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- 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
- TYPE NUM #1 #2 #3
- 1 CPP 1 1.967
- 2 INS 2 12.875
- 3 CPP 3 6.366
- 4 EQU 4 2
- 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
-
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- - 18 -