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Given this circuit I would like to know how can I find it's transfer function.

A method I've considered was to use Node Equations on nodes 1 and 2 , and I have a few things that I don't find clear :

  1. why do I have to introduce an additional equation with \$V2=0\$ ?
  2. Once I find the transfer function in form of \$H(s) = {\dfrac{R1}{R2+Ls+CR1R2s+CLR2s^2}}\$ I must calculate \$Vu(t)\$ with \$V1=10+cos(t)\$ , the solution states that \$Vu(t)= H(0)*10 + Re[H(j)*e^{(jt)}*1]=-10-{\dfrac{1}{2}}{sin(t)}\$ How does one achieve this result and first of all how does the \$Re\$ part of the transfer function is calculated ?

Thanks a lot for any insight provided.

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In this case a very good approximation to the transfer function can be seen from inspection. The output impedance of a TL082 is no match for a 1 Ω source. Therefore the output of this circuit will pretty much just follow the input regardless of anything the opamp tries to do.

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  • 1
    \$\begingroup\$ Yes, it is surprising how often instructors will ask their students to analyze an absurd circuit. One wonders if it was done on purpose as a red herring, by oversight, or because they didn't know themselves. \$\endgroup\$ – gsills Feb 8 '13 at 22:07
  • \$\begingroup\$ @gsills we had the 3rd kind of instructors in my college.. \$\endgroup\$ – abdullah kahraman Feb 10 '13 at 16:28
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To answer your first question: the negative feedback makes node 2 equal to the noninverting input, so that's 0 V. Then there flows a current of Vo/R1 through R1, and therefore also through L1 to node 1.

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