Signal and Systems
Prof. H. Sameti
Chapter 9: Laplace Transform

Motivation and Definition of the (Bilateral) Laplace Transform

Examples of Laplace Transforms and Their Regions of
Convergence (ROCs)
Properties of ROCs
Inverse Laplace Transforms
Laplace Transform Properties
The System Function of an LTI System
Geometric Evaluation of Laplace Transforms and Frequency
Responses





Book Chapter#: Section#
Motivation for the Laplace
Transform
 CT Fourier transform enables us to do a lot of things, e.g.
•
•
•
Analyze frequency response of LTI systems
Sampling
Modulation
 Why do we need yet another transform?
 One view of Laplace Transform is as an extension of the
Fourier transform to allow analysis of broader class of
signals and systems
 In particular, Fourier transform cannot handle large (and
important) classes of signals and unstable systems, i.e.
∞
when −∞ |𝑥(𝑡)| 𝑑𝑡 = ∞
Computer Engineering Department, Signal and Systems
2
Book Chapter#: Section#
Motivation for the Laplace
Transform (continued)

In many applications, we do need to deal with unstable
systems, e.g.
•
•
•
Stabilizing an inverted pendulum
Stabilizing an airplane or space shuttle
Instability is desired in some applications, e.g. oscillators and
lasers

How do we analyze such signals/systems? Recall from Lecture
#5, eigenfunction property of LTI systems:

𝑒 𝑠𝑡 is an eigenfunction of any LTI system
𝑠 = 𝜎 + 𝑗𝜔 can be complex in general

Computer Engineering Department, Signal and Systems
3
Book Chapter#: Section#
The (Bilateral) Laplace Transform
∞
−𝑠𝑡 𝑑𝑡
𝑥(𝑡)𝑒
−∞

𝑥(𝑡) ↔ 𝑋(𝑠) =

s = σ+ jω is a complex variable – Now we explore the full range
of 𝑠
absolute integrability needed
Basic ideas:

∞
= 𝐿{𝑥(𝑡)
𝑥(𝑡)𝑒 −𝜎𝑡 ]𝑒 −𝑗𝜔𝑡 𝑑𝑡 = 𝐹{𝑥(𝑡)𝑒 −𝜎𝑡
1.
𝑋(𝑠) = 𝑋(𝜎 + 𝑗𝜔) =
2.
A critical issue in dealing with Laplace transform is
convergence:—X(s) generally exists only for some values of s,
located in what is called the region of convergence(ROC):
∞
𝑅𝑂𝐶 = {𝑠 = 𝜎 + 𝑗𝜔 so that −∞ |𝑥 𝑡 𝑒 −𝜎𝑡 |𝑑𝑡 < ∞
If 𝑠 = 𝑗𝜔 is in the ROC (i.e. σ= 0), then
absolute
𝑋(𝑠)|𝑠=𝑗ω = 𝐹{𝑥(𝑡)
3.
−∞
integrability
condition
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4
Book Chapter#: Section#
Example #1:

𝑥1 (𝑡) = 𝑒 −𝑎𝑡 𝑢(𝑡) (a – an arbitrary real or complex number)

𝑋1 (𝑠) =
∞ −𝑎𝑡
−𝑠𝑡 𝑑𝑡
𝑒
𝑢(𝑡)𝑒
−∞
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=
∞ −(𝑠+𝑎)𝑡
𝑒
0
=−
1
[𝑒 −(𝑠+𝑎)∞
𝑠+𝑎
5
−
Book Chapter#: Section#
Example #2:
𝑥2 (𝑡) = −𝑒 −𝑎𝑡 𝑢(−𝑡)
∞ −𝑎𝑡
−𝑠𝑡
𝑒
𝑢(−𝑡)𝑒
𝑑𝑡
−∞
0
= − −∞ 𝑒 −(𝑠+𝑎)𝑡 𝑑𝑡
1
1
𝑒 −(𝑠+𝑎)𝑡 |0−∞ =
[1 − 𝑒 (𝑠+𝑎)∞
𝑠+𝑎
𝑠+𝑎
𝑋2 (𝑠) = −

