THEORETICAL INFORMATION

The insertion method can be used to characterise a filter response in microwave. It is

defined as the ratio of power available from source to power delivered to load. In this
program two common types of filter characteristics are used: maximally flat and equal
ripple (or Chebyshev) filters.

MAXIMALLY FLAT FILTER:
It provides the flattest possible pass band region for a given filter complexity or order (N).
it is specified by
(1)
Where wc is cut off frequency and N is the order of filter. At cut off it gives 3 dB power
loss. For wwc the attenuation increases monotically as shown in figure 1. Note that L(w) is
10logP
LR
where P
LR
is power loss.

Figure 1:Low-pass filter response for maximally flat filter.Frequency vs power loss
EQUAL RIPPLE (CHEBYSHEV):
It provides a sharper cut off , however the pass band has ripples of amplitude 1+k
2
. It is
specified by
(2)
T
N
(x) oscillates between +1 and 1. k
2
determines the pass band ripple level.




Figure 2: Low-pass filter response for equal ripple filter. Frequency vs power loss
LOW-PASS FILTER DESIGN
Ladder circuits, as shown below, gives the power loss characteristics of those filter
types. The lumped circuit elements are shunt capacitors and series inductors as shown in
figure 3. For maximally flat or equal ripple response in the pass band, the ladder network is
symmetric for odd number of elements. If we let Z
in
be the input impedance seen from Rs,
then reflection coefficient is
(3)
Then, power loss ratio becomes
(4)
At w=0, all capacitors open and all inductors short circuit , and hence Zin=1. P
LR
is zero for
maximally flat and equal ripple filter when N is odd. For equal ripple with N is even
P
LR
=1+k
2
at w=0.



Figure 3:Low-pass ladder prototype filter networks
For maximally flat filter with a power loss ratio
P
LR
=1+w
2N
(6)

the element values can be calculated by with R=1,
k=1,2,,N (7) where gk values are the value of inductance in
henries or the value of capacitance in farads.
Some numerical values of gk for maximally flat filter listed up to N =5 in table 1.

N
K 2 3 4 5
1 1.414 1.00 0.76 0.62
2 1.414 2.00 1.85 1.62
3 1.00 1.85 2.00
4 0.76 1.62
5 0.62
table 1: Values of gk for maximally flat filter for N=2,..,5

Equal ripple low pass filter element values can be calculated by
N even (8)
g
N+1
=1 N odd
when element g
N
is a capacitor g
N+1
=R, but when g
N
is an inductor g
N+1
=1/R.
(9)


Some numerical values of gk for equal ripple filter N up to 5 are listed in table 2.





N
K 2 3 4 5
1 0.84 1.03 1.11 1.15
2 0.84 1.15 1.35 1.37
3 1.03 1.77 1.97
4 0.82 1.37
5 1.15
table 2: Values of gk for equal ripple filter with k
2
=0.023 (0.1 dB ripple)
FILTER TRANSFORMATION:
The low pass prototype element values which are calculated for R=1 and wc=1 can be
modified for arbitrary source impedance and cut off frequency. In addition, by frequency
transformation, band pass and high pass filter element values can be obtained or vice
versa.
Impedance and Frequency Transformation: Element values calculated for wc=1 and
R=1 can be transformed for arbitrary cut off frequency wc and source impedance R
0
by
using following simple equations L =R
0
L/wc (10)
C= C/(R
0
wc)
Low pass to High pass Transformation: The frequency substitution wc/w yields high
pass filter. The point at w=0 is mapped to infinity, meaning that pass band of a low pass
filter is at infinity for a high pass filter. Then new element values can be found by




(11)



which shows that the series inductors must be replaced with capacitors C
k
and the shunt
capacitors must be replaced with inductors L
k
given by
(12)
Then new ladder network is shown in figure 4.



figure 4: high pass filter
ladder network

Low pass to Band pass Transformation:
where wo=(w1w2)
1/2
center frequency.
w=0 is mapped to w=wo, meaning that pass band of low pass filter is now centred around
wo. Cut off frequency 1 mapped to the edge frequencies w=w1 and w=w2 as desired for a
band pass filter. Similarly, new element values are determined by


(13)



Above equations show that the series inductor L
k
is transformed to a series LC circuit with
element values,
(14)

and the shunt capacitor C
k
is transformed to a shunt LC circuit with element values,
(15)
Then new circuit for band pass filter is as shown in figure 5

figure 5: band pass filter circuit.

