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}\$?


1 Answer 1


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.

  • \$\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\$
    – smollma
    May 12, 2016 at 3:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.