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Below shows the current generalized immittance converter. I am trying to derive the transfer function (V3/Vin) of the said filter but I am getting problems in obtaining the voltages at some of the nodes.

CGIC Diagram

These are the equations at the nodes I managed to obtain;

Node V1: (V3-V1)Y2 + (0-V1)Y6 + (Vin-V1)Y5 = 0

Node V2: (V4-V2)Y4 + (Vin-V2)Y7 + (0-V2)Y8 = 0

Node V3: (V1-V3)Y2 = 0

Node V4: (V2-V4)Y4 = 0

Node Vin: (V1-Vin)Y5 + (V2-Vin)Y7 = 0

I realized when I worked down with these equations I was not able to get the correct transfer function so I know some of the voltages into the nodes above are wrong. Can anyone help at least guide me to get the correct equations at each of the nodes so I can work out the transfer function. Any and all help will be greatly appreciated. Thank you.

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  • \$\begingroup\$ Dantec R.: What is your basic problem? Do you want to (1) KNOW the transfer function or to (2) DERIVE this function via node voltage analysis? I ask this question because there is another simpler way for finding the function (based on the classical GIC formula). This would allow to split the problem into two simpler tasks. \$\endgroup\$ – LvW Mar 28 '16 at 8:29
  • \$\begingroup\$ Hi, I want to know how to derive the function via node voltage analysis, but if there is another way (classical GIC formula you say) can you expand on that? \$\endgroup\$ – Dantec R. Mar 29 '16 at 1:07
  • \$\begingroup\$ OK - I will describe the way for finding the transfer function in a separate answer. \$\endgroup\$ – LvW Mar 29 '16 at 9:29
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  1. V4 is connected to more branches than just Y4. It is also connected to Y3 and the output of an amplifier.

  2. Similarly at V3 you didn't account for all the branches connected to the node.

  3. There is node in your circuit that you didn't label with a name or number. In order to get a complete set of nodal equations you need to have an equation for every node in the circuit except one. Typically the node with no equation is the ground node. So you need to add an equation for the unlabeled node.

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  • \$\begingroup\$ Hey, thanks for the advice. I will try to come up with new equations for V3 and V4. But I'm having problems understanding what is going at those nodes. Like, for V3, would that equation have V2 or V4 tied in somewhere? The jumped lines at the top to the diagram are slightly confusing. \$\endgroup\$ – Dantec R. Mar 29 '16 at 1:51
  • \$\begingroup\$ Really you can't make a node equation at V3 or V4 due to the voltage source. You just have V3 = A * (V2-Vx) where A is an unknown big number and Vx is the voltage at the unlabelled node. If the circuit feedback is done right, the gain factor (A) should end up not affecting the end result. \$\endgroup\$ – The Photon Mar 29 '16 at 1:54
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For my opinion, starting from the beginning (set of node voltages) seems to be a rather cumbersome way. Therefore, it is to be recommeded first to analyze the structure of the shown cicuit:

  • Without the elements Y5...Y8 the circuit resembles the classical unloaded GIC circuit as proposed by A. Antoniou in 1969. This circuit consists of two Negative-Impedance Converters (NIC) - one acting as a load for the other one.
  • The input impedances - looking into the non-inverting opamp input nodes of this GIC block - can readily be calculated; but the corresponding expressions can also be found in each relevant textbook because these blocks are used for realizing electronic inductances or Frequency-Dependent-Negative Resistors (FDNR), respectively.
  • Using the input impedance at both nodes it is not a problem to determine the input voltages at these two nodes as a function of the input signal voltage and the connected parts Y5...Y8 (simple voltage division).
  • Each input voltage (at the non-inv. input node) produces an output voltage - and the superposition of both voltages leads to the wanted transfer function V3/Vin.
  • The advantage of this method is (a) that the calculation starts at a rather low level (NIC input impedance) and (b) that the whole task is split into two independent (but very similar) portions.

(As a final result, the numerator as well as the denominator of the transfer function consists of 4 expressions like YaYbYc). Any further question?

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