How to find the impedance of this RLC circuit?

Question

I'm trying to find the impedance of this RLC circuit (sorry for awkward notation of Q instead of I and P instead of V due to that this scheme comes from an analogy of electrical circuits with 0D lumped models for blood flow simulations):

Due to the this circuit, we have:

$$Q_{0} - Q_{\ell} = \mathcal{C} \frac{d P_{\ell}}{dt}$$

$$P_{0} - P_{\ell} = \mathcal{L} \frac{d Q_{0}}{dt} + \mathcal{R} Q_{0}$$

Or in frequency domain:

$$\tilde{Q_{0}}(\omega) - \tilde{Q_{\ell}}(\omega) = j\omega \mathcal{C} \tilde{P_{\ell}}(\omega)$$

$$\tilde{P_{0}}(\omega) - \tilde{P_{\ell}}(\omega) = (\mathcal{R} + j\omega \mathcal{L})\tilde{Q_{0}}(\omega)$$

Or finally:

$$\tilde{P_{0}}(\omega) = (\mathcal{R} + j(\omega \mathcal{L} - \frac{1}{\omega \mathcal{C}}))\tilde{Q_{0}}(\omega) + \frac{j}{\omega\mathcal{C}} \tilde{Q_{\ell}}(\omega)$$

So, I'm stuck here cause I don't how to proceed and find the impedance. I'm not sure if it's correct to take $$\mathcal{R} + j(\omega \mathcal{L} - \frac{1}{\omega \mathcal{C}})$$ as impedance or not. Note that in my scheme I don't have any information about $$\tilde{Q_{\ell}}(\omega)$$ but I might assume that $$P_{\ell}$$ is just a constant value in time-domain (or probably a Dirac delta function in frequency domain).

Answer 1

Complex impedances that are purely series or parallel can be lumped similar to groups of resistors: $$ \begin{align} Z_{series} &= Z_1 + Z_2 + ... + Z_n\\ Z_{parallel} &= \biggl(Z_1^{-1} + Z_2^{-1} + ... + Z_n^{-1}\biggr)^{-1} \end{align} $$

The problem is that your circuit has 3 nets defining input and output, so it cannot be lumped as a single impedance; we can solve this by modeling a source across the input nets and/or a load across the output.

Borrowing my diagram from a similar question:

$$ \begin{align} Z_{COUT} &= \bigg({\frac{1}{j\omega C}}^{-1}+{Z_{LOAD}}^{-1}\bigg)^{-1} \\ &= \frac{Z_{LOAD}}{j\omega C Z_{LOAD}+1} \\ Z_{input} &= j\omega L + R + \frac{Z_{LOAD}}{j\omega C Z_{LOAD}+1} \end{align}$$

This gives the impedance seen by \$V_{IN}\$. To determine the impedance from the load's perspective, you have to account for the impedance of the source:

$$ Z_{output} = \biggl((Z_{VIN} + j\omega L + R)^{-1} + {\frac{1}{j\omega C}}^{-1}\biggr)^{-1} $$

Answer 2

^{simulate this circuit – Schematic created using CircuitLab}

Redrawn circuit with the specified condition applied. The impedance looking in is \$\frac{\Delta P_0}{\Delta Q_0} = R + j \omega L\$. The capacitor has no effect since the voltage at the top right node is now fixed by the constant voltage source \$P_l\$.

If the assumption of constant \$P_l\$ is wrong, then the impedance calculation shown above is invalid. It is better to add a general impedance \$Z_{nxt}\$ in parallel with the capacitance to represent the input impedance of whatever comes next in this circuit. And re-calculate the impedance (you wont need \$P_l\$ or \$Q_l\$ for that calculation since they have been modelled by \$Z_{nxt}\$).