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Circuit

I am trying to model this circuit in the s-plane, to prove that its transfer function is what I have written at the top.

The question I am asking the community is how to approach the first steps for this. I want to use Kirchhoff's current law, but I have forgotten how to express the different voltages and currents.

To be more specific, how can I express i1, i2 and i3 in terms of the voltages, resistances and capacitors? Could you explain your reasoning behind it as well?

I know that i1 is (Vi - Vx) / R1, because I am just taking the potential difference across the resistor, and dividing it by the resistors value, but I am unsure as to what to do next, for i2 and i3.

I know that it will depend on modelling the capacitor in the s-plane as sC, or 1/sC, depending on whether it is in parallel or in series, but again my circuit theory is rusty so when I looked at the solution I couldn't see how they derived it. I can post the solution if it helps.

Thank you

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You are on the right track.

Just remember that the impedance of a capacitor is

$$ Z_C = \dfrac{1}{sC} $$

These problems, are straight forward to setup but then a lot of algebra. Apply KCL at the intermediate node Vx as,

$$ \dfrac{ V_x - V_i}{R_1} + \dfrac{V_x}{1/(sC_1)} + \dfrac{V_x - V_o}{1/(sC_2)} = 0 $$

Do the same at the Vo node.

The solve the two equations as a function of Vi and Vo.

Eventually you will end up with something of the sort,

$$ Vo = H(s) Vi $$

Then divide each side by Vi.

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  • \$\begingroup\$ Thanks a lot for the reply, appreciated. With regards to the third term in the equation you posted, would it also be valid to just have Vx on the numerator, and then the addition of the impedance of the capacitor C2 with the resistor R2, in the denominator? \$\endgroup\$ – lgdl.y May 31 '17 at 13:46
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    \$\begingroup\$ Yes you could but you are trying to get a function of Vo & Vi not Vx, as you will see when you write out the second eqn you will get a function Vx=...Vo that you sub into the first I showed \$\endgroup\$ – sstobbe May 31 '17 at 13:49

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