# connectedness between the output voltage and input voltage

Given this circuit

I want to deduce the relationship between the output voltage $$U_A(t)$$ an d the input voltage $$U_E(t)$$ using Kirchhoff's current law. We're allowed to assume that $$R$$, $$L$$and $$C_1$$ and $$C_2$$ are given to us.

My solution so far: $$U_E = U_R + U_{C1}$$ $$U_{C1}=U_L+U_{C2}$$ $$U_A = U_{C2}$$ $$\rightarrow U_E = U_R+U_L+U_A$$ At point $$K$$, i want to use Kirchhoff's current law: $$I_R=I_{C1}+ I_L= I_{C1}+ I_{C2}$$ Is my last step correct? How to continue?

## 2 Answers

You can write two ODEs describing the relationship, and that would be in the time-domain. Alternatively you could describe things in the frequency domain and get the relationship as a transfer function. I will show the latter.

Using Kirchhoff's current law at the two nodes K and A, we get:

$$\frac{U_E-U_K}{R}=\frac{U_K-U_A}{L s}+C_1 s U_K$$ $$\frac{U_K-U_A}{L s}=C_2 s U_A$$

These two equations can be solved for the two unknowns $U_A$ and $U_K$. We are only interested in the former.

$$\frac{U_A}{U_E}=\frac{1}{C_1 C_2 L R s^3+C_2 L s^2+C_1 R s+C_2 R s+1}$$

One approach to solving this problem is to consider each part as a resistor with complex impedance. Now you can solve as you would if each part was just a resistor. You do the same series, parallel, or voltage divider arithmetic, except that it will be performed on complex numbers instead of the simple real numbers if the parts were all true resistors.

Another way is to think of R and C1 as forming a single-pole low pass filter, and L and C2 a two-pole low pass filter. You can get the Bode plot from inspection and a few seconds with a calculator. The tricky part with this method will be finding exact values near the rolloff frequencies when the impedances of the two filters interact. If you mostly need the response some distance below or above the rolloff frequencies, then this approach is simpler.