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)

EQUAL RIPPLE (CHEBYSHEV):

          (2)

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+k² at w=0.
 
 

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

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.

 
 
 

		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

FILTER TRANSFORMATION:

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 R₀ by using following simple equations L’ =R₀L/wc (10)

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

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

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
 
 

 Above equations show that the series inductor L_k is transformed to a series LC circuit with element values,

and the shunt capacitor C_k is transformed to a shunt LC circuit with element values,

 

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

Approximate Equivalent Circuits for short section transmission line:
Z-parameters for transmission line having Zo characteristic impedance given by,

the series elements of T equivalent circuit are given by

and shunt element value is Z12. The equivalent circuit (see figure 6) has the following element values:

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.
 
 

COUPLED LINE FILTERS

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. I₁, I₂, I₃, I₄ are the total port currents. i₁, i₃ are the currents driving the line in even mode, i₃, i₄ are in odd modes. Note that

V_A¹ ,V_B¹ voltages on line A and B due to the current i_1,

V_A² ,V_B² voltages on line A and B due to the current i_2,

V_A³ ,V_B³ voltages on line A and B due to the current i_3,

V_A⁴ ,V_B⁴ 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

The voltage on the line can be expressed as

The voltages at port 1 and 2 is for z=0;

Using (21) in (23),

In the same way voltage equation can be written in terms of i₃

Using (26) in (28),

In the same way voltage equation can be written in terms of i₄

Using the voltage equations the voltage at port 1 is found as (z=0)

V₁=-j0.5(Zoe(I₁+I₂ )+Zoo(I₁-I₂ ))cotq -j0.5(Zoe(I₃+I₄ )+Zoo(I₄-I₃ ))cscq                (33)

from symetry z-parameters can be found

  

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)

Also propagation constant is determined by

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.
 
 

Firstly, let us compute ABCD matrix for this circuit:

Choosing q =p /2, we can find . (39)
 

The propagation constant is from the matrix

Using (35) and (39)

         (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.

 
 

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 wo²=1/LC 

Using inductance and capacitance values following design equations can be derived

        k=2,3…N (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.

               (48)

Using (48) we can derive s/d and w/d ratios:

           where a²=Zoe/Zoo

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=59p²/(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

l₁ =0.64cm      w1/d= 22.65

l₂ =14cm         w2/d=0.33

l₃ =0.64cm      w3/d=22.65

For the substrate thickness of d=0.5 mm, I found the corresponding widths of lines.