More On the Laplace Transform
The Impulse Response
We have seen that we can solve the initial value problem
yÝ0Þ = y v Ý0Þ = 0
LßyÝtÞà = fÝtÞ,
by applying the Laplace transform to get
PÝs Þ ŷÝsÞ = !fÝsÞ
Then
ŷÝsÞ =
PÝsÞ = characteristic polynomial for L
where
1 !fÝsÞ,
PÝsÞ
and if we denote by YÝtÞ the inverse transform of
1 , then
PÝsÞ
t
yÝtÞ = Y D fÝtÞ = X YÝt ? bÞ fÝbÞ db.
0
Note that
0
yÝ0Þ = X YÝ0 ? bÞ fÝbÞ db = 0,
0
and
t
y v Ý0Þ = YÝ0ÞfÝtÞ + X Y v Ýt ? bÞ fÝbÞ db| t=0 = 0,
0
so this solution does satisfy the homogeneous initial conditions.
Now suppose that the forcing function f(t) is such that its Laplace transform is just 1. In
this case
!fÝsÞ = 1 implies ŷÝsÞ =
1
PÝsÞ
and yÝtÞ = YÝtÞ.
Evidently, Y(t) is the solution of the initial value problem in the special case that the forcing
has 1 for its Laplace transform. Then the question is, ”what is this f(t)?”
One approach to this question is to recall that
LßHÝtÞà = 1s
where
HÝtÞ =
0 if t ² 0
1 if t > 0
Then by property 3 of the Laplace transform,
LßH v ÝtÞà = 1 ? HÝ0Þ = 1,
which implies that the f(t) we are seeking is the derivative of HÝtÞ. It is usual to denote this
1
”function” by NÝtÞ but whatever notation is used, the fact remains that HÝtÞ has no derivative
in the setting of classical analysis. There is a well developed theory of generalized functions
in which the step function can be differentiated but we will not try to pursue this here.
Instead we will simply note that if HvÝtÞ did exist then the integration by parts formula would
imply that for any b > 0 and any smooth function dÝtÞ which tends to zero as t ¸ K,
K
K
v
X 0 H v Ýt ? bÞ dÝtÞ dt = H Ýt ? bÞ dÝtÞ| t=K
t=0 ? X 0 H Ýt ? bÞ d ÝtÞ dt
But HÝtÞ = 0, for t ² 0, and dÝtÞ = 0 at t = K, hence
H Ýt ? bÞ dÝtÞ| t=K
t=0 = 1 6 dÝKÞ ? HÝ?bÞdÝ0Þ = 0
K
K
X 0 H Ýt ? bÞ d v ÝtÞ dt = X b d v ÝtÞ dt = dÝKÞ ? dÝbÞ,
In addition
K
X 0 H v Ýt ? bÞ dÝtÞ dt = dÝbÞ.
and therefore
Then we take the defining property of NÝt ? bÞ = H v Ýt ? bÞ to be
K
X 0 NÝt ? bÞ dÝtÞ dt = dÝbÞ
for any smooth function dÝtÞ which tends to zero as t ¸ K. This suggests (but only
suggests, it is not true in the setting of classical analysis) that NÝt ? bÞ is very large near
t = b and is very small when t is not close to b. Physically we think of NÝt ? bÞ as an
idealization of something called an ”impulse”, an input of very high intensity concentrated on
a very small time interval. We repeat that this notion is only figurative and can not be
rigorously justified.
Now, using the definition given above for NÝt ? bÞ, we have
K
LßNÝt ? bÞà = X NÝt ? bÞ e ?st dt = e ?sb
0
and
K
LßNÝtÞà = X NÝtÞ e ?st dt = e ?sb | b=0 = 1.
0
Then YÝtÞ is the response to an impulse input of the system described by the initial value
problem. Y(t) is often called the ”impulse response”.
Evidently, Y(t) can be defined by either of two equivalent specifications, namely
ÝaÞ LßyÝtÞà = 0 and yÝ0Þ = 0,
y v Ý0Þ = 1
ÝbÞ LßyÝtÞà = NÝtÞ and yÝ0Þ = 0 = y v Ý0Þ
Once Y(t) is known, the initial value problem is completely solved. To see this, consider
2
LßyÝtÞà = y”ÝtÞ + by v ÝtÞ + cyÝtÞ = 0,
yÝ0Þ = A,
y v Ý0Þ = B
PÝsÞ ŷÝsÞ ? sA ? B ? bA = 0,
Then
and
ŷÝsÞ = ÝB + bAÞ
1 +A s .
PÝsÞ
PÝsÞ
yÝtÞ = ÝB + bAÞ YÝtÞ + A Y v ÝtÞ
It follows that
where we used property 3 again to show
L ?1
s
PÝsÞ
= Y v ÝtÞ ? YÝ0Þ = Y v ÝtÞ.
Note that
yÝ0Þ = ÝB + bAÞ YÝ0Þ + A Y v Ý0Þ = A
and
y v Ý0Þ = ÝB + bAÞ Y v Ý0Þ + A Y ”Ý0Þ
= ÝB + bAÞ + AÝ?bY v Ý0Þ ? cYÝ0ÞÞ
= B + bA ? Ab = B.
