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3.5 Impulses and the Delta Function 103

This is not a function in the conventional sense but is a more general object called a distribution.
For historical reasons, it continues to be known as the Dirac function after the Nobel laureate
physicist P.A.M. Dirac. The shifted delta function δ(t − a) is zero except for t = a, where it has
an infinite spike.
To take the Laplace transform of the delta function, begin with
1
δ (t − a) = [H(t − a) − H(t − a − )].

This has transform
1 1 −as 1 −(a+ )s

L[δ (t − a)]= e − e
s s
e −as (1 − e − s )
= ,
s
suggesting that we define

e −as (1 − e − s )
L[δ(t − a)]= lim = e −as .
→0+ s
In particular, we can choose a = 0 to get

L[δ(t)]= 1.

The following result is called the filtering property of the delta function. Suppose at time t = a a
signal is impacted with an impulse by mutliplying the signal by δ(t − a), and the resulting signal
is then summed over all positive time by integrating it from zero to infinity. We claim that this
yields exactly the value f (a) of the signal at time a.

THEOREM 3.6 Filtering Property of the Delta Function

∞
Let a > 0 and let f (t)dt converge. Suppose also that f is continuous at a. Then
0
∞

f (t)δ(t − a)dt = f (a).
0
A proof is outlined in Problem 9.
If we apply the filtering property to f (t) = e −st , we get

∞
e −st δ(t − a)dt = e −as ,
0
which is consistent with the definition of the Laplace transform of the delta function. Now change
notation in the filtering property, and write it as
∞

f (τ)δ(t − τ)dτ = f (t).
0
We recognize the convolution of f with δ . The last equation becomes

f ∗ δ = f.
The delta function therefore acts as an identity for the “product” defined by convolution.
In using the Laplace transform to solve an initial value problem involving the delta function,
proceed as we have been doing, except that now we must use the transform of the delta function.

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October 14, 2010 14:14 THM/NEIL Page-103 27410_03_ch03_p77-120