=

This converges only if Re(s+a) < 0, i.e. Re(s) < -Re(a)

𝑋2 (𝑠) =

Key Point (and key difference from FT): Need both X(s) and
ROC to uniquely determine x(t). No such an issue for FT.
1
, ℜ𝑒{𝑠}
𝑠+𝑎
< −ℜ𝑒{𝑎 Same as 𝑋1 (s), but different ROC
Computer Engineering Department, Signal and Systems
6
Book Chapter#: Section#
Graphical Visualization of the ROC

Example1:
1
, ℜ𝑒{𝑠}
𝑠+𝑎
−𝑎𝑡
𝑋1 (𝑠) =
𝑥1 (𝑡) = 𝑒
> −ℜ𝑒{𝑎
𝑢(𝑡) → 𝑟𝑖𝑔ℎ𝑡 − 𝑠𝑖𝑑𝑒𝑑
1

Example2:
𝑋2 (𝑠) = 𝑠+𝑎 , ℜ𝑒{𝑠} < −ℜ𝑒{𝑎
𝑥2 (𝑡) = −𝑒 −𝑎𝑡 𝑢(−𝑡) → 𝑙𝑒𝑓𝑡 − 𝑠𝑖𝑑𝑒𝑑
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7
Book Chapter#: Section#
Rational Transforms

Many (but by no means all) Laplace transforms of interest to us
are rational functions of s (e.g., Examples #1 and #2; in general,
impulse responses of LTI systems described by LCCDEs), where
X(s) = N(s)/D(s), N(s),D(s) – polynomials in s

Roots of N(s)= zeros of X(s)

Roots of D(s)= poles of X(s)

Any x(t) consisting of a linear combination of complex
exponentials for t > 0 and for t < 0 (e.g., as in Example #1 and
#2) has a rational Laplace transform.
Computer Engineering Department, Signal and Systems
8
Book Chapter#: Section#
Example #3
𝑥(𝑡) = 3𝑒 2𝑡 𝑢(𝑡) − 2𝑒 −𝑡 𝑢(𝑡)
𝑋(𝑠) =

𝑋(𝑠) = 3
𝑋(𝑠) =
∞
3𝑒 2𝑡 − 2𝑒 −𝑡 ]𝑒 −𝑠𝑡 𝑑𝑡
0
∞ −(𝑠−2)𝑡
∞
𝑒
𝑑𝑡 − 2 0 𝑒 −(𝑠+1)𝑡 𝑑𝑡
0
3
2
𝑠+7
−
= 2
, ℜ𝑒{𝑠} > 2
𝑠−2
𝑠+1
𝑠 −𝑠−2
Computer Engineering Department, Signal and Systems
9
Book Chapter#: Section#
Laplace Transforms and ROCs

Some signals do not have Laplace Transforms (have no ROC)

𝑎)𝑥(𝑡) = 𝐶𝑒 −𝑡 for all t since

𝑏)𝑥(𝑡) = 𝑒 𝑗𝜔0𝑡 for all t 𝐹𝑇: 𝑋(𝑗ω) = 2𝜋𝛿(𝜔 − 𝜔0 )
∞
−∞
|𝑥(𝑡)𝑒 −𝜎𝑡 |𝑑𝑡 =
∞
−𝜎𝑡
|𝑥(𝑡)𝑒
|𝑑𝑡
−∞
∞ −𝜎𝑡
𝑒
𝑑𝑡
−∞
= ∞ for all 𝜎
= ∞ for all 𝜎
X(s) is defined only in ROC; we don’t allow impulses in LTs
Computer Engineering Department, Signal and Systems
10
Book Chapter#: Section#
Properties of the ROC
 The ROC can take on only a small number of different
forms
1. 1) The ROC consists of a collection of lines parallel
to the jω-axis in the s-plane (i.e. the ROC only
depends on σ).Why?
∞
∞
−𝑠𝑡
−𝜎𝑡
|𝑥(𝑡)𝑒
|𝑑𝑡
=
|𝑥(𝑡)𝑒
| 𝑑𝑡 < ∞ depends
−∞
−∞
only on 𝜎 = ℜ𝑒{𝑠}
2. If X(s) is rational, then the ROC does not contain any
poles. Why?
Poles are places where D(s) = 0
⇒ X(s) = N(s)/D(s) = ∞ Not convergent.
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11
Book Chapter#: Section#
More Properties

