Tài liệu Signals And Systems With Matlab Applications P2 doc

Signals and Systems with MATLAB Applications, Second Edition 1-19
Orchard Publications
Summary
int(d) % Integrate the delta function
ans =
Heaviside(t-a)*k
1.8 Summary
• The unit step function that is defined as
• The unit step function offers a convenient method of describing the sudden application of a volt-
age or current source.
• The unit ramp function, denoted as , is defined as
• The unit impulse or delta function, denoted as , is the derivative of the unit step . It is also
defined as
and
• The sampling property of the delta function states that
or, when ,

• The sifting property of the delta function states that
• The sampling property of the doublet function states that
u
0
t()
u
0
t()
0t0<
1t0>
⎩
⎨
⎧
=

u
1
t()
u
1
t() u
0
τ()τd
∞–
t
∫
=
δ t() u
0
t()
δτ()τd
∞–
t
∫
u
0
t()=
δ t() 0 for all t0≠=
f
t()δta–()fa()δt()=
a0=
f
t()δt() f0()δt()=
ft()δt α–()td
∞–

∞
∫
f α()=
δ' t()
f
t()δ' ta–()fa()δ' ta–()f

' a()δta–()–=
Chapter 1 Elementary Signals
1-20
Signals and Systems with MATLAB Applications, Second Edition
Orchard Publications
1.9 Exercises
1. Evaluate the following functions:
a.
b.
c.
d.
e.
f.
2.
a. Express the voltage waveform shown in Figure 1.24, as a sum of unit step functions for
the time interval .
b. Using the result of part (a), compute the derivative of
, and sketch its waveform.
Figure 1.24. Waveform for Exercise 2
tδsin t
π
6
–

⎝⎠
⎛⎞
2tδcos t
π
4
–
⎝⎠
⎛⎞
t
2
δ t
π
2
–
⎝⎠
⎛⎞
cos
2tδtan t
π
8
–
⎝⎠
⎛⎞
t
2
e
t–
δ t2–()td
∞–
∞

∫
t
2
δ
1
t
π
2
–
⎝⎠
⎛⎞
sin
vt()
0t7 s<<
vt()
−10
−20
10
20
12345
6
7
0
vt()
ts()
e
2t–
V()
vt()
Signals and Systems with MATLAB Applications, Second Edition 1-21

Orchard Publications
Solutions to Exercises
1.10 Solutions to Exercises
Dear Reader:
The remaining pages on this chapter contain the solutions to the exercises.
You must, for your benefit, make an honest effort to solve the problems without first looking at the
solutions that follow. It is recommended that first you go through and solve those you feel that you
know. For the exercises that you are uncertain, review this chapter and try again. If your results do
not agree with those provided, look over your procedures for inconsistencies and computational
errors. Refer to the solutions as a last resort and rework those problems at a later date.
You should follow this practice with the exercises on all chapters of this book.
Chapter 1 Elementary Signals
1-22
Signals and Systems with MATLAB Applications, Second Edition
Orchard Publications
1. We apply the sampling property of the function for all expressions except (e) where we apply
the sifting property. For part (f) we apply the sampling property of the doublet.
We recall that the sampling property states that . Thus,
a.
b.
c.
d.
We recall that the sampling property states that . Thus,
e.
We recall that the sampling property for the doublet states that
Thus,
f.
2.
a.
or

δ t()
f
t()δta–()fa()δta–()=
tδsin t
π
6
–
⎝⎠
⎛⎞
t
t π 6⁄=
δ t
π
6
–
⎝⎠
⎛⎞
sin
π
6

