ABSTRACT
Pierre Simon Laplace (1749-1827) was a French astronomer and mathematician, whose work was important to the development of mathematical astronomy and statistics He also put the theory of mathematical probability on a sound footing He first discovered Laplace’s equation and Laplace transform, which is discussed in this chapter To Laplace, the universe is nothing but a giant problem in calculus
Laplace was born in Beaumont-en-Auge, Normandy, France on March 23, 1749, and died in Paris on March 5, 1827 Laplace became a professor of mathematics at the age of 20 He is remembered as one of the greatest scientists of all time He is sometimes referred to as the “Newton of France” He was widely known for his five-volume work, Celestial Mechanics, which supplemented the work of Newton on astronomy After the publication of this work, Laplace continued to apply his ideas of physics to other problems such as capillary action, double refraction, the velocity of sound, the theory of heat, rotation of the cooling earth, and elastic fluids He was born and died a Catholic
We saw in Chapter 2 that system analysis in the time-domain involves the evaluation of the convolution integral or sum since the response of the system is given by the convolution of the input and the impulse response This chapter introduces the Laplace transform, a very powerful, alternative tool for analyzing systems The Laplace transform is named after Pierre Simon Laplace (1749-1827), the French astronomer and mathematician In contrast to the time-domain models studied in the previous chapter, the Laplace transform is a frequency-domain representation that makes analysis and design of linear systems simpler
The Laplace transform is a well-established tool in analyzing continuous-time, linear systems It is important for a number of reasons First, it is applicable to a wider variety of inputs The Laplace transform of an unbounded signal can be found Second, Laplace transform is powerful for providing us in one single operation the complete response, that is, the steady-state plus transient or homogeneous, and particular Third, it automatically includes the initial conditions in the system analysis It allows us to convert ordinary differential equations into algebraic equations, which are easier to manipulate and solve It converts convolution into a simple multiplication Fourth, we can apply Laplace transform to generate the transfer function representation of a continuous-time LTI system
The chapter begins with the definition of the Laplace transform and uses that definition to derive the transform of some basic, important functions We will consider some properties of Laplace transform which are helpful in obtaining the Laplace transform of other functions We then consider the inverse Laplace transform We apply all these to solving integro-differential equations and system analysis, especially in circuit and control systems We finally demonstrate how MATLAB® can be used to do most of what we cover in this chapter
There are two types of Laplace transforms: 1 The two-sided (or bilateral) Laplace transform: The two-sided Laplace transform allows time functions to be nonzero for
negative time
L[ ( )] ( ) ( )x t X s x t e dtst= = - -¥
(31)
where s is the complex frequency given by
s j= +s w (32)
where σ (in Np/s) and ω (in rad/s) are the real and imaginary parts of s, respectively The advantage of the bilateral Laplace transform is that it can handle both causal and noncausal signals over −∞ to ∞
2 The one-sided (or unilateral) form: A one-sided Laplace transform is zero for negative time, that is, t < 0 It is
found from Equation 31 by setting the lower limit of the integral equal to zero
L[ ( )] ( ) ( )x t X s x t e dtst= = - ¥
(33)
where 0− denotes a time just before 0 The one-sided Laplace transform is a more commonly used transform
