Chapter 5
Laplace Transform Method
In this chapter, we introduce one of the crucial tools in engineering mathematics called
Laplace transform. This tool enables us to solve differential equations with discontin-
uous terms in an efficient and straightforward way. Besides solving ordinary differential
equations, with the aid of the Laplace tr a nsform, we are able to define the concept of the
transfer function of complex systems in engineering and the idea of the fundamental solu-
tion.
5.1 Defini tion of the transformation
The unilateral Laplace or L-transform of a function f (t) is defined by the following inte-
gral
L(f)(s) =
Z
0
1
e
¡st
f(t) dt; (5.1)
as long as the integral exists for some values of s. Note that (5.1) transforms a function of
t (that we can interpret as time) to a function of s (can be c o nsidered just as a para-
meter). In this chapter, we use notations f
^
(s), L(f) for the transform. It is called unilat-
eral because f(t) for t < 0 (the history of f ) does not affect the transformation.
Definition 5.1. A function f(t) is called L-admissible if there is an interval of s for
which the integral ( 5.1) converges.
Remark 5.1. The class of L-admissible functions is wide, however, the sub-exponential
functi ons are well-known to be
L
-adm is sible (see the problem set). A fun ction f (t) is
called sub-exponential if there a re a > 0; b > 0 such that jf(t)j< ae
bt
for all t 2[0; 1).
At the first glance, formula (5.1) looks peculiar, nevertheless, the transform possesses
nice properties that makes it desirable in engin eering mathematics. For example, it trans-
forms the derivative operation to an algebraic multiplication, that is, if f
0
(t) is the deriva-
tive of a function f (t), then
L(f
0
) = s L(f ) ¡ f(0):
The above property finds its importance to solve ordinary differential equations.
1
5.1.1 L-Transform of basic functions
1. The Heaviside or unit step function u(t) is defined by the following two valued
function
u(t) =
0 t < 0
1 t > 0
: (5.2)
In some texts, the value u(0) is considered
1
2
, or even 1, but u(0) does not affect its
transformation. We have
u^(s) =
Z
0
1
e
¡st
u(t) dt =
Z
0
1
e
¡st
dt =
1
s
h
lim
t!1
e
¡st
¡1
i
: (5.3)
Clearly, the above integral converges to
1
s
if s > 0. Note that L(1) =
1
s
since two
function u(t) and 1 are equal for t ≥0.
2. Consider the unit pulse
p(t) =
1 0 < t < 1
0 otherwise
: (5.4)
Its L-transform is
p^(s) =
Z
0
1
e
¡st
dt =
1 ¡e
¡s
s
: (5.5)
Note that the value of p(t) at the end points does not affect the transformation.
3. The transform of the ramp function r(t) = t is
L(t) =
Z
0
1
te
¡st
dt = ¡lim
t!1
t
s
e
¡st
+
1
s
Z
0
1
e
¡st
dt: (5.6)
If s > 0, we have
lim
t!1
t
s
e
¡st
= 0;
and thus L(t ) =
1
s
2
.
4. The transform of the exponential function f (t) = e
at
is
L(e
at
) =
Z
0
1
e
¡st
e
at
dt =
Z
0
1
e
¡(s¡a)t
dt: (5.7)
Clearly the integral converges for s > a and then L(e
¡at
) =
1
s ¡a
.
5. The L-transform of the function f(t) = sin(!t) is simply derived by the aid of the
Euler formula
sin(!t) =
e
i!t
¡e
¡i!t
2i
:
A simple algebraic simplification g ives
L(sin(!t)) =
1
2i
fL(e
i!t
) ¡L(e
¡i!t
)g=
!
s
2
+ !
2
: (5.8)
A similar argument applies for the transform of f (t) = cos(!t)
L(cos(!t)) =
s
s
2
+ !
2
:
2 Laplace Transform Method
6. Remember the formula sinh(!t) =
e
!t
¡e
¡!t
2
. For s > j!j, we have
L(sinh(!t)) =
1
2
fL(e
!t
) ¡L(e
¡!t
)g=
!
s
2
¡!
2
: (5.9)
A similar argument applies for the transform of cosh(!t)
L(cosh(!t)) =
s
s
2
¡!
2
: (5.10)
5.1.2 The inverse of Laplace transform
We saw that L(u(t)) =
1
s
for s > 0, and it implies that functions u(t) in t-domain has the
representation
1
s
in s-domain through the Laplace transform; see the figure (5.1)
1 2 3
t
u
(
t
)
L-Transfo rm
1 2 3
s
4
8
12
ˆ
u
(
s
)
Figure 5.1.
An inverse transformation from s-domain to t-domain is also can be defined through
the following relation
L(f(t)) = f
^
(s) , f(t) = L
¡1
(f
^
(s)):
However, sinc e the transform is un ilateral, we just write
L
¡1
(f
^
(s)) = f (t) for t > 0.
Accordingly, since L(t) =
1
s
2
, we write L
¡1
¡
1
s
2
= t; t > 0. There is still a minor problem
here. Let f
~
(t) be the function
f
~
(t) =
t t =/ 1
0 t = 1
:
It is simply seen that L(t) = L(f
~
) =
1
s
2
, and therefore it is not clear that L
¡1
¡
1
s
2
how
should be defined. In fact the transformation L is not one-to-one. Here we make the fol-
lowing convention.
Convention . The function L
¡1
(f
^
(s)) refers to the most possibly continuous function f (t)
for t > 0 such that L(f (t)) = f
^
(s).
Accordi ng to the above conventi on, we define
L
¡1
¡
1
s
= 1,
L
¡1
¡
1
s ¡ a
= e
at
,
L
¡1
¡
s
s
2
+ !
2
= cos(t), and L
¡1
¡
!
s
2
+ !
2
= sin(t) all for t > 0.
5.1 Definition of the transformation 3
Proposition 5.1. Assume that f (t) is sub-exponential, then
lim
s!1
f
^
(s) = 0: (5.11)
Proof. We have
lim
s!1
f
^
(s) = lim
s!1
Z
0
1
e
¡st
f(t) dt: (5.12)
Since the function f (t) is sub-exponential, the function je
¡st
f(t)j is bounded. By the
dominant convergence theorem, we can pass the limit inside the integral and write it as
lim
s!1
f
^
(s) =
Z
0
1
lim
s!1
e
¡st
f(t) dt = 0; (5.13)
and this completes the proof.
We have also the following fact.
Proposition 5.2. Assume th at f (t) is sub-exponential, then f
^
(s) is continuous in its
domain of definition.
Proof. Since the domain of f
^
(s) is open, let us consider the fraction
f
^
(s + h) ¡ f
^
(s) =
Z
0
1
e
¡st
f(t)(e
¡ht
¡1) dt:
Now, for h !0, we have
lim
h!0
(f
^
(s + h) ¡f
^
(s)) = lim
h!0
Z
0
1
e
¡st
f(t)(e
¡h t
¡1) dt:
Now, if jf(t)j < ae
bt
for some a; b > 0 holds, then the condition of dominant convergence
theorem (see the appendix to this book) holds for the above integral since
je
¡st
f(t)(e
¡ht
¡1)j≤ae
¡(s¡b)t
je
¡ht
¡1j≤ae
¡(s¡b+h)t
< a
if s is sufficiently large. Therefore
lim
h!0
Z
0
1
e
¡st
f(t)(e
¡ht
¡1) dt =
Z
0
1
e
¡st
f(t)lim
h!0
(e
¡h t
¡1) dt = 0 ;
and this completes the proof.
Problems
Problem 5.1. Assume that the function f (t) is sub-exponential, that is, there are constants a; b such
that jf(t)j< ae
bt
for all t 2[0; 1). Show that f(t) is L-admissible.
Problem 5.2. Assume that f(t) is a sub-exponential function and f
0
(t) is L-admissible.
a) Use the definition and show
L(f
0
) =
Z
0
1
e
¡st
f
0
(t) dt + s
Z
0
1
e
¡st
f(t) dt:
4 Laplace Transform Method
b) Show the relation
lim
t!1
e
¡st
f(t) = 0;
and conclude
L(f
0
) = s L(f ) ¡ f (0):
Problem 5.3. Use the definition and find the L-transform of the following functions
i.
f(t) =
1 0 < t < 1
¡1 1 < t < 2
:
ii.
f(t) =
(
e
t
¡1 0 < t < 1
0 t > 1
:
iii. f(t) = e
t
¡e
¡2t
.
Problem 5.4. Use the definition and calculate the transform of the following functions. For each
function determine the domain of s for which the transform exists
a)
f(t) =
(
e
2t
0 ≤t < 1
t t > 1
b) f(t) = cos
2
(t)
c) f (t) = sinh
2
(t)
d)
f(t) =
1 1 < t < 2
¡1 2 < t < 3
Problem 5.5. If f
^
(s) is given we can recover the function f (t) such that L(f (t)) = f
^
(s). For the fol-
lowing functions, find f (t) such that L(f ) = f
^
.
i. f
^
(s) =
3
s
2
+ 1
ii. f
^
(s) =
2 ¡ 3s
s
2
+ 4
iii. f
^
(s) =
3
s ¡ 2
iv. f
^
(s) =
1
(s + 1)(s + 2)
Problem 5.6. Is there any L-transform of the function f(t) = e
t
2
? Is it possible the function f
^
(s) = 1
to be the transform of a sub-exponential function?
