence on pathological functions :) – Laurent Duval Nov 17 '19 at 10:01
• should not be more surprising to see some uniform convergence of Cauchy series on a set of real numbers where between each two rational members existing are infinitely many non-rational numbers as well ;-) – Fat32 Nov 17 '19 at 12:29
• I hope you'll like this one. For $p=1,2$, the smooth plane curve $x^p+y^p=1$ (a line or a circle) has infinitely many couples of rational solutions. Interestingly (for me), when $p>2$, Wiles-Fermat says that the same parametric smooth curves avoid ALL rational points $(r_1,r_2)$ that are dense in the $[0\,,1]^2$ – Laurent Duval Nov 17 '19 at 12:43

First of all it seems useful to establish what we mean by an equation like

$$\delta(at+b)=\frac{1}{|a|}\delta(t+b/a),\qquad a\neq 0\tag{1}$$

Since the Dirac impulse $$\delta(t)$$ is a distribution, Eq. $$(1)$$ only makes sense if interpreted as

$$\int_{-\infty}^{\infty}\delta(at+b)\phi(t)dt=\frac{1}{|a|}\int_{-\infty}^{\infty}\delta(t+b/a)\phi(t)dt\tag{2}$$

where $$\phi(t)$$ is a so-called test function, by which we generally mean that it has derivatives of any order, and that for $$|t|\to\infty$$ it tends to zero sufficiently rapidly.

In order to prove $$(1)$$, it is sufficient to show that

$$\delta(at)=\frac{1}{|a|}\delta(t),\qquad a\neq 0\tag{3}$$

because substituting $$t+t_0$$ for $$t$$ in $$(3)$$ results in

$$\delta(a(t+t_0))=\frac{1}{|a|}\delta(t+t_0),\qquad a\neq 0\tag{4}$$

which is equal to $$(1)$$ for $$t_0=b/a$$.

Proving $$(3)$$ means that we need to show that

$$\int_{-\infty}^{\infty}\delta(at)\phi(t)dt=\frac{1}{|a|}\int_{-\infty}^{\infty}\delta(t)\phi(t)dt,\qquad a\neq 0\tag{5}$$

Eq. $$(5)$$ can be shown in a straightforward manner by the variable substitution $$at=\tau$$, which results in $$a\,dt=d\tau$$:

\begin{align}\int_{-\infty}^{\infty}\delta(at)\phi(t)dt=\begin{cases}\displaystyle\frac{1}{a}\int_{-\infty}^{\infty}\delta(\tau)\phi(\tau/a)d\tau,&a>0\\\displaystyle\frac{1}{a}\int_{\infty}^{-\infty}\delta(\tau)\phi(\tau/a)d\tau,&a<0\end{cases}\tag{6}\end{align}

Since

$$\frac{1}{a}\int_{\infty}^{-\infty}\delta(\tau)\phi(\tau/a)d\tau=-\frac{1}{a}\int_{-\infty}^{\infty}\delta(\tau)\phi(\tau/a)d\tau\tag{7}$$

the result $$(6)$$ can be summarized as

$$\int_{-\infty}^{\infty}\delta(at)\phi(t)dt=\frac{1}{|a|}\int_{-\infty}^{\infty}\delta(t)\phi(t/a)dt,\qquad a\neq 0\tag{8}$$

which is equivalent to $$(5)$$ because

\begin{align}\frac{1}{|a|}\int_{-\infty}^{\infty}\delta(t)\phi(t/a)dt&=\frac{1}{|a|}\int_{-\infty}^{\infty}\delta(t)\phi(t)dt\\&=\frac{1}{|a|}\phi(0),\qquad a\neq 0\end{align}\tag{9}

Consequently, $$(8)$$ and $$(9)$$ prove $$(3)$$, which in turn proves $$(4)$$ and $$(1)$$.

Thanks to Laurent Duval for commenting and motivating me to rewrite this answer with a bit more (engineering) rigor.