than the two-sided Laplace transform, because we encounter only positivetime signals in practice and it is defined for positive-time signals
According to Equation 33, the t-domain signal x(t) is changed into the s-domain function X(s)
The Laplace transform of a signal x(t) is the integration of the product of x(t) and e−st over the interval from 0 to +∞
In inverse Laplace transform the s-domain X(s) is changed back to t-domain x(t) and is given as
L-
= = ò1 12 1
[ ( )] ( ) ( )X s x t j X s e ds st
p s
(34)
where the integration is performed along a straight line (σ1 + jω, −∞.< ω.< ∞) as shown in Figure 31 The Laplace transform pair x(t) and X(s) are represented as
x t X s( ) ( )Û (35)
because there is one-to-one correspondence between x(t) and X(s)
A signal x(t) is Laplace transformable if the integral in Equation 33 exists; that is, in order for x(t) to have a Laplace transform, the integral in Equation 33 must converge The integral converges when
x t e dtj t( ) ( )- + ¥
(36)
This can be simplified as
x t e dt x t e dt
x t e dt
( ) ( )
( )
£ < ¥
(37)
Since |ejωt| = 1 for any value of t,
x t e dtt( ) - ¥
< ¥ - ò s 0
(38)
for some real value of σ = σc The inequality in Equation 38 states that the Laplace transform exists Hence, the region of convergence (ROC) for Laplace transform is Re(s) = σ > σc, as shown in Figure 31, the ROC means, the range of s for which the Laplace transform converges We cannot define X(s) outside the region of convergence The portion of the s plane in which the integral converges is found by the value of σ All signals that we come across within the system analysis fulfill the convergence criterion in Equation 38 and have their corresponding Laplace transforms
Example 3.1
Example 3.2
The properties of the Laplace transform are summarized below, which are useful to obtain the transform of many signals Some of the properties are linearity, scaling, time shifting, frequency shifting, time differentiation, time convolution, time integration, frequency differentiation, time periodicity, modulation, and initial and final values
Let the Laplace transform of two signals x1(t) and x2(t) be X1(s) and X2(s), respectively Then the linearity property states that the Laplace transform of the sum of two functions of time is equal to the sum of the transforms of each function and is given as
L a x t a x t a X s a X s1 1 2 2 1 1 2 2( ) ( ) ( ) ( )+êë úû = + (39)
where a1 and a2 are constants The linearity property is explained as follows:
L[ ( ) ( )] [ ( ) ( )]
( )
a x t a x t a x t a x t e dt
a x t e d
+ = +
=
t a x t e dt
a X s a X s
at+
= +
( )
( ) ( )
(310)
For example, consider
x t e e t t( ) .= + -3 54 6 (311)
By using linearity property, X(s) = L[3e4t + 5e−6t] = L[3e4t] + L[5e−6t]
X s
s s ( ) =
- +
+ 3
4 5
6 (312)
Let X(s) be the Laplace transform of the signal x(t) Let a be any positive real constant, then Laplace transform of x(at) is given as
L[ ( )] ( )x at x at e dtst= - ¥
(313)
Let λ = at, dλ = a dt, we obtain
L[ ( )] ( ) ( )x at x e e d a a
x e d s
ò òl l l ll l l
1 (314)
Thus, the scaling property is obtained as
L[ ( )] ,x at a
X s a
a= æ è ç
ö ø ÷ >
1 0 (315)
The unilateral Laplace transform is valid only for causal signals and not valid for noncausal signal (when a is negative ie, a < 0)
For example, consider
L[ ( )]r t
s = 12 (316)
By applying the scaling property,
L[ ( )] ( )r t s s2
1 2
1 2
2 2 2= =/
(317)
Often we can see that there is a signal delay in control systems due to transmission delays or other effects Let X(s) be the Laplace transform of signal x(t) Then time shifting is given as
L x t a u t a e X ss( ) ( ) ( )- -éë ùû = -a (318)
If a function is delayed in time by a, then the Laplace transform of the function is multiplied by e−αs The time shifting property can be proved as follows:
L[ ( ) ( )] ( ) ( ) ,x t a u t a x t a u t a e dt ast-- = - - ³- ¥
0 (319)
If u(t − a) = 0 for t < a and u(t − a) = 1 for t > a then,
L[ ( ) ( )] ( )( )x t a u t a e dt x t a e dtst st a
- - = + -- - ¥¥
òò0 1 0
(320)
by substituting λ = t − a, dλ = dt, and t = λ + a As t→a, λ→0, and as t→∞, λ→∞ thus,
L[ ( ) ( )] ( )
( )
( )x t a u t a x e
e x e d
e X
- - =
=
=
l
l l
( )s
(321)
For example, consider
g t u t u t( ) ( ) ( ).= - - -4 6 (322)
We know that L[ ( )]u t s
= 1
Using linearity property and the time shifting property,
G s u t u t
s e
s e
s e es s s s( ) [ ( ) ( )]= = = ( )L - - - - -- - - -4 6 1 1 14 6 4 6 (323)
Let X(s) be the Laplace transform of the signal x(t) Then the frequency shifting is given as
L e x t u t e x t e dt x t e dt X s aat at st s a t-- - ¥
- +éë ùû = = = +ò( ) ( ) ( ) ( ) ( )( ) 0 0
L e x t u t X s aat-éë ùû = +( ) ( ) ( )
The Laplace transform turns multiplication by e−at in the time-domain into s + a in the transform X(s)
For example, consider
cos ( )w
w tu t
s
s Û
+2 2 (325)
By applying the frequency property and replacing every s by s + a, the Laplace transform of the cosine function is given by
L[ cos ( )] ( )e tu t
s a
s a
at-w w
= + + +2 2
(326)
Time differentiation property is important for solving differential equations Let X(s) be the Laplace transform of the signal x(t) The Laplace transform of the derivative of x(t) is given as
L dx dt
u t dx dt
e dtst( ) ( )é ëê
ù ûú
= -
(327)
By differentiation by parts, U = e−st, dU = −se−stdt, and dV dxdt dt dx= æ
è ç
ö ø ÷ = , V = x(t),
then
L dx dt
u t x t e x t se dt
x s
st st( ) ( )
( )
( )é ëê
ù ûú
= - -éë ùû
= - +
0 0 x t e dt
sX s x
st( )
( ) ( )
= -
(328)
L ¢éë ùû = -
-x t sX s x( ) ( ) ( )0
By differentiating the above equation, we get
L L d x dt
s x t x s sX s x x
s
2 0 0 0 é
ë ê
ù
û ú = éë ùû - = -éë ùû - ¢
=
¢ ¢ - - -( ) ( ) ( ) ( ) ( )
2 0 0X s sx x( ) ( ) ( )- - ¢- - (329)
L ¢¢éë ùû = - - ¢
- -x t s X s sx x( ) ( ) ( ) ( )2 0 0
In the same way, for the nth derivative, we get
L d x dt
s X s s x s x s x n
ë ê
ù
û ú = - - - -¢
For example, consider x(t) = cos ωt u(t) Then at t = 0, we have x(0) = 1 By applying time differentiation, the Laplace transform of the derivative of x(t) is
L d dt
tu t s s
s u
s
s
cos ( ) cos( ) ( )w w
-
w
- -é ëê
ù ûú
= +
æ è ç
ö ø ÷
= +
0 0
(331)
Let X(s) and H(s) be the Laplace transform of the signal x(t) and h(t) Then the convolution property is given as
L[ ( )* ( )] ( ) ( )x t h t X s H s= (332)
In time-domain the convolution property,
x t h t x t h d( )* ( ) = -( ) ( ) ¥
ò t t t 0
Considering the Laplace transform on both sides
L
L
x t h t h x t d e dt
x t h t
st( ) ( )éë ùû = - é
ë ê ê
ù
û ú ú
é
òò* ( ) ( )
( )* ( )
t t t 00
ë ùû = ( ) - é
ë ê ê
ù
û ú ú
òòh x t e dt dstt t t( ) 00
(333)
Let λ = t − λ so that t = τ + λ and dt = dλ
L x t h t h x e d d
h e
( )* ( ) ( )éë ùû = ( ) é
ë ê ê
ù
û ú ú
= ( )
ò òt l l t
t
d x e d H s X sst l ll
ò ò =( ) ( ) ( )
(334)
For example, consider
g t u t u t( ) ( ) ( ).= - - -3 4 By applying the Laplace transform on both sides
G s
s e es s( ) = -( )- -1 3 4
Considering the convolution of g(t)*g(t), which is G2(s),
G s
s e e
2 6 7 81 1 2( ) = -( ) = - +( )- - - - - (335)
By applying inverse Laplace transform, we get
x t g t g t t u t t u t t u t
r t
( ) ( )* ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
= = - - - - - + - -
= -
6 6 2 7 7 8 8
6 - - + -2 7 8r t r t( ) ( )
(336)
Let X(s) be the Laplace transform of the signal x(t) Then Laplace transform of the integral is given as
L x t dt s
X s t
( ) é
ë ê ê
ù
û ú ú
1 ( )
From the convolution property, we get
L L Lx d x u t d u t x t t
( ) ( ) ( ) ( )* ( )l l l l l 0 0 ò ò
é
ë ê ê
ù
û ú ú
= - é
ë ê ê
ù
û ú ú
= éë ùû
=
é
ë ê ê
ù
û ú ú
s X s
x t dt s
X s t
( )
( ) ( ) L (337)