Problem 5.7. Determine which one of the following functions are sub-exponential in [0; 1)
i. f (x) = t sin(t)
ii. f (x) = t sin(1/t)
iii. f (x) = 1 + e
t
+ e
2t
+ ···+ e
nt
.
iv. f (x) = 1 + e
t
+ e
2t
+ ···
v. f (x) = u(t) + u (t ¡1) + u(t ¡2) + ···
Problem 5.8. Show L( t
p
) =
π
p
2
s
¡3/2
. Hint: Remember the integral
Z
0
1
e
¡u
2
du =
π
p
2
:
5.1 Definition of the transformation 5
5.2 Properties of L aplace transform
Here we study only important properties of L-transform that we need for solving initial
value problems.
Linearity.
L-transform is linear in the sense that for any pair of admissible functions f(t); g(t) and
any arbitrary constants c
1
; c
2
, it satisfies the relation
L(c
1
f + c
2
g) = c
1
L(f) + c
2
L(g) (5.14)
The property is directly proved by the definition (5.1):
L(c
1
f + c
2
g) =
Z
0
1
e
¡st
(c
1
f(t) + c
2
g(t))dt = c
1
Z
0
1
e
¡st
f(t) dt +
+c
2
Z
0
1
e
¡st
g(t) dt = c
1
L(f) + c
2
L(g):
The linearity of L-transform implies the linearity of L
¡1
as well. In fact, by the relation
c
1
f + c
2
g = L
¡1
L(c
1
f + c
2
g) = L
¡1
(c
1
f
^
+ c
2
g^);
we obtain
L
¡1
(c
1
f
^
(s) + c
2
g^(s)) = c
1
L
¡1
(f
^
(s)) + c
2
L
¡1
(g^(s)):
Example 5.1. Let us find L
¡1
1
s(s + 1)
. According to the relation
1
s(s + 1)
=
1
s
¡
1
s + 1
;
we can write
L
¡1
1
s(s + 1)
= L
¡1
1
s
¡L
¡1
1
s + 1
= (1 ¡e
¡t
); t > 0:
Multiplication by t.
If f (t) is an sub-exponential functions then
d
ds
f
^
(s) =
Z
0
1
d
ds
e
¡st
f(t) dt = ¡
Z
0
1
e
¡st
tf (t) dt (5.15)
In other word we can write
d
ds
f
^
(s) = ¡L(tf (t)); (5.16)
or equivalently
L(tf (t)) = ¡
d
ds
f
^
(s): (5.17)
It is left as an exercise to the reader to verify the following formula by the mathematical
induction
L(t
n
f(t)) = (¡1)
n
d
n
ds
n
f
^
(s): (5.18)
6 Laplace Transform Method
Example 5.2. We calculate the inverse transform o f the fo llowing function
f
^
(s) = ln
s ¡1
s
: (5.19)
Taking derivative of f
^
(s) gives
df
^
ds
(s) =
1
s(s ¡1)
=
1
s ¡1
¡
1
s
: (5.20)
According to the linearity of
L
¡1
, we have
L
¡1
(
df
^
ds
) = e
t
¡ 1. Now use the property
L
¡1
(
df
^
ds
) = ¡tf(t) and write
f(t) = L
¡1
(f
^
(s)) =
1 ¡e
t
t
; t > 0: (5.21)
Note that f(t) is a sub-exponential function.
Division by t.
Let f (t) be a sub-exponential function that satisfies the relation
lim
t#0
f(t)
t
<1: (5.22)
Consider the function g(t) =
f(t)
t
. By (5.16), we have
d
ds
g^( s) = ¡L (tg(t)) = ¡L(f (t)) = ¡f
^
(s): (5.23)
Integrating both sides of the above formula in [s; 1) gives
¡g^(s) + lim
s!1
g^(s) = ¡
Z
s
1
f
^
(v) dv: (5.24)
By (5.11) we have
lim
s!1
g^(s) = 0;
and thus
L
f(t)
t
=
Z
s
1
f
^
(v) dv: (5.25)
Example 5.3. Some definite integrals can be calculated by the aid of the above property.
For example, let us find the fo llowing integral
I =
Z
0
1
sin(t)
t
dt: (5.26)
We can write
Z
0
1
sin(t)
t
dt =
Z
0
1
lim
s!0
e
¡st
sin(t)
t
dt: (5.27)
5.2 Propert ies of L aplace transfo rm 7
Since
sin(t)
t
is sub-exponential, we can take the limit out of the integral and write
Z
0
1
sin(t)
t
dt = lim
s!0
Z
0
1
e
¡st
sin(t)
t
dt = lim
s!0
L
sin(t)
t
:
But, we have
L
sin(t)
t
=
Z
s
1
dv
v
2
+ 1
=
π
2
¡tan
¡1
(s): (5.28)
Finally for s !0, we obtain
Z
0
1
sin(t)
t
dt =
π
2
¡lim
s!0
tan
¡1
(s) =
π
2
: (5.29)
Shift in t-domain.
Suppose that f (t) is an admissible function and g(t) = f(t ¡t
0
)u(t ¡t
0
) is the time shift of
f by t
0
≥0; see the figure (5.2).
t
0
f (t ¡a)u(t ¡a)
f (t)
Figure 5.2. f (t) and its shifted graph f (t ¡t
0
).
We have
L(g) =
Z
0
1
e
¡st
f(t ¡t
0
)u(t ¡t
0
) dt =
Z
t
0
1
e
¡st
f(t ¡t
0
) dt: (5.30)
For τ = t ¡t
0
, we obtain
Z
t
0
1
e
¡st
f(t ¡t
0
) dt =
Z
0
1
e
¡s(τ +t
0
)
f(τ ) dτ = e
¡t
0
s
Z
0
1
e
¡sτ
f(τ ) dτ ; (5.31)
and then
L(f(t ¡t
0
) u(t ¡t
0
)) = e
¡t
0
s
f
^
(s): (5.32)
Example 5.4. The above property is used to calculate the inverse transform of some s-
functions of the form g^(s) = e
¡t
0
s
f
^
(s). For example,
L
¡1
e
¡2s
s
s
2
¡1
= L
¡1
s
s
2
¡1
j
t!t¡2
u(t ¡2): (5.33)
But,
L
¡1
s
s
2
¡1
= cosh(t): (5.34)
and hence
L
¡1
e
¡2s
s
s
2
¡1
= cosh(t ¡2) u(t ¡2): (5.35)
8 Laplace Transform Method
Example 5.5. Let g(t) be the following function:
g(t) =
X
k=0
1
f(t ¡kt
0
)u(t ¡kt
0
); (5.36)
and assume that f (t) is sub-exponential. We have
jg(t)j≤
X
k=0
1
jf (t ¡kt
0
)ju(t ¡kt
0
) ≤a
X
k=0
1
e
b(t¡kt
0
)
u(t ¡kt
0
) ≤
≤ae
bt
X
k=0
1
e
¡bkt
0
u(t ¡kt
0
):
Since
X
k=0
1
e
¡bkt
0
u(t ¡kt
0
) ≤
1
1 ¡e
¡bt
0
; (5.37)
we conclude
jg(t)j≤
a
1 ¡e
¡bt
0
e
bt
: (5.38)
The above argument guarantees that g(t) is a sub-exponential function. Let us define
g
n
(t) =
X
k=0
n
f(t ¡kt
0
)u(t ¡kt
0
); (5.39)
and then
g^(s) = lim
n!1
g^
n
(s) = lim
n!1
Z
0
1
e
¡st
g
n
(t) dt: (5.40)
But, we have
Z
0
1
e
¡st
g
n
(t) dt =
X
k=0
n
L(f(t ¡kt
0
) u(t ¡kt
0
)) =
X
k=0
n
e
¡kt
0
s
f
^
(s); (5.41)
and finally
g^(s) = f
^
(s) lim
n!1
X
k=0
n
e
¡kt
0
s
=
1
1 ¡e
¡t
0
s
f
^
(s); (5.42)
or equivalently
L
¡1
1
1 ¡e
¡t
0
s
f
^
(s)
=
X
k=0
1
f(t ¡kt
0
) u(t ¡kt
0
): (5.43)
In particular, if f(t) is a T -periodic function, that is, f (t + T ) = f (t), then
L(f(t)) =
1
1 ¡e
¡sT
Z
0
T
e
¡st
f(t) dt:
Shift in s-domain.
