Transcript Analog Filters

Analog Filters: Doubly-Terminated
LC Ladders
Franco Maloberti
Basic Formulation

The basic arrangement is
R1
Es
I1
+
E1
I2
Lossless
Twoport
+
E2
R2
Z 11(s)


The voltage or the current are no more meaningful
What matter is the power delivered to the load
Franco Maloberti
Analog Filters: Doubly-Terminated LC Ladders
2
Power Transmission

Maximum transmission
R1
Es
I1
I2
+
E1
+
E2
Es
E2  E1 
R1  R 2
E2
P2 
R2
R2
2
Es
P2 
R2
2
(R1  R2 )
2
R1  R2
Max for
2
P2,MAX

Es

4R1
If the two resistances are not equal maximum
power by using a transformer
Franco Maloberti
Analog Filters: Doubly-Terminated LC Ladders
3
Transmission and Reflection

The transmission coefficient is defined by
t (j) 
2

P2
P2,MAX

4R1 E2
R 2 E1
1
The reflection coefficient is
( j)  1 t ( j )
2

2
For a filter
t (j)  1
2
in the pass-band
t (j)  1 in the stop-band
Franco Maloberti
Analog Filters: Doubly-Terminated LC Ladders
4
Transmission and Reflection (ii)

Define the input impedance at port 1 with port 2
terminated with R2
R1
Es
Es
I1 
R1  Z11 (s)
I1
+
E1
2
P1 ()  I1 R11 ( ) 
2
Z 11(s)

The two-port is loss-less
P1 ()  P2 ( )
t (j) 
2
Franco Maloberti
E s R11 ()
R1  Z11 ()
2
Es R11 ()
2
R1  Z11 ()
2
2

Es
4 R1
Analog Filters: Doubly-Terminated LC Ladders
4 R1 R11 ( )
R1  Z11 ()
2
5
Transmission and Reflection (iii)

The reflection coefficient becomes
 ( j)  1 t ( j )  1
2
( j )  

2
R1  Z11 ()
R1  Z11 ()
4R1R11 ()
R1  Z11 ()
2

R1  Z11 ()
2
R1  Z11 ()
2
Z11 (s)  R1
1  (s)
1  (s)
Realizing a given transmission coefficient becomes
realizing a corresponding Z11(s)
Franco Maloberti
Analog Filters: Doubly-Terminated LC Ladders
6
Ladder Filter Design: Procedure




Given a certain transmission coefficient |t(j)|2
Estimate the reflection coefficient |(j)|2
Determine (s)
Chose the upper or lower sign
1  (s)
Z11 (s)  R1
1  (s)
1  (s)
Z11 (s)  R1
1  (s)

1  (s)
Z11 (s)  R1
1  (s)
Realize Z11(s)
Franco Maloberti
Analog Filters: Doubly-Terminated LC Ladders
7
Example
1
1
t (j) 

1  6 1 s 6 s j
s3
 (s)  3
s  2s 2  2s 1
2
6
s6
 ( j ) 
6  6
1 
s  1 s j 
2
den=[-1 0 0 0 0 0 1];
»b=roots(den)
b=
-1.0000
-0.5000 + 0.8660i
-0.5000 - 0.8660i
0.5000 + 0.8660i
0.5000 - 0.8660i
1.0000
»c=poly([b(1:3)])
c = 1.0000 2.0000 2.0000
Franco Maloberti
1 (s) 2s 2  2s 2  2s 1
Z11 (s) 

1 (s)
2s2  2s 1
Long division or other
Implementation methods
1.0000
Analog Filters: Doubly-Terminated LC Ladders
8
Unequal terminations


Three methods
Implement the problem with equal terminations
R1
Lossless
Twoport
Es
R2R3
R2
R3
R2+R3
= R1
R1
Es


Twoport
R3
R2
Use a transformer
Account for the gain reduction in t(j)
Franco Maloberti
Analog Filters: Doubly-Terminated LC Ladders
9
Use in Reverse

Assume to exchange the role of the two ports
I1
R1
+
E1
I2
Lossless
Twoport
+
E2
R2
Es
The transmission coefficient in reverse is the same
as the one in direct configuration (no demonstration)

We can flip the network
Franco Maloberti
Analog Filters: Doubly-Terminated LC Ladders
10