analog filters: singly-terminated lc ladders franco maloberti

Analog Filters: Singly-Terminated LC Ladders

Franco Maloberti

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


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


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


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)


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

€

Z21(s) =

KN1(s)

M2(s)

1+N2(s)

M2(s)

→ z22 =N2(s)

M2(s)and z21 =

KN1(s)

M2(s)


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

€

Z21(s) =P(s)

Q3(s)

Z1Z2Z3Z1Z2Z3

Series Lzero @ ∞

Series CZero @ 0

Shunt CZero @ ∞

Shunt LZero @ 0


Intuitive view (Example Q3(s))

∞∞∞

000

∞00

∞∞0


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


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


Example 1 (iii)

Network realization

Evaluation of k

I1 E2Ω1F23

F2

1

H3

4

KZI

E=== )0(1 21

1

2


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


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


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


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

€

Z21(s) =P(s)

Q(s)=

P(s)

M2(s) + N2(s)


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


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( )


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


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


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


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


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


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)

analog filters: singly-terminated lc ladders franco maloberti

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