There is a beautiful duality between the shift in t dom a in and the shift in s domain some-
times called phase shift. For g(t) = e
at
f(t), we have
L(g) =
Z
0
1
e
¡st
e
at
f(t) dt =
Z
0
1
e
¡(s¡a)t
f(t) dt = f
^
(s ¡a); (5.44)
5.2 Propert ies of L aplace transfo rm 9
or equivalently
L
¡1
(f
^
(s ¡a)) = e
at
L
¡1
(f
^
(s)): (5.45)
Example 5.6. Let us use the above property to find
L
¡1
e
¡2s
s ¡1
(s ¡1)
2
¡4
: (5.46)
We have
L
¡1
e
¡2s
s ¡1
(s ¡1)
2
¡4
= L
¡1
s ¡1
(s ¡1)
2
¡4
j
t!t¡2
u(t ¡2): (5.47)
Since
L
¡1
s ¡1
(s ¡1)
2
¡4
= e
t
cosh(2t); (5.48)
we obtain
L
¡1
e
¡2s
s ¡1
(s ¡1)
2
¡4
= e
(t¡2)
cosh2(t ¡2)u(t ¡2): (5.49)
Derivative.
If f (t) is a sub-exponential differentiable function, then
Z
0
1
e
¡st
f
0
(t) dt = lim
t!1
e
¡st
f(t) ¡ lim
t!0
+
f(t) + s
Z
0
1
e
¡st
f(t) dt: (5.50)
Since f(t) is sub-exponential, we have lim
t!1
e
¡st
f(t) = 0. If f(t) is continuous at t = 0,
then lim
t!0
+
f(t) = f (0), and then
L(f
0
) = s f
^
(s) ¡f (0): (5.51)
Example 5.7. Consider the following problem
y
0
+ y = u(t ¡1)
y(0) = y
0
:
Taking the L-transform to the both sides of the equati on gives
s y^(s) ¡ y
0
+ y^(s) =
e
¡s
s
;
or equivalently
(s + 1)y^(s) =
e
¡s
s
+ y
0
:
This yields y^(s) as
y^(s) =
1
s + 1
y
0
+
e
¡s
s(s + 1)
:
To obtain y(t), we should take inverse transform of y^(s). Note that
L
¡1
e
¡s
s(s + 1)
= (1 ¡e
¡(t¡1)
) u(t ¡1);
and
L
¡1
1
s + 1
y
0
= y
0
e
¡t
; t > 0:
10 Laplace Transform Method
Here we should take a notice. The term L
¡1
¡
1
s + 1
y
0
is associated to the initial condition
of the equation at time t = 0 and L
¡1
e
¡s
s(s + 1)
is assoc iated to the external term that is
exercised for t > 1. For this the homogeneous solution y
0
e
¡t
is not only defined for t > 0
but also for all t 2(¡1; 1). Therefore, we can write the s o luti o n to the equation a s
y(t) = y
0
e
¡t
+ (1 ¡e
¡(t¡1)
) u(t ¡1); (5.52)
for all t 2 (¡1; 1). Note also that the solution is continuous everywhere and differen-
tiable, but it is not differentiabl e at t = 1 due t the term u(t ¡1).
The following formula is simply proved by the mathematical induction:
L(f
(n)
(t)) = s
n
f
^
(s) ¡s
n¡1
f(0) ¡···¡ f
(n¡1)
(0): (5.53)
Integration.
Let us d erive the L-transform of the integral function
g(t) =
Z
0
t
f(τ) dτ ; t > 0: (5.54)
Since g
0
(t) = f (t) and lim
t!0
+
g(t) = 0, we obtain
f
^
(s) = L(g
0
) = s g^(s) (5. 55)
and thus
L
Z
0
t
f(τ ) dτ
=
1
s
f
^
(s); (5.56)
or equivalently
Z
0
t
f(τ ) dτ = L
¡1
1
s
f
^
(s)
; (5.57)
for t > 0.
Example 5.8. By the aid of the integral property, we are able to calculate the inverse of
some s-functions of the form g^(s) =
1
s
f
^
(s). Consider the function
f
^
(s) =
1
s(s
2
+ 1)
: (5.58)
We have
L
¡1
(f
^
(s)) = L
¡1
1
s
1
(s
2
+ 1)
=
Z
0
t
L
¡1
1
s
2
+ 1
=
Z
0
t
sin (v) dv = (1 ¡cos(t))u(t): (5.59)
Example 5.9. Let us calculate L
¡1
f
^
(s)
s + a
. According to the phase shift property, we
have
L
e
¡at
Z
0
t
f(τ )
= L
Z
0
t
f(τ )
j
s!s+a
=
f
^
(s)
s
j
s!s+a
=
f
^
(s + a)
s + a
:
5.2 Propert ies of L aplace transfo rm 11
Since we need f
^
(s) in the top, we write
L
e
¡at
Z
0
t
e
aτ
f(τ )
= L
Z
0
t
e
aτ
f(τ )
j
s!s+a
:
But
L
Z
0
t
e
aτ
f(τ )
=
1
s
L(e
at
f(t)) =
f
^
(s ¡a)
s
;
and therefore
L
e
¡at
Z
0
t
e
aτ
f(τ )
=
f
^
(s ¡a)
s
j
s!s+a
=
f
^
(s)
s + a
;
or equivalently
L
¡1
f
^
(s)
s + a
!
=
e
¡at
Z
0
t
e
aτ
f(τ ) dτ
u(t)
5.2.1 Dirac delta function and its Laplace tr ansform
The Dirac delta functio n δ, i ntroduced by the E nglish physicist Paul Dirac (1902-
1984) is a specific type of singularity used in advanced physics. There are different ways
to define the Dirac delta function. One way that we adopt here is to define it as the limit
of a sequence of functions called the delta-sequence functions. Consider the following
sequence
δ
n
(t) =
(
n
2
¡
1
n
< t <
1
n
0 otherwise
; (5.60)
for n = 1; 2; 3; ···. The graph is shown below.
δ
n
(t)
n
2
¡1
n
1
n
t
Figure 5.3. An example of δ
n
-sequence functions.
Assume that f (t) is a continu o us function. We can write
min
t2
h
¡1
n
;
1
n
i
f(t) ≤
Z
¡1
1
f(t)δ
n
(t) dt ≤ max
t2
h
¡1
n
;
1
n
i
f(t): (5.61)
12 Laplace Transform Method
Since
lim
n!1
min
t2
h
¡1
n
;
1
n
i
f(t) = lim
n!1
max
t2
h
¡1
n
;
1
n
i
f(t) = f(0);
then by the squeeze theorem, we can write
lim
n!1
Z
¡1
1
f(t)δ
n
(t) dt = f(0):
As it is seen, δ
n
(t) converges to a spike at t = 0 with infinite height and zero width with
the property
Z
¡1
1
δ
n
(t) dt = 0; 8n ≥1:
This limiting function w hich is a singularity at t = 0 is called the Dirac delta function.
The above defined sequence is not unique and there are other sequences with the same
property. For example, consider the sequence of function G
n
(t) =
n
π
p
e
¡n
2
t
2
for n = 1; 2; ···
shown below. Note that all G
n
(x) are derived from the Gaussian function G (t) =
1
π
p
e
¡t
2
by the relation G
n
(t) = nG(nt).
−
2 2
1
2
3
4
n
= 1
n
= 8
Figure 5.4.
When n approaches infinity, the function approaches to a spike with the width 0 a nd
the height infinity at t = 0. Let f(t) be a continuous and bounded function. We have
Z
¡1
1
f(t) G
n
(t) dt =
Z
¡1
1
f(t)
n
π
p
e
¡n
2
t
2
dt;
and by taking τ = nt, we obtain
Z
¡1
1
f(t)G
n
(t) dt =
1
π
p
Z
¡1
1
f
τ
n
e
¡τ
2
dτ :
5.2 Propert ies of L aplace transfo rm 13
For n !1, we can write
lim
n!1
Z
¡1
1
f(t)G
n
(t) dt =
1
π
p
Z
¡1
1
lim
n!1
f
τ
n
e
¡τ
2
dτ =
1
π
p
Z
¡1
1
f(0)e
¡τ
2
dτ = f (0):
Note that in the above calculations, we took the limit inside the integral which is j ustifi-
able by the assumption on f . Note also the relation
Z
¡1
1
G
n
(t) dt = 1; 8n ≥1:
Definition 5.2. A sequence of function (δ
n
(x)) is called a δ-sequence function if the fol-
lowing relation holds for any bounded and continuous function f
lim
n!1
Z
¡1
1
f(t) δ
n
(t)dt = f (0):
The limit function of a δ-sequence function is called the Dirac delta function and is
denoted by δ(t).
An alternative way for defining δ is by the following equality
Z
¡a
a
f(t) δ(t)dt = f (0); (5.62)
for any a > 0 and any continuous function f. The formula (5.62) gives immediately the
following relation
Z
t¡a
t+a
f(τ )δ(t ¡τ )dτ = f (t); (5.63)
or equivalently
Z
¡1
1
f(τ ) δ(t ¡τ ) dτ = f(t):
That δ(t) i s not a function in the usual sense is seen from the following observation. If
δ was a classical function then it would had to satisfy the following condition
lim
a!0
Z
¡a
a
f(t) δ(t) d t = 0;
which contradicts the relation (5.62). In engineering literature, δ(t) is called an impulse
and is usually denoted by a unit arrow; see Fig.5.5. For this rea son, δ(t) is som etime
defined (and it is not technically correct) by the following relation
δ(t) =
1 t = 0
0 otherwise
: (5.64)
14 Laplace Transform Method
δ(t)
t
Figure 5.5. The graph of δ function.