For example, let x(t) = u(t) We know that X(s) = 1/s By applying the time integration property,
L 0
2 1 1 1
u t dt L t s s sò
é
ë ê ê
ù
û ú ú
= = =( ) [ ]
Let X(s) be the Laplace transform of the signal x(t) Then Laplace transform of the frequency differentiation is given as
L tx t
dX s ds
( ) ( )éë ùû = -
X(s) is given as
X s x t e dtst( ) = ( ) - ¥
Differentiating both sides, we get
dX s ds
x t te dt tx t e dt tx tst st( ) ( )( ) ( ) ( )= - = -( ) = -éë ùû-¥
L
The frequency differentiation property is given as
L tx t
dX s ds
( ) ( )éë ùû = - (338)
Similarly, the general form of frequency differentiation property is given as
L t x t
d X s ds
n ( ) ( )éë ùû = -
( )1 (339)
For example,
L t t
d ds
s
s
s
s cos ( )w - w
- w w
( ) = +
é ëê
ù ûú
= +2 2
L t t
d ds s
s
s sin ( )w -
w w
w w
( ) = +
é ëê
ù ûú
(341)
Consider a periodic signal x(t) and it can be expressed as a sum of time-shifted functions as (Figures 33 and 34)
x t x t x t x t
x t x t T u t T x t T u t
( ) ( ) (
( ) ( ) ( ) ( ) ( ) ( )
= + + +
= + - - + - -
T ) +
(342)
where 0 < t < T By applying the Laplace transform and the time-shifting property, we get
X s X s X s e X s e X s e
X s e e
( ) ( ) ( ) ( ) ( )
( )
= + + + +
= + +
1 1
- -+ +éë ùû 2 3Ts Tse
(343)
But
1 1
1 2 3+ + + + =
- a a a
a (344)
if |a| < 1
X s X s
e Ts ( ) ( )=
1 (345)
For example, consider
x t t t
t
X s x t e dt t est st
0 1 0 1 2
( )
( ) ( )
sin , ,
sin
= < < < <
ì í î
= =- -
p
p dt
e bt dt e a b
a bt b btat at
= +
-sin ( sin cos )
(346)
where a = −s b = π
X s e s
s t t e
s T
2 2 1 2( ) sin cos ( ) ,=
+ - -éë ùû =
+ +
p p p p p
p (347)
X s X s
e
e
s e
( )
( ) ( )( )
= ( )
-
= + + -
1 1
p p
(348)
Let X(s) be the Laplace transform of the signal x(t), then for any real number ω,
L
L
x t t X s j X s j
x t t j X s
( )cos
( )sin [ (
w w w
w
éë ùû = +( ) + -( )éë ùû
éë ùû = +
1 2
2 j X s jw w) ( )]- -
(349)
The signals x(t) cos ωt and x(t) sin ωt can be expressed as
x t t x t e e
x t t j x t e e
( )cos ( )
( )sin ( )
w
w
= +éë ùû
= -é
1 2
1 2 ë
ùû
(350)
From Equation 324,
L e x t X s jj t± ( )éë ùû = ±( )
w w (351)
For example, we know that
u t s
( )Û 1 (352)
u t t s j s j
s
s
u t t j
( )cos( )
( )sin( )
w w w w
w
Û +
+ -
é
ë ê
ù
û ú = +
Û
1 2
1 1
1 2
1 1 2 2s j s j so o
o+ -
- é
ë ê
ù
û ú = +w w
w w
(353)
The initial value x(0) and final value x(∞) are used to relate frequency domain expressions to the time-domain as time approaches zero and infinity, respectively These properties can be derived by using the differentiation property as
sX s x dx dt
dx dt
e dtst( ) ( )- = é ëê
ù ûú
L (354)
By taking limits, we get
lim ( ) ( ) s
sX s x ®¥
-éë ùû =0 0
Since x(0) is not dependent on s,
x sX s
s ( ) lim ( )0 =
®¥ (355)
This is called the initial-value theorem x(0) can be found using X(s), without using the inverse transform of x(t)
If we consider s→0, then
sX s x dx dt
e dt dx x xt( ) ( ) ( )- ( ) = = = ¥ -- - - ¥¥
-- òò0 00 00
x sX s
s ( ) lim ( )¥ =
®0 (356)
This is known as the final-value theorem. In the final-value theorem, all poles of X(s) must be located in the left half of the s-plane
For example, consider
x t e t u t X s
s
s
s
t( ) cos ( ) ( ) ( )= + Û = + +
+ + -4 3 4 2
2 3 2
By applying the initial-value theorem,
x sX s s s s s
s
s
( ) lim ( )0 4 2 4 13
4 1 2
1 4
2= = + +
+ +
= + +
+
lim
lim +
= + + + +
=
13 4 1 0
1 0 0
0 5
2s
x( )
(358)
As another example, consider
x t e u t X s
s s
t( ) ( ( ) ( )= + Û = + +
-5 5 1 2
2 ) (359)
Using the final-value theorem,
x sX s s
ss s ( ) lim ( ) lim¥ = = +
+ æ è ç
ö ø ÷ = + =® ®0 0
5 2
5 0 5 (360)
The final-value theorem cannot be used to find the value of x(t) = cos 2t, because X(s)
has poles at s j= ± 2 , which are not in the left half of the s plane For convenience, Table 31 provides the derived list of properties of Laplace
