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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:

enter image description here

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):

enter image description here

How can I get my slope, that is 40dB/decade, to match his 20dB/decade? Thanks.

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  • \$\begingroup\$ What are the units in the xaxis of the plot? This is important as it affects the scaling of the roll off. What are the units in your xaxis of your plot? \$\endgroup\$
    – Voltage Spike
    Jan 28, 2016 at 0:03

2 Answers 2

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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.

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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: -

enter image description here

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  • \$\begingroup\$ Thanks for your response. You are correct. I changed the equation in my original question. In my Matlab code I actually plotted 1-TF. My apologies. For the error function then, that has a high pass response like that given in my question, how do I change the slope to 20dB/decade? \$\endgroup\$
    – Joe
    Jan 27, 2016 at 23:40
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    \$\begingroup\$ @Joe it's the difference between a 2nd order and a first order filter - if you are trying to mimic something then use a first order filter. \$\endgroup\$
    – Andy aka
    Jan 28, 2016 at 8:26

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