δ t
π
6
–
⎝⎠
⎛⎞
sin 0.5δ t
π
6

–
⎝⎠
⎛⎞
===
2tδcos t
π
4
–
⎝⎠
⎛⎞
2t
t π 4⁄=
δ t
π
4
–
⎝⎠
⎛⎞
cos
π
2

δ t
π
4
–
⎝⎠
⎛⎞
cos 0===
t

2
δ t
π
2
–
⎝⎠
⎛⎞
cos
1
2

12tcos+()
t π 2⁄=
δ t
π
2
–
⎝⎠
⎛⎞
1
2

1 πcos+()δt
π
2
–
⎝⎠
⎛⎞
1
2

11–()δt
π
2
–
⎝⎠
⎛⎞
0====
2tδtan t
π
8
–
⎝⎠
⎛⎞
2t
t π 8⁄=
δtan t
π
8
–
⎝⎠
⎛⎞
π
4

δ t
π
8
–
⎝⎠

⎛⎞
tan δ t
π
8
–
⎝⎠
⎛⎞
===
ft()δt α–()td
∞–
∞
∫
f α()=
t
2
e
t–
δ t2–()td
∞–
∞
∫
t
2
e
t–
t2=
4e
2–
0.54===
f

t()δ' ta–()fa()δ' ta–()f

' a()δta–()–=
t
2
δ
1
t
π
2
–
⎝⎠
⎛⎞
sin t
t π 2⁄=
2
δ
1
t
π
2
–
⎝⎠
⎛⎞
sin
d
dt

t
t π 2⁄=

2
δ t
π
2
–
⎝⎠
⎛⎞
sin–=
1
2

12tcos–()
t π 2⁄=
δ
1
t
π
2
–
⎝⎠
⎛⎞
2t
t π 2⁄=
δ t
π
2
–
⎝⎠
⎛⎞
sin–=

1
2

11+()δ
1
t
π
2
–
⎝⎠
⎛⎞
πδ t
π
2
–
⎝⎠
⎛⎞
sin– δ
1
t
π
2
–
⎝⎠
⎛⎞
==
vt() e
2t–
u
0

t() u
0
t2–()–[]10t 30–()u
0
t2–()u
0
t3–()–[]+=
+ 10– t50+()u
0
t3–()u
0
t5–()–[]10t 70–()u
0
t5–()u
0
t7–()–[]+
Signals and Systems with MATLAB Applications, Second Edition 1-23
Orchard Publications
Solutions to Exercises
b.
(1)
Referring to the given waveform we observe that discontinuities occur only at , ,
and . Therefore, and . Also, by the sampling property of the delta
function
and with these simplifications (1) above reduces to
The waveform for is shown below.
vt() e
2t–
u
0

t() e
2t–
u
0
t2–()10tu
0
t2–()30u
0
t2–()10tu
0
t3–()30u
0
t3–()+––+–=
10tu
0
t3–()– 50u
0
t3–()10tu
0
t5–()50u
0
t5–()10tu
0
t5–()+–++
70u
0
t5–()10tu
0
t7–()70u
0

t7–()+––
e
2t–
u
0
t() e
2t–
10t 30–+–()u
0
t2–()20t 80+–()u
0
t3–()20t 120–()u
0
t5–()+++=
+ 10t 70+–()u
0
t7–()
dv
dt
2e
2t–
u
0
t() e
2t–
δ t() 2e
2t–
10+()u
0
t2–()e

2t–
10t 30–+–()δt2–()++ +–=
20u
0
t3–() 20t– 80+()δt3–()20u
0
t5–()20t 120–()δt5–()+++–
10u
0
t7–() 10t– 70+()δt7–()+–
t2= t3=
t5= δ t() 0= δ t7–()0=
e
2t–
10t 30–+–()δt2–() e
2t–
10t 30–+–()
t2=
δ t2–()= 10δ t2–()–≈
20t– 80+()δt3–() 20t– 80+()
t3=
δ t3–()= 20δ t3–()=
20t 120–()δt5–()20t 120–()
t5=
δ t5–()= 20– δ t5–()=
dv dt⁄ 2e
2t–
u
0
t() 2e

2t–
u
0
t2–()10u
0
t2–()10δ t2–()–++–=
20u
0
t3–()20δ t3–()20u
0
t5–()20δ t5–()10u
0
t7–()––++–
2e
2t–
u
0
t() u
0
t2–()–[]10δ t2–()10 u
0
t2–()u
0
t3–()–[]20δ t3–()++––=
10 u
0
t3–()u
0
t5–()–[]– 20δ t5–()10 u
0

t5–()u
0
t7–()–[]+–
dv dt⁄
dv dt⁄
20
10
Vs⁄()
t s()
20–
10–
1
2
3
4
5
6
7
10δ t2–()–
20δ t 3–()
20δ t 5–()–
2e
2t–
–
Chapter 1 Elementary Signals
1-24
Signals and Systems with MATLAB Applications, Second Edition
Orchard Publications
NOTES
Signals and Systems with MATLAB Applications, Second Edition 2-1

