Scaling property of the impulse

The scaling property of the unit impulse says that scaling the argument by a nonzero constant $a$ scales the impulse by $1/|a|$:

$\delta (a(t-t_0))=\frac{1}{|a|}\delta (t-t_0).$

This is one of those things that catches you. With ordinary functions, scaling the argument by $a$ doesn’t change the function’s value at any individual point — $f(2t)$ takes the same value at $t=1$ that $f(t)$ takes at $t=2$; it’s just relabeled. For $\delta$, scaling the argument changes the strength by a factor $1/|a|$. This is one of those moments that remind us $\delta$ is not a function.

Where the $1/|a|$ comes from

Approximate $\delta$ as a tall narrow rectangle of width $w$ and height $1/w$, so its area is $1$. If we replace the argument $t$ by $at$, the rectangle gets narrower by a factor $|a|$ — its new width is $w/|a|$. But its height stays $1/w$. So its area becomes $(1/w)\cdot (w/|a|)=1/|a|$, not $1$.

To restore unit area in the limit, we have to multiply by $|a|$. In equation form, the impulse $\delta (at)$ has strength $1/|a|$, hence $\delta (at)=\delta (t)/|a|$.

Worked example

What is $\delta ((t-1)/2)$? Apply the scaling property with $a=1/2$ and $t_0=1$:

$\delta \left(\frac{1}{2}(t-1)\right)=\frac{1}{|1/2|}\delta (t-1)=2\delta (t-1).$

So integrating $\delta ((t-1)/2)$ over any interval containing $t=1$ gives $2$, not $1$. Slowing down the argument has made the impulse “fatter,” and its strength has doubled.

This shows up whenever you change variables inside an integral involving an impulse — a substitution like $u=at$ rescales $dt$ to $du/a$, and the scaling property of the impulse keeps everything consistent.

For the other two properties, see Equivalence property of the impulse (the pointwise statement) and the sifting property (the integrated form).