Analog Filters: Singly-Terminated LC
Ladders
Franco Maloberti
Introduction

The purpose of this part is to design a LC ladder
network that:
 Is a two-port network
 It contains inductors and capacitors
 Has a resistive termination at the output
 The source is a voltage or a current generator
Or a voltage source
E 1  z11 I1  z12 I 2
I2
I1
LC Ladder
1
E2
E 2  z 21 I1  z 22 I 2
Franco Maloberti
Analog Filters: Singly-Terminated LC Ladders
2
LC Ladder with Current source

Consider a singly-terminated filter with a current
source and a normalized resistive load
I2
I1
1
LC Ladder
E2
E 2  z 21 I 1  z 22 I 2
 z 21 I 1  z 22 E 2
 Z 21 ( s ) 
Franco Maloberti
E2
I1

z 21
1  z 22
Analog Filters: Singly-Terminated LC Ladders
3
LC Ladder with Current source (ii)
Z 21 ( s) 

E2
I1

z 21
1  z 22
How to realize an LC network for a given Z21(s)?
Properties
of z21 and z22

 z21 = (even poly)/(odd poly) or vice versa
 Zeros of Z21 (transmission zeros) are zeros of z21
 z22 is a lossless function
P22/Q22 = (even poly)/(odd poly) or vice versa
Franco Maloberti
Analog Filters: Singly-Terminated LC Ladders
4
Implementing Z21(s)

Consider the transfer function
Z 21 ( s ) 

P (s)

Q (s)
KN 1 ( s )
M 2 (s)  N 2 (s)
or
M ( s)  even terms
KM 1 ( s )
N ( s)  odd terms
M 2 (s)  N 2 (s)
Q ( s)  Hurwitz polynomial
If P(s) is an even polynomial

KM 1 ( s)
Z 21 ( s) 
N 2 ( s)
M 2 ( s)
1
N 2 ( s)

z 22 
M 2 ( s)
N 2 ( s)
and
z 21 
KM 1 ( s)
N 2 ( s)

Franco Maloberti
Analog Filters: Singly-Terminated LC Ladders
5
Implementing Z21(s) (ii)

If P(s) is an odd polynomial
KN 1 (s)
Z 21 ( s) 



M 2 ( s)
N ( s)
1 2
M 2 ( s)

z 22 
N 2 (s)
M 2 ( s)
and
z 21 
KN 1 ( s)
M 2 ( s)
For a given Z21(s) we have to design an LC network
that realizes z21(s) and z22(s) simultaneously
Proceed from left to right instead of going from right
to left
Franco Maloberti
Analog Filters: Singly-Terminated LC Ladders
6
Transmission zeros at 0 or infinite


Use of Cauer’s realizations to remove zeros at the
origin or infinite completely.
Use an intuitive view.
Series L
Example
Z 21 ( s) 
P ( s)
Q 3 ( s)
Three non-dissipative elements
Input is a parallel element

Z2
Z1
Z3
Franco Maloberti
Z2
Z1
Z3
Analog Filters: Singly-Terminated LC Ladders
zero @ ∞
Series C
Zero @ 0
Shunt C
Zero @ ∞
Shunt L
Zero @ 0
7
Intuitive view (Example Q3(s))
Franco Maloberti
Analog Filters: Singly-Terminated LC Ladders
8
Example 1
Z 21 ( s ) 
K
 even numerator
3
2
s  2s  2s  1
K
3
s
 2s
Z 21 ( s ) 
2s 2  1
1 3
s  2s
2s  1
 three transmission
zeros at s  
2
z 22 
Franco Maloberti
s  2s
3
;
z 21 
K
s  2s
3
Analog Filters: Singly-Terminated LC Ladders
9
Example 1 (ii)

Three transmission zeros at s = ∞
Applying the long division to
get circuit parameters
1
3
s 3  2 s  (2 s 2  1)  s  s
2
2
3 4
2s 2  1  s  s  1
2 3
3
3
s  1 s  0
2
2
Franco Maloberti
Analog Filters: Singly-Terminated LC Ladders
10
Example 1 (iii)

Network realization
4
H
3
I1
3
F
2
1
F
2
1
E2
Evaluation of k
E2
I1
 1  Z 21 ( 0 )  K
Franco Maloberti
Analog Filters: Singly-Terminated LC Ladders
11
Example 1 (iv)

Use of Matlab for removing transmission zeros
2s  1
2
z 22 

s  2s
3
clear all
num=[1 0 2 0];
den=[2 0 1];
[c1,r1]=deconv(num,den);
c1
r1=r1(3:4);
[l,r2]=deconv(den,r1);
l
r2=r2(3);
[c2,r3]=deconv(r1,r2)
Franco Maloberti
ex6_1
c1 = 0.5000
l = 1.3333
c2 =1.5000
r3 = 0
Analog Filters: Singly-Terminated LC Ladders
0
0
0
0
12
Example 2