transform, while Table 32 gives the Laplace transform of some common functions For signals in Table 32, we have omitted the factor u(t) except where it is necessary because we are only interested in functions that exist for t > 0
TABLE 3.1 Properties of the Laplace Transform
TABLE 3.2 Laplace Transform Pairsa
Example 3.3
Example 3.4
Example 3.5
Example 3.6
(d) g t x d t
( ) ( )= ò t t 0
Example 3.7
For a given X(s), we can obtain the corresponding x(t) by using Equation 34, but this requires contour integration, which is beyond the scope of this text We will use an easier approach by employing the partial fraction expansion of X(s) and matching entries in Table 32
In most cases, we can express X(s) in the standard form of
X s N s
D s ( ) ( )( )= (361)
where N(s) is the numerator polynomial D(s) is the denominator polynomial
Equation 361 can be expressed as
X s N s
D s b s b s b s b s b a s a s
( ) ( )( )= = + + + + + +
+ + + +- -a s a s an n2 2 1 0
(362)
We can express X(s) in terms of factors of N(s) and D(s) as
X s N s
D s k s z s z s z
s p s p s p m
( ) ( )( ) ( )( ) ( ) ( )( ) ( )= =
+ + + + + +
(363)
where k = bm/an The roots of N(s) = 0 are called the zeros of X(s), that is, the values −z1,−z2,…,zm are the zeros of X(s)
Similarly, the roots of D(s) = 0 are the poles of X(s), that is, the values −p1,−p2,…,pn are the poles of X(s) We assume that X(s) is proper (m < n), that is, the degree of N(s) is less than the degree of D(s) (In case X(s) is not proper (m > n), we use long division to reduce it proper fraction) We use partial fraction to break X(s) down into simple terms whose inverse transform is obtained by matching entries in Table 32 Partial fraction expansions are useful in finding the inverse Laplace, inverse Fourier, or inverse z-transforms Thus, to find the inverse Laplace transform of X(s) requires two steps:
1 Use partial fraction expansion to break X(s) into simple terms 2 Refer to Table 32 to obtain the inverse Laplace transform of each term
We will now consider the three common forms X(s) may take and how we apply the two steps to each form
A simple pole is a first-order pole If the poles s = −p1,−p2,…,−pn are simple and distinct (pi ≠ pj for all i ≠ j), then D(s) in Equation 363 becomes a product of factors, so that
X s N s
s p s p s pn ( ) ( )( )( ) ( )= + + +1 2
(364)
Using partial fraction expansion, we decompose X(s) in Equation 364 as
X s k
s p k
s p k
s p n
( ) = +
+ +
+ + +
2 (365)
To get the expansion coefficients k1, k2,…, kn (known as the residues of X(s)), we multiply both sides of Equation 365 by (s + pi) for each pole For example, if we multiply by (s + p1), we obtain
( ) ( ) ( ) ( )s p X s k s p k
s p s p k s p
+ = + + +
+ + + +1 1
1 (366)
Setting s = −p1 in Equation 366 leaves only k1 on the right-hand side of Equation 366 By isolating k1 this way, we obtain
( ) ( )s p X s k
s p + =
In general,
k s p X si i s pi= + =-( ) ( ) (368)
This is referred to as Heaviside’s formula From Table 32, the inverse transform of each term in Equation 365 is
L - -1[ / ( )] ( )k s a ke u tat+ = (369)
Applying Equation 369 to each term in Equation 365 leads to the inverse Laplace transform of X(s) as
x t k e k e k e u tp t p t n p tn( ) ( )= + + +( )- - -1 21 2
(370)
If X(s) has repeated poles, we employ a different strategy Let X(s) have n repeated poles at s = −p, then we may represent X(s) as
X s k
s p k
s p k
s p k
s p X sn
n ( ) ( ) ( ) ( ) ( )= + + + + + + + + +
where X1(s) stands for the remaining part of X(s) that does not contain the repeated poles We find the coefficient kn as
k s p X sn n
s p = +
=- ( ) ( ) (372)
as in Equation 368 To determine kn−1, we multiply each term in Equation 371 by (s + p)n, differentiate, and then set s = −p to get rid of the other coefficients except kn−1 Thus, we obtain
k d ds
s p X sn n s p
= +éë ùû1 ( ) ( ) (373)
We repeat the same process to get
k d ds
s p X sn n s p
= +éë ùû2 2
2 1 2!
