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I have been trying to find the transfer function for this circuit, and I think I may have to use the voltage divider twice.

enter image description here

What I have tried:

I have used the voltage divider formula to find the Voltage of C1. enter image description here

How can I use the voltage divider formula again to find Vout?

This is supposed to be the right transfer function: enter image description here

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  • \$\begingroup\$ 1st formula <-- how can a voltage equal an impedance? \$\endgroup\$
    – Andy aka
    Oct 10, 2023 at 20:52

2 Answers 2

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I'll just replicate results using my own approach.

Let's call the unlabeled node you have in the schematic as \$V\$. Then the output is \$V_{_\text{OUT}}=V\cdot\frac{R}{R+Z_{C_2}}\$. That's a simple voltage divider.

But... What's \$V\$ here? Well, it's the divided output given \$V_{_\text{IN}}\$, right? But note here that \$C_1\$ is in parallel with \$R+C_2\$! So in this case we can find that \$V=V_{_\text{IN}}\cdot\frac{Z_{C_1}\,\mid\mid\,\left(R+Z_{C_2}\right)}{Z_L+Z_{C_1}\,\mid\mid\,\left(R+Z_{C_2}\right)}\$.

So putting those together find \$V_{_\text{OUT}}=V_{_\text{IN}}\cdot\frac{Z_{C_1}\,\mid\mid\,\left(R+Z_{C_2}\right)}{Z_L+Z_{C_1}\,\mid\mid\,\left(R+Z_{C_2}\right)}\cdot\frac{R}{R+Z_{C_2}}\$ or that:

$$\frac{V_{_\text{OUT}}}{V_{_\text{IN}}}=\frac{Z_{C_1}\,\mid\mid\,\left(R+Z_{C_2}\right)}{Z_L+Z_{C_1}\,\mid\mid\,\left(R+Z_{C_2}\right)}\cdot\frac{R}{R+Z_{C_2}}$$

Rather than do a bunch of algebra I don't want to do, I'll use SymPy:

def par(a,b): return a*b/(a+b)
L,C1,C2,R=symbols('L,C1,C2,R',real=True,positive=True)
simplify(par(C1,C2+R)/(L+par(C1,C2+R))*R/(C2+R))
C1*R/(C1*(C2 + R) + L*(C1 + C2 + R))

I think you can re-arrange that to get the right result.

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The analysis is very simple and is done just as you say.

enter image description here

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