LECTURE 22: DIRAC DELTA AND IMPULSE RESPONSE
MINGFENG ZHAO
October 27, 2014
Definition 1. Let f (t) be a function on [0, ∞), then
I. The Laplace transform of f , denoted by L[f ](s), is defined as:
Z
∞
L[f ](s) =
f (t)e−st dt,
for all s > 0.
0
II. If F (s) = L[f ](s), the inverse Laplace transform of F , denoted by L−1 [F ](t), is defined as:
L−1 [F ](t) = f (t),
∀for all t > 0.
Definition 2. Let f (t) and g(t) be two functions on [0, ∞), the convolution of f and g is defined as:
Z
(f ∗ g)(t) =
t
f (τ )g(t − τ ) dτ.
0
Definition 3. For any continuous function f (t) on (−∞, ∞), we have
Z
∞
δ(t)f (t) dt = f (0).
−∞
Proposition 1. There holds that
I. Linearity:
L[af (t) + bg(t)](s) = aL[f (t)](s) + bL[g(t)](s).
That is,
L−1 [aF (s) + bG(s)](t) = aL−1 [F (s)](t) + bL−1 [G(s)](t).
II. First Shifting Property:
L e−at f (t) (s) = L[f (t)](s + a).
That is,
L−1 [F (s + a)] (t) = e−at L−1 [F (s)](t).
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MINGFENG ZHAO
III. Transforms of derivatives:
L[f 0 ](s)
=
sL[f ](s) − f (0)
L[f 00 ](s)
=
s2 L[f ](s) − sf (0) − f 0 (0).
IV. Second Shifting Property:
L[ua (t)f (t − a)] = e−as L[f (t)](s),
for all s > 0.
That is,
L−1 e−as G(s) (t) = ua (t)L−1 [G(s)](t − a).
V. Transform of Integrals:
t
Z
L
0
L[f (t)](s)
.
f (τ ) dτ (s) =
s
That is,
L−1
Z t
F (s)
=
L−1 [G(s)](τ ) dτ.
s
0
V. Transform of Convolution:
L[(f ∗ g)(t)] = L[f (t)] · L[g(t)].
That is,
L−1 [L[f (t)] · L[g(t)]] = f ∗ g(t)
VI. Transform of Dirac delta:
L[δ(t − a)] = e−as .
That is,
L−1 [e−as ] = δ(t − a).
Example 1. Find L
−1
s+1
.
s
In fact, we have
L−1
s+1
1
1
= L−1 1 +
= L−1 [1] + L−1
.
s
s
s
By looking up the table, we have
L−1 [1] = δ(t),
and L−1
1
= 1.
s
Then we know that
L
−1
s+1
= δ(t) + 1. .
s
LECTURE 22: DIRAC DELTA AND IMPULSE RESPONSE
Example 2. Let ω0 > 0, solve x00 + ω02 x = δ(t), x(0) = x0 (0) = 0.
Let X(s) = L[x(t)], apply the Laplace transform on the both sides of x00 + ω02 x = δ(t), then
L[x00 ] + ω02 L[x] = L[δ(t)].
By the transform of derivatives, we have
L[x00 ] = s2 L[x] − sx(0) − x00 (0) = s2 X(s).
By looking up the table, we have
L[δ(t)] = 1.
So we have
s2 X(s) + ω02 X(s) = 1.
That is,
X(s) =
1
.
s2 + ω02
By looking up the table, we get
x(t) = L−1 [X(s)] = L−1
1
sin(ω0 t)
=
.
2
2
s + ω0
ω0
Therefore, the solution to x00 + ω02 x = δ(t), x(0) = x0 (0) = 0 is:
x(t) =
sin(ω0 t)
.
ω0
Example 3. Solve y 00 + 4y 0 + 5y = δ(t − 3), y(0) = 1, y 0 (0) = 0.
Let Y (s) = L[y(t)], apply the Laplace transform on the both sides of y 00 + 4y 0 + 5y = δ(t − 3), then
L[y 00 ] + 4L[y 0 ] + 5L[y] = L[δ(t − 3)].
By the transform of derivatives, we have
L[y 0 ] = sL[y] − y(0) = sY (s) − 1,
and L[y 00 ] = s2 L[y] − sy(0) − y 0 (0) = s2 Y (s) − s.
By looking the table, we have
L[δ(t − 3)] = e−3s .
So we get
s2 Y (s) − s + 4[sY (s) − 1] + 5Y (s) = e−3s .
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MINGFENG ZHAO
Then
Y (s)
=
=
=
=
e−3s + s + 4
s2 + 4s + 5
s+4
e−3s
+ 2
2
s + 4s + 5 s + 4s + 5
s+4
e−3s
+
2
(s + 2) + 1 (s + 2)2 + 1
2
e−3s
s+2
+
+
.
(s + 2)2 + 1 (s + 2)2 + 1 (s + 2)2 + 1
By looking up the table, we have
s+2
L−1
= e−2t cos(t),
(s + 2)2 + 1
and L−1
1
= e−2t sin(t).
(s + 2)2 + 1
By the Second Shifting Property, we have
1
e−3s
−1
−1
= µ3 (t) · L
(t − 3) = µ3 (t)e−2(t−3) sin(t − 3).
L
(s + 2)2 + 1
(s + 2)2 + 1
Then we get
y(t) = L−1 [Y (s)] = e−2t cos(t) + 2e−2t sin(t) + µ3 (t)e−2(t−3) sin(t − 3).
Therefore, the solution to y 00 + 4y 0 + 5y = δ(t − 3), y(0) = 1, y 0 (0) = 0 is:
y(t) = e−2t cos(t) + 2e−2t sin(t) + µ3 (t)e−2(t−3) sin(t − 3) .
Example 4. Find the Laplace transform Y (s) = L[y(t)] of the solution of y 00 + y = g(t), y(0) = 1 and y 0 (0) = 2, where

