Signals and Systems
Fall 2003
Lecture #19
18 November 2003
1. CT System Function Properties
2. System Function Algebra and
Block Diagrams
3. Unilateral Laplace Transform and
Applications
CT System Function Properties
x(t)
h(t)
y(t)
Y ( s)  H ( s) X ( s)
1) System is stable 



h(t ) dt  
H(s) = “system function”
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
Properties of CT Rational System Functions
a)
However, if H(s) is rational, then
The system is causal
b)
⇔ The ROC of H(s) is to the
right of the rightmost pole
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
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

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 )
s 0
Final value
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
 No poles at s  0
s 0
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)
bs

where H ( s ) 
s
a

M
k
k 0 k
N
roots of numerator ⇒ zeros
k
roots of denominator ⇒ poles
k 0
k
Rational
ROC =?
Depends on: 1) Locations of all poles.
2) Boundary conditions, i.e.
right-, left-, two-sided signals.
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)
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
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 )
2
dt
dt
Notation:
1/s—an integrator
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.
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
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 ) 
 

dt 2
dt  dt 
dx(t )
UL
dt

s( sX ( s )  x(0  ))  x' (0  )

 s X ( s)  sx(0 )  x' (0 )
2


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

)(
s 
2)


1

ZIR
ZSR
ZIR — Response for
zero input x(t)=0
ZSR — Response for zero state,
β=γ=0, initially at rest
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