Network Analysis and Synthesis
Chapter 3
Elements of Realizability Theory
Introduction
• In the last chapter we were concerned with
the problem of identifying the response given
the excitation and network.
• When we discuss about synthesis we are
concerned with the problem of constructing a
network given the excitation and response.
• The starting point for any synthesis is the
system function
H (s) 
R (s)
E (s)
• The first step in synthesis procedure is
determining whether the system function can
be realized with a physical passive network.
• There are two considerations
– Causality and
– Stability
1. Causality
• By causality we mean that a voltage doesn’t
appear between any terminals in the network
before a current/voltage is applied.
• In other words, the impulse response of the
network must be zero for t<0.
h ( t )  0 for
t0
Example
1. h(t)=etu(t)
•
Is causal because for t<0 u(t)=0, hence, h(t)=0.
2. h(t)=e|t|
•
Is not causal because for t<0, h(t) is not zero.
• In certain cases, the network can be made
causal by delaying the impulse response by a
certain time delay.
• If we delay h(t) by T, then h(t-T) will be causal.
• In the frequency domain causality is implied
when the Paley-Wiener criterion is satisfied.
• The Paley-Wiener criterion states that a
necessary and sufficient condition for
causality is



log H ( jw )
w 1
2
dw  
• The physical implication of the Paley-Wiener
criterion is that the amplitude response of a
causal network can’t be zero over a finite band
of frequency.
Example
1. The ideal low pass filter
– Is not causal because it is zero
for all frequencies w>wc.
2. The Gaussian filter
H ( jw )  e
w
2
– Is not causal because



log H ( jw )
w 1
2
is not finite.

dw 
w

w
2
2
1
dw
3. The amplitude function
1
H ( jw ) 
w 1
2
– Is causal because



log H ( jw )
w 1
2

dw 
 w

1
2
1

2
dw  
2. Stability
• If a network is stable, then for a bounded
excitation e(t) the response will also be
bounded.
| e ( t ) | C 1
0t
| r ( t ) | C 2
0t
• Where C1 and C2 are real, positive and finite
numbers.
• If a linear system is stable, then we get from the
convolution integral and the above definition of
stability
| r ( t ) | C  h ( ) d   C

1
2
0
• The above equation implies that the impulse
response be absolutely integrable.

 h ( ) d 

0
• One important requirement for h(t) to be absolutely
integrable is h(t) approach zero as t increases to
infinity.
• Note that our definition of stability implies systems
with sinwt impulse response are not stable because

 sin wt dt  
• However, a simple L-C network has such an impulse
response.
• Since we don't want to call these systems unstable,
we call them marginally stable if they satisfy the
following criterion.
0
lim
t 
h ( t )  0 and
h (t )  C
all t
• Stability in the frequency domain implies that
the system function should only have poles on
the left had side of the ‘s’ plane or simple
poles on the jw axis.
• This is because if we have a pole on the right
hand side, then the impulse response will
have an exponentially increasing term, eαt.
• Hence, our response will not be bounded.
• If there is a double pole on the jw axis, then
the impulse response of the network will have
a term tsin(wt).
• This will not be bounded.
• If H(s) is given as
a n s  a n 1 s
n
H (s) 
n 1
b m s  b m 1 s
m
m 1
 ...  a1 s  a 0
 ...  b1 s  b0
• Due to the requirement of simple poles on the
jw axis, the order of the numerator shouldn’t
exceed the order of the denominator by more
than 1. That is n  m  1
• If n  m  1 then there would be multiple poles
on the s=jw=infinity.
• To summarize, for a network to be stabile the
following three conditions must be satisfied
1. H(s) can’t have poles on the right side of the ‘s’
plane.
2. H(s) can’t have multiple poles on the jw axis.
3. The degree of the numerator of H(s) can’t
exceed that of the denominator by more than 1.
3. Hurwitz polynomials
• We mentioned in the previous section that in
order for a system to be stable, its poles must
lie in the left side of the ‘s’ plane; moreover
the poles on the jw axis must be simple.
• The denominator polynomial of a system
function H(s) that satisfies these criteria
belongs to a class of polynomials called
Hurwitz polynomials.
• In these section, we will discuss the properties
of these types of polynomials.
• A polynomial P(s) is said to be Hurwitz if it
satisfies
1. P(s) must be real if s is real.
2. The real part of its roots must be negative or zero.
• As a result of these conditions, if P(s) is a
Hurwitz polynomial given by
P ( s )  a n s  a n 1 s
n
n 1
 ...  a1 s  a 0
• Then all coefficients an must be real and if
si=α+jβ is root of P(s), then α must be negative.
Example
1. The polynomial
Hurwitz because
•
•

 is
For real s P(s) is real, P(s)=(s+1)(s2+3s+2)
None of the roots lie on the right hand side of the ‘s’ plane.
2. The polynomial
Hurwitz
•

