Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!usc!apple!kchen From: kchen@Apple.COM (Kok Chen) Newsgroups: sci.electronics Subject: Re: A question about a PLL synth loop filter Message-ID: <46225@apple.Apple.COM> Date: 2 Nov 90 04:02:56 GMT References: <1990Oct31.210242.20619@aplcen.apl.jhu.edu> <12890@chaph.usc.edu> Organization: Apple Computer Inc., Cupertino, CA Lines: 54 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.