Laplace Transform in signal and system slide

Laplace Transform with full explanation

1. Laplace Transform
Representation of continuous time aperiodic signals using general
complex exponentials
(General form of Fourier Transform)

2. In Chapter 3, we saw that the response of a linear time-invariant system with impulse
response /(t) to a complex exponential input of the form e” is

y(t) = He", (9.1)

where

H(s) = | hide" dt. (9.2)

For s imaginary (i.e., s = jw), the integral in eq. (9.2) corresponds to the Fourier trans-
form of h(t). For general values of the complex variable s, it is referred to as the Laplace
transform of the impulse response h(t).

The Laplace transform of a general signal x(t) is defined as!

x) à | "ge dt, (9.3)

and we note in particular that it is a function of the independent variable s corresponding
to the complex variable in the exponent of e~'. The complex variable s can be written as
s = 0 + jw, with o and o the real and imaginary parts, respectively. For convenience,
we will sometimes denote the Laplace transform in operator form as £{x(r)} and denote
the transform relationship between x(t) and X(s) as

£
x(t) —> X(5). (9.4)

3. Relationship of Laplace Transform and Fourier Transform
3

4. Example Let the signal x(t) = 67000).
Le * 1
X(jw) = | e “une (dt = | ete dt = — , a>0 (9.9)
u lo Jo +a
the Laplace transform is
X(s) = | e-tu(the-sdt -| els + arar, (9.10)
-= 0
or, with s = 9 + jw,
X(0 + jw) = | ete jai dt, (9.11)
0

By comparison with eq. (9.9) we recognize eq. (9.11) as the Fourier transform of
e-(7* y(t), and thus,

1

X(œ + jw) = Grarie o+a>0, (9.12)

5. since s = o + jw and & = Refs},

X(s) = E - 7 Res} > -a.

We note, in particular, that just as the Fourier transform does not converge for all
signals, the Laplace transform may converge for some values of Re{s} and not for others.
In eq. (9.13), the Laplace transform converges only for = Re{s} > —a. If a is positive,

then X(s) can be evaluated at o = 0 to obtain
X(0 + jw) =

Tore (9.15)

fora = Othe Laplace transform is equal to the Fourier transform,

6. Example x(t) = —e “u(-2).

X(s) = -| ee Su(-t)dt

0
= -| en ta gy,
1
X(s) = a

For convergence in this example, we require that Re{s + a} < 0, or Re{s} < —a; that is,

= £ 1
ーe ツー の 一， 一 一 一
sta

Refs} < —a.

(9.19)

7. Comparing 605. (9.14) and (9.19), we see that the algebraic expression for the
Laplace transform is identical for both of the signals considered in Examples 9.1 and 9.2.
However, from the same equations, we also see that the set of values of s for which the
expression is valid is very different in the two examples. This serves to illustrate the fact
that, in specifying the Laplace transform of a signal, both the algebraic expression and
the range of values of s for which this expression is valid are required. In general, the
range of values of s for which the integral in eq.(9.3) converges is referred to as the region
of convergence (which we abbreviate as ROC) of the Laplace transform. That is, the
ROC consists of those values of s = d + jw for which the Fourier transform of x(rJe”*"
converges.

8. A convenient way to display the ROC is shown in Figure 9.1. The variable s is a
complex number, and in the figure we display the complex plane, generally referred to as
the s-plane, associated with this complex variable. The coordinate axes are Re{s} along
the horizontal axis and Sr{s} along the vertical axis. The horizontal and vertical axes are
sometimes referred to as the d-axis and the jw-axis, respectively. The shaded region in
Figure 9.1(a) represents the set of points in the s-plane corresponding to the region of
convergence for Example 9.1. The shaded region in Figure 9.1(b) indicates the region of
convergence for Example 9.2.

im Im

レレ
Y splane

S

_

«plane

=a Ge

Re

(a) fb)

Figure 9.1 (a) ROC for Example 9.1; (b) ROC for Example 9.2.