P ( s )  ( s  1) s  1  j 2 s  1  j 2
G ( s )  ( s  1)( s  2 )( s  3 )
is not
The root s=1 lies on the positive ‘s’ plane.
Properties of Hurwitz polynomial
1. All the coefficients of the polynomial are non
negative.
• This is readily seen by examining the types of
terms P(s) can have

P ( s )  s   i  s   i
Simple
real pole
2
2
 s  
Simple pole
on the jw axis
i
2   i 2 
Complex
conjugate roots
• The multiplication of these non negative
coefficients can only give non negative
coefficients.
2. The even and odd parts of P(s) have roots on
the jw axis only.
• If we denote the even and odd parts of P(s)
as n(s) and m(s)
P (s)  n(s)  m (s)
• Then both n(s) and m(s) have roots on the jw
axis only.
3. The continued fraction expansion of n(s)/m(s)
or m(s)/n(s) of a Hurwitz polynomial yields
only positive quotient terms.
 (s) 
n(s)
m (s)
1
 q1 s 
1
q2s 
q3s 
1
q4s 
1
.....
1
qns
• All the q’s are positive.
Example
• Obtain the continued fraction expansion of
F ( s )  s  s  5s  3s  4
4
3
2
• Solution:
n ( s )  s  5 s  4 and
4
2
m ( s )  s  3s
3
– Since the order of n(s) is higher than m(s), we
start with n(s)/m(s).
n(s)
s  5s  4
4

m (s)
2
s  3s
3
2s  4
2
 s
 s
s  3s
3
1
s  3s
3
2s  4
2
 s
1
1
s
2
 s
s
2s  4
2
1
1
s
2
1
2s  4
2
s
 s
1
1
1
s
2
2s 
4
s
 s
1
1
2
s
1
2s 
1
s
4
Note that all the
coefficients of the
quotients are positive
4. Positive Real Functions
• These functions are important because they
represent physically realizable passive driving
point immitances.
• A function is positive real if
– F(s) is real for real s, that is F(σ) is real.
– The real part of F(s) is greater or equal to zero
when the real part of s is greater than or equal to
zero. That is
• In other words, the right half of the ‘s’ plane
maps with the right half of F(s) plane.
• In addition, the real axis of ‘s’ plane maps with
real axis of F(s) plane.
• A further restriction is that F(s) be rational.
Example
1. F(s)=Ls (where L is positive real number), is
positive real by definition. Inductor
2. F(s)=R (where R is positive real number), is
positive real by definition. Resistor
3. F(s)=K/s (K real and positive) is positive real
because when s is real F(s) is real and when
the real part of s is positive the real part of
F(s) is also positive. Capacitor
F (s) 
1
  j

1
  j
*
  j
  j

  j
 
2
2
• The necessary and sufficient condition for F(s)
to be a positive real function is
– F(s) must have no poles on the right side of s
plane.
– F(s) may have only simple poles on the jw axis
with real and positive residues.
– Re (F(jw))  0 for all w.
Example
1. Is
F (s) 
s2
s  3s  2
2
F (s) 
positive real function?
s2
 s  1( s  2 )

1
s 1
– Its pole s=-1 lies on the left of s plane
– No multiple poles on the jw axis
– Its real part is
 1
Re  F ( jw )   Re 
 1  jw
is always positive.

1
 1  jw 
  Re 


2
2
1 w  1 w

2. Is F ( s )  ss 12
Solution:
2
F (s) 
positive real?
s2
s  j 2 s  j 2 
– No poles on the right hand side.
– No multiple poles on jw axis.
– The real part of F(jw) is
1
 jw  1 
Re  F ( jw )   Re 


2
2
w 2 2w
this can be a negative number. For example for
w=2.
Exercise
3. Is
F (s) 
s4
s  2s  1
2
positive real?