• Could be useful to integrate $\delta$ against some generic "good function" – Laurent Duval Nov 16 '19 at 14:32
• @LaurentDuval: Well, yes, I guess you could say so ... I'll add some explanation later on. Anyway, the only way Eq. (4) can make sense is if interpreted as applied to a test function, but my impression was that the OP would prefer to be spared with that kind of stuff ... – Matt L. Nov 16 '19 at 15:00
• @LaurentDuval: You're right, I'll add that later on. – Matt L. Nov 16 '19 at 15:41
• @LaurentDuval: I believe that my edited answer makes more sense now, while hopefully still being accessible for engineers. Thanks for pointing out the hand-waviness of the previous version. – Matt L. Nov 17 '19 at 13:35
• It does, yet I already upvoted it before the update (trust). On such a Q&A forum, it is difficult to answer efficiently (from the OP point of view) while keeping sufficient rigor (for next generations) on enough places of the reasoning. I do have my own share of approximations in answers as you have noticed. Here, I thought (future) trouble could come from the $\int f = \int g$, thus $f=g$ simplification. Witnessing how SE pals like you do is very instructive. – Laurent Duval Nov 17 '19 at 13:50

For those who prefer an approach using limits instead of integrals, it follows like this:

Consider the following definition of the unit-impulse located at the origin $$t=0$$ :

$$\delta(t) = \lim_{\Delta \to 0} \delta_{\Delta}(t) \tag{1}$$

where the classical function $$\delta_{\Delta}(t)$$ is defined as

$$\delta_{\Delta}(t) = \begin{cases} {\frac{1}{\Delta} ~~~, ~~~ 0 < t < \Delta \\ ~ 0 ~~~~,~~~ \text{otherwise} } \end{cases} \tag{2}$$

Observe that the area $$A$$ of this pulse is always unity for any value of $$\Delta > 0$$. This area becomes the weight of the impulse as we take the limit.

Now consider the following pulse $$\delta_{\Delta}( a t)$$ whose limit is (let $$a >0$$) :

$$\lim_{\Delta \to 0} \delta_{\Delta}(a t) = \delta(a t) \tag{3}$$

But the new pulse is defined as: $$\delta_{\Delta}(at) = \begin{cases} {\frac{1}{\Delta} ~~~, ~~~ 0 < at < \Delta \\ ~ 0 ~~~~,~~~ \text{otherwise} } \end{cases} = \begin{cases} {\frac{1}{\Delta} ~~~, ~~~ 0 < t < \Delta/a \\ ~ 0 ~~~~,~~~ \text{otherwise} } \end{cases} \tag{4}$$

This new time-scaled pulse has an area of $$A_a = 1/a$$. Then using the unit-area definition of the basic pulse we can re-write the new pulse as :

$$\delta_{\Delta}(at) = \begin{cases} {\frac{1}{a\Delta} ~~~, ~~~ 0 < t < \Delta \\ ~ 0 ~~~~~,~~~ \text{otherwise} } \end{cases} ~~ = ~~ \frac{1}{a} \delta_{\Delta}(t) \tag{5}$$

Finally taking the limits we see that

$$\lim_{\Delta \to 0} \delta_{\Delta}(a t) = \delta(a t) = \lim_{\Delta \to 0} \frac{1}{a} \delta_{\Delta}(t) = \frac{1}{a} \delta(t) \tag{6}$$

The shift can also be shown on a similar basis and it's no surprise to see that

$$\lim_{\Delta \to 0} \delta_{\Delta}(t-b) = \delta(t-b) \tag{7}$$

is a unit impulse located at $$t=b$$ instead of $$t=0$$.

Then combining the time-scale and time-shift we can argue that

$$\delta_{\Delta}(a t + b) = \frac{1}{a} \delta_{\Delta}(t + b/a) \tag{8}$$

and the limit yields: $$\lim_{\Delta \to 0} \delta_{\Delta}(a t + b) = \frac{1}{a} \delta(t+b/a) \tag{9}$$

For $$a<0$$ we use the absolute value on the weight.

• I've messed around with this sort of thing; it's useful at times to consider $\delta_\Delta(t)$ to be any function that is (A) centered on $t=0$, (B) has a total area of 1, and (C) tends to zero as $\Delta t$ gets sufficiently small (and, I'm not going to try to explain what I mean by "sufficiently small" here -- I'm in the deep end and furiously treading water!) – TimWescott Nov 18 '19 at 20:32