Orchard Publications
Chapter 2
The Laplace Transformation
his chapter begins with an introduction to the Laplace transformation, definitions, and proper-
ties of the Laplace transformation. The initial value and final value theorems are also discussed
and proved. It concludes with the derivation of the Laplace transform of common functions
of time, and the Laplace transforms of common waveforms.
2.1 Definition of the Laplace Transformation
The two-sided or bilateral Laplace Transform pair is defined as
(2.1)
(2.2)
where denotes the Laplace transform of the time function , denotes the
Inverse Laplace transform, and is a complex variable whose real part is , and imaginary part ,
that is, .
In most problems, we are concerned with values of time greater than some reference time, say
, and since the initial conditions are generally known, the two-sided Laplace transform
pair of (2.1) and (2.2) simplifies to the
unilateral or one-sided Laplace transform defined as
(2.3)
(2.4)
The Laplace Transform of (2.3) has meaning only if the integral converges (reaches a limit), that is, if
(2.5)
To determine the conditions that will ensure us that the integral of (2.3) converges, we rewrite (2.5)
T
L ft(){}Fs()= ft()
∞–
∞
∫
e
st–

dt=
L
1–
Fs(){}ft()=
1
2πj

Fs()
σ jω–
σ jω+
∫
e
st
ds=
L ft(){}
f
t() L
1–
Fs(){}
s σω
s σ jω+=
t
tt
0
0==
L ft(){}F= s() ft()
t
0
∞
∫

e
st–
dt f t()
0
∞
∫
e
st–
dt==
L
1–
Fs(){}f= t()
1
2πj

Fs()
σ jω–
σ jω+
∫
e
st
ds=
ft()
0
∞
∫
e
st–
dt ∞<

Chapter 2 The Laplace Transformation
2-2
Signals and Systems with MATLAB Applications, Second Edition
Orchard Publications
as
(2.6)
The term in the integral of (2.6) has magnitude of unity, i.e., , and thus the condition
for convergence becomes
(2.7)
Fortunately, in most engineering applications the functions are of exponential order
*
. Then, we
can express (2.7) as,
(2.8)
and we see that the integral on the right side of the inequality sign in (2.8), converges if .
Therefore, we conclude that if is of exponential order, exists if
(2.9)
where denotes the real part of the complex variable .
Evaluation of the integral of (2.4) involves contour integration in the complex plane, and thus, it will
not be attempted in this chapter. We will see, in the next chapter, that many Laplace transforms can
be inverted with the use of a few standard pairs, and therefore, there is no need to use (2.4) to obtain
the Inverse Laplace transform.
In our subsequent discussion, we will denote transformation from the time domain to the complex
frequency domain, and vice versa, as
(2.10)

2.2 Properties of the Laplace Transform
1. Linearity Property
The linearity property states that if
have Laplace transforms
* A function is said to be of exponential order if .
ft()e
σt–
0
∞
∫
e
jωt–
dt ∞<
e
jωt–
e
jωt–
1=
ft()e
σt–
0
∞
∫
dt ∞<
f
t()
f
t() ft() ke
σ
0

t
for all t0≥<
ft()e
σt–
0
∞
∫
dt ke
σ
0
t
e
σt–
0
∞
∫
dt<
σσ
0
>
f
t() L ft(){}
Re s{} σ σ
0
>=
Re s{} s
f
t() Fs()⇔
f
1

t()f
2
t()…f
n
t(),,,
Signals and Systems with MATLAB Applications, Second Edition 2-3
Orchard Publications
Properties of the Laplace Transform
respectively, and
are arbitrary constants, then,
(2.11)
Proof:
Note 1:
It is desirable to multiply by to eliminate any unwanted non-zero values of for .
2. Time Shifting Property
The time shifting property states that a right shift in the time domain by units, corresponds to mul-
tiplication by in the complex frequency domain. Thus,
(2.12)
Proof:
(2.13)
Now, we let ; then
, and . With these substitutions, the second integral
on the right side of (2.13) becomes
3. Frequency Shifting Property
The frequency shifting property states that if we multiply some time domain function by an
exponential function where
a is an arbitrary positive constant, this multiplication will produce a
shift of the
s variable in the complex frequency domain by units. Thus,
F