If x(t) is of finite duration and is absolutely integrable, then the ROC is
the entire s-plane.
∞

𝑋(𝑠) =
𝑥(𝑡)𝑒 −𝑠𝑡 𝑑𝑡 =
−∞
𝑇2
−𝑠𝑡 𝑑𝑡
𝑥(𝑡)𝑒
𝑇1
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< ∞ 𝑖𝑓
𝑇2
|𝑥(𝑡)| 𝑑𝑡
𝑇1
12
<∞
Book Chapter#: Section#
ROC Properties that Depend on
Which Side You Are On - I
 If x(t) is right-sided (i.e. if it is zero before some time), and
if Re(s) = 𝜎0 is in the ROC, then all values of s for which
Re(s) > 𝜎0 are also in the ROC.
ROC is a right half plane (RHP)
Computer Engineering Department, Signal and Systems
13
Book Chapter#: Section#
ROC Properties that Depend on
Which Side You Are On -II
 If x(t) is left-sided (i.e. if it is zero after some time), and
Re(s) = 𝜎0 is in the ROC, then all values of s for which
Re(s) < 𝜎0 are also in the ROC.
ROC is a left half plane (LHP)
Computer Engineering Department, Signal and Systems
14
if
Book Chapter#: Section#
Still More ROC Properties
 If x(t) is two-sided and if the line Re(s) = 𝜎0
is in the ROC,
then the ROC consists of a strip in the s-plane
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15
Book Chapter#: Section#
Example:
 𝑥(𝑡)
= 𝑒 −𝑏|𝑡|
Intuition?
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
Okay: multiply by
constant (𝑒 𝜎𝑡 ) and
will be integrable

Looks bad: no 𝑒 𝜎𝑡
will dampen both
sides
16
Book Chapter#: Section#
Example (continued):

𝑥(𝑡) = 𝑒 𝑏𝑡 𝑢(−𝑡) + 𝑒 −𝑏𝑡 𝑢(𝑡) ⇒ −
 Overlap if 𝑏
>0⇒𝑋 𝑠 =
 What if b < 0?
1
, ℜ𝑒{𝑠}
𝑠−𝑏
−2𝑏
𝑠 2 −𝑏2
<𝑏+
1
, ℜ𝑒{𝑠}
𝑠+𝑏
> −𝑏
, with ROC:
⇒No overlap ⇒ No Laplace Transform
Computer Engineering Department, Signal and Systems
17
Book Chapter#: Section#
Properties, Properties

If X(s) 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.
Suppose X(s) is rational, then
a) If x(t) is right-sided, the ROC is to the right of the rightmost pole.
b) If x(t) is left-sided, the ROC is to the left of the leftmost pole.

If ROC of X(s) includes the jω-axis, then FT of x(t) exists.

Computer Engineering Department, Signal and Systems
18
Book Chapter#: Section#
Example:
 Three possible ROCs
Fourier
Transform
exists?
x(t) is right-sided
x(t) is left-sided
x(t) extends for all time
Computer Engineering Department, Signal and Systems
ROC:
ROC:
ROC:
III
I
II
No
No
Yes
19
Book Chapter10: Section 1
The z-Transform
xn  X ( z ) 

 xnz
n
 Z xn
n  


ROC   z  re j at which


•

| xnr
n  
n


|  


-depends only on r = |z|, just like the ROC in s-plane
only depends on Re(s)
Last time:
•
Unit circle (r = 1) in the ROC ⇒DTFT X (e j  ) exists
•
Rational transforms correspond to signals that are linear
combinations of DT exponentials
Computer Engineering Department, Signal and Systems
20
Book Chapter10: Section 1
Some Intuition on the Relation
between ZT and LT
x (t )  X (s ) 
Let t=nT



