Equivalence property of the impulse

Multiply any function $g(t)$ continuous at $t_0$ by an impulse located at $t_0$ and you get a scaled impulse, scaled by the function’s value at $t_0$:

$g(t)\delta (t-t_0)=g(t_0)\delta (t-t_0).$

The reasoning is direct. The impulse $\delta (t-t_0)$ is zero everywhere except at $t=t_0$, so the product is zero everywhere except possibly at that one point. At $t=t_0$ the multiplier $g(t)$ takes the value $g(t_0)$, so we have a scaled-by-$g(t_0)$ impulse at $t_0$.

The right-hand side has $g(t_0)$, a constant, with no $t$ in it. The function $g$ is evaluated at the impulse’s location, and only that one value of $g$ matters. Everything else gets discarded.

Why this is useful

This is what makes convolutions with impulses trivial: the impulse picks out one value of the function. The sifting property (sometimes called the sampling property) is a direct corollary, integrating both sides over $t$:

${\int }_{-\infty }^{\infty }g(t)\delta (t-t_0)dt=g(t_0){\int }_{-\infty }^{\infty }\delta (t-t_0)dt=g(t_0).$

Equivalence is the pointwise statement; sifting is its integrated version. In a derivation you’ll often write the pointwise form first to manipulate algebraically, then integrate at the end.

Worked example

Compute $e^{-3t}\delta (t)$. By equivalence with $g(t)=e^{-3t}$ and $t_0=0$:

$e^{-3t}\delta (t)=e^{-3\cdot 0}\delta (t)=e^0\delta (t)=\delta (t).$

This kind of simplification appears in every verification of an impulse-response calculation: when you plug $h(t)=e^{-\alpha t}u(t)$ back into the ODE, $e^{-\alpha t}\delta (t)$ simplifies to $\delta (t)$ exactly via this property.

The third member of the trio is Scaling property of the impulse.