Transform of the Unit-Impulse Function

Consider the following example.

EXAMPLE 10.6.1   Find the Laplace transform, $D_h(s)$ , of

$$\delta_h(t) = \left\{\begin{array}{ll} 0 & t < 0 \\ \frac{1}{h} & 0 \leq t < h \\ 0 & t > h. \end{array}\right.$$

Solution: Note that $\delta_h(t) = \displaystyle\frac{1}{h}\bigl( U_0(t) - U_h(t) \bigr).$ By linearity of the Laplace transform, we get

$$D_h(s) = \frac{1}{h} \bigl( \frac{ 1 - e^{-hs} }{s} \bigr).$$

Remark 10.6.2  

1. Observe that in Example 10.6.1, if we allow $h$ to approach 0 , we obtain a new function, say $\delta(t).$ That is, let
   $$\delta(t) = \lim_{h \longrightarrow 0} \delta_h(t).$$
   This new function is zero everywhere except at the origin. At origin, this function tends to infinity. In other words, the graph of the function appears as a line of infinite height at the origin. This new function, $\delta(t)$ , is called the UNIT-IMPULSE FUNCTION (or Dirac's delta function).

2. We can also write
   $$\delta(t) = \lim_{h \longrightarrow 0} \delta_h(t) = \lim_{h \longrightarrow 0} \frac{1}{h}\bigl( U_0(t) - U_h(t) \bigr).$$

3. In the strict mathematical sense $\lim\limits_{h \longrightarrow 0} \delta_h(t)$ does not exist. Hence, mathematically speaking, $\delta(t)$ does not represent a function.

4. However, note that
   $$\int_0^\infty \delta_h(t) d t = 1, \;\; {\mbox{ for all }} \; \; h.$$

5. Also, observe that ${\mathcal L}(\delta_h(t)) = \displaystyle \frac{ 1 - e^{- h s}}{hs}.$ Now, if we take the limit of both sides, as $h$ approaches zero (apply L'Hospital's rule), we get
   $${\mathcal L}(\delta(t)) = \lim_{h \longrightarrow 0} \frac{ 1 - e^{- h s}}{hs}= \lim_{h \longrightarrow 0} \frac{ s e^{- h s}}{s} = 1.$$