STEP-IMPEDANCE LOW-PASS AND HIGH-PASS FILTER DESIGN
Low pass filter and high pass filter can be realised by using alternating sections of very
high and very low characteristic impedance lines. They can be implemented either in
microstrip or stripline. Impedance values are limited by fabrication. Typical value is 150W
for Zh (highest impedance) and 10W for Zl (lowest impedance). Because of the
approximations involved the electrical performances are not perfect but they are easier to
design and take less place compared to other designs.
Approximate Equivalent Circuits for short section transmission line:
Z-parameters for transmission line having Zo characteristic impedance given by,
(16)
the series elements of T equivalent circuit are given by
(17)

and shunt element value is Z12. The equivalent circuit (see figure 6) has the following
element values:
X/2=Zotan(b l/2) (18)
B=sin(b l )/Zo
Assuming length is short and the characteristic impedance is large then,
X ~ Zob l, (19)
B~0,
The result implies that it is a series inductor. Similarly, assuming very small characteristic
impedance results in a shunt capacitor.(see figure 6)
X ~ 0 (20)
B~Yob l,





figure 6: Approximate equivalent circuits for
transmission lines
a) T equivalent circuit for a TLb) Equivalent
circuit for smallbl
and large Zo c) Equivalent circuit for small b l
and small Zo
From (19) b l=LR
0
/ Zh for inductor, and from Eq 20 b l=CZl/R
0
for capacitor can be
obtained. The required lengths of the lines are determined from
l = v
p
gkR
0
/(Zh wc) for inductor,
l = v
p
gkZh/(R
0
wc) for inductor, where v
P
is the phase velocity of the
propagating wave on the line.
A typical to view of the stepped impedance low pass filter can be seen in figure 2. For high
pass filter it is just replacing high impedance lines with low impedance transmission line
section and element values transformed.



figure 7: Microstrip layout of low pass filter
COUPLED LINE FILTERS
Coupled transmission lines can be used to implement various filter circuit.



figure 8: a)a parallel coupled line b) even and odd mode currents

Filter characteristics of a single quarter wave coupled line:
To find the open circuit impedance matrix we will consider even and odd mode excitations
separately, then find the result by superposition. I
1
, I
2
, I
3
, I
4
are the total port currents. i
1
,
i
3
are the currents driving the line in even mode, i
3
, i
4
are in odd modes. Note that
V
A
1
,V
B
1
voltages on line A and B due to the current i
1,

V
A
2
,V
B
2
voltages on line A and B due to the current i
2,

V
A
3
,V
B
3
voltages on line A and B due to the current i
3,

V
A
4
,V
B
4
voltages on line A and B due to the current i
4.



At the even mode, the impedance seen from ports 1 and 2 is
Z
e
in=-jZ
oe
cot b l. (21)
The voltage on the line can be expressed as
V
A
1
(z)= V
B
1
(z) =Ve
+
[e
-jb (z-l)
+ e
jb (z-l)
]=2Ve
+
cosb (l-z), (22)



The voltages at port 1 and 2 is for z=0;
V
A
1
(0)= V
B
1
(0)=2Ve
+
cos b l=i
1
Zin
e
.(23)



Using (21) in (23),
V(z)=-jZ
oe
. (24)
In the same way voltage equation can be written in terms of i
3

V
A
3
(z)= V
B
3
(z)=-jZ
oe
. (25)
In the odd mode, the line is driven by i
2
. The impedance seen from 1 and 2
is
Z
o
in=-jZ
oe
cot b l. (26)
The voltage on the line A can be expressed as, the voltage on B is just negative of the
following,
V
A
2
(z)=- V
B
2
(z)=Vo
+
[e
-jb (z-l)
+ e
jb (z-l)
]=2Vo
+
cosb (l-z), (27)
The voltages at port 1 and 2 is for z=0;
V
A
2
(0)=- V
B
2
(0)=2Vo
+
cos b l=i
2
Zin
o
.
Using (26) in (28),
V
A
2
(z)=- V
B
2
(z)=-jZ
oo
. (28)
In the same way voltage equation can be written in terms of i
4

V
A
4
(z)=- V
B
4
(z)=-jZ
oo
. (29)
Using the voltage equations the voltage at port 1 is found as (z=0)
V
1
= V
A
1
(0)+V
A
2
(0) + V
A
3
(0)+V
A
4
(0) (30)
=-j(Zoe i
1
+ Zoo i
2
)cotq - j(Zoe i
3
+ Zoo i
4
)cscq .
Also total currents on the line can be expressed in terms of even and odd
mode currents:
I
1
=i
1
+i
2
,
I
2
=i
1
-i
2
,
I
3
=i
3
-i
4
,
I
4
=i
3
+i
4
. (31)
From (31) w can drive,
i
1
=0.5(I
1
+I
2
),
i
2
=0.5(I
1
-I
2
),
i
3
=0.5(I
3
+I
4
),
i
4
=0.5(I
3
-I
4
),
Using the above results in (30):
V
1
=-j0.5(Zoe(I
1
+I
2
)+Zoo(I
1
-I
2
))cotq -j0.5(Zoe(I
3
+I
4
)+Zoo(I
4
-I
3
))cscq (33)
from symetry z-parameters can be found
Z11=Z22=Z33=Z44=-j0.5(Zoe+Zoo)cotq ,
Z12=Z21=Z34=Z43=-j0.5(Zoe-Zoo)cotq ,
Z13=Z31=Z24=Z42=-j0.5(Zoe-Zoo)cscq ,
Z14=Z41=Z32=Z23=-j0.5(Zoe+Zoo)cscq ,
(34)

