Band Pass Filter Design & L C Realization
By Fu, Law
Chain
EE 175
Abstract To
understand the Band Pass Filter & L C Realization. The following circuit is
an example of a band pass filter:
First we will consider a qualitative analysis of the circuit. Recall that the impedance of the inductor and capacitor are:
and
.
Hence if the frequency is zero (i.e. D.C.) the impedance of the inductor is zero (i.e. a short circuit) and the impedance of the capacitor is infinite (i.e. an open circuit), this is shown in the circuit below:
Now if the frequency is infinite, the impedance of the inductor is infinite (i.e. an open circuit) and the impedance of the capacitor is zero (i.e. a short circuit), this is shown in the circuit below:
Now we will consider the quantitative analysis.
Using Kirchoffs' voltage law gives:
and ohm's law:
we can calculate the gain of the circuit by:
The following graph is of the gain of the band pass filter circuit shown above:
The gain of the circuit is:
L C
realization with load 50 Ω
Odd = S3 + 4S; Even = S4+10S2+9
If N(s) is even function, where
S3+4S S4+10S2+9
=
Test Z2LC for an open and short circuit @ s = 0 and s = ∞ respectively
Z2LC ( s → 0 ) = 0 ; Z2LC ( s → ∞ ) = 1/ s = 0
Note next that Z (s) has four poles: therefore the network realization contains four elements. Since the number is even, there are two inductors and two capacitors. Next we can use foster I, foster II, cauer I and cauer II.network. Before we use it, we have to check the first equation first .
When
H(s) / s = 0 = H S2 = 0
H (s) / s = ∞ = 
Therefore, it match with cauer I network. For L C network, Cauer I has inductors as series elements and capacitor as shunt elements.
Cause I network
Check the
validity of Cauer I Impedance
When
s = 0, L1 and L2 will short out the signal, resulting zero impedance.
When s = ¥, L1 will yield an infinite impedance.
Algorithm
|
Y1=1s |
1 |
10 |
9 |
|
Z2=1/6s |
1 |
4 |
0 |
|
Y3=12/5s |
6 |
9 |
0 |
|
Z4=5/18s |
5/2 |
0 |
0 |
|
|
9 |
0 |
0 |
L1= 
, C2 = 
, L2 =
, C1 = 1
Once the LC network is synthesized by using the above mentioned two cases approach, the elements’ values can be scaled accordingly by using the scaling formulas given below.
: Rn
= New, or scaled resistance value
R = Old, or un-scaled resistance value
Km= Magnitude scaling factor
: Km= Magnitude scaling
factor
Kf= Frequency scaling factor
Ln = New, or scaled inductance value
L = Old, or un-scaled inductance value
: Km= Magnitude scaling factor
Kf= Frequency scaling factor
Cn = New, or scaled capacitance value
C = Old, or un-scaled capacitance value
Cauer I elements’ values can now be scaled by using scaling formulas, as shown in next
\
: Rn
= New, or scaled resistance value
R = Old, or un-scaled resistance value
Km= Magnitude scaling factor
: Km= 50Kf = Frequency scaling factor = 1
Ln = New, or scaled inductance value
L = Old, or un-scaled inductance value
Ln1= 
= 50 * 
H = 
H = L3
Ln2= 
= 50 * 
H = 
H = L4
: Km= 50
Kf = Frequency scaling factor = 1
Cn = New, or scaled capacitance value
C = Old, or un-scaled capacitance value
Cn1= 
= 
= C4
Cn2= 
= 
= 
= C3
Now, we have to find the H(s) value. Which is
H(s) = 
/s=∞ = 
= 
Therefore
H = 
Appendix:
Reference: Strasilla Lecture note
Programs were used: Words for Windows, PSICE.