9. Example x(t) = 3e 700) — 2e 'u(t). (9.20)

The algebraic expression for the Laplace transform is then

X(s) = | Fe u(t) — 2e so dt
> ー (9.21)
| ee “u(t)dt 一 2] ee “u(t)dt.
3 2
Xi) =; ;
6) s+2 stl (9.22)

To determine the ROC we note that x(r) is a sum of two real exponentials, and
from eq. (9.21) we see that X(s) is the sum of the Laplace transforms of each of the
individual terms. The first term is the Laplace transform of 3e7*u(r) and the second
term the Laplace transform of —2e ‘u(r). From Example 9.1, we know that

1

e u(t) 5 st Res>-—l,

un > = Rels)> -2

10. The set of values of Re{s} for which the Laplace transforms of both terms converge is
@Rets} > —1, and thus, combining the two terms on the right-hand side of eq. (9.22), we
obtain

*ー1

> _ 1 2 4 u Pr
3e “u(t) — 2e ‘u(t) — rer) ets} ラー1. (9.23)
_ NG)
X(s) = Do’

where N(s) and D(s) are the numerator polynomial and denominator polynomial,

Im

s-plane

Nk
AA
-?

Re

10

11. For rational Laplace transforms, the roots of the numerator polynomial are com-
monly referred to as the zeros of X(s), since, for those values of s, X(s) = 0. The roots
of the denominator polynomial are referred to as the poles of X(s), and for those values
of s, X(s) is infinite. The poles and zeros of X(s) in the finite s-plane completely char-
acterize the algebraic expression for X(s) to within a scale factor. The representation of
X(s) through its poles and zeros in the s-plane is referred to as the pole-zero plot of X(s).

Also, while they are not needed to specify the algebraic form of a rational transform
X(s), it is sometimes convenient to refer to poles or zeros of X(s) at infinity. Specifically,
if the order of the denominator polynomial is greater than the order of the numerator poly-
nomial, then X(s) will become zero as s approaches infinity. Conversely, if the order of the
numerator polynomial is greater than the order of the denominator, then X(s) will become
unbounded as s approaches infinity. This behavior can be interpreted as zeros or poles at
infinity.

12. x(t) = 60) 一 Se un + Leu. (9.32)
4 1 11
X(s) = 1- 3571 + 3.27 Res}>2, (9.34)
6
XO = nea Rels) > 2, (9.35)
sm
|
A + Be a

13. 13
Fourier Transform Convergence and Laplace Transform ROC

14. THE REGION OF CONVERGENCE FOR LAPLACE TRANSFORMS

Property 1: The ROC of X(s) consists of strips parallel to the jw-axis in the s-plane.

Property 2: For rational Laplace transforms, the ROC does not contain any poles.

Property 3: If x(¢) is of finite duration and is absolutely integrable, then the ROC is
the entire s-plane.

Decaying exponential
N

Growing exponential

15. Property 4: If x(r) is right sided, and if the line Re{s} = go is in the ROC, then all
values of s for which Re{s} > oo will also be in the ROC.

xt)

T t Figure 9.6 Right-sided signal.

suppose that the Laplace transform converges for some value of o, which we denote by
go. Then

ㅣ lle dr < (9.44)
or equivalently, since x( is right sided,

[ Inle ou dt <=. (9.45)
n

Then if O, > oo, it must also be true that x(1)e ”'' is absolutely integrable,

16. Property 5: If x( is left sided, and if the line Refs} = oo is in the ROC, then all
values of s for which Re(s) < oo will also be in the ROC.

x(t)

Tt Figure 9.8 Left-sided signal.

Property 6: If x(1) is two sided, and if the line Re{s} = on is in the ROC, then the
ROC will consist of a strip in the s-plane that includes the line Re{s} = oo.

x(t)

OA

Tp, t
@

xn xt)

Dune

To

17. Im Im

이 Be on

© 加

Figure 9.10 (a) ROC for xq(t) in Figure 9.9; (b) ROC for x.(2) in Figure
9.9; (c) the ROC for x(t) = xa(t) + x(t), assuming that the ROCs in (a) and
(b) overlap.

