Impulse function (Delta Dirac function) strength 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.
 A: 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: 
$$
\int a\delta(x)f(x)dx = a\int \delta(x)f(x)dx = af(0)
$$
A: 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".   