x (t )e st dt  L {x (t )}

 lim
T 0

n  
x(nT )(e sT ) n  T



x n 

 lim T
T 0

xn(e sT )  n
n  
The (Bilateral) z-Transform
xn  X ( z ) 


xnz n  z{xn}
n  
Can think of z-transform as DT
version of Laplace transform with
z  e sT
Computer Engineering Department, Signal and Systems
21
Book Chapter10: Section 1
More intuition on ZT-LT, s-plane - z-plane
relationship
e sT  z



jT
j axis in s - plane ( s  j )  z  e  1 a unit circle in z - plan
LHP in s-plane, Re(s) < 0⇒|z| = | esT| < 1, inside the |z| = 1 circle.
Special case, Re(s) = -∞ ⇔|z| = 0.
RHP in s-plane, Re(s) > 0⇒|z| = | esT| > 1, outside the |z| = 1 circle.
Special case, Re(s) = +∞ ⇔|z| = ∞.
A vertical line in s-plane, Re(s) = constant⇔| esT| = constant, a circle in z-plane.
Computer Engineering Department, Signal and Systems
22
Book Chapter 9 : Section 2
Inverse Laplace Transform

X ( s )   x (t )e  st dt , s    j  ROC

 F {x (t )e t }
Fix σ ∈ ROC and apply the inverse Fourier
1 
jt
x (t )e 
X
(


j

)
e
d



2
1 
(  j ) t
x (t ) 
X
(


j

)
e
d

2  
t
But s = σ + jω (σ fixed)⇒ ds =jdω
1   j
st
x (t ) 
X
(
s
)
e
ds



j

2j
Computer Engineering Department, Signal and Systems
23
Book Chapter 9 : Section 2
Inverse Laplace Transforms Via Partial Fraction
Expansion and Properties
Example:
x( s) 
s3
A
B


( s  1)( s  2) s  1 s  2
2
5
A , B 
3
3
Three possible ROC’s — corresponding to three different signals
Recall
1
, e{s}  a  e  at u( t ) left - sided
sa
1
, e {s }  a  e at u (t ) right-sided
s a
Computer Engineering Department, Signal and Systems
24
Book Chapter 9 : Section 2
ROC I:
— Left-sided signal.
x (t )   Ae t u( t )  Be 2 t u( t )
5
2

  e t  e 2 t  u ( t )
3 
3
ROC II:
Diverg es as t  
— Two-sided signal, has Fourier Transform.
x(t )  Ae t u(t )  Be 2 t u( t )
5 2t
 2 t

   e u ( t )  e u ( t ) 
3
3

Div erges as t  
ROC III: — Right-sided signal.
x (t )  Ae t u(t )  Be 2 t u(t )
5 
 2
  e t  e2 t u(t)
3 
 3
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Div erges as t  
25
Book Chapter 9 : Section 2
Properties of Laplace Transforms


Many parallel properties of the CTFT, but for Laplace transforms we need
to determine implications for the ROC
For example:
Linearity
ax1 (t )  bx2 (t )  aX 1 ( s)  bX 2 ( s)
ROC at least the intersection of ROCs of X1(s) and X2(s)
ROC can be bigger (due to pole-zero cancellation)
E.g.
Then
x1 (t )  x2 (t ) and a  b
ax1 (t )  bx2 (t )  0  X ( s )  0
⇒ ROC entire s-
Computer Engineering Department, Signal and Systems
26
Book Chapter 9 : Section 2
Time Shift
x (t  T )  e
 sT
X ( s), same ROC as X ( s)
Example :
e 3s
,
s2
e{s}  2  ?
e  sT
, e{s}  2  e  2t u(t ) | t  t  T
s2
 T  3
e 3s
,
s2
e{s}  2 
Computer Engineering Department, Signal and Systems
e  2 ( t  3) u(t  3)
27
Book Chapter 9 : Section 2
Time-Domain Differentiation
x (t ) 
1   j
dx(t )
1   j
st
st
X
(
s
)
e
ds
,

sX
(
s
)
e
ds




j



j

2j
dt
2j

dx (t )
 sX ( s ), with ROC containing the ROC of X ( s )
dt
ROC could be bigger than the ROC of X(s), if there is pole-zero
cancellation. E.g.,
x (t )

dx ( t )

dt
u(t ) 
1
,
s
 (t )  1  s 
e{s}  0
1
s
ROC  entire s - plane
s-Domain Differentiation
dX ( s )
 tx(t ) 
, with same ROC as X(s)
ds
E.g. te at u(t )  
(Derivation is
similar to
d  1 
1

