Dirac Delta Function (δ(t))
Lecture Notes: Monday, June 1
We want a function which represents an “instanteous” force. For example when you hit a
billard ball with a cue stick, you impart a force on the system that immediately disappears
after the hit.
Definition 1. (Delta Function) In this class, we define the delta function as δ(t) =
u00 (t) where
(
0 t<0
u0 (t) =
.
1 0≤t<∞
Remark:
(a) u0 (t) is NOT differentiable at 0. Everywhere, but at 0, u00 (t) = 0, and so we define
u00 (0) = ∞.
(b) Because u0 (t) is not differentiable, δ(t) is not a function. In fact, it is called a
generalized function.
*A generalized function inputs functions and outputs a real value. We will see in the
case of the δ-function that δ(f ) = f (0).
With this definition in mind, we think of the δ function as
(
0 t 6= 0
δ(t) ”=”
.
∞ t=0
Remark: This ”definition” agrees with the previous one for the derivative of the heavyside
function except at 0 is 0 and we could define the derivative at 0 to be ∞.
Properties of the δ function
Property 1:
Z
∞
δ(t) dt = 1.
−∞
Proof.
Z
∞
Z
∞
δ(t) dt =
−∞
−∞
u00 (t) dt
= u0 (∞) − u0 (−∞)
definition of heavyside
1
= 1 − 0 = 1.
Observe that if δ(t) were a function and equaled to 0 everywhere except at 0 and ∞ at 0,
how could the area under this function equal to 1? This is one reason why δ(t) is not a
function.
Property 2 (Most Important):
Suppose f (t) is a function such that f (∞) = f (−∞) = 0. Then
Z ∞
f (t)δ(t) dt = f (0).
−∞
Most people define the δ function based on this property.
Proof.
Z
∞
Z
∞
u00 (t)f (t) dt
−∞
Z
∞
= u0 (t)f (t) −∞ −
δ(t)f (t) dt =
−∞
IBP
∞
u0 (t)f 0 (t) dt
Z ∞
= 1 · f (∞) − 0 · f (−∞) −
f 0 (t) dt
0
∞
= −f (t)0 = −f (∞) + f (0) = f (0).
−∞
Example 1. Evaluate
Z
∞
(t − 1)3 δ(t) dt.
−∞
Solution:
Z
∞
(t − 1)3 δ(t) dt
Prop. 2
=
(0 − 1)3 = −1.
−∞
Definition 2. (The Shifted Delta Function) The shifted delta function δc (t) or δ(t − c)
is
δ(t − c) = u0c (t).
Like the δ function, one can think of the shifted delta function as
(
0
t 6= c
δ(t − c) =
.
∞ t=c
Properties of the shifted delta function:
Property 1:
Z ∞
δ(t − c) dt = 1.
−∞
2
Property 2:
Suppose f is a function such that f (−∞) = f (∞) = 0. Then
Z ∞
f (t)δ(t − c) dt = f (c).
−∞
Example 2. Evaluate
Z
∞
(t − 1)3 δ(t − 2) dt.
−∞
Solution:
Z
∞
(t − 1)3 delta(t − 2) dt = (2 − 1)3 = 1.
−∞
Example 3. Evaluate
Z
∞
cos(t)δ(t − π) − t2 δ(t − 4) dt.
−∞
Solution:
Z
∞
cos(t)δ(t − π) − t2 δ(t − 4) dt = cos(π) − (4)2 = −17.
−∞
Laplace Transform
Suppose c ≥ 0. Then
L δ(t − c) = e−sc .
Proof.
L{δ(t − c)} = L u0c (t)
by def.
Derivative rule = sL uc (t) + uc (0)
#12 on Table
=s·
3
e−sc
− 0 = e−sc .
s