For any τ > 0, the relation (5.63) implies
L(δ(t ¡a)) =
Z
0
1
e
¡st
δ(t ¡a) dt = e
¡as
:
We accept this definition for a = 0 as well and write
L(δ) = 1; and L
¡1
(1) = δ(t): (5.65)
By this de finition, the following relation is immediately followe d
L(f(t)δ(t ¡a)) = f(a)e
¡as
: (5.66)
Problems
Problem 5.9. Find the transform of the following functions
i. f (t) = e
t
u(t ¡1)
ii. f (t) = te
¡t
u(t ¡1)
iii. f (t) = u(t) ¡e
t
u(t¡1)
iv. f (t) = e
sin(πt)
δ(t ¡2)
v. f (t) = (1 ¡u(t ¡2)) u(t ¡1).
Problem 5.10. Compare the transformation of the function
f(t) =
1 0 < t < 1
0 otherwise
;
and the function
g(t) =
t 0 < t < 1
0 otherwise
;
and conclude g^(s) = ¡
d
ds
f
^
(s).
Problem 5.11. Use the definition and derive the Laplace transform of the function
f(t) =
1 0 < t < 1
0 otherwise
; g(t) =
1 1 < t < 2
0 otherwise
;
and conclude g^(s) = e
¡s
f
^
(s).
Problem 5.12. Use the Laplace transform properties to calculate
i. L(te
at
)
ii. L(t sin(!t))
5.2 Propert ies of L aplace transfo rm 15
iii. L(t cos(!t))
iv. L(t sinh(!t))
v. L(t cosh(!t))
Problem 5.13. Find the transformation of the following functions
i. L(t
3/2
).
ii. L(t
¡1/2
)
iii. L(e
¡t
t
¡1/2
)
Problem 5.14. Use mathematical induction to prove the identities
L(t
n
f(t)) = (¡1)
n
d
n
ds
n
f
^
(s);
and conclude
i. L(t
n
) =
n!
s
n+1
,
ii. L(t
n
e
¡at
) =
n!
(s + a)
n+1
.
Problem 5.15. Use partial fraction to find the inverse transformation o f the following functions:
i. f
^
(s) =
s
s
2
¡3s + 2
ii. f
^
(s) =
1
s(s
2
+ 4)
iii. f
^
(s) =
s + 2
s
3
¡s
2
+ s ¡ 1
.
Problem 5.16. Use the properties of Laplace transform to obtain the following integrals:
i.
R
0
1
1 ¡e
¡t
t
e
¡t
dt,
ii.
R
0
1
1 ¡cos(t)
t
e
¡t
dt,
iii.
R
0
1
sin(t)
t
e
¡t
dt
iv.
R
¡1
1
e
¡jtj
1 ¡ cos(t)
ln(2) jtj
dt
Problem 5.17. Find the inverse transformation of the following functions:
i. f
^
(s) =
s
s
2
¡4s + 5
ii. f
^
(s) =
s + 3
s
2
+ 2s + 5
iii. f
^
(s) =
s
(s ¡ 1)
2
+ 3
iv. f
^
(s) = e
¡s
log
¡
s + 1
s
v. f
^
(s) = e
¡2s
s
(s
2
+ 3)
2
,
vi. f
^
(s) = e
¡s
s
s
2
+ 3s + 2
.
Problem 5.18. Find the inverse transformation of following functions:
i. f
^
(s) =
1
1 ¡ e
¡s
1
s
2
¡1
ii. f
^
(s) =
1 + e
¡s
1 ¡ e
¡2s
1
s
2
+ 1
.
Problem 5.19. If we define
g(t) =
X
k=0
1
(¡1)
k
f(t ¡ka)u(t ¡ka)
16 Laplace Transform Method
show that
g^(s) =
1
1 + e
¡as
f
^
(s):
Using the above argument calculate the following inverse transformation
L
¡1
1
1 + e
¡2s
3!
(s ¡3)
4
:
Problem 5.20. Use mathematical induction to prove
L(f
(n)
(t)) = s
n
f
^
(s) ¡s
n¡1
f(0) ¡···¡ f
(n¡1)
(0):
Problem 5.21. Use Laplace tra nsform of tsin(!t) and tcos(!t) to find the inverse transformation of
the following functions:
i. f
^
(s) =
2
(s
2
+ 1)
2
,
ii. f
^
(s) =
2
((s ¡ 2)
2
+ 1)
2
Problem 5.22. Find the transformation of the following periodic functions:
i. f (t) = t; 0 < t < 1; f (t + 1) = f(t),
ii. f (t) =
1 0 < t < 1
¡1 1 < t < 2
, f(t + 2) = f (t).
Problem 5.23. Show the following relation
L
¡1
f
^
(s)
(s + a)
2
!
= e
¡at
Z
0
t
Z
0
τ
e
as
f(s) ds
dτ :
Use integration by part formula and rewrite the right hand side of the above formula as
e
¡at
Z
0
t
Z
0
τ
e
as
f(s) ds
dτ =
Z
0
t
(t ¡τ )e
¡a(t¡τ )
f(τ) dτ :
Problem 5.24. Even though the derivative of the unit step function u(t ¡ a) is not defined at t = a
(it is n ot even continuous at th is point), show that δ(t ¡ a) and u(t ¡ a) are related in the following
way
L(δ(t ¡a)) = s L(u(t ¡a));
that is δ(t ¡a) can be considered as the generalized derivative of u(t ¡a).
Problem 5.25. Consider the sequence of functions
f
n
(t) =
8
<
:
nt 0 ≤t ≤
1
n
1 t ≥
1
n
:
i. Find f
^
n
(s) and then obtain lim
n!1
f
^
n
(s).
ii. What is the limit function lim
n!1
f
n
(t)?
iii. Compare lim
n!1
f
^
n
(s) with the the L-transform of the limiting function you obtained in part
(ii).
Problem 5.26. In the proposition (5.2) we proved that f
^
(s) is continuous in its domain of definition.
i. For f (t) = e
¡t
sin(t) find
R
0
1
f(t)dt.
5.2 Propert ies of L aplace transfo rm 17
ii. Find f
^
(s) and let lim
s!0
f
^
(s) and explain why it gives the same result in part (i).
iii. Now for f (t) = sin(t) we know f
^
(s) =
1
s
2
+ 1
and lim
s!0
f
^
(s) = 1. Does it imply that
Z
0
1
sin(t) dt =
1? why?
Problem 5.27. It is not always allowed to pas s the limit inside the integral.
i. Let f
n
(t) =
n
1 + n
2
t
2
. Find
R
0
1
f
n
(t).
ii. Compare lim
n!1
Z
0
1
f
n
(t)dt and
Z
0
1
lim
n!1
f
n
(t)dt and explain why they are not equal.
Problem 5.28. Let f
n
(t) = sin(nt).
i. Find lim
n!1
f
^
n
(s).
ii. Can we pass the limit in side the integral of L-transform?
5.3 In itial value probl ems and L-tr ans form
In this section, we solve initial value problems by the aid of the Laplace transform. As
we will see, this method offers considerable advantag e over the classical methods for
solving initial value problem, specially when the fo rcing terms are discontinuous. Consider
the following initial value problem
y
00
+ a y
0
+ by = f(t)
y(0) = y
0
; y
0
(0) = y
1
; (5.67)
where a; b are constants. The L-transform of the equation transforms the equation into the
following algebraic one
L(y
00
) + a L(y
0
) + b L(y) = f
^
(s): (5.68)
Using the relations
L(y
00
) = s
2
y^(s) ¡sy
0
¡y
1
; and L(y
0
) = sy^(s) ¡ y
0
;
we obtain
s
2
y^(s) ¡sy
0
¡y
1
+ as y^(s) ¡ay
0
+ b y^(s) = f
^
(s); (5.69)
and thus we reach
(s
2
+ as + b) y^(s) = (s + a)y
0
+ y
1
+ f
^
(s): (5.70)
Note that the coefficient of y^(s) i s the characteristic polynomial of the differential equa-
tion. Therefore,
y^(s) =
f
^
(s)
s
2
+ as + b
+
s + a
s
2
+ as + b
y
0
+
1
s
2
+ as + b
y
1
; (5.71)
and finally
y(t) = y
0
L
¡1
s + a
s
2
+ as + b
+ y
1
L
¡1
1
s
2
+ as + b
+ L
¡1
f
^
(s)
s
2
+ as + b
!
