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The Dirac Delta Function and Impulse Response

and

lim f (t) = 0, t ̸= 0, lim f (0) = ∞. (3.4)
ε
ε
ε→0 + ε→0 +
Our ultimate interest is the behaviour of the solution to equation (3.1) with forcing
+
function f (t) in the limit ε → 0 . That is, what happens to the system output as
ε
we make the applied force progressively ”sharper” and ”stronger?”.
Let y (t) be the solution to equation (3.1) with f(t) = f (t). Then the unique
ε
ε
solution is given by
t
∫
y (t) = ϕ(t − s)f (s)ds.
ε
ε
0
For t ≥ ε the last equation becomes

∫ ε
1
y (t) = ϕ(t − s)ds.
ε
ε 0
Since ϕ(t) is continuous for all t ≥ 0 we can apply the mean value theorem for
integrals and write

y(t) = ϕ(t − ψ)

+
for some 0 ≤ ψ ≤ ε. Letting ε → 0 and using the continuity of ϕ we find

y(t) = lim y (t) = ϕ(t).
ε
ε→0 +

We call y(t) the impulse response of the linear system.

The Dirac Delta Function

The problem with the integral

t
∫
ϕ(t − s)f (s)ds
ε
0
is that lim f (0) is undefined. So it makes sense to ask the question of whether
ε
ε→0 +
we can find a function δ(t) such that

t t
∫ ∫
lim y (t) = lim ϕ(t − s)f (s)ds == ϕ(t − s)δ(s)ds = ϕ(s)
ε
ε
ε→0 + ε→0 + 0 0
where the role of δ(t) would be to evaluate the integrand at s = 0. Note that
because of (3.4), we cannot interchange the operations of limit and integration in
the above limit process. Such a function δ exist in the theory of distributions and
can be defined as follows:

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