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1986-08-29
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56 lines
I. EXAMPLES
a. LOWPASS FILTER
Let's begin the examples with a simple five-pole lowpass
filter, having a cutoff frequency of 1 GHz and a maximally flat
passband response. Using commonly found design tables, we de-
rive the following circuit:
ELEMENT PARAMETER
12.875nH 12.875nH TYPE NUM #1 #2 #3
O-o-MMM--o--MMMM-o-O 1 CPP 1 1.967
|1.967p| |1.967pf 2 INS 2 12.875
=== === === 3 CPP 3 6.366
| |6.366pf| 4 EQU 4 2
O-o------o-------o-O 5 EQU 5 1
6 CAX 6 1 5
7 END 6
Note that the EQU (equivalent) command has been used for
the last two elements. We can do this as the filter is symmet-
rical. This saves execution time as we are using the previous-
ly calculated S-parameters of elements 1 and 2 rather than cal-
culate them again. Now, let's see how this filter performs
from .1 to 2 GHz in .1 GHz steps (see the chapter on response
for instructions on how to enter analysis frequencies).
FREQ S11 S21 S12 S22
in GHz in dB in dB in dB in dB
0.1 -99.81 0 0 -99.81
0.2 -69.85 0 0 -69.85
0.3 -52.27 0 0 -52.27
0.4 -39.79 0 0 -39.79
0.5 -30.1 0 0 -30.1
0.6 -22.21 -0.03 -0.03 -22.21
0.7 -15.61 -0.12 -0.12 -15.61
0.8 -10.13 -0.44 -0.44 -10.13
0.9 -5.88 -1.3 -1.3 -5.88
1 -3.01 -3.01 -3.01 -3.01
1.1 -1.42 -5.55 -5.55 -1.42
1.2 -0.65 -8.57 -8.57 -0.65
1.3 -0.3 -11.7 -11.7 -0.3
1.4 -0.15 -14.76 -14.76 -0.15
1.5 -0.07 -17.68 -17.68 -0.07
1.6 -0.04 -20.45 -20.45 -0.04
1.7 -0.02 -23.06 -23.06 -0.02
1.8 -0.01 -25.54 -25.54 -0.01
1.9 -0.01 -27.89 -27.88 -0.01
2 0 -30.1 -30.1 0
è - 18A -