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From: "Christopher R. Carlen"
Subject: Re: How to increase PLL order?
Date: Mon, 09 Dec 2002 16:37:56 -0800
Organization: Sandia National Laboratories, Albuquerque, NM USA
NNTP-Posting-Date: Mon, 9 Dec 2002 23:36:08 +0000 (UTC)
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Jason Rosinski wrote:
> (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).
Heck, that's the book I've got! Unfortunately, this is rather difficult
material for a non-EE with only a chemistry background, so I have been
wading my way through it all very slowly, and have been detoured for a
few months by a course in EM. After that, hopefully I can begin looking
at the Wolaver book in depth once again, as well as a text on control
> I'm assuming your system is of the form:
> with the 1/s from the VCO, etc.
> Reduces to:
> So a proportional loop gives:
> Putting an integrator in parallel gives:
> G(s)=K1+K2/s leaving
> Xfer(s)= K1*s + K2
> s^2 + K1*s + K2
The problem is that if you look at my VCO transfer function, I've
already got a third order denomiator polynomial, as opposed to the
typical first order Kvco/s. It seems I made an error and wrote:
Hmotor(s)=(1/(LC))/(s2+(R/L)s+1/(LC)s)*N/Kv in (rad/V)
when it should have been:
Hmotor(s)=(1/(LC))/((s^2+(R/L)s+1/(LC))s)*N/Kv in (rad/V)
One of the motor/VCO poles is at quite a high frequency (the one from
the winding inductance) and can be almost ignored for reasonable unity
gain frequency choices, so there is effectively only one pole at some
finite frequency near the PLL's desired bandwidth, and a pole at DC.
It is because of this stuff in my motor/VCO H(s) that the typical PI
loop filter doesn't work with motors. Thus, my present "zero-pole"
filter has a response that looks like:
Unlike the typical "pole-zero" filter used in PLLs with Kvco/s VCO
Putting in an integrator *factor* in the filter transfer function that I
have creates a response that looks something like:
Without the increasing slope of the zero-pole filter through the unity
gain frequency to compensate for the steep open loop gain response
rolloff (the open loop excluding the filter, that is) passing through
the unity gain frequency, I don't know if this integrating factor will
allow the total open loop response to pass through unity gain at the
required <6 dB/octave required for stability.
So does that mean I need to add another zero to compensate for the
integrator's pole at DC?
> 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
I think I just have to hack through a whole pile of math starting in
January to figure all this out to where I can understand it and model
everything. But I'll try some experiments with integrating terms and
factors in the mean time.
Thanks for the input.
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
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