We can analyze the filter characteristic by calculating the image impedance (Zi) and the
propagation constant.
for q =pi/2 Zi=0.5(Zoe-Zoo) (35)
When q goes to 0 or pi, Zi goes to infinity; meaning that stop band. Cut off frequency from
(35)
cosq
1
=-cosq
2
=(Zoe-Zoo)/(Zoe+Zoo) (36)
Also propagation constant is determined by
cosb = . (37)




Design of Coupled Line Band pass filters:
A coupled line can be modelled by two transmission line and an admittance inverter in
between as shown in figure 8.



figure9:Equivalent circuit for coupled line

Firstly, let us compute ABCD matrix for this circuit:
(38)
Choosing q =p /2, we can find . (39)

The propagation constant is from the matrix
. (40)
Using (35) and (39)
JZo
2
= (Zoe-Zoo)/2, (41)
And using (40) and (37), also assuming sinq» 1
(42)
Solving (41) and (42) we obtain even and odd impedances
Zoe=Zo[1+JZo+(JZo)
2
],
Zoo=Zo[1-JZo+(JZo)
2
],
Now, we cascade N+1 coupled line as in figure 9


figure 10:Cascaded coupled lines






figure 11: Equivalent circuit for a transmission line with length l /2.




When we look at the equivalent circuit of two adjacent pair of N+1 cascaded coupled
line, there are a transmission line with l /2 length and a J inverter. Using ABCD matrices we
can show that a transmission line with l /2 length can be modelled by a shunt LC circuit
(Fig 11) and an admittance inverter between these LC circuits which transforms shunt LC
to series circuit (Fig 12). In this way, we can use low pass filter prototype element values
to determine J
n
.

figure 12:Equivalent circuit for admittance inverter
the ABCD matrix for T network folowed by transformer is given by

If we equate it to the ABCD matrix of transmission line with length l/2 we can find the
following
Z12=jZo/sin2q
Z11=-jZocot2q
Z11-Z12=-jZocotq
the transformer in the equivalent circuit (Fig 11) shifts the filter response without changing
the amplitude so we can eliminate it. Then we have left with T network. For w=wo+Dw
where wo is the center frequency
2q=bl=( wo+Dw )p/wo (45)
for q~p/2 the series arms impedance of T network become zero, parallel arm impedance is
Z12=jZo/(sinp(1+Dw/wo))~-jZowo/(p(w-wo)) (46)
The impedance of parallel LC circuit near resonance is given by
Z= when we equate these impedances
Z= (46)


Using wo
2
=1/LC
Using inductance and capacitance values following design equations can be derived
k=2,3N (47)


By using (43), even and odd impedances of each coupled line can be found


Finding geometry ratios by using even and odd impedances:
Cascaded coupled lines can be realised in microstrip or stripline form.

figure 13: Coupled stripline
Stripline: For the geometry in figure 13, even and odd impedances are given by
(48)
Using (48) we can derive s/d and w/d ratios:
where a
2
=Zoe/Zoo

figure 14: Coupled microstrip
Microstrip: to find the ratios for microstrip is more laborious to carry out. Here I used
Akhazard method.
Firstly, we should determine equivalent single microstrip shape ratios (w/d)
s
. Then we can
realte coupled line ratios to single line ratios.
Zose=Zoe/2
Zoso=Zoo/2 for single microstrip
Using even and odd impedances for a single line we can find w/d ratio,
For narrow strips Zos {44-2e
r
}



for wide strips Zos {44-2e
r
}
where d=59p
2
/(Zoe
r
1/2
)



A DESIGN EXAMLE AND COMPARISON:
I run the program for a low pass filter power loss of 15 dB at 2 GHz and cut off frequency
at 1 GHz. The filter impedance is chosen as 50 ohm. The highest and the lowest practical
impedance values are 150 and 10 ohm respectively.
The output of the program for e=2.2:
Lengths of lines w/d ratios
l
1
=0.64cm w1/d= 22.65
l
2
=14cm w2/d=0.33
l
3
=0.64cm w3/d=22.65
For the substrate thickness of d=0.5 mm, I found the corresponding widths of lines.
Then I designed the circuit by using SONNET the results was satisfactory. The graph is
drawn according to SONNET, as it can be seen from the graph at 1 GHz 3.6 and 2 GHz 17.7
dB power loss is obtained.




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