(5.72)
18 Laplace Transform Method
Example 5.10. Consider the following problem
y
00
+ y = u(t ¡2) ¡u( t ¡8)
y(0) = y
0
(0) = 0
: (5.73)
Here a pulse in the period (2; 8) is applied to a harmonic oscillator. By the method of
Laplace transform, the problem reduces to the following algebraic one
(s
2
+ 1)y^(s) =
e
¡2s
¡e
¡8s
s
;
and thus
y(t) = L
¡1
e
¡2s
¡e
¡8s
s(s
2
+ 1)
= L
¡1
e
¡2s
s(s
2
+ 1)
¡L
¡1
e
¡8s
s(s
2
+ 1)
:
By the property
L
¡1
(e
¡as
f
^
(s)) = f (t ¡a) u(t ¡a);
we have
L
¡1
e
¡2s
s(s
2
+ 1)
= L
¡1
1
s(s
2
+ 1)
j
t=t¡2
u(t ¡2):
Since
L
¡1
1
s(s
2
+ 1)
=
Z
0
t
L
¡1
1
s
2
+ 1
=
Z
0
t
sinvdv = 1 ¡cost;
we conclude
y(t) = [1 ¡cos(t ¡2)] u(t ¡2) ¡[1 ¡cos(t ¡8)] u(t ¡8): (5.74)
The figure (5.6) shows the graph o f this solution.
2 8 16
t
1
2
y
(
t
)
Figure 5.6. The graph of the solution y(t).
Observe from the figure that y(t) ≡ 0 for t < 2 th a t confirms our expectation because
the system is at rest for t < 2. The solution y(t) is smooth of order 1 at t = 2 since y
00
(t)
has a finite jump at that time. A similar argument holds at t = 8 where f (t) jumps from 1
to 0. By the classical method of previous chapters, to solve the problem, we had to split
the problem into the three dub-domains t < 2, 2 < t < 8 and t > 8 as follow s. For t < 0 the
problem is
y
00
+ y = 0;
y(0) = y
0
(0) = 0
; (5.75)
5.3 Initial value pro blems and L-transform 19
and thus the unique solution is y(t) = 0. In the interval (2; 8), the problem reads
y
00
+ y = 1;
y(2) = y
0
(2) = 0
; (5.76)
with the solution y(t) = 1 ¡cos(t ¡2). In the interval (8; 1), we have
y
00
+ y = 0; for t > 8
y(8) = 1 ¡cos(6); y
0
(8) = sin(6)
; (5.77)
with the solution y = cos(t ¡ 2) ¡ cos(t ¡ 8). Note the initial conditions in the above
problem.
Example 5.11. Let us solve the following problem
y
00
+ y = u(t ¡1)
y(1) = 1; y
0
(1) = 0
:
Here the initial data is given at t = 1 instead of t = 0 . To solve the problem, we pro cess as
if conditions y(0) = y
0
and y
0
(0) = y
1
are known. Taking L-transform we obtain
y^(s) =
e
¡s
s(s
2
+ 1)
+ y
0
s
s
2
+ 1
+ y
1
1
s
2
+ 1
:
This implies
y(t) = (1 ¡cos(t ¡1))u(t ¡1) + y
0
cos(t) + y
1
sin(t):
Now, we apply the data y(1) and y
0
(1) and get y
0
= cos(1) and y
1
= sin(1) and thus
y(t) = (1 ¡cos(t ¡1))u(t ¡1) + cos(t ¡1) =
cos(t ¡1) t ≤1
1 t ≥1
:
The graph o f the solution for t > 1 is shown in the figure (5.7).
Figure 5.7.
Note that two branches of the solution for t ≤ 1 and for t ≥ 1 connect together
smoothly of first order. This is because y
00
has a finite jump at t = 1 due to the term u(t ¡
1) and therefore y
0
is c o ntinuous at this point. Note that the initial condition at t = 1
determines y(t) for t < 1.
20 Laplace Transform Method
Example 5.12. Let f (t) be the function
f(t) = u(t) ¡2u
t ¡
π
2
+ u(t ¡π);
and let g(t) be the π-periodic extension of f in (0; 1); see th e figure (5.8).
π
2
¡1
1
π
f(t)
2π
t
3π
2
Figure 5.8.
Note that g has the representation
g(t) =
X
k=0
1
f(t ¡kπ) u(t ¡kπ):
Now consider the problem
y
00
+ y = f (t)
y(0) = y
0
(0) = 0
: (5.78)
We have
L(f ) =
X
k=0
1
L(g(t ¡kπ)u(t ¡kπ)) = g^(s)
X
k=0
1
e
¡kπs
=
1
1 ¡e
¡πs
g^(s):
By the L-transform method, we have
y^(s) =
1
1 ¡e
¡πs
g^(s)
s
2
+ 1
;
and thus
y(t) = L
¡1
1
1 ¡e
¡πs
g^(s)
s
2
+ 1
: (5.79)
We use the formula (5.42) to write y(t) as
y(t) =
X
k=0
1
L
¡1
g^(s)
s
2
+ 1
j
t!t¡kπ
u(t ¡kπ): (5.80)
But
g^(s) =
1 ¡2e
¡πs/2
+ e
¡πs
s
;
and therefore
L
¡1
g^(s)
s
2
+ 1
= (1 ¡cos(t)) u (t) ¡2 (1 ¡sin(t)) u
t ¡
π
2
+ (1 + co s(t)) u(t ¡π):
The figure (5.9) shows the solution to the problem.
5.3 Initial value pro blems and L-transform 21
-1
0
1
2
4 8 12 16 20
Figure 5.9. The graph of the solution y(t).
Example 5.13. Consider the initial value problem
(
y
00
+ y = e
sin(log(t/π))
δ(t ¡π)
y(0) = y
0
(0) = 0
: (5.81)
The transform reduces the equation to the following algebraic one
(s
2
+ 1)y^(s) = L(e
sin(log(t/π))
δ(t ¡π)): (5.82)
It is simply seen that
L(e
sin(log(t/π))
δ(t ¡π)) = e
sin(log(π/π))
e
¡πs
= e
¡πs
; (5.83)
and thus y^(s) =
e
¡πs
s
2
+ 1
which gives
y = sin(t ¡π) u(t ¡π): (5.84)
Observe that the solution starts at t = π, the time when the external source is applied.
Problems
Problem 5.29. Use the Laplace transform method to solve the following initial value problems
i.
y
0
+ y = sin(t ¡1) u(t ¡1)
y(0) = 0:
ii.
(
y
00
+ 4y
0
+ 3y = e
t
y(1) = 0; y
0
(1) = 1
iii.
y
00
+ y = sin(t)
y(0) = 0; y
0
(0) = 1
iv.
y
00
+ y = u(t ¡1)
y(0) = 0; y
0
(1) = 1
v.
(
y
00
+ 3y
0
+ 2y = 2u(t ¡1) + e
t
δ(t ¡2)
y(0) = y
0
(0) = 0
22 Laplace Transform Method
vi.
(
y
00
+ y = u(t ¡1) ¡u(t ¡2);
y(0) = y(
π
2
); y
0
(0) = 1:
vii.
y
00
+ y = sin(t) u(t ¡π)
y(0) = y
0
(0) = 0
;
viii.
(
y
000
¡3y
00
+ 3y
0
+ y = e
t
(1 + t)
y(0) = y
0
(0) = y
00
(0) = 0
:
Problem 5.30. Let us solve the prob lem (5.11) as follows. Let τ = t ¡ 1 and defined z(τ ) = y(t).
Then we have
z
00
(τ) + z(τ ) = u(τ )
z(0) = 1; z
0
(0) = 0
:
The L-transform of the problem gives
z^(s) =
1
s(s
2
+ 1)
+
s
s
2
+ 1
=
1
s
:
Therefore z(τ) = u(τ) and y(t) = u(t ¡1) (!) What is wrong here?
Problem 5.31. Solve the f ollowing equations and use a computer software to draw the solutions
i. y
00
+ 2y
0
+ y = f (t); y(0) = y
0
(0) = 0 and f(t) is a period ic function as follows
f(t) = u(t) ¡u(t ¡1), f(t + 2) = f (t)
ii. y
00
+ y = f (t), f (t) = t for 0 < t < 1 and f(t + 1) = f(t)
iii. y
00
+ y = f (t) where f (t) is given in the following figure
1
2 3
t
2
1
3
Problem 5.32. Assume that s(t) is a continuous function such that
Z
¡1
1
s(t) dt = 1: (5.85)
Prove that the sequence s
n
(t) = ns(n t) is a δ-sequence functions for the class of continuous and
bo unded functions f (t). Use this property to conclude that
Z
¡1
1
n cos(t)
π(1 + n
2
t
2
)
dt
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
n!1
1:
Problem 5.33. If L
¡1
(F (s)) = f (t), show that
L
¡1
(e
¡2s
sF (s)) = f
0
(t ¡2)u(t ¡2) + f (0) δ(t ¡2):
Problem 5.34. Solve the f ollowing problem and draw the solution y(t) in the interval [0; 10]
(
y
00
+ y =
P
k=1
1
δ(t ¡k)
y(0) = y
0
(0) = 0
:
5.3 Initial value pro blems and L-transform 23
Problem 5.35. Solve the f ollowing problem and draw the solution y(t) in the interval [0; 10]
(
y
(4)
+ 2y
00
+ y =
P
k=1
1
2
¡k
δ(t ¡k)
y
k
(0) = 0; k = 0; 1; 2; 3
:
5.4 Impulse respo nse and convolu tion
5.4.1 Convolution
Definition 5.3. Assume that f(t) and g(t) are two integrable functions defined on (¡1;
1). The convolution f ∗g is defined by the following relation integral relation
(f ∗ g) (t) =
Z
¡1
1
f(τ ) g(t ¡τ ) dτ : (5.86)
Although it looks odd at first glance, we will see the importance of convolu tion in real
applications. Before that, let us solve a few e xamples.
