# Informations deduced from the Impedance curve

#### Gomez_T

Joined Nov 16, 2015
2
Dear all,
I would like to know how I can distinguish between series resonance characteristics and the parallel ones only from the impedance curves(Real and imaginary parts), in order to build an equivalent circuit using R,L and C elements.
Have a good day!

#### MrAl

Joined Jun 17, 2014
7,659
Hello there,

The parallel circuit has a peaked response with frequency, while the series circuit has a dip (min) with frequency.
The parallel circuit phase decreases and then levels off with frequency, the series circuit phase increases and then levels off with frequency.

But what quantities did you really intend to measure, the real and imag parts of the impedance or the magnitude and phase?

If really the real and imag parts, then we have the following...

For the parallel case:
Z=(w^2*L^2*R-j*w^3*C*L^2*R^2+j*w*L*R^2)/(w^4*C^2*L^2*R^2-2*w^2*C*L*R^2+R^2+w^2*L^2)

and for the series case:
Z=R+j*w*L-j/(w*C)

and from those you can deduce the real and imag parts and solve for whatever you want to solve for.
For example, note that in the series case the R is the real part so knowing the real part means you already know the resistance. The parallel case will be harder to solve, however if we assume R non-zero:
R=(b^2+a^2)/a
where a=real part of impedance, b=imag part of impedance,
then that gets us that far already.
The solution for L then is:
L=(b^2+a^2)/(C*(b^2+a^2)*w^2+b*w)

which basically says that the value of L depends on the value of C, so there is no unique solution for L and C. So we either choose C and calculate L, or we choose L and calculate C.

This happens in the series case too, where we know the real part is R, and then we have for L:
L=(b*w*C+1)/(w^2*C)

so again we have multiple solutions, so we choose C and calculate L or the other way around again.

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