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enter image description here

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.

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3 Answers 3

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[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 $1$.

Dirac as a limit of rectangles or triangles

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.

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    $\begingroup$ honestly I didn't read your answer before posting mine :-)) $\endgroup$
    – Fat32
    Nov 17, 2019 at 1:06
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    $\begingroup$ Good to see some uniform convergence on pathological functions :) $\endgroup$ Nov 17, 2019 at 10:01
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    $\begingroup$ 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 ;-) $\endgroup$
    – Fat32
    Nov 17, 2019 at 12:29
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    $\begingroup$ 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 $[0\,,1]^2$ $\endgroup$ Nov 17, 2019 at 12:43
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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.

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    $\begingroup$ Could be useful to integrate $\delta$ against some generic "good function" $\endgroup$ Nov 16, 2019 at 14:32
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    $\begingroup$ @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 ... $\endgroup$
    – Matt L.
    Nov 16, 2019 at 15:00
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    $\begingroup$ @LaurentDuval: You're right, I'll add that later on. $\endgroup$
    – Matt L.
    Nov 16, 2019 at 15:41
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    $\begingroup$ @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. $\endgroup$
    – Matt L.
    Nov 17, 2019 at 13:35
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    $\begingroup$ 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. $\endgroup$ Nov 17, 2019 at 13:50
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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.

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  • $\begingroup$ 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!) $\endgroup$
    – TimWescott
    Nov 18, 2019 at 20:32

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