 1,
if 0 ≤ t < 2,
g(t) =
 cos(t − 2), if t ≥ 2.
Apply the Laplace transform on the both sides of y 00 + y = g(t), then
L[y 00 ] + L[y] = L[g(t)].
By transform of derivatives, we have
L[y 00 ] = s2 L[y] − sy(0) − y 0 (0) = s2 Y (s) − s − 2.
Then
s2 Y (s) − s − 2 + Y (s) = L[g(t)].
LECTURE 22: DIRAC DELTA AND IMPULSE RESPONSE
So we get
Y (s) =
L[g(t)] + s + 2
s2 + 1
For L[g(t)], it’s easy to see that
g(t) = 1 − u2 (t) + u2 (t) cos(t − 2)
By the Second Shifting Property, we have
L[g(t)] = L[1] − L[u2 (t)] + L[u2 (t) cos(t − 2)] = L[1] − e−2s L[1](s) + e−2s L[cos(t)](s).
By looking up the table, we know that
L[1] =
1
,
s
and L[cos(t)] =
s2
s
.
+1
Then we get
L[g(t)] =
se−2s
1 e−2s
−
+ 2
.
s
s
s +1
So
Y (s)
=
=
1
s
−
e−2s
s
+
se−2s
s2 +1
+s+2
s2
+1
s+2
1
e−2s
se−2s
+
−
+ 2
.
2
2
2
s + 1 s(s + 1) s(s + 1) (s + 1)2
Therefore, we know that
Y (s) =
s+2
1
e−2s
se−2s
.
+
−
+ 2
2
2
2
s + 1 s(s + 1) s(s + 1) (s + 1)2
In general, let 0 ≤ a1 < a2 < a3 < · · · < an , if



f1 (t),







f2 (t),





 f3 (t),
f (t) =
..


.,







fn (t),





 fn+1 (t),
if 0 ≤ t < a1 ,
if a1 ≤ t < a2 ,
if a2 ≤ t < a3 ,
..
.,
if an−1 ≤ t < an ,
if t ≥ an ,
then,
f (t) = f1 (t) +
n
X
k=2
fk (t)[uk−1 (t) − uk (t)] + fn+1 (t)un (t).
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MINGFENG ZHAO
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca