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I have derived the stage 1 and stage 2 transfer function separately (both have their own feedback and schematics). I tried just to assume that the transfer function of the two stage as multiplication of both TF1*TF2.

My problem I don't know how can I get the full transfer function with the feedback coming from second stage output to first stage input through R1 as I don't know how I calculate the current flowing through R3. Is there a technique I can think of?

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    \$\begingroup\$ This is a problem in feedback control. Have you been exposed to any material on that subject? \$\endgroup\$ – TimWescott Nov 19 '19 at 17:13
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    \$\begingroup\$ When TF1 has internal feedback it is necessary to see the complete circuit ......otherwise we do not know the value of the input resistance (in series to R3) \$\endgroup\$ – LvW Nov 19 '19 at 17:54
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    \$\begingroup\$ What exactly are you trying to achieve? \$\endgroup\$ – Chu Nov 19 '19 at 23:56
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    \$\begingroup\$ What about TF2 ? Open inverting terminal? No internal feedback? \$\endgroup\$ – LvW Nov 20 '19 at 9:18
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    \$\begingroup\$ Related: electronics.stackexchange.com/questions/282826/… \$\endgroup\$ – Voltage Spike Nov 21 '19 at 19:22
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When two circuits - isolated from each other - form the forward path of a system with feedback, you can multiply the two transfer functions TF=TF1*TF2.

As a second step, the general feedback formula (H. Black) can be applied H(s)=TF/(1-LG). LG is the loop gain (which must be negative for negative feedback).

However, in your circuit, there is no overall feedback at all (the feedback signal is shorted in the signal source Vin)

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    \$\begingroup\$ What is LG? And what if the the feedback connection was at the right pin of R1 that is the non inverting pin. \$\endgroup\$ – Navaro Nov 20 '19 at 12:56
  • \$\begingroup\$ I did exactly as you said. but when it comes to R1, I have no idea how this must be calculated to the general feedback formula. I tried tow assumption:1- that current flowing to the TF1 is zero.2- assuming Voltage at node (R1, C1 and Vin) is zero. the bode plot of the second assumption give little bit of similarity to the bode plot given in LTspice \$\endgroup\$ – Yaakov Nov 20 '19 at 13:14
  • \$\begingroup\$ @navaro...As I have mentioned, LG is the loop gain (a fixed and defined terminus in feedback theory). \$\endgroup\$ – LvW Nov 20 '19 at 14:28
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    \$\begingroup\$ @Yaakov...It is hard to answer because we do not know what your task is. Is there any transfer function you want to realize? What is the purpose of C1 (in paralle to a voltage source). When Rfeedback is connected to the inv. opamp input, you have a summing operation at this input. \$\endgroup\$ – LvW Nov 20 '19 at 14:32
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    \$\begingroup\$ ....because the inverting input of an opamp (wired as an inverer) acts of a summing junction for several input signals (virtual ground principle). This is the basic principle of a summing amplifier.... \$\endgroup\$ – LvW Nov 20 '19 at 15:20

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