18. Property 7: Ifthe Laplace transform X(s) of x(t) is rational, then its ROC is bounded
by poles or extends to infinity. In addition, no poles of X(s) are contained in the ROC.

Property 8: Ifthe Laplace transform X(s) of x(1) is rational, then if x(t) is right sided, |
the ROC is the region in the s-plane to the right of the rightmost pole. If x(r) is left sided,
the ROC is the region in the s-plane to the left of the leftmost pole.

19. (9.52)

1
(s + Dis + 2)'

X(s)

Im

Im

s-plane

Re

s-plane

Re

s-plane

Re

s-plane

X-X-

(c)

20. THE INVERSE LAPLACE TRANSFORM
X(o + jo) = Hxtde””) = | ADE ist qi

3006 = F'{X(a + jo) = x] X(o + jue!" do,

on

ar, multiplying both sides by e”', we obtain

x(t) = >| Ko + jo)e or dw.

1 마쓰
x(t) = | X(s)e" ds.

£ 7 一 jz

This equation states that x(1) can be represented as a weighted integral of complex
exponentials. The contour of integration in eq. (9.56) is the straight line in the s-plane
corresponding to all points s satisfying Re{s} = a. This line is parallel to the jw-axis.
Furthermore. we can choose anv such line in the ROC—i.e., we can choose any value
of o such that X(@ + jw) converges.

21. Alternate method to find the inverse Laplace
21

22. 1

* の = で Te+ 2

Res} > -1.
1 A B

o srl 542

A = [(s + DAS) --ı = 1,
B = [(s + 2)X(5)]|,..-2 = 一 1.

Thus, the partial-fraction expansion for X(s) is

X(s) =

s+l o s+2

22

23. > 2 1 u
ecu < 一 > SET Rels) > —1,
eun eos => Res} > -2.

We thus obtain

a) > Refs} > -1.

- 1
le G+ Its +2)

24. Let us now suppose that the algebraic expression for X(s) is again given by eq. (9.58),
but that the ROC is now the left-half plane Re{s} < —2.”

1

£
u = _
e ul ee 호 Refs} = —1,
-e ツー が es +, Res} < -2
s+2 i "
50 that
= = £ 1
x(t) = [me +e “lu — Res} < 一 2.

(s + 1)(s + 2)

24

25. GEOMETRIC EVALUATION OF THE FOURIER TRANSFC Im
FROM THE POLE-ZERO PLOT s-plane

As we saw in Section 9.1, the Fourier transform of a signal is the Laplace transform evalu-
ated on the jw-axis. In this section we discuss a procedure for geometrically evaluating the
Fourier transform and, more generally, the Laplace transform at any set of values from the
pole-zero pattern associated with a rational Laplace transform. To develop the procedure,
let us first consider a Laplace transform with a single zero [i.e., X(s) = s — a], which we
evaluate at a specific value of s, say, s = sı. The algebraic expression sı — a is the sum
of two complex numbers, 51 and —a, each of which can be represented as a vector in the
complex plane, as illustrated in Figure 9.15. The vector representing the complex number
si — ais then the vector sum of 51 and —a, which we see in the figure to be a vector from
the zero ats = ato the point sı. The value of X(s) then has a magnitude that is the length
of this vector and an angle that is the angle of the vector relative to the real axis.

26. If X(s)

instead has a single pole at s = a [i.e., X(s) = 1/(s — a)], then the denominator would be
represented by the same vector sum of sı and —a, and the value of X(sı) would have a!
magnitude that is the reciprocal of the length of the vector from the pole to s = sı and an
angle that is the negative of the angle of the vector with the real axis.