High-pass Butterworth filter
Z 21 ( s) 
Ks
K
3
s  2s  2s  1
3
s  2s
2
3
z 22 



2s  1
2
;
z 21 
Z 21 ( s) 
;
1
Ks
s
3
2s  1
3
s  2s
2
2s  1
2
3
2s  1
2
One zero at 0 and
two zero at ∞
Franco Maloberti
Analog Filters: Singly-Terminated LC Ladders
13


Example 2 (ii)

Remove the pole of z22 at infinite
s  2s
3
z1 
z 21 
2s  1
2
Ks

2
2
I1

4s  2
2
1
3/2 H
s
s1

3
1/2 H
I1
2
 Ks
3
z 21 
z1 
s;
3s
3
2s  1
IL
1
IL
3/4 F
s  Ks
2
2
Franco Maloberti
Analog Filters: Singly-Terminated LC Ladders
14
Zeros at finite frequency

Finite zeros are zeros of P(s) (or z21)
Z 21 ( s) 
P ( s)
Q ( s)

P ( s)
M 2 ( s)  N 2 ( s)
We need to create the finite zeros of Z12(s) while
realizing z22.
z22 does not have the zeros of Z12!


Partial (and complete) removal of poles shifts the
zeros
Franco Maloberti
Analog Filters: Singly-Terminated LC Ladders
15
Zero Shifting
The partial removal of poles from a function DOES
affect the zeros of the remainder  Zero shifting is
necessary
( s  1)( s  9 )
Consider Z ( s )  s ( s  4 )

2

2
2
X ( )  Im[Z ( j )]
X ( )  Im[Z ( j )]
k' 
0
1
2

3
1
2
3

0
 k ' / 
When part of the pole at  (left) and at the origin (right) is removed
Franco Maloberti
Analog Filters: Singly-Terminated LC Ladders
16
Zero Shifting (ii)




A partial pole removal shits all the finite (and nonzero) zeros toward the affected pole
The larger part is removed the more the zeros are
shifted toward the pole
Zeros cannot be shifted beyond adjacent poles
Shifting a zero in a given desired position is not
always possible
X()

Franco Maloberti
Analog Filters: Singly-Terminated LC Ladders
17

Example 6.4
K (s  4 )
2
Z 21 ( s) 
2
z 22 ( s) 
s  2s  2s  1
3
2
Zeros of z22 is at  
Poles of z22 are at
2s  1
1
s  2s
3
We can move the
zeros only in the range
2 
  0;
2
0;
2
 y !!
Work with
22

3
y 22 ( s) 
Franco Maloberti
s  2s
2s  1
2
Analog Filters: Singly-Terminated LC Ladders
18
Example 6.4 (ii)

Zeros of y22 is at
Poles of y22
are at  

2
1
; 
2
2

y 22 ( s)  C 1 s s
2


2
Franco Maloberti
1/ 2;
y 22  C 1 s  0 s

1
We can move the
zeros only in the range

j4
7
j2

j2
s 3  2 s

  2
 C 1 s
2 s  1

 j 2C 1  0
Analog Filters: Singly-Terminated LC Ladders
C1 
2
7
19
Example 6.4 (iii)
y 22 ( s) 
2
3s( s  4 )
2
s
7
7(2 s  1)
7(2 s  1)
2
2
z1 
3s(s  4 )
2
 z2 
k1 s
s 4
2
7(2 s 2  1) s 2  4 
49
k1  



2
s s 2  4 12
3s(s  4 )
49
s
7
12
z 2  z1  2

s  4 12 s
Franco Maloberti
Analog Filters: Singly-Terminated LC Ladders
20
Example 6.4 (iv)





Another option:
Remove completely the pole of z22 at infinite
This produces a zero at infinite
Produce the required pair of zeros by partially
removing the zero at infinite
The partial removal leave the two zeros only
Franco Maloberti
Analog Filters: Singly-Terminated LC Ladders
21
LC Ladder with Voltage Source

Consider a singly-terminated filter with a voltage
source and a normalized resistive load
I 2  y 21 E 1  y 22 E 2
 y 21 E 1  y 22 I 2
 Y 21 ( s) 
Franco Maloberti
I2
E1

y 21
1  y 22
Analog Filters: Singly-Terminated LC Ladders
22
LC Ladder with Voltage Source (ii)

The y parameters and the z parameters of a
lossless two-port have the same properties.
Use the same procedure studied for current source

Transmission poles (of y) at the origin and infinite

Non-zero transmission poles (of y)

Franco Maloberti
Analog Filters: Singly-Terminated LC Ladders
23