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DC gain of Type-1 PLL

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Finite DC gain of Type-1 PLL

I suppose Type-1 PLL also has NCO/VCO as integrator , so Why Type-1 PLL has finite DC gain while Type-2 has infinite DC gain ?

I have found some similar claim below, but there is no mathematical proof. Could anyone help ?

Section 2.3.3 of Floyd Gardner book

**broken link removed**

VKlG32T.png
 

I understand that there is only one pole in the open-loop transfer function (which is G(s)) of type-1 PLL , therefore the use of final value theorem to obtain DC gain will cancel out the only pole (which is the denominator 's' within G(s))

However, according to the screenshot below obtained from https://www.egr.msu.edu/classes/me451/jchoi/2007/handouts/ME451_S07_lecture12.pdf#page=5 , if we apply a step input (which is represented as 1/s in laplace domain), then we still have a denominator 's' within G(s) after the use of final value theorem to obtain DC gain.

Could anyone help ?

mfRB0Er.png
 

Re: Finite DC gain of Type-1 PLL

There is an ideal integrator in the type2 PLL which has got infinite gain for DC signals. See what is a Bode diagram for such a loop filter.
 

The problem is in using ambiguous terms. "DC gain" in the quoted lecture is referring to the phase detector + loop filter, not the VCO transfer function respectively the product of both, the loop gain. The VCO pole doesn't cancel phase errors, you need a finite control voltage to maintain the operation frequency which corresponds to a respective phase error in a first order PLL.
 

The answer is actually : A type 1 PLL has only one integrator in the loop, that from the VCO , therefore the use of final value theorem to obtain DC gain resulted in finite value for Type-I PLL

Note: Screenshots below are from Floyd Gardner book.

xiGzJ5T.png


fGf1bRj.png
 

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