A more general rational Laplace transform consists of a product of pole and zero
terms of the form discussed in the preceding paragraph; that is, it can be factored into the
form

X) = met Bo. (9.70)
[26 — aj)

To evaluate X(s) at 5 = sı, each term in the product is represented by a vector from the
zero or pole to the point s;. The magnitude of X(s¡) is then the magnitude of the scale
factor M, times the product of the lengths of the zero vectors (i.e., the vectors from the
zeros to 51) divided by the product of the lengths of the pole vectors (i.e., the vectors from
the poles to sı). The angle of the complex number X(s) is the sum of the angles of the zero
vectors minus the sum of the angles of the pole vectors.

27. it is readily evident that the length of the pole
vector monotonically increases with
increasing w, and thus, the magnitude of the
Fourier transform will monotonically
decrease with increasing w.
27

28. PROPERTIES OF THE LAPLACE TRANSFORM

Linearity of the Laplace Transform
If

with a region of convergence that

£
AO > Xi) be denoted as R,

and

with a region of convergence that

£
Xx) X) will be denoted as Ro,

then

£
axı() + bxal) + aXi(s) + bX2(s), with ROC
containing Ry N Ro.

As indicated, the region of convergence of X(s) is at least the intersection of Rı and Ra,
which could be empty, in which case X(s) has no region of convergence—i.e., x(1) has

no Laplace transform. 28

29. Time Shifting

x(t) > XG) with ROC = R,

X(t — 10) > 0906) with ROC = R.

Shifting in the s-Domain

£
x(t) — X(s), with ROC = R,

0090) > X(s—5), with ROC = R + Refs}.

£
ex) <—> X(s — jwo), with ROC = R.

30. Time Scaling

£
x(t) — X(s),

with ROC = R,

2 1
xa) > コテ
lal

} with ROC Rı =aR.

Im

s-plane

s-plane

Re

A A

Re

al

xD) = X(=s) with ROC = —R.

31. Conjugation

£
x(t) — X(s), with ROC = R,

xo > x), with ROC = R.

X(s) = X*(s*) when x(t) is real.

Consequently, if x(t) is real and if X(s) has a pole or zero at s = so (i.e., if X(s) is un-
bounded or zero at s = so), then X(s) also has a pole or zero at the complex conjugate
points = 5.

32. Convolution Property

£
xi の — Xi(s),

£
x) — Xa(s),

with ROC = Rı,

with ROC = Ro,

nm の Or の ~ XOX),

with ROC containing Rı N Ra.

Differentiation in the Time Domain

£
x(t) X),

with ROC = R,

040

2, 5x
ao. (s),

with ROC containing R.

32

33. Differentiation in the s-Domain

x(t) 4 X(s), with ROC = R,

with ROC = R.

math of, 0,
ds

Integration in the Time Domain

e

x(t) ÁS X(s), with ROC = R,

]

m 7
| x(t)dtT — Iyo, with ROC containing
= 8 RN {Refs} > 0).

34. The Initial- and Final-Value Theorems

Under the specific constraints that x(t) = 0 for + < 0 and that x(D contains no impulses
or higher order singularities at the origin, one can directly calculate, from the Laplace
transform, the initial value x(0*)—i.e., x(t) as 7 approaches zero from positive values of
t. Specifically the initial-value theorem states that

x(0*) = lim sX(s), (9.110)

Also, if x(t) = 0 for r < 0 and, in addition, x(t) has a finite limit as 7 > +, then the final-
value theorem says that lim x(t) = limsX(s). (9.111)
to $0

34

35. TABLE 9.2 LAPLACE TRANSFORMS OF ELEMENTARY FUNCTIONS
Transform
pair Signal Transform ROC
1 60) 1 Alls
2 u(t) ı Refs} >0
3 un) : Refs} < 0
a 1
4 ao = Refs} > 0
5 o = ets} <0
=D s
= 1 =
6 eut) == Rels)> a
PE 1 一
7 eu) oa Res} <-a
no. 1 _
8 won’ u(t) Gray Reis} > —a
9 a Lt ー の an Refs} < -
Gen ur) Gray ets a

