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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

( ) ( ) ( ) E s H s  R s

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 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    





dw w

H jw 1

log ( ) 2

• 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.

3. The amplitude function
   • Is causal because

1 ( )^1 H jw  (^) w (^2) 

      

 

 

log wH 2 ( jw 1 ) dw w 21 12 dw

2. Stability

• If a network is stable, then for a bounded

excitation e(t) the response will also be

bounded.

• Where C 1 and C 2 are real, positive and finite numbers.

  

   | ( )| 0 t

| ( )| 0 t 2

1 r t C

e t C

• Note that our definition of stability implies systems with sin wt impulse response are not stable because
• 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

sin wtdt

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 H(s) is given as
• 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
• If then there would be multiple poles

on the s=jw=infinity.

1 1 1 0

1 1 1 0 ... ( ) ... b s b s bs b H s a s a s a s a m m m m

n n n n           

 

n  m  1 n  m  1

• 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.

• 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
• Then all coefficients an must be real and if si=α+jβ is root of P(s), then α must be negative.

P ( s ) an sn  an  1 sn ^1 ... a 1 s  a 0

Example

1. The polynomial is Hurwitz because - For real s P(s) is real, P(s)=(s+1)(s^2 +3s+2) - None of the roots lie on the right hand side of the ‘s’ plane.
2. The polynomial is not Hurwitz - The root s=1 lies on the positive ‘s’ plane.

P ( s ) ( s  1 ) s^  1  j 2  s^  1  j 2 

G ( s ) ( s  1 )( s  2 )( s  3 )