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 one 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 $t=2$, just relabeled. For $\delta$, scaling the argument changes the strength by a factor $1/|a|$. Another reminder that $\delta$ 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 $1$. Replace the argument $t$ by $at$ and 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 keeps everything consistent.