36. Transform
pair

12
13
14

15

ô@-T)

[cos wot]u(r)
[sin wot}u(r)
[e° cos egg の

le“ sinworlu(t)

d"5(t)
dt"

0) =

und) = u(t) +" ul)
m

n times

aT

e
s

2

2 + wè
w

2 2

2+ 08
Sta

(s + a} + w

we

(s + a} +0

All
Re

Re:

Rel

All

Re

ROC
s
s}>0
s}>0
s}>-a

s}> -a

s}>0

36

37. ANALYSIS AND CHARACTERIZATION OF LTI SYSTEMS USING
THE LAPLACE TRANSFORM

Y(s) = H(s)X(s).

where X(s), Y(s), and H(s) are the Laplace transforms of the input, output, and
impulse response of the system, respectively.

If the ROC of H(s) includes the imaginary axis, then for s = jw, H(s) is the
frequency response of the LTI system. In the broader context of the Laplace
transform, H(s) is commonly referred to as the system function or, alternatively, the

transfer function. Many properties of LTI systems can be closely associated with the
characteristics of the system function in ‘the s- plane.

38. Causality

For a causal LTI system, the impulse response is zero for ¢ < O and thus is right sided.

The ROC associated with the system function for a
causal system is a right-half plane.

|

38

39. Consider a system with impulse response
A(t) = e u(t). (9.113)

Since h(t) = O for r < 0, this system is causal, Also, the system function can be obtained
from Example 9.1:

1
s+]

H(s) = , Res}> -1. (9.114)
Tn this case, the system function is rational and the ROC in eq. (9.114) is to the right of
the rightmost pole, consistent with our statement that causality for systems with rational
system functions is equivalent to the ROC being to the right of the rightmost pole.

39

40. Consider a system with impulse response
h(t) = eh.

Since h(t) # 0 for £ < 0, this system is not causal. Also, from Example 9.7, the system
function is

ー1 < Refs} < +1.

Thus, H(s) is rational and has an ROC that is not to the right of the the rightmost pole,
consistent with the fact that the system is not causal.

40

41. Example Consider the system function

e
H(s) = 아고 Reis} > 一 1. (9.115)

For this system, the ROC is to the right of the rightmost pole. Therefore, the impulse
response must be right sided. To determine the impulse response, we first use the result

£
e 'ult) — = : i Res} > -1.
eur + D > oS Rels} > -1,

so that the impulse response associated with the system is
A(t) = ent + 1),
which is nonzero for —1 < t < 0. Hence, the system is not causal. This example serves

as a reminder that causality implies that the ROC is to the right of the rightmost pole,
but the converse is not in general true, unless the system function is rational.

42. In an exactly analogous manner, we can deal with the concept of anticausality. A
system is anticausal if its impulse response A(t) = 0 for £ > 0. Since in that case A(r)
would be left sided, we know from Section 9.2 that the ROC of the system function H(s)
would have to be a left-half plane. Again, in general, the converse is not true. That is, if
the ROC of H(s) is a left-half plane, all we know is that A(t) is left sided. However, if H(s)
is rational, then having an ROC to the left of the leftmost pole is equivalent to the system
being anticausal.

43. Stability

Let us consider an LTI system with system function

=1

Ss
AC) = n=

44. Im

s-plane

Re

s-plane

四

45. For one particular and very important class of systems, stability can be characterized
very simply in terms of the locations of the poles. Specifically, consider a causal LTI system
with a rational system function H(s). Since the system is causal, the ROC is to the right of
the rightmost pole. Consequently, for this system to be stable (i.e., for the ROC to include
the jw-axis), the rightmost pole of H(s) must be to the left of the jw-axis. That is,

A causal system with rational system function H(s)
is stable if and only if all of the poles of H(s) lie in
the left-half of the s-plane—i.e., all of the poles have
negative real parts.

46. LTI Systems Characterized by Linear Constant-Coefficient

Differential Equations

20

+ 3y(t) = 30).

sY(s) + 3¥(s) = X(s).