1
s()F
2
s()…F
n
s(),,,
c
1
c
2
… c
n
,, ,
c
1
f
1
t() c
2
f
2
t() … c
n
f
n
t()+++ c
1
F
1
s() c

2
F
2
s() … c
n
F
n
s()+++⇔
L c
1
f
1
t() c
2
f
2
t() … c
n
f
n
t()+++{}c
1
f
1
t() c
2
f
2
t() … c
n

f
n
t()+++[]
t
0
∞
∫
dt=
c
1
f
1
t()
t
0
∞
∫
e
st–
dt c
2
f
2
t()
t
0
∞
∫
e
st–

dt … + c
n
f
n
t()
t
0
∞
∫
e
st–
dt++=
c
1
F
1
s() c
2
F
2
s() … c
n
F
n
s()+++=
f
t() u
0
t()
f

t() t0<
a
e
as–
ft a–()u
0
ta–()e
as–
Fs()⇔
L ft a–()u
0
ta–(){}0
0
a
∫
e
st–
dt f t a–()
a
∞
∫
e
st–
dt+=
ta– τ= t τ a+= dt dτ=
f τ()
0
∞
∫
e

s τ a+()–
dτ e
as–
f τ()
0
∞
∫
e
sτ–
dτ e
as–
Fs()==
f
t()
e
at–
a
Chapter 2 The Laplace Transformation
2-4
Signals and Systems with MATLAB Applications, Second Edition
Orchard Publications
(2.14)
Proof:
Note 2:
A change of scale is represented by multiplication of the time variable by a positive scaling factor
. Thus, the function after scaling the time axis, becomes .
4. Scaling Property
Let be an arbitrary positive constant; then, the
scaling property states that
(2.15)

Proof:
and letting , we get
Note 3:
Generally, the initial value of is taken at to include any discontinuity that may be present
at . If it is known that no such discontinuity exists at , we simply interpret as .
5. Differentiation in Time Domain
The
differentiation in time domain property states that differentiation in the time domain corresponds
to multiplication by in the complex frequency domain, minus the initial value of at .
Thus,
(2.16)
Proof:
e
at–
ft() Fs a+()⇔
L e
at–
ft(){}e
at–
ft()
0
∞
∫
e
st–
dt f t()
0
∞
∫
e

sa+()t–
dt F s a+()== =
t
a
f
t()
f
at()
a
fat()
1
a

F
s
a

⎝⎠
⎛⎞
⇔
L fat(){} fat()
0
∞
∫
e
st–
dt=
t τ a⁄=
L fat(){} f τ()
0

∞
∫
e
s τ a⁄()–
d
τ
a

⎝⎠
⎛⎞
1
a

f τ()
0
∞
∫
e
sa⁄()τ–
d τ()
1
a

F
s
a

⎝⎠
⎛⎞
== =

f
t() t0
−
=
t0= t0
−
=
f
0
−
()
f
0()
s
f
t() t0
−
=
f

' t()
d
dt

ft()= sF s() f0
−
()–⇔
L f

' t(){} f

' t()
0
∞
∫
e
st–
dt=
Signals and Systems with MATLAB Applications, Second Edition 2-5
Orchard Publications
Properties of the Laplace Transform
Using integration by parts where
(2.17)
we let and . Then, , , and thus
The time differentiation property can be extended to show that
(2.18)
(2.19)
and in general
(2.20)
To prove (2.18), we let
and as we found above,
Then,
Relations (2.19) and (2.20) can be proved by similar procedures.
We must remember that the terms , and so on, represent the initial conditions.
Therefore, when all initial conditions are zero, and we differentiate a time function times,
this corresponds to multiplied by to the
power.
vud
∫
uv u vd

