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Referring to https://www.analog.com/media/en/technical-documentation/data-sheets/AD2S90.pdf page 9 states that RDC chip implements a type2 servo loop and the compensation is provided by the function $$\frac{K}{s}\cdot \frac{1+st_1}{1+st_2}.$$

Elsewhere in https://www.analog.com/media/en/technical-documentation/data-sheets/AD2S80A.pdf, page 11, the circuit appears different. Adding to confusion, https://www.ti.com/lit/an/slva662/slva662.pdf page 4, the Transfer function appears different too. Why is this so?

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    \$\begingroup\$ Please extract the useful information you need help with rather than asking us to browse all these documents. It is faster and more efficient to formulate an answer then. Thank you. \$\endgroup\$ Nov 5, 2017 at 9:39

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This is all about properly writing a transfer function in a low-entropy form. In a type II compensator, you have a pole at the origin to provide infinite gain in dc (actually bounded by the op amp open-loop gain \$A_{OL}\$) plus one pole and one zero.

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The original equation obtained by rearranging the division of the impedance in the feedback path with \$R_1\$ leads to \$G(s)=-\frac{1+sR_2C_1}{sR_1(C_1+C_2)(1+sR_2\frac{C_1C_2}{C_1+C_2})}\$. However, you have absolutely 0 insight in this expression. The best is to reorganize it by factoring \$sR_2C_1\$ in the numerator and highlight what is called an inverted zero:

\$G(s)=-G_0\frac{1+\frac{s_z}{s}}{1+\frac{s}{s_p}}\$ with \$G_0=\frac{R_2C_1}{R_1(C_1+C_2)}\$, \$\omega_z=\frac{1}{R_2C_1}\$ and \$\omega_z=\frac{1}{R_2\frac{C_1C_2}{C_1+C_2}}\$. In this expression, you first adjust the zero and the pole to create a phase boost at crossover and then, you adjust the mid-band gain \$G_0\$ as described in this document. The expression I gave is similar to that given in the first doc as \$G(s)=\frac{k_1}{s}\frac{1+s\tau_1}{1+s\tau_2}\$ however, you now clearly see the leading term \$G_0\$ as a gain. In page 11 of the second document, the compensator transfer function is lumped in the closed-loop equation (bottom right) so it is difficult to judge what expression they have adopted. Then in the TI document, the expression in (4) is mathematically correct and is that of the type 2 but gives absolutely 0 insight on poles and gain as it does not stick to the low-entropy format I used.

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  • \$\begingroup\$ Please look at the link ti.com/lit/an/slyt688/slyt688.pdf , page 4. What could be the transfer function of this PI controller? It seems to be different that what is described in AD and TI app notes \$\endgroup\$
    – vishnu
    Nov 12, 2017 at 13:16
  • \$\begingroup\$ They say (type II) so it is the above op-amp-based transfer function. \$\endgroup\$ Nov 12, 2017 at 16:20
  • \$\begingroup\$ Again please see the link folk.ntnu.no/skoge/prost/proceedings/ifac2014/media/files/… (page 4). It says standard PI controller instead of the above mentioned compensator. What really is confusing me is that there are multiple configurations for type 2 compensator. Now which one to adopt. Are all of them right or is it situational. Does PI alone would suffice as suggested in the link \$\endgroup\$
    – vishnu
    Nov 12, 2017 at 17:01
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    \$\begingroup\$ It depends on the amount of phase boost you need at crossover. The thing is to compensate for the total plant lag and build phase margin at \$f_c\$. A type II compensator boosts the phase up to 90° and a type III up to 180°. This is something you need to assess based on the plant dynamic response. Check out the link I gave in the response, it is all detailed there. \$\endgroup\$ Nov 12, 2017 at 19:30

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