¥(s)

H(s) = XO)’

we obtain, for this system,

47. This, then, provides the algebraic expression for the system function, but not the
region of convergence.

If. in addition to the

differential equation, we know that the system is causal, then the ROC can be inferred
to be to the right of the rightmost pole, which in this case corresponds to Re{s} > —3. If
the system were known to be anticausal, then the ROC associated with H(s) would be
Re{s} < —3. The corresponding impulse response in the causal case is

h(t) = e*u(t), (9.129)
whereas in the anticausal case it is

h(t) = -e "(一 の. (9.130)

48. N k M ik
d*y(t) _ d* x(t)
TS = >, qe (9.131)

Applying the Laplace transform to both sides and using the linearity and differenti-
ation properties repeatedly, we obtain

N m
E ask po = E he (9.132)
k=0

k=0

M
| > bys* |
H(s) = 3 (9.133)
ayst
ze]

Thus, the system function for a system specified by a differential equation is always ratio-
nal, with zeros at the solutions of

a
SN bist = (9.134)
k=0

and poles at the solutions of

(9.135)

48

49. if the values of R, L, and C are all positive, the poles of this system
function will have negative real parts, and consequently, the system will be
stable.
49

50. Examples Relating System Behavior to the System Function
Suppose we know that if the input to an LTI system is
x(t) = e "ult),

then the output is

4

y(t) = [e *—e "Jun.

Taking Laplace transforms of x(t) and y(1), we get

X(s) = Rels) > -3,

1
s+3
and

YO = ara

Refs} > -1. Refs} > -1.

51. rs) _ s+3 _ s+3
- 6) (s+DG65+2) 9243942

Furthermore, we can also determine the ROC for this system. In particular, we
know from the convolution property set forth in Section 9.5.6 that the ROC of Y(s) must
include at least the intersections of the ROCs of X(s) and H(s). Examining the three
possible choices for the ROC of H(s) (i.e., to the left of the pole ats = —2, between the
poles at -2 and —1, and to the right of the pole ats = — 1), we see that the only choice
that is consistent with the ROCs of X(s) and Y(s) is Re{s} > —1. Since this is to the
right of the rightmost pole of H(s), we conclude that H(s) is causal, and since both poles
of H(s) have negative real parts, it follows that the system is stable. Moreover, from the
relationship between eqs. (9,131) and (9.133), we can specify the differential equation
that, together with the condition of initial rest, characterizes the system:

04270) ,dy(t)
3 34)
de dt

+ 2y = zo + 3300.

52. SYSTEM FUNCTION ALGEBRA AND BLOCK DIAGRAM REPRESENTATIONS

x(t)

hi⑪
Hi($)

hatt)
Haz(s)

(a)

x(t) 一 一

ht) hatt)

Hi(s) Pa] Hals) > 0

Figure 9.30

y(t)

53. System Functions for Interconnections of LTI Systems

Consider the parallel interconnection of two systems, as shown in Figure 9.30(a). The
impulse response of the overall system is

A(t) = hy(t) + M), (9.155)
and from the linearity of the Laplace transform,
H(s) = Hı(s) + HA). (9.156)
Similarly, the impulse response of the series interconnection in Figure 9.30(b) is
A(t) = hy(t) * h2(t), (9.157)
and the associated system function is

H(s) = Hy(s)H2s). (9.158)

54. x(t) y(t)
Y(s) = Hi(s)E(s),

E(s) = X(s) - Z(s),

and
Z(s) = Ha(s)Y(s),
from which we obtain the relation
Y(s) = Hi(s)[X(s) — H2(s)¥(s)],

or
Y(s) _ _ Hi(s)
xo ~ 2 = Toomey

55. Block Diagram Representations for Causal LTI Systems
Described by Differential Equations and Rational

System Functions

Consider the causal LTI system with system function

の + 3» = x,

Vs

H(s) = 143%

KLUB CA E > y
1 -3
H(s) = 7
s+3
(a)
x⑩ el 2 von
= = 1 . 3
+3

(o)
(a) Block diagram representation of the causal LTI system in
(b) equivalent block diagram representation.