Example 5.14. If f (t) = δ(t), then f o r arbitrary integrable function g(t), we have
(δ ∗ g) (t) =
Z
¡1
1
δ(τ) g(t ¡τ ) dτ = g(t): (5.87 )
If f (t) = u(t) then for arbitrary integrable function g(t) we have
(u ∗ g)(t) =
Z
¡1
1
u(τ) g(t ¡τ ) dτ =
Z
0
1
g(t ¡τ ) dτ =
Z
¡1
t
g(τ) dτ ; (5.88)
and in particular, if g(t) = 0 for t < 0, then
(u ∗ g)(t) =
Z
0
t
g(τ )dτ:
Assume f (t) = e
¡t
u(t), then
(f ∗f )(t) =
Z
¡1
1
e
¡τ
u(τ ) e
¡(t¡τ)
u(t ¡τ ) dτ =
Z
0
t
e
¡t
dτ = te
¡t
:
Remark 5.2. If f ; g are zero for t < 0 (as we usually assume for the Laplace transform)
the integral reads
(f ∗ g )(t) =
Z
0
t
f(τ) g(t ¡τ ) d τ : (5.89)
Remark 5.3. It is simply verified that the convolution relation is commutative, that is,
f ∗ g = g ∗ f
(f ∗ g)(t) =
Z
¡1
1
f(τ )g(t ¡τ) dτ =
Z
¡1
1
f(t ¡τ ) g(τ ) dτ = (g ∗ f )(t):
24 Laplace Transform Method
The equality is verified simply by the change of variable v = t ¡τ .
Theorem 5.1. For any admissible function f (t) and g(t) we have
L(f ∗ g) = f
^
(s)g^(s): (5.90)
Proof. By de finition, we can write
L(f ∗g) =
Z
0
1
Z
0
t
e
¡st
f(τ ) g(t ¡τ ) dτdt: (5.91)
Remember the follow ing change of order of the integration from calculus
Z
0
b
Z
0
t
h(t; τ ) dτdt =
Z
0
b
Z
τ
b
h(t; τ ) dtdτ : (5.92)
The following figure justifies the above relation
b
τ
τ = t
t
Figure 5.10.
Using the above formula for b = 1, we can write
Z
0
1
Z
0
t
e
¡st
f(τ ) g(t ¡τ ) dτdt =
Z
0
1
Z
τ
1
e
¡st
f(τ ) g(t ¡τ ) dtdτ =
=
Z
0
1
f(τ )
Z
τ
1
e
¡st
g(t ¡τ ) dt
dτ:
By the change of variable t ¡τ = v, we reach
Z
0
1
Z
0
t
e
¡st
f(τ ) g(t ¡τ ) dτdt =
Z
0
1b
e
¡sτ
f(τ ) dτ
Z
0
1
e
¡sv
g(v) dv
= f
^
(s) g^(s);
and this completes the proof.
As it is seen, the Laplace transform, transforms the convolution relation to a simple
algebraic multiplication. As we will see below, this relation offers significant a simplifica-
tion in the calculations of linear systems. But before that, let us solve a few examples.
Example 5.15. Let us start with the following inverse transformation
L
¡1
1
s
f
^
(s)
:
5.4 Impulse response and convolution 25
As we know, it is equal to
L
¡1
1
s
f
^
(s)
=
Z
0
t
f(t) dτ :
By the convolution relation, we can write
L
¡1
1
s
f
^
(s)
= L
¡1
1
s
∗L
¡1
ff
^
(s)g= u(t) ∗ f(t) =
Z
0
t
f(τ )u(t ¡τ ) dτ =
Z
0
t
f(τ )dτ ;
for t > 0. Similarly, we can write
L
¡1
1
s + a
f
^
(s)
= L
¡1
1
s + a
∗L
¡1
ff
^
(s)g= e
¡at
∗f (t) =
Z
0
t
f(τ) e
¡a(t¡τ)
dτ = e
¡at
Z
0
t
f(τ ) e
aτ
dτ :
The above relation is the solution of the familiar initial value problem
y
0
+ ay = f(t )
y(0) = 0
;
which is by the Laplace transform is equivalent to
y^(s) =
f
^
(s)
s + a
:
Example 5.16. Let us solve the following integral equation
y ¡
Z
0
t
e
t¡τ
y(τ) dτ = δ(t ¡π): (5.93)
Since the integral term in the equation is equal to e
t
∗ y, the Laplace transform of the
equation is
y^(s) ¡L(e
t
∗y) = e
¡πs
; (5.94)
and therefore
y^(s) ¡
y^(s)
s ¡1
= e
¡πs
: (5.95 )
The above equation is solved for y^ as
y^(s) =
s ¡1
s ¡2
e
¡πs
; (5.96)
and finally
y(t) = L
¡1
s ¡1
s ¡2
e
¡πs
= δ(t ¡π) + e
2(t¡π)
u(t ¡π): (5.97)
5.4.2 System approach and transfer function
The word system is a prevalent term in the whole of applied sciences. Almost all physical
systems, incl uding man-machine and natural systems, are formulated in te rms of three
fundamental terms: input, output, and transaction.
26 Laplace Transform Method
Transaction
input
output
For example, the movement of a mass connected to a spring of stiffness k under the influ-
ence o f an external force f(t), or the change of inflation rate in a n economy, or the reac-
tion of a body to some virus all of them can be interpreted in terms o f the above simple
block-diagram. The advantage of block-diagram representation is find out the exis tent
functional similarities between seemingly different systems. For ex a mple, the dynamic of
a mass-spring system is describ ed by the following differential equation
m
d
2
x
dt
2
+ kx = f (t); (5.98)
where k is the stiffness of the spring, m is the mass of the body, and f(t) is th e external
force exerting to the mass. In the block-diagram representation, we can show the above
system as follows
1
ms
2
+ k
x^(s)
f
^
(s)
In the above block-diagram, the transaction is represented by the Laplace transform of
the differential equation
x^(s) =
1
ms
2
+ k
f
^
(s): (5.99)
The expression h
^
(s) =
1
ms
2
+ k
is als o called the transfer function of the mass-spring system.
Now, let us consider the following LC circuit and assume that the switch S connects at
t = 0.
According to the voltage-current relationships of the capacitor C and the inductor L,
we can write
LC
d
2
V
c
dt
2
+ V
c
= v
s
(t) u(t): (5.100)
Assuming that the system is at rest, the Laplace transform of the equation is
(LCs
2
+1)V
^
c
(s) = v^
s
(s); (5.101)
or equivalently
V
^
c
(s) =
1
LCs
2
+ 1
v^
s
(s); (5.102)
which has the following block-diagram.
5.4 Impulse response and convolution 27
1
LCs
2
+ 1
V
c
^
(s)
v
s
^(s)
As it is seen, two systems have the completely similar transfer function and thus they
behave completely similar from the functional point of view.
Note that if the input in the above examples is δ(t), then the Laplace transform of
the output will be h
^
(s). In fact, fo r the mass-spring functional equation (5.99), if f (t) =
δ(t), then f
^
(s) = 1 and thus x^(s) =
1
ms
2
+ k
:= h
^
(s).
Definition 5.4. The response of a system to the input δ(t) is called the impulse response
of the system and is usually denoted by h(t). The Laplace transform of the impulse
response is called the transfer function of a system and is denoted by h
^
(s).
For example, for the impulse response of the mass-spring system is
h(t) = L
¡1
(h
^
(s)) =
1
km
r
sin
k
m
r
t
u(t); (5.103)
and the impulse response of the above LC circuit is
h(t) =
1
LC
p
sin
t
LC
p
u(t): (5.104)
Example 5.17. Consider the following initial value prob lem
y
0
+ y = δ(t ¡τ ); τ > 0
y(0) = 0
:
By Laplace transform method, the response of the system to the input δ(t ¡τ ) is
h(t ¡τ ) = e
¡(t¡τ)
u(t ¡τ ):
Note that since y
0
(t) is a Dirac singularity at t = τ , the response y(t) is of a finite jump
at that time. For τ !0, we take
h(t) =
e
¡t
t > 0
0 t ≤0
:
On the other hand, the response of the system to input r(t) is
y^(s) =
r^(s)
s + 1
;
and hence,
y(t) = h(t) ∗r(t) = e
¡t
Z
0
t
e
τ
r(τ) dτ :
5.4.3 LTI systems
General systems are usually classified into the fo llowing classes:
a) Causal system: The output y at time t depends only on the input x(τ) for τ ≤ t
and is independent of τ > t.
