🚀 go-pugleaf

kjh@aludra.usc.edu (Kenneth J. Hendrickson) writes:

>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, you forgot the ( K/s ) term of the VCO.  I believe that in 
Marshall's original posting, he gave the correct closed-loop transfer 
function

	H(s) = A(s)B(s)/( 1 + A(s)B(s) )

where A(s) is the 2-pole filter above and B(s) is the VCO transfer
function.

I have always avoided 2nd-order loop filter for precisely this
reason.  Very hard to estimate what a 3rd-order system would do.
One of the poles is necessarily real, and it's locus better not 
fall on the wrong side of the plane.

73,

Kok Chen, AA6TY				kchen@apple.com
Apple Computer, Inc.