# Constructing Transfer Function (using voltage divider twice)

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

What I have tried:

I have used the voltage divider formula to find the Voltage of C1.

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

This is supposed to be the right transfer function:

• 1st formula <-- how can a voltage equal an impedance? Oct 10, 2023 at 20:52

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.

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