# Emulate impedance response using RLC components

I need to emulate measured impedance response of some blackbox using only RLC components. Measurements:

[Hz]    [Ohms]
--------------
20      14.5
40      15
70      15.5
100     15.75
200     16
300     16
400     16.75
500     17
700     18.5
1000    20
2000    26
3000    31
4000    35
5000    39
6000    42.5
8000    50
10000   57
20000   85


The closest I found is circuitry like this:

But it has major error around 2kHz (5 ohms of mismatch):

Blue plot is impedance response I want to model.

Red plot is impedance response of cuircuitry I found.

Previously I thought it is quite easy task, but unfortunately it found to be quite challenging for me as I do not have much experience with modeling impedance responses. Do you have any idea how to make the circuitry more accurate to measurements I have?

BTW. Impedance I want to emulate is impedance response of simplified closed speaker cabinet. It is simplified, becasue currently I ignored bass resonant peak (I already perfectly modelled it). The only problem I have is shape of this treble cuircut.

EDIT: I tweaked as suggested with resonance above 20kHz, but adding additional capacitor on L1. Still cannot cover the knee: I tweaked with various cobination of values of resonance components values, but still it does not fit in any configuration.

• Try adding a capacitor across L1 - start low to see the effect. I think the blue graph has resonance. 5 ohms in 26 ohms is only 1.8 dB in audio terms so not much of a problem given that most folk wouldn't notice if something suddenly got louder by 1 dB. – Andy aka Jul 4 at 13:25
• Try .stepping each component and see what comes closer. By the looks of it, unless there is some resonance above 20 kHz, you have a greater than 20 dB/dec slope, and an overdamped corner frequency, so another inductor (+ resistance?) might be needed. But first, try stepping the values; too often a simple tweak can come close enough for an approximation. – a concerned citizen Jul 4 at 13:43
• Thank you for your suggestions. I added additional capacitor across L1 as suggested to intoduce resonance above 20kHz. But still cannot fit with knee of measured impedance. I updated description with latest results. – Madras Jul 4 at 15:09

The expression for impedance with one inductor and two resistors is

$$\ R1 + j\omega\ R2L1 / (R2 + j\omega\ L1) \$$

At low frequency the impedance is $$\\approx R1\$$ and at very high frequency the impedance is $$\\approx\ R1 + R2\$$. With the three tunable parameters (R1,R2,L1) you can match only the low frequency impedance, high frequency impedance and the corner frequency. As @a concerned citizen mentioned in the comment, you may need more components for closer match.

The mismatch at 2kHz can be bridged by adding another impedance in series which has low values at frequencies below and above 2kHz. A parallel LC circuit can be used for this purpose. A sample circuit with sample values (not tuned for best match) is shown below. LC was chosen so that

$$\2*\pi* 2kHz \approx 1/\sqrt{LC}\$$.

The parallel LC circuit will have infinite impedance at 2kHz. However, we want to bridge only the 5ohms difference seen in your graph. Hence, a 5ohm resistor is also kept in parallel. The above calculations are all thumb rules / approximate formulae.

simulate this circuit – Schematic created using CircuitLab

The plot of the R|L|C section's impedance is shown below.

The net impedance is shown below.

You can independently verify if the given scheme is suitable. For a better match, use the tuning method you have used for the calculation of (R1, R2, L1).