, e{s}  a
2


ds  s  a  ( s  a )
Computer Engineering Department, Signal and Systems
d
 s)
dt
28
Book Chapter 9 : Section 2
Convolution Property
x(t)
h(t)
y(t)=h(t)＊x(t)
x (t )  X ( s ), y (t )  Y ( s ), h(t )  H ( s )
For
Y ( s)  H ( s)  X ( s)
Then
• ROC of Y(s) = H(s)X(s): at least the overlap of the ROCs of H(s) & X(s)
• ROC could be empty if there is no overlap between the two ROCs
E.g.
x(t)=etu(t),and
h(t)=-e-tu(-t)
• ROC could be larger than the overlap of the two.
x (t ) * h (t )   (t )
Computer Engineering Department, Signal and Systems
29
Book Chapter 9 : Section 2
The System Function of an LTI System
h(t)
x(t)
y(t)
h(t )  H ( s )  the system function
The system function characterizes the system
⇓
System properties correspond to properties of H(s) and its ROC
A first example:

System is stable   h(t ) dt  

Computer Engineering Department, Signal and Systems

ROC of H(s)
includes the jω axis
30
Book Chapter 9 : Section 2
Geometric Evaluation of Rational Laplace Transforms
Example #1:
X 1 ( s)  s  a
Computer Engineering Department, Signal and Systems
A first-order zero
31
Book Chapter 9 : Section 2
Example #2:
A first-order pole
X 2 ( s) 
 X 2 ( s) 
1
1

sa
X 1 ( s)
1
X 1 ( s)
(or log |X 2(s)|   log | X 1 ( s ) |)
X 2 ( s )   X 1 ( s ) Still reason with vector, but
remember to "invert" for poles
Example #3:
A higher-order rational Laplace transform
X ( s) 
X ( s) 
 iR1 ( s   i )
M
 Pj1 ( s   j )
M
 iR1 s   i
 Pj1 s   j
R
p
i 1
j 1
X ( s )  M   ( s   i )   ( s   j )
Computer Engineering Department, Signal and Systems
32
Book Chapter 9 : Section 2
First-Order System
h(t ) 
1

H ( s) 
1
1 /
1

, e{s}  
s  1 s  1 / 


Graphical evaluation of H(jω)

s (t )  1  e  t /  u(t )
e  t / u (t )
H ( j ) 
Computer Engineering Department, Signal and Systems
1 /
1
1
 
j  1 / 
 j  1 / 
33
Book Chapter 9 : Section 2
Bode Plot of the First-Order System
H ( j ) 
H ( j ) 
1/
j  1/
1 /
 2  (1 /  ) 2
0
1

 1 / 2   1/
1 /    1/

H ( j )      tan 1 ( )
0
0

   / 4   1/
  / 2   1/

Computer Engineering Department, Signal and Systems
34
Book Chapter 9 : Section 2
Second-Order System
n2
H ( s)  2
s  2 n s  n2
0   1

ROC e{s}  e(pole)
complex poles
— Underdamped
 1

double pole at s = −ωn
— Critically damped
 1

2 poles on negative real axis
— Overdamped
Computer Engineering Department, Signal and Systems
35
Book Chapter 9 : Section 2
Demo Pole-zero diagrams, frequency response, and step
response of first-order and second-order CT causal systems
Computer Engineering Department, Signal and Systems
36
Book Chapter 9 : Section 2
Bode Plot of a Second-Order System
Top is flat when
ζ= 1/√2 = 0.707
⇒a LPF for
ω < ωn
Computer Engineering Department, Signal and Systems
37
Book Chapter 9 : Section 2
Unit-Impulse and Unit-Step Response of a SecondOrder System
No oscillations when
ζ≥1
⇒ Critically (=) and
over (>) damped.
Computer Engineering Department, Signal and Systems
38
Book Chapter 9 : Section 2
First-Order All-Pass System
sa
H ( s) 
, e{s}  a ( a  0)
sa
1. Two vectors have
the same lengths
2. H ( j )  1   2
Computer Engineering Department, Signal and Systems