56. Consider now the causal LTI system with system function

Ho = £720 je +2). (9.164)

s+3

As suggested by eq. (9.164), this system can be thought of as a cascade of a system
with system function 1/(s + 3) followed by a system with system function s + 2, and

we have illustrated this in Figure 9.33(a), in which we have used the block diagram in
Figure 9.32(a) to represent 1/(s + 3).

vn

57. equivalent block diagram representation.

y(t)

ae

It is also possible to obtain an alternative block diagram representation for the
system in eq. (9.164). Using the linearity and differentiation properties of the Laplace
transform, we know that y(r) and z(D in Figure 9.33 (a) are related by

020

YN = di

+ 220).

However, the input e(f) to the integrator is exactly the derivative of the output z(t), so
that

y(t) = e(t) + 220),

58. Consider next a causal second-order system with system function

1 il
HO AD TT 2.169)
The input x(+) and output y(z) for this system satisfy the differential equation
ya) ¿00
I 43 + 270) = x. À
de af pr 十 2y(1) = x(t). (9.166)

vit)

59. |
:
|

二 ーー
(b)
T
3
=
x(t)
1 1
H(s) = -
때 6571 6+2'

y(t)

60. THE UNILATERAL LAPLACE TRANSFORM

In the preceding sections of this chapter, we have dealt with what is commonly called
the bilateral Laplace transform. In this section, we introduce and examine a somewhat
different transform, the unilateral Laplace transform, which is of considerable value in
analyzing causal systems and, particularly, systems specified by linear constant-coefficient
differential equations with nonzero initial conditions (i.e., systems that are not initially at
rest).

The unilateral Laplace transform of a continuous-time signal x(1) is defined as

800) à [ xen" dt, (9.170)
に

where the lower limit of integration, 0”, signifies that we include in the interval of
integration any impulses or higher order singularity functions concentrated at r = 0.
Once again we adopt a convenient shorthand notation for a signal and its unilateral
Laplace transform:

x(t) > Xs) = ULL xt}. (9.171)

60

61. Since the unilateral transform of x(r) is identical to the bilateral transform of the
signal obtained from x(t) by setting its value to 0 for all + < 0, many of the insights,
concepts, and results pertaining to bilateral transforms can be directly adapted to the
unilateral case. For example, using Property 4 in Section 9.2 for right-sided signals,
we see that the ROC for eq. (9.170) is always a right-half plane. The evaluation of
the inverse unilateral Laplace transforms is also the same as for bilateral transforms,
with the constraint that the ROC for a unilateral transform must always be a right-half
plane.

Consider the signal

(n- 1)!

x(0) = e “ult.

Since x(t) = 0 for t < 0, the unilateral and bilateral transforms are identical.

1

X(s) = Gta”

Rels) > —a.

61

62. Consider next
x(t) = eur + 1). (9.174)

The bilateral transform X(s) for this example can be obtained from Example 9.1 and the
time-shifting property (Section 9.5.2):

X(s) = = Res} > —a. (9.175)

By contrast, the unilateral transform is

X(s) = | eur + le dt

0-

- [ A (9.176)
o
1

一 rara Reis} > —a.

62

63. Consider the signal
x(t) = S(t) + 2uy(t) + e'ult). (9.177)

Since x(t) = 0 fort < 0, and since singularities at the origin are included in the interval
of integration, the unilateral transform for x(r) is the same as the bilateral transform.

