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!kjhkchen@Apple.COM (Kok Chen) (11/02/90)
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.
mark@mips.COM (Mark G. Johnson) (11/02/90)
O---VVVVV----+-------+------O
R1 | |
| >
| > R2
| >
--- |
--- C1 |
| |
| ---
| --- C2
| |
O------------+-------+------O
For a first cut at a design, you can save some effort by using simplifying
approximations:
Pole at (approx) T = ((R1+R2) * C2)
Zero at (approx) T = (R2 * C2)
Pole at (approx) T = (R1 * C1)
--
-- Mark Johnson
MIPS Computer Systems, 930 E. Arques M/S 2-02, Sunnyvale, CA 94086
(408) 524-8308 mark@mips.com {or ...!decwrl!mips!mark}wang@motcid.UUCP (Jerry Wang) (11/03/90)
kchen@Apple.COM (Kok Chen) writes: >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. Agree. There is another way to look at this. 1. By eliminating C1 or making it negligibly small, this becomes a first order 'lag-lead' filter. The resulting second order PLL is stable by a properly chosen R2. 2. The above system is characterized by its damping factor d, and undamped natural frequency Wn. Wn = sqrt ( K / (T1+T2) ) -------- (1) d = Wn * (T2 + 1/K) / 2 -------- (2) where K is the loop gain determined by Kvco and N, N being the divide ratio in the feedback loop. T1 = R1*C2 T2 = R2*C2 The system has optimum unit (phase) step response when d =1. 3. Theoretically we can choose an arbitary Wn (which determines the frequency response of the system), calculate T2 by setting d = 1 in equation 2, then solve T1 by equation 1. Note: If the PLL is programable, K will vary based on the operating point of VCO and N value for the feedback counter. d will vary according to K. Under all situations, d >= 1 must be maintained in order to maintain the stability of the loop. 4. But PLL is unique in a way that the phase detector output is best modeled by a bipolar pulse train at the frequency of it's refernce input. The amount of 'glitch' attenuation priveded by the above lag-lead filter is roughly R1/(R1+R2), which is poor. The poorly attenuated phase detector output glitches will be translated into phase jitter by the VCO. 5. So how do we deal with this? a. Choose a Wn much much less than the reference frequency (this produces large T1 + T2). b. Follow the steps in #3 to determine T1 and T2. c. Now let's introduce a C1 to attenuate the phase detector output glitches at the reference frequency. Since the Wn is much much less than the reference frequency, R1*C1 can be much less than T1+T2 and becomes negligible as assumed in #1. d. The best strategy is still a clean VCO and a clean refernce input, which will generate pulses with nearly zero pulse width. The nearly zero energy associated with nearly zero pulse width will reduce the R1*C1 requirement. Jerry Wang (Motorola - Radio Telephone System Group)