28 Laplace Transform Method
b) Memory-less system: The output y at time t depends on ly on the in put x at time t,
not on the previous or future values of t.
c) Time Invariant: The output y is not sensitive to the time of applying input x . In
other word, if y(t) is the response to the input x(t), then y(t ¡ τ ) will be the
response of x(t ¡τ ).
d) Linear system: If y
1
; y
2
are respectively the response to inputs x
1
; x
2
, then the
response of the system to αx
1
+ βx
2
for arbi trary constants α; β is αy
1
+ βy
2
.
Example 5.18. Suppose that the transaction T is a derivative operator, that is,
y(t) = T fx(t)g= x
0
(t):
The system is not causal and memory-less b ecause of the definition of the derivative:
x
0
(t) = lim
h!0
x(t + h) ¡x(t)
h
:
The system is linear due to the relation T fax
1
+ bx
2
g= ax
1
0
+ bx
2
0
= ay
1
+ by
2
. In addition,
this system is time invariant y(t ¡t
0
) = x
0
(t ¡t
0
): S imilarly, an integrator
y(t) =
Z
¡1
t
x(τ ) dτ ;
is a linear, time invariant and causal but not memory-less.
Definition 5.5. Consider a system (S) with input x(t) and output y(t). The system is
called linear if it satisfies the following two conditions:
1. for arbitrary λ 2R, the response to input λx(t) is λy(t), ( homogeneity property)
2. if y
1
, y
2
are responses t o inputs x
1
and x
2
respectively, the response to the input
x
1
(t) + x
2
(t) is y
1
(t) + y
2
(t) ( additive property)
This concept is shown in the figure (5.11) schematically.
S
λyλx
x
1
y
S
y
1
y
2
x
2
+
+
Figure 5.11. A linear system.
Definition 5.6. A system (S) with input x(t) and output y(t) is called Time Invariant
(TI) if response to x(t ¡τ ) is y(t ¡τ ) for arbitrary τ ≥0.
5.4 Impulse response and convolution 29
Example 5.19. Consider the mass-spring system (5.99). It is almost straightforward to
verify that the system is linear. In fact, if f(t) is replaced by λf (t) (for arbitrary constant
λ), the response changes to λx(t). Furthermore, if f(t) is replaced by f
1
(t) + f
2
(t), then
by superposition principle, we can write
x(t) = x
1
(t) + x
2
(t): (5.105)
In addition, if f(t)u(t) is replaced by f(t ¡τ)u(t ¡τ ) for s o me τ ≥0, then the response of
the system changes to
L
¡1
e
¡sτ
f
^
ms
2
+ k
!
= L
¡1
f
^
ms
2
+ k
!
j
t!t¡τ
u(t ¡τ ) = x(t ¡τ ) u(t ¡τ ): (5.106)
Therefore, the mass spring system is an LTI sy stem. The reader can check that the LC
circuit in the previo us example is LTI as well.
Example 5.20. Consider the following initial value prob lem
(
y
(n)
+ a
1
y
(n¡1)
+ ···+ a
n
y = x(t)
y
(k)
(0) = 0; for k = 0; :::; n ¡1
; (5.107)
where a
k
are some constants. We write the equation as L[y] = x(t), for L the op erator
L := D
n
+ a
1
D
n¡1
+ ···+ a
n
: (5.108)
It is simply verified that the equations represents an LTI system with x(t); y(t) as its
input and output respectively. In fact, for arbitrary λ
1
; λ
2
2 R, the solution to the input
x(t) = λ
1
x
1
(t) + λ
2
x
2
(t), is y(t) = λ
1
y
2
(t) + λ
2
y
2
(t), where L[y
1
] = x
1
(t) and L[y
2
] = x
2
(t).
Furthermore, according to the relation
y
(n)
(t ¡τ ) + a
1
y
(n¡1)
(t ¡τ ) + ···+ a
n
y(t ¡τ ) = x(t ¡τ ); (5.109)
we have L[y](t ¡τ ) = x(t ¡τ ). The transfer function of this system is:
h
^
(s) =
1
s
n
+ a
1
s
n¡1
+ ···+ a
n
: (5.110)
In this section, we study causal linear time invariant (LTI) systems described by linear
differential equations with constant coefficients
y
00
+ ay + by = r(t)
y(t
0
) = 0; y
0
(t
0
) = 0
: (5.111)
Or in the operator form as
L[y](t) = r(t);
where L stands for the differential operator
d
2
dt
2
+ a
d
dt
+ b. Accordingly, we can write
y(t) = L
¡1
[r(t)];
where L is the inverse of the differential operator L.
r(t)
y(t)
y^(s)
1
s
2
+ as + b
r^(s)
L
¡1
30 Laplace Transform Method
5.4.4 Convolution and LTI control systems
The importance of convolution lies in the f o llowing theorem.
Theorem 5.2. Assume that h(t) is the impulse response of an LTI system. Then the
response of the sy stem to an arbitrary input x(t) is determined by the following convolu-
tion
y(t) =
Z
0
t
h(t ¡τ ) x(τ ) dτ : (5.112)
Proof. Note that we can write x(t) as the convolution as
x(t) =
Z
0
t
x(τ) δ(t ¡τ ) dτ : (5.113)
Let us write the above integral as the following Riemann sum
Z
0
t
x(τ) δ(t ¡τ ) dτ = lim
n!1
X
k=1
n
x(τ
k
) δ(t ¡τ
k
) ∆τ
k
; (5.114)
and define x
n
(t) by the following partial sum
x
n
(t) =
X
k=1
n
x(τ
k
)δ(t ¡τ
k
) ∆τ
k
: (5.115)
Since the system is linear time-invariant, y
n
(t) is determined by the following summation
y
n
(t) =
X
k=1
n
x(τ
k
) h(t ¡τ
k
) ∆τ
k
; (5.116)
where h(t ¡τ
k
) is the response to the impulse δ(t ¡τ
k
). Now, we can write
y(t) = lim
n!1
y
n
(t) = lim
n!1
X
k=1
n
x(τ
k
) h(t ¡τ
k
) ∆τ
k
=
Z
0
t
x(τ) h(t ¡τ ) dτ ; (5.117)
and this completes the proof.
Example 5.21. For a LTI system, we know that the impulse response function h(t) is
h(t) = e
¡t
sin(t) u(t):
We would like to find the response of the system to the input function r(t) = e
¡t
u(t).
According to the above theorem, we can write
y(t) = r(t) ∗h(t) =
Z
0
t
e
¡(t¡τ)
e
¡τ
sin(τ) dτ = e
¡t
(1 ¡cos(t)) u(t):
Problem 5.36. Following problems:
a) If we know that the impulse response of an LTI system is sin(t), find the response to the input
x(t) = cos(t).
5.4 Impulse response and convolution 31
b) If we know that the response of an LTI system to input u(t) is sin(t), find the response to the
input x(t) = cos(t).
Working with convolution integrals is not always as simple as above, specially for con-
trol systems. For example consider a simple feedback control system shown below.
+
c(t)
y(t)
x(t)
h(t)
¡
Figure 5.12.
By a a straightforward calculation, we can write y(t) as the following convolution:
y(t) = (x(t) ¡c(t) ∗ y(t)) ∗h(t); (5.118)
or equivalently in the integral form as
y(t) =
Z
0
t
x(τ) ¡
Z
0
τ
c(τ ¡v)y(v) dv
h(t ¡τ ) dτ :
As it is seen, the formula looks very complicated even for such a simple feedback system.
Here we s ee how the Laplace transform simplifies the calculations. Remember the re la-
tion
Lff ∗ gg= f
^
(s) g^(s);
and thus fo r (5.118), we can write
y^(s) = h
^
(s)[x^(s) ¡c^(s) y^(s)];
which is solved simply for y^(s) as
y^(s) =
h
^
(s)
1 + c^(s) h
^
(s)
x^(s):
Therefore, the transfer function of the above control system is
h
^
c
(s) =
h
^
(s)
1 + c^(s) h
^
(s)
:
Problem 5.37. Find the transfer function of the control system shown in the following diagram
c
1
(t)
h
1
(t)
¡
+
¡
x(t)
h
2
(t)
c
2
(t)
+
y(t)
32 Laplace Transform Method
Problems
Problem 5.38. Show that an integrator
y(t) =
Z
¡1
t
x(τ ) dτ ;
is time invariant system.
Problem 5.39. Consider the discrete system y[n] = x[2n] for n 2Z. Give and example that shows the
system is not time invariant.
Problem 5.40. Consider the equation
y
0
+ ay = r(t);
where r(t) = 0 for t < 0. Sh ow that the transaction L
¡1
defined by
L
¡1
[r](t) =
Z
0
t
e
¡a(t¡τ)
r(τ)dτ ;
is a system representation of the given equation. Show that the system L
¡1
is linear t ime invariant.