Xs) = X(s) = 14254 2. = s(2s— 1)

— 一 Reis} >1. (9.178)

63

64. PROPERTIES OF THE UNILATERAL LAPLACE TRANSFORM

Property Signal Unilateral Laplace Transform
30) 609]
xO X)
(1) X)

Linearity axı() + bx2(t) aXi(s) + bXr(s)

Shifting in the s-domain

Time scaling

ee の

x(at) a>0

Us — 50)

|

Conjugation

x(t)

x*(s)

Convolution (assuming
that xi and x(t)
are identically zero for
t<0)

x(t) * (の

POR の O)

64

65. Property Signal Unilateral Laplace Transform

Differentiation in the time Lx sX(s) — x(0 )
domain
A 6 d
Differentiation in the IMM) — X(s)
. ds
s-domain
x 1
Integration in the time [ x(r)dr —X(s)
domain ° ;

Initial- and Final-Value Theorems
Tf x(t) contains no impulses or higher-order singularities at ¢ = 0, then
x(0*) = lim sX(s)

lim x(7) = lim s (5)
1 q

66. Properties of the Unilateral Laplace Transform

A particularly important difference between the properties of the unilateral and bi-
lateral transforms is the differentiation property. Consider a signal x(t) with unilateral
Laplace transform X(s). Then, integrating by parts, we find that the unilateral transform
of dx(1)/dt is given by

* dx) st ap = mens
[ 고 e “dt = x(De

MM + 中 x( の De "gd (9.193)
= sX(s) - (0).

Similarly, a second application of this would yield the unilateral Laplace transform of
dx( の 7 だ 、iLe.。

SCH) — sx(07) — x (05), (9.194)

where x'(07) denotes the derivative of x(r) evaluated att = 0”. Clearly, we can continue
the procedure to obtain the unilateral transform of higher derivatives.

67. Solving Differential Equations Using the Unilateral
Laplace Transform

A primary use of the unilateral Laplace transform is in obtaining the solution of linear
constant-coefficient differential equations with nonzero initial conditions. We illustrate
this in the following example:

4090) | dy(t)
+ 3 + 270) = x(t 9.189
Se + 35S +20 = (の (9.189)
Consider the system characterized by the differential equation (9.189) with initial con-
ditions
yO") = B, y@)=y¥. (9.195)

Let x(t) = au(t). Then, applying the unilateral transform to both sides of eq. (9.189),
we obtain

SUS) - Bs — y + 38968) — 36 + 245) = で (9.196)

68. or

6) = _ Bis +3) _ Y 。 a
(+ 1D(s+2) | + 1) + 2) GADEA (9.197)

we see that the last

term on the right-hand side of eq. (9.197) is precisely the unilateral Laplace transform
of the response of the system when the initial conditions in eq. (9.195) are both zero
(B = y = 0). That is, the last term represents the response of the causal LTI system
described by eq. (9.189) and the condition of initial rest. This response is often referred

to as the zero-state response—i.e., the response when the initial state (the set of initial
conditions in eq. (9.195)) is zero.

68

69. An analogous interpretation applies to the first two terms on the right-hand side of
eq. (9.197). These terms represent the unilateral transform of the response of the system
when the input is zero (a = 0). This response is commonly referred to as the zero-
input response. Note that the zero-input response is a linear function of the values of
the initial conditions (e.g., doubling the values of both B and y doubles the zero-input
response). Furthermore, eq. (9.197) illustrates an important fact about the solution of
linear constant-coefficient differential equations with nonzero initial conditions, namely,
that the overall response is simply the superposition of the zero-state and the zero-input
responses. The zero-state response is the response obtained by setting the initial condi-
tions to zero—i.e., it is the response of an LTI system defined by the differential equa-
tion and the condition of initial rest. The zero-input response is the response to the initial

conditions with the input set to zero.

69

70. Finally, for any values of a, B, and y, we can, of course, expand Y(s) in eq. (9.197)
in a partial-fraction expansion and invert to obtain y(t). For example, fa = 2, B = 3,
and y = —5, then performing a partial-fraction expansion for eq. (9.197) we find that

Ws) = 1 _ . 4 3 (9.198)

Applying Example 9.32 to each term then yields

y(t) = [l-e'+3eJu(t) for 1>0. (9.199)