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The Dirac Delta Function and Impulse Response

differential equations. The next theorem finds the Laplace transform of the delta

function.
Theorem 3.1È

With δ(t) defined as above, if a ≤ t < b
0

b
∫
(t)δ(t − t )dt = f(t ).
0
0
a
. . . . . . .
PROOF. We have

b b 1 t 0 +ε
∫ ∫ ∫
f(t)δ(t − t )dt = lim f(t)f (t − t )dt = lim f(t)dt =
0
ε
0
+
a ε→0 + a ε→0 ε t 0
1
= lim f(t + βε)ε = f(t )
0
0
ε→0 ε
+
. . . . where 0 < β < 1 and the mean-value theorem for integrals has been used 2
Remark 3.1E Since p ε (t − t0) = 1 for t 0 ≤ t ≤ t 0 + ε and 0 otherwise we see that
ε
∫ b f(t)δ(t − a)dt = f(a) and ∫ b
a a f(t)δ(t − t 0 )dt = 0 for t 0 ≥ b.
. . . . .
It follows immediately from the above theorem that

∫
∞
L[δ(t − t )] = e −st δ(t − t )dt = e −st 0 , t > 0.
0
0
0
0
In particular, if t = 0 we find
0
L[δ(t)] = 1.
.
The following example illustrates the formal use of the delta function.

Example 3.1, A spring-mass system with mass 2, damping 4, and
spring constant 10 is subject to a hammer blow at time t = 0.
The blow imparts a total impulse of 1 to the system, which was

initially at rest. Find the response of the system.
. . . . .
.
Solution. The situation is modelled by the initial value problem

2y + 4y + 10y = δ(t), y(0) = 0, y (0) = 0.
′
′
′′
Taking Laplace transform of both sides we find
2
2s Y (s) + 4sY (s) + 10Y (s) = 1.

Solving for Y (s) we find
1
Y (s) = .
2
. . . . 2s + 4s + 1

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