For example,
y”ÝtÞ + I 2 yÝtÞ = 0,
leads to
y v Ý0Þ = B
PÝsÞ = s 2 + I 2 , and
YÝtÞ = L ?1
so
yÝ0Þ = A,
1
PÝsÞ
= 1 L ?1 2 I 2
I
s +I
= 1 sin It
I
yÝtÞ = B YÝtÞ + A Y v ÝtÞ = B sin It + A cos It.
I
Additionally, the solution of
LßyÝtÞà = y”ÝtÞ + by v ÝtÞ + cyÝtÞ = fÝtÞ,
yÝ0Þ = A,
y v Ý0Þ = B
can be expressed entirely in terms of the impulse response by writing
t
yÝtÞ = ÝB + bAÞ YÝtÞ + A Y v ÝtÞ + X YÝt ? bÞ fÝbÞ db.
0
Periodic Forcing Functions
Suppose fÝtÞ is a periodic function of period T; i.e.,
fÝtÞ = fÝt ? TÞ
for all t > T.
3
Here are two examples. The first, the square wave is piecewise continuous, and is given by
SqÝtÞ =
1
if 2k < t < 2k + 1
k = 0, 1, ...
?1 if 2k + 1 < t < 2k + 2
Periodic Square wave
The second function, the sawtooth wave is continuous but only piecewise differentiable and
is given by
WÝtÞ =
1?t
if
0<t<2
t?3
if
2<t<4
_
2k ? 1 ? t
k = 1, 2, ...
_
if
2k ? 2 < t < 2k
t ? Ý2k + 1Þ if
2k < t < 2k + 2
Periodic Sawtooth Wave
Now, for f(t) periodic of period T, define
f F ÝtÞ = fÝtÞ ? fÝt ? TÞ HÝt ? TÞ =
fÝtÞ if 0 < t < T
0
if t > T
4
That is, f F ÝtÞ is ju;st one fundamental period of fÝtÞ. Then by the result for the transform of a
shift, we have
!f F ÝsÞ = !fÝsÞ ? !fÝsÞ e ?Ts = !fÝsÞÝ1 ? e ?Ts Þ
!fÝsÞ =
and hence
!f F ÝsÞ
Ý1 ? e ?Ts Þ
where
!f F ÝsÞ = X T fÝtÞ e ?st dt.
0
For example,
2
LßSq F ÝtÞà= ŜqÝsÞ = X S qF ÝtÞ e ?st dt
0
1
2
= X e ?st dt ? X e ?st dt = 1s ße ?2s ? 2e ?s + 1à.
0
1
This is the transform of the fundamental period of the square wave.
Now we recall a result from high school algebra, namely for |r| < 1, the sum of a
geometric series is given by,
K
1 + r + r2 + ` = > rn =
n=0
1 .
1?r
K
1
= > e ?Tsn = 1 + e ?Ts + e ?2Ts + `
Ý1 ? e ?Ts Þ
n=0
Then
Now consider the problem
LßyÝtÞà = fÝtÞ,
yÝ0Þ = y v Ý0Þ = 0
where fÝtÞ is periodic of period T. Then
ŷÝsÞ =
1 !f ÝsÞ ß1 + e ?Ts + e ?2Ts + `à,
PÝsÞ F
so, if we let y F ÝtÞ = L ?1
implies,
1 !f F ÝsÞ , then the shifting property of the Laplace transform
PÝsÞ
yÝtÞ = y F ÝtÞ + y F Ýt ? TÞ HÝt ? TÞ + y F Ýt ? 2TÞ HÝt ? 2TÞ + `
Example:
y vv ÝtÞ + 16 yÝtÞ = WÝtÞ,
yÝ0Þ = 0, y v Ý0Þ = 0,
5
where WÝtÞ denotes the period 4 sawtooth wave defined earlier. Then, letting
T
2
4
Ŵ F ÝsÞ = X WÝtÞ e ?st dt = X Ý1 ? tÞ e ?st dt + X Ýt ? 3Þ e ?st dt
0
0
2
= 2e
?2s
s + e ?2s + s ? 1 ? e ?2s e ?2s s + e ?2s + s ? 1
s2
s2
= 2 1s ? 12
s
we have
and
+ e ?2s 1s + 32
s
? e ?4s 1s + 12
s
.
Ýs 2 + 16ÞŷÝsÞ = Ŵ F ÝsÞß1 + e ?4s + e ?8s + `à
Ŵ F ÝsÞ
ß1 + e ?4s + e ?8s + `à.
s 2 + 16
ŷÝsÞ =
Then
yÝtÞ = y F ÝtÞ + y F Ýt ? 4ÞHÝt ? 4Þ + y F Ýt ? 8ÞHÝt ? 8Þ + `
where
y F ÝtÞ = L ?1
=
1
8
Ŵ F ÝsÞ
s 2 + 16
1 ? t ? cosÝ4tÞ + 14 sinÝ4tÞ
1
+ 16
HÝt ? 2Þ 1 + 3Ýt ? 2Þ ? cos 4Ýt ? 2Þ ? 34 sin 4Ýt ? 2Þ
1
? 16
HÝt ? 4Þ 1 ? Ýt ? 4Þ ? cos 4Ýt ? 4Þ + 14 sin 4Ýt ? 4Þ
6