(   2 )   2

  2 2

0


 / 2   a
~ 0   a

39
CT System Function Properties
h(t)
x(t)
Y ( s)  H ( s) X ( s)
1) System is stable 



y(t)
H(s) = “system function”
h(t ) dt    ROC of H(s) includes jω axis
2) Causality => h(t) right-sided signal => ROC of H(s) is a right-half plane
Question:
If the ROC of H(s) is a right-half plane, is the system causal?
E.x.
e eT
H ( s) 
, e{e}  1  h(t ) right - sided
s 1
 e eT 
1 
1 
h (t )  L 
 e t u(t ) |t t T
L 

 s  1  t t  T
 s  1
Non-causal
 e  ( t T ) u(t  T )  0 at t  0
1
40
Properties of CT Rational System Functions
a)
However, if H(s) is rational, then
The system is causal
⇔ The ROC of H(s) is to the
right of the rightmost pole
b) If H(s) is rational and is the system function of a causal
system, then
The system is stable
⇔
jω-axis is in ROC
⇔ all poles are in
41
Checking if All Poles Are In the Left-Half Plane
N ( s)
H ( s) 
D( s )
Poles are the root of D(s)=sn+an-1sn-1+…+a1s+a0
Method #1: Calculate all the roots and see!
Method #2: Routh-Hurwitz – Without having to solve for roots.
Polynomial
First - order
s  a0
Second - order
s 2  a1 s  a0
Third - order
s 3  a 2 s 2  a1 s  a0

Condition so that all
roots are in the LHP
a0  0
a1  0, a0  0
a 2  0, a1  0, a0  0
and a0  a1a 2

42
Initial- and Final-Value Theorems
If x(t) = 0 for t < 0 and there are no impulses or higher order
discontinuities at the origin, then
x(0 )  lim sX ( s)
Initial value
s 
If x(t) = 0 for t < 0 and x(t) has a finite limit as t → ∞, then
x ( )  lim sX ( s )
Final value
s 0
43
Applications of the Initial- and Final-Value Theorem
X ( s) 
For
n-order of polynomial N(s),
•
N ( s)
D( s )
d – order of polynomial D(s)
Initial value：
d  n 1
0


x (0 )  lim sX ( s )  finite  0 d  n  1
s 

d  n 1

E.g.
•
X ( s) 
1
s 1
x (0  )  ?
Final value：
If x()  lim sX ( s)  0  lim X ( s)  
s 0
s 0
 No poles at s  0
44
LTI Systems Described by LCCDEs
d k y (t ) M d k x (t )
ak
  bk

k
dt
dt k
k 0
k 0
N
d
dk
k
Repeated use of differenti ation property ：  s,

s
dt
dt k
N
M
 a s Y ( s)   b s
k
k 0
k
k 0
k
k
X ( s)

Y ( s)  H ( s) X ( s)
k
b
s
k  0 k
M
where H ( s ) 
k
a
s
k 0 k 

N
roots of numerator ⇒ zeros
roots of denominator ⇒ poles
Rational
ROC =?
Depends on: 1) Locations of all poles.
2) Boundary conditions, i.e.
right-, left-, two-sided signals.
45
System Function Algebra
Example:
A basic feedback system consisting of causal blocks
E( s)  X ( s)  Z ( s)  X ( s)  H 2 ( s)Y ( s)
Y ( s)  H1 ( s)E( s)  H1 ( s)X ( s)  H 2 ( s)Y ( s)

H ( s) 
ROC:
H1 ( s)
Y ( s)

