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For the second order PLL with equation:
$$1-\dfrac{2\zeta\omega s + \omega^{2}}{s^{2}+2\zeta\omega s + \omega^{2}}$$
I used Matlab and obtained a Bode Magnitude (with zeta = 0.3, 0.5) like:
In the book "Phase Lock Loops and Frequency Synthesis" by V. F. Kroupa, the author has a plot that looks like this (for a 2nd order system found on page 14 of the book):
How can I get my slope, that is 40dB/decade, to match his 20dB/decade? Thanks.
You have a 2nd order system, it will always be 40dB/decade. The figure is for a 1st order system (it says so in the caption), so it is 20 dB/decade.
This answer was given before the question detail was corrected!
Take your formula: -
\$\dfrac{2\zeta\omega s + \omega^{2}}{s^{2}+2\zeta\omega s + \omega^{2}}\$
Now, put s to zero. You then get unity gain at DC. This is not what your graph shows. Your graph shows a high pass filter response: -