∫
–=
du f

' t()= ve
st–
= uft()= dv se
st–
–=
Lf

' t(){}ft()e
st–
0
−
∞
sft()
0
−
∞
∫
e
st–
dt+ ft()e
st–
0
−
a
a ∞→
lim sF s()+==

e
sa–
fa() f0
−
()–[]
a ∞→
lim sF s()+ 0f0
−
()– sF s()+==
d
2
dt
2

ft() s
2
Fs() sf 0
−
()– f

' 0
−
()–⇔
d
3
dt
3

ft() s
3

Fs() s
2
f0
−
()– sf

' 0
−
()– f

'' 0
−
()–⇔
d
n
dt
n

ft() s
n
Fs() s
n1–
f0
−
()– s
n2–
f

' 0
−

()– … f–
n1–
0
−
()–⇔
gt() f

' t()
d
dt

ft()==
L
g

' t(){}sL g t(){}g0
−
(
)
–=
L f

'' t(){}sL f

' t(){}f

' 0
−
()– ssL ft()[]f0
−

()–[]f

' 0
−
()–==
s
2
Fs() sf 0
−
()– f

' 0
−
()–=
f
0
−
() f

' 0
−
() f

'' 0
−
(),,
f
t() n
Fs() snth
Chapter 2 The Laplace Transformation

2-6
Signals and Systems with MATLAB Applications, Second Edition
Orchard Publications
6. Differentiation in Complex Frequency Domain
This property states that differentiation in complex frequency domain and multiplication by minus one,
corresponds to multiplication of by in the time domain. In other words,
(2.21)
Proof:
Differentiating with respect to s, and applying Leibnitz’s rule
*
for differentiation under the integral, we
get
In general,
(2.22)
The proof for follows by taking the second and higher-order derivatives of with respect
to .
7. Integration in Time Domain
This property states that
integration in time domain corresponds to divided by plus the initial
value of at
, also divided by . That is,
(2.23)
* This rule states that if a function of a parameter is defined by the equation where f is some
known function of integration x and the parameter , a and b are constants independent of x and , and the par-
tial derivative exists and it is continuous, then .
f
t() t
tf t()
d
ds

– Fs()⇔
L ft(){}Fs() ft()
0
∞
∫
e
st–
dt==
α F α() fxα,()xd
a
b
∫
=
αα
f∂α∂⁄
dF
dα

x α,()∂
α()∂

xd
a
b
∫
=
d
ds

Fs()

d
ds

ft()
0
∞
∫
e
st–
dt
s∂
∂
0
∞
∫
e
st–
ft()dt t–
0
∞
∫
e
st–
ft()dt===
tf t()[]
0
∞
∫
e
st–

dt– L tf t()[]–==
t
n
ft() 1–()
n
d
n
ds
n

Fs()⇔
n2≥ Fs()
s
Fs() s
f
t() t0
−
= s
f τ()
∞–
t
∫
dτ
Fs()
s

f0
−
()
s

+⇔
Signals and Systems with MATLAB Applications, Second Edition 2-7
Orchard Publications
Properties of the Laplace Transform
Proof:
We express the integral of (2.23) as two integrals, that is,
(2.24)
The first integral on the right side of (2.24), represents a constant value since neither the upper, nor
the lower limits of integration are functions of time, and this constant is an initial condition denoted
as . We will find the Laplace transform of this constant, the transform of the second integral
on the right side of (2.24), and will prove (2.23) by the linearity property. Thus,
(2.25)
This is the value of the first integral in (2.24). Next, we will show that
We let
then,
and
Now,
(2.26)
and the proof of (2.23) follows from (2.25) and (2.26).
f τ()
∞–
t
∫
dτ f τ()
∞–
0
∫
dτ f τ()
0
t

∫
dτ+=
f
0
−
()
L f0
−
(){} f0
−
()
0
∞
∫
e
st–
dt f 0
−
() e
st–
0
∞
∫
dt f 0
−
()
e
st–
s–