X ( s) 1  H1 ( s) H 2 ( s)
More on this later
in feedback
Determined by the roots of 1+H1(s)H2(s), instead of H1(s)
46
Block Diagram for Causal LTI Systems with Rational
System Functions
Example：
Y ( s)  H ( s) X ( s)
— Can be viewed
1
 2
2s 2  4s  6  
 2
 ( 2 s  4 s  6)
H ( s)  2
as cascade of
s

3
s

2


s  3s  2
two systems.
1
Define ：
W ( s)  2
X ( s)
s  3s  2
d 2 w(t )
dw(t )

3
 2 w(t )  x (t ), initially at rest
2
dt
dt
d 2 w(t )
dw(t )
or

x
(
t
)

3
 2 w(t )
2
dt
dt
Similarly
Y ( s )  ( 2 s 2  4 s  6)W ( s )
d 2 w(t )
dw(t )
y (t )  2

4
 6w(t )
2
dt
dt
47
Example (continued)
H ( s)
Instead of
x (t )
s
2
1
 3s  2
 2s
2
 4s  6
 y (t )
2
We can construct H(s) using: d w(t )  x (t )  3 dw(t )  2 w(t )
2
dt
dt
d 2 w(t )
dw(t )
y (t )  2

4
 6w(t )
dt 2
dt
Notation:
1/s—an integrator
48
Note also that
H ( s) 
PFE
 2(s  1 )   s  3   s  3   2( s  1) 
 s  2   s  1    s  2   s  1 
 2
6
8

s  2 s 1
 Cascade
 parallel connection
Lesson to be learned：There are many different ways to construct a
system that performs a certain function.
49
The Unilateral Laplace Transform
(The preferred tool to analyze causal CT systems described by
LCCDEs with initial conditions)
Note:
X ( s) 
1) If x(t) = 0 for t < 0,


0
x (t )e  st dt  ULx (t )
X ( s)  X ( s)
2) Unilateral LT of x(t) = Bilateral LT of x(t)u(t-)
3) For example, if h(t) is the impulse response of a causal LTI
system then,
H ( s)  H ( s)
4) Convolution property: If x1(t) = x2(t) = 0 for t < 0,
ULx1 (t )  x2 (t ) X1 ( x)X2 ( s)
Same as Bilateral Laplace transform
50
Differentiation Property for Unilateral Laplace Transform
x (t )
 X ( s)

Initial condition!
dx (t )
 sX ( s )  x (0  )
dt
integration by parts
∫f．dg=fg－∫g．df
Derivation：
 dx (t ) 
UL 

 dt 

dx (t )  st
e dt

0
dt




s  x (t )e  st dt  x (t )e  st | 
0
0


X (s)
 sX ( s)  x(0 )
Note：
d 2 x (t ) d  dx (t ) 
 

2
dt
dt  dt 
dx (t )
UL
dt

s( sX ( s )  x (0  ))  x ' (0  )

 s X ( s)  sx(0
2

)  x ' (0  )
51
Use of ULTs to Solve Differentiation Equations
with Initial Conditions
Example：
d 2 y (t )
dy (t )
3
 2 y (t )  x (t )
2
dt
dt
y (0  )   , y ' (0  )   , x (t )  u(t )

2
( s )  s    3(Y( s )   )  2Y( s ) 
Take ULT： sY
 
s
2
 d y 
UL  2 
 dt 
Y( s ) 
 dy 
UL  
 dt 
 ( s  3)




( s  1)( s  2) ( s  1)( s  2) s( s  1)( s  2)


 

ZIR
ZSR
ZIR — Response for
zero input x(t)=0
ZSR — Response for zero state,
β=γ=0, initially at rest
52
Example (continued)
• Response for LTI system initially at rest (β = γ = 0 )

Y( s )
1
H ( s) 

 H ( s)
 ( s ) ( s  1)( s  2)
• Response to initial conditions alone (α = 0).
example:
x(t )  0( no input ),
y(0 )  1,
For
y ' (0  )  0
(   1,   0)

s3
2
1
Y( s ) 


( s  1)( s  2) s  1 s  2
y ( t )  2e  t  e  2 t ,
t0
53

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