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From: Jason Rosinski
X-Mailer: Mozilla 4.7 [en] (X11; U; SunOS 5.8 sun4u)
Subject: Re: How to increase PLL order?
Date: Mon, 09 Dec 2002 17:35:05 -0500
NNTP-Posting-Date: Mon, 09 Dec 2002 17:35:15 EST
Organization: Bell Sympatico
"Christopher R. Carlen" wrote:
> It seems I could put the integrator in series with the existing loop
> filter, in which case I'd change the total filter transfer function to
> something like:
> Hf(s)=Kf*(1+s/wz)/(1+s/wp) * Ki/s
> Or I could split the phase detector signal, and sent it to both the
> existing loop filter, and an integrator, then sum the result, yielding a
> PI controller within the PLL:
> Hf(s)=Kf*(1+s/wz)/(1+s/wp) + Ki/s
> Which one of these methods is the right one, or am I heading out to left
> field here, and to increase the order of the loop I must do something
> substantially different than what I'm thinking?
> Suggestions appreciated.
> Good day!
(finally something I can contribute to the group!)
One way to form a 2nd order loop in a PLL is to place the integrator in
parallel with the proportional term and sum them at the end. This is
really well covered in Dan Wolaver's book, 'Phase-locked loop circuit
design'. He's got some nice closed-form solutions to show damping
constrains and passband gain (essential for telecom PLL's).
I'm assuming your system is of the form:
with the 1/s from the VCO, etc.
So a proportional loop gives:
Putting an integrator in parallel gives:
Xfer(s)= K1*s + K2
s^2 + K1*s + K2
So to put your mind at ease, yes there are lots of PLL's that use the
method of an integrator in parallel with a summation at the filter
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