0
∞
===
f0
−
()0
f0
−
()
s
–
⎝⎠
⎛⎞
–×
f0
−
()
s
==
f τ()
0
t
∫
dτ
Fs()
s

⇔
gt() f τ()
0

t
∫
dτ=
g' t() f τ()=
g0() f τ()
0
0
∫
dτ 0==
L g' t(){}Gs() sL gt(){}g0
−
()– Gs() 0–== =
sL gt(){}Gs()=
L gt(){}
Gs()
s
=
L f τ()
0
t
∫
dτ
⎩⎭
⎨⎬
⎧⎫
Fs()
s
=
Chapter 2 The Laplace Transformation
2-8

Signals and Systems with MATLAB Applications, Second Edition
Orchard Publications
8. Integration in Complex Frequency Domain
This property states that integration in complex frequency domain with respect to corresponds to
division of a time function by the variable , provided that the limit exists. Thus,
(2.27)
Proof:
Integrating both sides from to , we get
Next, we interchange the order of integration, i.e.,
and performing the inner integration on the right side integral with respect to , we get
9. Time Periodicity
The
time periodicity property states that a periodic function of time with period corresponds to
the integral divided by in the complex frequency domain. Thus, if we let
be a periodic function with period , that is, , for we get the trans-
form pair
(2.28)
s
f
t() t
ft()
t

t0→
lim
ft()
t

Fs()sd
s

∞
∫
⇔
Fs() ft()
0
∞
∫
e
st–
dt=
s ∞
Fs()sd
s
∞
∫
ft()
0
∞
∫
e
st–
dt sd
s
∞
∫
=
Fs()sd
s
∞
∫

e
st–
s
∞
∫
sdft()td
0
∞
∫
=
s
Fs()sd
s
∞
∫
1
t
– e
st–
s
∞
ft()td
0
∞
∫
ft()
t

e
st–

td
0
∞
∫
L
ft()
t

⎩⎭
⎨⎬
⎧⎫
===
T
ft()
0
T
∫
e
st–
dt 1 e
sT–
–()
f
t()
T
f
t() ft nT+()= n123…,,,=
ft nT+()
ft()
0

T
∫
e
st–
dt
1e
sT–
–

⇔
Signals and Systems with MATLAB Applications, Second Edition 2-9
Orchard Publications
Properties of the Laplace Transform
Proof:
The Laplace transform of a periodic function can be expressed as
In the first integral of the right side, we let , in the second , in the third ,
and so on. The areas under each period of are equal, and thus the upper and lower limits of
integration are the same for each integral. Then,
(2.29)
Since the function is periodic, i.e., , we can write
(2.29) as
(2.30)
By application of the binomial theorem, that is,
(2.31)
we find that expression (2.30) reduces to
10. Initial Value Theorem
The
initial value theorem states that the initial value of the time function can be found
from its Laplace transform multiplied by and letting .That is,
(2.32)

Proof:
From the time domain differentiation property,
or
L ft(){} ft()
0
∞
∫
e
st–
dt f t()
0
T
∫
e
st–
dt f t()
T
2T
∫
e
st–
dt f t()
2T
3T
∫
e
st–
dt …+++==
t τ= t τ T+= t τ 2T+=
f

t()
L ft(){} f τ()
0
T
∫
e
sτ–
dτ f τ T+()
0
T
∫
e
s τ T+()–
dτ f τ 2T+()
0
T
∫
e
s τ 2T+()–
dτ…++ +=
f
τ() f τ T+()f τ 2T+()…f τ nT+()== ==
L f τ(){} 1e
sT–
e
2sT–
…++ +()f τ()
0
T
∫

e
sτ–
dτ=
1aa
2
a
3
…++++
1
1a–
=
L f τ(){}
f τ()
0
T
∫
e
sτ–
dτ
τ e
sT–
–
=
f
0
−
()
f
t()
ss∞→

