# Deriving the Impedance equation of a RLC circuit (with 2 Inductor) simulate this circuit – Schematic created using CircuitLab

During this lockdown I had some time in hand and decided to learn more about impedance. So I gave myself a project, trying to create a impedance graph in excel base on a the schematic seen above. The passive components will be a variable.

To do that, I first need to derive the impedance equation for the schematic seen above. When I cross check it with a simulation tool and a proven graph, I could not get the graph to match. Hence, I believe I must have done something wrong. I had doubled, tripled checked and corrected all the mistakes.

I do not know where I went wrong. I am hoping if I can get a 2nd or 3rd eye to look to see where I might have gone wrong.

This is my first post here, I had done the search but to no avail of information I can obtain anymore. Do let me know if I did violate any rules by posting this.

Edit: I have removed my derivation, as it is wrong and not working. This is to not confuse other readers as User Verbal Kint has provided a really good solution.

• That's a cranky way of doing things. – Andy aka Apr 28 at 14:25
• This would all be easier for you if you worked with a numerical analysis tool that can handle complex numbers instead of using Excel and having to do all that algebra yourself. For example: Matlab, Octave, Numpy, Maple, Mathematica, ... – The Photon Apr 28 at 14:31
• I am not sure you can obtain a tractable result using this approach. I would highly recommend the FACTs to get there. You have a 3rd-order denominator and a 2nd-order numerator as a first observation. – Verbal Kint Apr 28 at 14:35
• Or use an algebraic tools like Maple, Maxima, ... – Huisman Apr 28 at 14:37
• Z1 (first half page) seems right to me. Why do conjugate this afterwards? – Stefan Wyss Apr 28 at 15:56

There are several ways to look at this type of 3rd-order transfer function: brute-force analysis - which often leads to algebraic paralysis - and the fast analytical techniques or FACTs.

The brute-force expression is not that complicated to write but good luck to develop it and express the result as a meaningful transfer function. Fortunately, it is easy to plot it with a mathematical solver like Mathcad: With the FACTs (or any other approach), you determine this transfer function (yes, an impedance is a transfer function) by installing a test generator $$\I_T\$$ - the stimulus - across the connecting terminals and determine the response $$\V_T\$$ that it generates. To determine the denominator, you turn the excitation off (the current source is zeroed or open-circuited) and you "look" through the connecting terminals of the 3 energy-storing elements to determine the resistance $$\R\$$ that drives them. They are temporarily removed during this exercise. This is to write the denominator $$\D(s)=1+sb_1+s^2b_2+s^3b_3\$$.

For the zeroes, you null the response across the current source $$\I_T\$$ meaning that you replace it by a short circuit (this is a degenerate case). Again, you "look" through the connecting terminals of the 3 energy-storing elements to determine the resistance $$\R\$$ that drives them. This is to write the numerator $$\N(s)=1+sb_1+s^2b_2\$$. There is no third-order coefficient as $$\C_3\$$ is shorted in this exercise.

If you draw intermediate sketches and write carefully all the time constants, without a single line of algebra, then you assemble the transfer function at the end: Finally, you can plot this expression and compare its response versus that of the brute-force formula used a reference here: If you now want to extract the magnitude and phase expressions, I let you replace $$\s\$$ by $$\j\omega\$$ and collect real and imaginary terms. I'm done for tonight : ) 