Impulse function (Delta Dirac function) strength

Question

The strength of the delta function by definition is infinity, so how come in some questions a number is assigned to the strength of the impulse function? Although these numbers are not necessarily big.

Answer 1

A "delta function" is not really a function, it's what is called a "generalized function", or distribution. So, when you see people write $\delta(0) = \infty$, this is just a "formality", and a very misleading one at that. The only correct way to think about a delta function is under an integral sign with another ("normal") function $f(x)$:

$$ \int \delta(x)f(x)dx $$ This expression makes perfect sense, and as long as $f(x)$ is "nice enough" (e.g. continuous at 0), it will evaluate to $f(0)$. If someone wants to "scale" a delta function, they will usually write $a\delta(x)$, but again, this isn't a function, so it doesn't make sense to write

$$ a\delta(x) = \left\{\begin{array}{cc} 0 & x\neq 0 \\ a\cdot \infty & x = 0 \end{array}\right. $$ so the only way to make sense of $a\delta$ is under an integral sign:

$$ \int a\delta(x)f(x)dx = a\int \delta(x)f(x)dx = af(0) $$

Answer 2

The Dirac function is not really a function, rather a distribution representing a point mass or impulse. It is helpful to view the Dirac function as a limit of functions $$ \delta_a(t) = \left\{ \begin{array}{cl} \dfrac{1}{2a}, \ \ & |t| < a \\[10pt] 0, \ \ & \text{otherwise} \end{array} \right. , $$ as $a \to 0^+.$ Notice that $\int^\infty_{-\infty} \delta_a (t) \ dt = 1$ regardless of the value of $a.$ If we scaled by positive constant, say $ m \delta_a(t),$ then we would increasing the "impulse".