ft()
t0→
lim sF s()
s ∞→
lim f 0
−
()==
d
dt

ft() sF s() f0
−
()–⇔
Chapter 2 The Laplace Transformation
2-10
Signals and Systems with MATLAB Applications, Second Edition
Orchard Publications
Taking the limit of both sides by letting , we get
Interchanging the limiting process, we get
and since
the above expression reduces to
or
11. Final Value Theorem
The
final value theorem states that the final value of the time function can be found from
its Laplace transform multiplied by
s, then, letting . That is,
(2.33)
Proof:
From the time domain differentiation property,

or
Taking the limit of both sides by letting , we get
L
d
dt

ft()
⎩⎭
⎨⎬
⎧⎫
sF s() f0
−
()–
d
dt

ft()
0
∞
∫
e
st–
dt==
s ∞→
sF s()f0
−
()–[]
s ∞→
lim
d

dt

ft()
ε
T
∫
e
st–
dt
T ∞→
ε 0→
lim
s ∞→
lim=
sF s()f0
−
()–[]
s ∞→
lim
d
dt

ft()
ε
T
∫
e
st–
s ∞→
lim dt

T ∞→
ε 0→
lim=
e
st–
s ∞→
lim 0=
sF s()f0
−
()–[]
s ∞→
lim 0=
sF s()
s ∞→
lim f 0
−
()=
f
∞()
f
t()
s0→
ft()
t ∞→
lim sF s()
s0→
lim f ∞()==
d
dt

ft() sF s() f0
−
()–⇔
L
d
dt

ft()
⎩⎭
⎨⎬
⎧⎫
sF s() f0
−
()–
d
dt

ft()
0
∞
∫
e
st–
dt==
s0→
Signals and Systems with MATLAB Applications, Second Edition 2-11
Orchard Publications
Properties of the Laplace Transform
and by interchanging the limiting process, we get
Also, since

the above expression reduces to
and therefore,
12. Convolution in the Time Domain
Convolution
*
in the time domain corresponds to multiplication in the complex frequency domain,
that is,
(2.34)
Proof:
(2.35)
We let ; then,
, and . By substitution into (2.35),
* Convolution is the process of overlapping two signals. The convolution of two time functions and is
denoted as , and by definition, where is a dummy variable. We will
discuss it in detail in Chapter 6.
sF s()f0
−
()–[]
s0→
lim
d
dt

ft()
ε
T
∫
e
st–
dt

T ∞→
ε 0→
lim
s0→
lim=
sF s()f0
−
()–[]
s0→
lim
d
dt

ft()
ε
T
∫
e
st–
s0→
lim dt
T ∞→
ε 0→
lim=
e
st–
s0→
lim 1=
sF s()f0
−

()–[]
s0→
lim
d
dt

ft()
ε
T
∫
dt
T ∞→
ε 0→
lim f t()
ε
T
∫
T ∞→
ε 0→
lim==
fT() f ε()–[]
T ∞→
ε 0→
lim f ∞()f0
−
()–==
sF s()
s0→
lim f ∞()=
f

1
t()
f
2
t()
f
1
t()*f
2
t()
f
1
t()*f
2
t() f
1
τ()f
2
t τ–()
∞–
∞
∫
dτ= τ
f
1
t()*f
2
t() F
1
s()F

2
s()⇔
L f
1
t()*f
2
t(){}L f
1
τ()f
2
t τ–()
∞–
∞
∫
dτ f
1
τ()f
2
t τ–()
0
∞
∫
dτ
0
∞
∫
e
st–
dt==
f

1
τ() f
2
t τ–()
0
∞
∫
e
st–
dt
0
∞
∫
dτ=
t τ– λ= t λτ+= dt dλ=
Chapter 2 The Laplace Transformation
2-12
Signals and Systems with MATLAB Applications, Second Edition
Orchard Publications
13. Convolution in the Complex Frequency Domain
Convolution in the complex frequency domain divided by , corresponds to multiplication in the
time domain. That is,
(2.36)
Proof:
(2.37)
and recalling that the Inverse Laplace transform from (2.2) is
by substitution into (2.37), we get
We observe that the bracketed integral is ; therefore,
For easy reference, we have summarized the Laplace transform pairs and theorems in Table 2.1.
2.3 The Laplace Transform of Common Functions of Time