Problem 5.41. Consider a system that its response is described by the following differential equation
y
0
+ ay = r(t):
i. Find the im pulse response of the system.
ii. Write the solution in t he convolution form.
The solution is the same as we obtained through solving a linear first order equation in previous chap-
ters.
Problem 5.42. Consider a system that its response is described by the following differential equation
y
00
+y = r(t):
i. Find the im pulse response of the system.
ii. Write the solution in t he convolution form.
iii. Verify that the obtained formula is a solution to the problem
y
00
+ y = r(t)
y(0) = y
0
(0) = 0
:
Problem 5.43. Consider the following initial value problem
y
00
+ a
1
y
0
+ a
2
y = r(t)
y(0) = y
0
(0) = 0
:
i. If the characteristic polynomial of the equation has two distinct roots λ
1
=/ λ
2
, use convolution
theorem to s how that the solution to the equation is
φ(t) =
1
λ
1
¡λ
2
(e
λ
1
t
∗r(t) ¡e
λ
2
t
∗r(t)):
ii. If the characteristic polynomial has a repeated root λ, use convolution theorem to show that
φ(t) = te
λt
∗r(t):
Problem 5.44. For the RC circuit shown in the figure (5.13), find the impulse response h(t) and the
transfer function h
^
(s) if v
o
(t) is considered as the response of the s ystem. Find v
o
(t) if v(t) = 1 and
the switch S connect s at time t = 1.
5.4 Impulse response and convolution 33
Figure 5.13.
Problem 5.45. Consider the circuit shown in the figure (5.14) where v
i
(t) and v
o
(t) are the input
and output of the system respectively. Find the transfer function and the impulse response of the
system. If R = L = C = 1, find the response v
o
(t) provided that v
i
(t) = sin(t).
Figure 5.14.
Problem 5.46. Use the definition of convolution t o calculate the co nvolution of following function.
Use Laplace transform method for convolution and compare the results.
i. f (t) = tu(t) , g(t) = u(t)
ii. f (t) = e
¡t
, g(t) = u(t) ¡u(t ¡1)
iii. f (t) = u(t), g (t) = t[u(t) ¡u(t ¡1)]
iv. f (t) = u(t) ¡u(t ¡1), g(t) = u(t ¡1) ¡u(t ¡2)
Problem 5.47. Use convolution property to calculate the inverse transformation of the following
function
i.
L
¡1
1
s(s
2
+ 1)
ii.
L
¡1
1
(s
2
+ 1)
2
iii.
L
¡1
e
¡s
(s ¡1)(s
2
+ 2s + 2)
Problem 5.48. Show that convolution is commutative, that is, f ∗ g = g ∗ f .
Problem 5.49. Us Laplace transform method to solve the following integro-differential equations.
The initial condition(s) for all equations is assumed zero.
i. y + 3
R
0
t
y(v) sin(t ¡v) dv = u(t ¡1),
ii. y
0
+ e
t
R
0
t
e
¡τ
y(τ) dτ = cos(t) δ(t ¡π)
iii. y +
R
0
t
y
0
(τ) e
t¡τ
dτ = u(t ¡1).
iv. y
0
0
+
R
0
t
y(v) u(t ¡v) dv = δ(t ¡1) + u(t ¡1)
5.5 Systems of differential equations
In this section we see how the method is employed to solve the linear systems of differen-
34 Laplace Transform Method
tial equations with constant coefficients. Consider the following system
8
>
>
<
>
>
:
y
1
0
= a
11
y
1
+ a
12
y
2
+ b
1
(t)
y
2
0
= a
21
y
1
+ a
22
y
2
+ b
2
(t)
y
1
(0) = y
2
(0) = 0
: (5.119)
The Laplace transform of the system reduces it to the following algebraic system
(
(s ¡a
11
)y^
1
(s) ¡a
12
y^
2
(s) = b
^
1
(s)
(s ¡a
22
)y^
2
(s) ¡a
21
y^
1
(s) = b
^
2
(s)
; (5.120)
that can be put in the following matrix form in turn
s ¡a
11
¡a
12
¡a
21
s ¡a
22
y^
1
(s)
y^
2
(s)
=
b
^
1
(s)
b
^
2
(s)
!
: (5.121)
The system (5.121) is solvable if the coefficient matrix is invertible. If so, we can write
y^
1
(s)
y^
2
(s)
=
1
p(s)
s ¡a
22
a
12
a
21
s ¡a
11
b
^
1
(s)
b
^
2
(s)
!
; (5.122)
where p(s) is the characteri sti c polynomial of the matrix
p(s) = s
2
¡(a
11
+ a
22
)s + a
11
a
22
¡a
12
a
21
: (5.123)
Now, y
1
(t) and y
2
(t) can be restored by the inverse transform of y^
1
(s) and y^
2
(s) respec-
tively.
Example 5.22. Consider the following system
8
>
>
<
>
>
:
y
1
0
= y
1
¡y
2
y
2
0
= y
1
+ δ(t ¡1)
y
1
(0) = y
2
(0) = 0
: (5.124)
By Laplace transform, we write the above system as the following algebraic one
(s ¡1)y^
1
(s) + y^
2
(s) = 0
sy^
2
(s) ¡ y^
1
(s) = e
¡s
;
or in the matrix form
s ¡1 1
¡1 s
y^
1
(s)
y^
2
(s)
=
0
e
¡s
:
The above system is solved for the vector (y^
1
; y^
2
) by the formula
y^
1
(s)
y^
2
(s)
=
1
s
2
¡s + 1
s ¡1
1 s ¡1
0
e
¡s
=
1
s
2
¡s + 1
¡e
¡s
(s ¡1) e
¡s
:
Thus y^
1
(s) =
¡e
¡s
s
2
¡s + 1
and therefore
y
1
(t) =
¡2
3
p
e
t¡ 1
2
sin
3
p
2
(t ¡1)
u(t ¡1):
5.5 Systems of differential equations 35
Likewise, we have y^
2
(s) =
(s ¡1)e
¡s
s
2
¡s + 1
and therefore
y
2
(t) = e
(t¡1)/ 2
cos
3
p
2
(t ¡1)
u(t ¡1) ¡
1
3
p
e
t¡ 1
2
sin
3
p
2
(t ¡1)
u(t ¡1):
The method is applied to higher order systems as well. The f o llowing example presents
the method for a second order system.
Example 5.23. Consider the following system
8
>
>
<
>
>
:
y
1
00
= 2 y
2
+ u(t)
y
2
00
= 8y
1
y
1
(0) = y
2
(0) = y
1
0
(0) = y
2
0
(0) = 0
:
The system reduces to the algebraic one
(
s
2
y
1
^ (s) = 2y^
2
(s) +
1
s
s
2
y
^
2
(s) = 8y
^
1
(s)
:
Solving the system for y^
1
(s) and y^
2
(s) gives
y^
1
(s) =
s
s
4
¡16
, y^
2
(s) =
8
s(s
4
¡16)
:
The inverse transform yields y
1
(t) and y
2
(t) as follows
y
1
(t) =
1
8
cosh(2t) ¡
1
8
cos(2t) and y
2
(t) =
1
4
cosh(2t) +
1
4
cos(2t) ¡
1
2
:
Problems
In the following problems, use Laplace transform method to solve the following systems.
Problem 5.50.
(
y
1
0
= y
2
+ δ(t ¡1)
y
2
0
= ¡y
1
+ u(t)
; y
1
(0) = y
2
(0) = 0
Problem 5.51.
(
y
1
0
= ¡y
1
¡ y
2
+ δ(t ¡1)
y
2
0
= ¡2y
1
; y
1
(0) = 0; y
2
(0) = 1
Problem 5.52.
(
y
1
00
= 2y
1
+ y
2
y
2
00
= 12y
1
¡2y
2
; y
1
(0) = 1; y
2
(0) = y
1
0
(0) = y
2
0
(0) = 0
Problem 5.53.
(
y
1
00
= 2 y
2
¡ y
1
0
+ y
2
0
y
1
0
¡y
2
0
= ¡y
1
; y
1
(0) = y
2
(0) = 1; y
1
0
(0) = y
2
0
(0) = 0
Problem 5.54.
(
y
1
00
+ 3y
2
00
= y
1
+ δ(t ¡1)
y
1
0
+ 3y
2
0
= 2y
2
; y
1
(0) = y
2
(0) = y
1
0
(0) = y
2
0
(0) = 0
Problem 5.55. Use the Laplace transform method to prove tha t the following two systems have
the same solution
36 Laplace Transform Method
1. y~
0
= A
2×2
y~ ; y~(0) = y~
0
2. y~
0
= A
2×2
y~ + δ(t)y~
0
; y~(0) = 0.
Here y~ stands for the vector
y
1
y
2
and the matrix A
2×2
is a constant matrix
A
2×2
=
a
11
a
12
a
21
a
22
:
5.5 Systems of differential equations 37