mjj@stda.jhuapl.edu (Marshall Jose) (11/01/90)
I have been trying to understand a passive loop filter I have twice seen used
in ham radio construction articles. It looks like this:
O---VVVVV----+-------+------O
R1 | |
| >
| > R2
| >
--- |
--- C1 |
| |
| ---
| --- C2
| |
O------------+-------+------O
I have been having a great deal of trouble trying to factor out at least
one zero in the denominator of H(s) = F(s)G(s)/[1 + F(s)G(s)], where
F(s) = K/s and G(s) = response of above filter. I get something like
C(as + 1)
H(s) = ---------------------
3 2
s + ps + qs + r
where p, q, & r are expressions involving R1, R2, C1, and C2. It looks
like it ought to be a good performer as long as I could move the poles
to the right place. When I try to cheat and have an expression manipulator
(such as Mathematica) find the roots, I (deservedly) get an enormous,
intuitively opaque result.
Has anybody out there dealt with this at the design level, or at least
seen a reference to it? The Egan, Manassewitz, and Gardner books seem
not to address it, but I haven't checked the journals yet.
Thanks in advance,
Marshall Jose WA3VPZ
mjj%stda@aplcen.apl.jhu.edu || ...mimsy!aplcen!aplvax!mjjkjh@aludra.usc.edu (Kenneth J. Hendrickson) (11/02/90)
In article <1990Oct31.210242.20619@aplcen.apl.jhu.edu> @aplvax.jhuapl.edu:mjj@stda.jhuapl.edu (Marshall Jose) writes:
%I have been trying to understand a passive loop filter I have twice seen used
%in ham radio construction articles. It looks like this:
%
%
% O---VVVVV----+-------+------O
% R1 | |
% | >
% | > R2
% | >
% --- |
% --- C1 |
% | |
% | ---
% | --- C2
% | |
% O------------+-------+------O
%
%
% C(as + 1)
% H(s) = ---------------------
% 3 2
% s + ps + qs + r
%
Your first clue, is that since you have two energy storage devices, that
you should have two poles. When I solve this, I get:
c1r1 * s + 1
H(s) = ---------------------------------------------
c1c2r1r2 * s^2 + (c2r2 + c1r1 + c2r1) * s + 1
Ken Hendrickson N8DGN/6 kjh@usc.edu ...!uunet!usc!pollux!kjh