In this section, we will present several examples for finding the Laplace transform of common func-
tions of time.
Example 2.1
Find
L f
1
t()*f
2
t(){}f
1
τ() f
2
λ()
0
∞
∫
e
s λτ+()–
dλ
0
∞
∫
dτ f
1
τ()e
sτ–
dτ
0
∞
∫

f
2
λ()
0
∞
∫
e
sλ–
dλ==
F
1
s()F
2
s()=
12πj⁄
f
1
t()f
2
t()
1
2πj

F
1
s()*F
2
s()⇔
L f
1

t()f
2
t(){}f
1
t()f
2
t()
0
∞
∫
e
st–
dt=
f
1
t()
1
2πj

F
1
σ jω–
σ jω+
∫
µ()e
µt
dµ=
L f
1
t()f

2
t(){}
1
2πj

F
1
σ jω–
σ jω+
∫
µ()e
µt
dµ f
2
t()
0
∞
∫
e
st–
dt=
1
2πj

F
1
σ jω–
σ jω+
∫
µ() f

2
t()
0
∞
∫
e
s µ–()t–
dt dµ=
F
2
s µ–()
Lf
1
t()f
2
t(){}
1
2πj

F
1
σ jω–
σ jω+
∫
µ()F
2
s µ–()dµ
1
2πj

F
1
s()*F
2
s()==
L u
0
t(){}
Signals and Systems with MATLAB Applications, Second Edition 2-13
Orchard Publications
The Laplace Transform of Common Functions of Time
TABLE 2.1 Summary of Laplace Transform Properties and Theorems
Property/Theorem Time Domain Complex Frequency Domain
1 Linearity
2 Time Shifting
3 Frequency Shifting
4 Time Scaling
5 Time Differentiation
See also (2.18) through (2.20)
6 Frequency Differentiation
See also (2.22)
7 Time Integration
8 Frequency Integration
9 Time Periodicity
10 Initial Value Theorem
11 Final Value Theorem
12 Time Convolution
13 Frequency Convolution
c
1

f
1
t() c
2
f
2
t()+
+ … c
n
f
n
t()+
c
1
F
1
s() c
2
F
2
s()+
+ … c
n
F
n
s()+
ft a–()u
0
ta–()
e

as–
Fs()
e
as–
ft()
Fs a+()
f
at()
1
a

F
s
a

⎝⎠
⎛⎞
d
dt

ft()
sF s() f0
−
()–
tf t() d
ds
– Fs()
f τ()
∞–
t

∫
dτ
Fs()
s

f0
−
()
s
+
f
t()
t

Fs()sd
s
∞
∫
f
tnT+()
ft()
0
T
∫
e
st–
dt
1e
sT–
–

ft()
t
0→
lim
sF s()
s ∞→
lim f 0
−
()=
ft()
t
∞→
lim
sF s()
s
0→
lim f ∞()=
f
1
t()*f
2
t()
F
1
s()F
2
s()
f
1

t()f
2
t() 1
2πj

F
1
s()*F
2
s()
Chapter 2 The Laplace Transformation
2-14
Signals and Systems with MATLAB Applications, Second Edition
Orchard Publications
Solution:
We start with the definition of the Laplace transform, that is,
For this example,
Thus, we have obtained the transform pair
(2.38)
for .
*
Example 2.2
Find
Solution:
We apply the definition
and for this example,
We will perform integration by parts recalling that
(2.39)
We let
then,

* This condition was established in (2.9).
L
ft(){}Fs() ft()
0
∞
∫
e
st–
d
t
==
L u
0
t(){} 1
0
∞
∫
e
st–
dt
e
st
–
s

0
∞
0
1
s

–
⎝⎠
⎛⎞
–
1
s

====
u
0
t()
1
s

⇔
Re s{} σ 0>=
L u
1
t(){}
L ft(){}Fs() ft()
0
∞
∫
e
st–
dt==
L u
1
t(){}L t{} t
0

∞
∫
e
st–
dt==
uvd
∫
uv v ud
∫
–=
ut and dv e
st–
==
du 1 and v
e
st–
–
s
==