Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 2
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
LC LadderI1
I2
E2Ω1
Or a voltage source
€
E1 = z11I1 + z12I2E2 = z21I1 + z22I2
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 3
LC Ladder with Current source
Consider a singly-terminated filter with a current source and a normalized resistive load
22
21
1
221
222121
2221212
1)(
z
z
I
EsZ
EzIz
IzIzE
+==∴
−=+=
LC LadderI1
I2
E2Ω1
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 4
LC Ladder with Current source (ii)
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
€
Z21(s) =E2
I1=z21
1+ z22
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 5
Implementing Z21(s)
Consider the transfer function
If P(s) is an even polynomial
)()(
)(or
)()(
)(
)(
)()(
22
1
22
121 sNsM
sKM
sNsM
sKN
sQ
sPsZ
++==
€
M(s) = even terms
N(s) = odd terms
Q(s) = Hurwitz polynomial
€
Z21(s) =
KM1(s)
N2(s)
1+M2(s)
N2(s)
→ z22 =M2(s)
N2(s)and z21 =
KM1(s)
N2(s)
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 6
Implementing Z21(s) (ii)
If P(s) is an odd polynomial
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
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Z21(s) =
KN1(s)
M2(s)
1+N2(s)
M2(s)
→ z22 =N2(s)
M2(s)and z21 =
KN1(s)
M2(s)
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 7
Transmission zeros at 0 or infinite
Use of Cauer’s realizations to remove zeros at the origin or infinite completely.
Use an intuitive view.
Example
Three non-dissipative elementsInput is a parallel element
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Z21(s) =P(s)
Q3(s)
Z1Z2Z3Z1Z2Z3
Series Lzero @ ∞
Series CZero @ 0
Shunt CZero @ ∞
Shunt LZero @ 0
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 8
Intuitive view (Example Q3(s))
∞∞∞
000
∞00
∞∞0
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 9
Example 1
numeratoreven 122
)(2321 ←
+++=
sssK
sZ
ss
sss
K
sZ
2
121
2)(
3
2
3
21
++
+
+=∞=
←s at zeros
ion transmissthree
€
z22 =2s2 +1
s3 + 2s; z21 =
K
s3 + 2s
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 10
Example 1 (ii)
Three transmission zeros at s = ∞
sssss2
3
2
1)12(2 23 +⋅+=+
134
23
12 2 +⋅=+ sss
023
123
+⋅= ss
Applying the long division toget circuit parameters
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 11
Example 1 (iii)
Network realization
Evaluation of k
I1 E2Ω1F23
F2
1
H3
4
KZI
E=== )0(1 21
1
2
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 12
Example 1 (iv)
Use of Matlab for removing transmission zeros
€
z22 =2s2 +1
s3 + 2s
clear allnum=[1 0 2 0];den=[2 0 1];[c1,r1]=deconv(num,den);c1r1=r1(3:4);[l,r2]=deconv(den,r1);lr2=r2(3);[c2,r3]=deconv(r1,r2)
ex6_1c1 = 0.5000 0l = 1.3333 0c2 =1.5000 0r3 = 0 0
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 13
Example 2
High-pass Butterworth filter
One zero at 0 and
two zero at ∞
€
Z21(s) =Ks3
s3 + 2s2 + 2s+1; Z21(s) =
Ks3
2s2 +1
1+s3 + 2s
2s2 +1
€
z22 =s3 + 2s
2s2 +1; z21 =
Ks3
2s2 +1∞∞0
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 14
Example 2 (ii)
Remove the pole of z22 at infinite
€
z1 =s3 + 2s
2s2 +1−
1
2s; z1 =
3s
4s2 + 2 ∞∞0
1/2 H
3/4 F3/2 H
€
z21 =ILI1
=
3
2s
1
2s+1
→3
2s =Ks
I1 IL
€
z21 =Ks3
2s2 +1→Ks
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 15
Zeros at finite frequency
Finite zeros are zeros of P(s) (or z21)
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
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Z21(s) =P(s)
Q(s)=
P(s)
M2(s) + N2(s)
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 16
Zero Shifting
The partial removal of poles from a function DOES affect the zeros of the remainder Zero shifting is necessary
Consider
When part of the pole at (left) and at the origin (right) is removed
)4(
)9)(1()(
2
22
+++
=ss
sssZ
ω
)](Im[)( ωω jZX =
1 2 3
ω'∞k
0ω
)](Im[)( ωω jZX =
1 2 3
ω/'∞− k
0
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 17
Zero Shifting (ii)
A partial pole removal shits all the finite (and non-zero) 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 possibleX( )
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 18
Example 6.4
€
Z21(s) =K(s2 + 4)
s3 + 2s2 + 2s+1
€
z22(s) =2s2 +1
s3 + 2s
Zeros of z22 is at
Poles of z22 are at
€
ω =1
2
€
ω =0; 2
We can move thezeros only in the range
€
0; 2
Work with y22 !!
€
y22(s) =s3 + 2s
2s2 +1
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 19
Example 6.4 (ii)
Zeros of y22 is at
Poles of y22 are at
€
ω =1
2; ∞
€
ω = 2We can move thezeros only in the range
€
1/ 2; ∞
€
1
2€
2
€
2€
y22 −C1s = 0s= j 2
€
y22(s) −C1s s= j2 =s3 + 2s
2s2 +1−C1s
⎡
⎣ ⎢
⎤
⎦ ⎥=
=j4
7− j2C1 = 0
€
C1 =2
7
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 20
Example 6.4 (iii)
€
y22(s) −2
7s =
3s(s2 + 4)
7(2s2 +1)
€
z1 =7(2s2 +1)
3s(s2 + 4)= z2 +
k1s
s2 + 4
€
k1 =7(2s2 +1)
3s(s2 + 4)⋅s2 + 4
s
⎡
⎣ ⎢
⎤
⎦ ⎥s2 =−4
=49
12
€
z2 = z1 −
49
12s
s2 + 4=
7
12s
1Ω2/7 Fz1( )sI1( )s
1Ω2/7 Fz2( )sI1( )s48/49 F49/12 H
1Ω2/7 FI1( )s48/49 F49/12 H12/7 F
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 21
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 22
LC Ladder with Voltage Source
Consider a singly-terminated filter with a voltage source and a normalized resistive load
€
I2 = y21E1 + y22E2
= y21E1 − y22I2
∴Y21(s) =I2E1
=y21
1+ y22
LC LadderI1 I2E2Ω1E1
Franco Maloberti Analog Filters: Singly-Terminated LC Ladders 23
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)