Tutorial 6 Network Synthesis EEL2186 Circuits and Signals

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Questions
1. Check the positive realness of the following function
s 3
F ( s) 
( s  1)

2. Identify the given impedance functions whether they are of RL, RC or LC

(s 2  0.5)(s 2  1.5)(s 2  3)
i. s(s 2  1)(s 2  2)

(s  2)(s  5)
ii. (s  1)(s  3)

3. Is it possible to realize the following function using canonical forms?

s 2  2s  6
s(s  3)

4. Find the network for the given functions below by:

s(s 2  4)
Z(s) 
i. s2 1

s(s 2  2)(s 2  4)
Y(s) 
ii. (s 2  1)(s 2  3)(s 2  5)

2s 2  5
Z s   2
s s  1
5. Determine whether the function is positive real or not.

6. Determine the range of  such that the polynomial

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Tutorial 6 Network Synthesis EEL2186 Circuits and Signals
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P ( s ) s 4  s 3  4 s 2   s  3 is Hurwitz.

Solutions
1. Check the positive realness of the following function
s 3
F ( s) 
( s  1)

F(s) has all the poles lying on the left half of the s – plane. The other
condition required for prf is Re [F (j)]  0 for all values of,
(since, no pole falls on j axis second necessary and sufficient
condition need not to be checked).

 j  3   ( j  3)(1  j ) 
ReF ( j )  Re   Re  
 j  1   12 

 j  3   2  3 j   3   2  2 j  3   2
 Re   Re  
 1   2
  1   2
 1
2

Thus, Re [F (j)]  0 for all values of. Hence F(s) is a positive

2. Identify the given impedance functions whether they are of RL, RC or LC

(s 2  0.5)(s 2  1.5)(s 2  3)
2 2
.a. s (s  1)( s  2)
 Find the poles and zeros
Zeros are: ± j√ 0 .5 , ± j√ 1. 5 , ± j√ 3

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Poles are:0, ± j1, ± j√ 2

 Since all the poles and zeros interlace on “j” axis, it is a

(s  2)(s  5)
b. (s  1)(s  3)
 Find the poles and zeros
Zeros are: -2, -5
Poles are: -1, -3
 Critical frequency far away from the origin is zero (-5)
 All the poles and zeros are simple and interlaced on
negative real axis of the s-plane
 Hence it is an RC network.

s 2  2s  6
3.
s(s  3)
 Find the poles and zeros
Zeros are: -1 + j 2.236 and -1 – j 2.236
Poles are: 0, - 3
 Since it is not a ratio of even to odd or odd to even
function, it is not a LC function
 Since the roots are complex [ not simple ], it is not a RC/RL
function

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Since it is neither a LC nor a RC/RL network function, it
can not be realized by using any canonical form.

s(s 2  4)
Z(s)  2
4. a. i. s 1
Since it is a ratio of even to odd function, it is a LC function.
Foster – I form:

s(s 2  4)
Z(s)  2
s 1
Partial fraction of the above function leads to
3s
2
Z(s) = s + s  1
Since it is a series function, it has a series of two impedance
terms.

Foster – I network
Foster II form:
2
1 s 1
2
Y(s) = Z ( s ) = s ( s  4)

Partial fraction of the function leads to

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1 ( s 2  1)( s 13/)(4s  35s) / 4
2 2

Y ( s) s ( s 2  2)( s 2  
4) 2
Y(s) = s s 4

Foster – II network
4. a.ii.

s(s 2  2)(s 2  4)
Y(s)  2
(s  1)(s 2  3)(s 2  5)

Foster – I form:

1 ( s 2  1)( s 2  3)( s 2  5)
 2 2
Z(s) = Y ( s ) s ( s  2)( s  4)

Partial fraction of the above function leads to

3s / 4 3s / 8 15 / 8
2
 2 s
Z(s) = s  2 s  4 s

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Foster – I network
Foster- II form:

s(s 2  2)(s 2  4)
Y(s)  2
(s  1)(s 2  3)(s 2  5)
Partial fraction leads to

3s / 8 s/4 3s / 8
2
 2
 2
Y(s) = s  1 s  3 s 5

Foster – II network
4. b.i
Cauer – I form:

s(s 2  4)
Z(s)  2
s 1
The CFE of the function leads to

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Cauer – I network
Cauer – II form:

4s  s 3
2
Z(s) = 1  s [Rearrange both numerator and denominator in an
ascending order]

1 1 s2
 3
Y(s) = Z ( s ) 4 s  s
By taking CFE of Y(s),
4 s  s 3 )1  s 2 (1 / 4 s
1 s2 / 4
   
3s 2 / 4)4 s  s 3 (16 / 3s
4s
   
s 3 )3s 2 / 4(3 / 4 s
3s 2 / 4
  
0
  

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Cauer – II network
4b.ii.
Cauer-I form:

( s2 +1 )( s 2 +3 )( s 2 +5 ) s 6 + 9 s 4 + 23 s 2 +15
2 2
=
Z(s) = s ( s + 2)( s + 4 ) s5 +6 s 3 + 8 s
s5 + 6 s 3 + 8 s ) s 6 + 9 s 4 + 23 s 2 +15( s
s6 +6 s4 +8 s2
−−−−−−−−
3 s4 +15 s2 + 15) s 5 +6 s 3 +8 s( s/ 3
s5 +5 s 3 +5 s
−−−−−−−−
s3 +3 s ) 3 s 4 +15 s 2 +15 (3 s
3 s4+ 9 s2
−−−−−
6 s 2 +15 ) s3 +3 s ( s / 6
s 3 +15 s /6
−−−−−−−
2
3 s /6 ) 6 s +15 (12 s
2
6s
−−−−−−
15 ) 3 s / 6( s/ 30
3s/6
−−−−
0
−−−−−

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Cauer – I network

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Cauer – II form:
Rearrange the numerator and denominator in ascending order
15+ 23 s 2 +9 s 4 + s6
Z ( s )=
8 s+ 6 s 3 + s5
Cauer – II network

3 5 2 4 6 15
8 s+6 s +s ) 15+23 s +9 s + s (
8s
45 2 15 4
15+ s+ s
4 8
47 2 57 4 6 32
s + s + s ) 8 s+ 6 s 3 + s5 (
4 8 47 s
228 3 32 5
8 s+ s + s
47 47
54 3 15 5 47 2 57 4 6 2209
s + s ) s + s +s (
47 47 4 8 216 s
47 2 235 4
s + s
4 72
139 4 6 54 3 15 4 1944
s +s ) s + s (
36 47 47 6533 s
54 3 1944 5
s + s
47 6533
3 139 4 6 19321
s5 ) s +s (
139 36 108 s
139 4
s
36
3 3
s6 ) s5(
139 139 s
3
s5
139
⋅

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5.

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6.

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