simulate this circuit – Schematic created using CircuitLab

For a problem I am tasked with finding the transfer function \$G(s) = \frac{V_o(s)}{V_i(s)}\$. I am having trouble defining the 2nd loop in my node voltage analysis since there is no circuit element between the 2 nodes.

These are the current equations that I am working with \$V_1(s)[\frac{1}{sL+R_1}+\frac{1}{sc_1}] = \frac{V_i(s)}{sL+R_1}\$ and then \$\frac{V_o(t)}{\frac{1}{sc_2}+R_2}=0\$

  • \$\begingroup\$ Just note that C1 is in parallel with the series combination of C2 and R2, so that the impedance Z of those three elements is Z1 || (R2 + Z2). \$\endgroup\$
    – Null
    Jan 28, 2022 at 13:47
  • \$\begingroup\$ Can you show us what you have tried? It will help us pinpoint where exactly you went wrong. \$\endgroup\$ Jan 28, 2022 at 15:10
  • \$\begingroup\$ i have updated. \$\endgroup\$ Jan 28, 2022 at 21:30
  • \$\begingroup\$ @thejacobdaniels18 Do you want to know how to include the nodes that are just straight through connections between two devices, in series? Or do you want to avoid that trouble? You can go either way and both work. Just one generates more equations than another. The solution is the same. \$\endgroup\$
    – jonk
    Jan 29, 2022 at 1:50

1 Answer 1


Well, when we have the circuit:


simulate this circuit – Schematic created using CircuitLab

We know that the transfer function is given by:


And in your case it is not hard to see that:



\begin{equation} \begin{split} \underline{\text{Z}}_{\space2}&=\frac{1}{\text{j}\omega\text{C}_1}\space\text{||}\space\left(\text{R}_2+\frac{1}{\text{j}\omega\text{C}_2}\right)\\ \\ &=\frac{\frac{1}{\text{j}\omega\text{C}_1}\left(\text{R}_2+\frac{1}{\text{j}\omega\text{C}_2}\right)}{\frac{1}{\text{j}\omega\text{C}_1}+\text{R}_2+\frac{1}{\text{j}\omega\text{C}_2}}\\ \\ &=\frac{\frac{\text{j}\omega\text{C}_1}{\text{j}\omega\text{C}_1}\left(\text{R}_2+\frac{1}{\text{j}\omega\text{C}_2}\right)}{\frac{\text{j}\omega\text{C}_1}{\text{j}\omega\text{C}_1}+\text{j}\omega\text{C}_1\text{R}_2+\frac{\text{j}\omega\text{C}_1}{\text{j}\omega\text{C}_2}}\\ \\ &=\frac{\text{R}_2+\frac{1}{\text{j}\omega\text{C}_2}}{1+\text{j}\omega\text{C}_1\text{R}_2+\frac{\text{C}_1}{\text{C}_2}}\\ \\ &=\frac{\text{j}\omega\text{C}_2}{\text{j}\omega\text{C}_2}\cdot\frac{\text{R}_2+\frac{1}{\text{j}\omega\text{C}_2}}{1+\text{j}\omega\text{C}_1\text{R}_2+\frac{\text{C}_1}{\text{C}_2}}\\ \\ &=\frac{\text{j}\omega\text{C}_2\text{R}_2+\frac{\text{j}\omega\text{C}_2}{\text{j}\omega\text{C}_2}}{\text{j}\omega\text{C}_2\cdot1+\text{j}\omega\text{C}_2\text{j}\omega\text{C}_1\text{R}_2+\frac{\text{j}\omega\text{C}_2\text{C}_1}{\text{C}_2}}\\ \\ &=\frac{\text{j}\omega\text{C}_2\text{R}_2+1}{\text{j}\omega\text{C}_2-\omega^2\text{C}_1\text{C}_2\text{R}_2+\text{j}\omega\text{C}_1}\\ \\ &=\frac{1+\omega\text{C}_2\text{R}_2\text{j}}{\left(\text{C}_1+\text{C}_2\right)\omega\text{j}-\omega^2\text{C}_1\text{C}_2\text{R}_2} \end{split}\tag3 \end{equation}

  • \$\begingroup\$ Jan, you can reduce the size of the schematic a bit by adding a period (text) to the left (and also worth experimenting with is another one to the right) of the schematic. The system will measure the image, find it wider than before, and resize it better so that it doesn't take up your entire post. Something to consider when you have the time and inclination. \$\endgroup\$
    – jonk
    Jan 28, 2022 at 17:40
  • \$\begingroup\$ im a little bit confused by this answer, check out my edit which may highlight my thinking process. \$\endgroup\$ Jan 28, 2022 at 21:31

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