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Suppose I am trying to solve the following circuit, where \$V_i\$ is the potential seen from the top branch to the bottom branch (which is 'grounded') and \$V_o\$ is the potential across \$C_2\$:

Simple RC circuit with parallel combination in the top branch

If I want to find the transfer function \$\frac{V_o(t)}{V_i(t)}\$, I can of course go through the laborious method of obtaining the voltage drop across the top branch and calling it \$V(s)\$ and then substituting this into an equation in terms of \$V_i\$ and \$V_o\$, respective of the whole circuit, finally obtaining the Laplacian transfer function:

$$\frac{V_o[s]}{V_i[s]}=\frac{RC_1s+1}{RC_2s + RC_1s + 1}\ \ \ \ \ (\text{Eq 1})$$

I have been told that I can also do this by considering impedances though?

How would I do this?

I guess I would consider the top branch parallel combination of R and C as:

$$Z_\text{total} = (1/R+1/X_c)^{-1} \equiv (1/R + j\omega C_1)^{-1} = \frac{R}{1+j\omega RC_1}$$ And then consider this in series with the second capacitor?


Further work:

The voltage drop of the top branch must be: $$V_s(t)=i(t) \times \frac{R}{1+j\omega RC_1}$$ Laplace transform gives: $$V_s(s)=I(s) \times \frac{R}{1+j\omega RC_1}$$

\$V_i(s)=V_s(s)+V_o(s)\$, just by KVL. We know that \$V_o(t)=i(t)\times1/j\omega C_2\$ and Laplace transforming natürlich gives: \$V_o(s)=I(s)\times1/j\omega C_2\$. Therefore, the transfer function is something like:

$$\frac{V_o}{V_i}(s)=\frac{\frac{I(s)}{j\omega C_2}}{\frac{I(s)R}{1+j\omega R C_1}+\frac{I(s)}{j \omega C_2}}$$

Is this somewhere along the right lines? I don't think it can be, because to me, there is no way to now cancel out \$j\$ and leave myself with \$\text{Eq 1}\$?

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You are mixing Laplace transforms (\$s\$-variable) with phasors (\$j\omega\$-variable). Please see: Under what conditions does jw equal the laplace variable s in an electrical circuit?. Both are called impedances, but you either use one or the other at the same time.

I share what I would consider the easy way of solving that circuit. If we call \$Z_1\$ to the parallel impedance of \$R\$ and \$C_1\$, and \$Z_2\$ to the impedance of \$C_2\$, we get a simple voltage divider with the following transfer function:

$$H(s) =\frac{V_o}{V_i}= \frac{Z_2}{Z_1 + Z_2}$$

The impedances are: $$Z_1 = R \parallel (C_1s)^{-1} = \frac{R \times \frac{1}{C_1s}}{R + \frac{1}{C_1s}} = \frac{R}{1 + RC_1s}$$ $$Z_2 = (C_2s)^{-1}$$

Resulting in: $$H(s) = \frac{\frac{1}{C_2s}}{\frac{R}{1 + RC_1s} + \frac{1}{C_2s}}$$

As you see, this equation is identical to the last one you wrote, but with \$s = j\omega\$ \$(I(s)\$ can be cancelled\$)\$.

Simplifying we get: $$\boxed{H(s) = \frac{RC_1s + 1}{R(C_1 + C_2)s + 1}}$$ and, depending on the ROC of the transfer function: $$\boxed{H(j\omega) = \frac{RC_1j\omega + 1}{R(C_1 + C_2)j\omega + 1}}$$


Edit: All this is valid if the circuit is initially relaxed. In this case if the capacitors are initially discharged. See s-domain equivalent circuits and impedances.

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  • \$\begingroup\$ Great answer and the relationship between the \$j\omega\$ variable and the s-domain is something I momentarily overlooked! Thank you for making the clear to me! \$\endgroup\$ – lmsavk May 12 '16 at 3:27

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