# Converting impedence response to electrical conductivity

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1. Jul 13, 2018

### Alba19

Hi,

I have a curve, impedance magnitude vs frequency. i would like to convert this curve to electrical conductivity vs time. how can i do that?

2. Jul 13, 2018

### f95toli

You can't.
Impedance vs frequency is an AC measurement; meaning the capacitance and inductance (i.e. the reactive contribution) also plays a role. Hence, if you only have the magnitude there is no way to figure out what the conductivity is.
Also, "conductivity vs time" does not make sense; the frequency dependence of the impedance comes from the reactance, not the conductivity.

3. Jul 13, 2018

### Staff: Mentor

Perhaps you mean period of the sinusoidal input?

4. Jul 13, 2018

### Staff: Mentor

What units do you have in mind for "electrical conductivity"?

5. Jul 14, 2018

### Alba19

S/m(unit)

6. Jul 14, 2018

### Staff: Mentor

Conductivity is the inverse of resistance. It is measures in mhos (reciprocal of ohms). You might also express it as current/voltage, or amps/volts.

Is that what you mean?

We are having trouble understanding your question.

7. Jul 14, 2018

### K Murty

Hi.
Impedence is the complex number that captures the magnitude and phase of the quotient of the complex numbers of a sinusoidal voltage over a sinusoidal current. The units of impedence are Ohms.

$$v(t) = A \cos( \omega t) \,\, V \rightarrow A \angle{0} \,\, V \\ i(t) = B \cos(\omega t + \phi) \,\, A \rightarrow B \angle{\phi} \,\, A \\ Z_{L} = \dfrac{ A \angle{0} }{ B \angle{\phi} } \,\, \Omega = \dfrac{A}{B} \angle{ - \phi } \,\, \Omega$$
The inverse of this complex number, is admittance. Admittance has the same units as conductance: Siemens $$\,\, S$$
You can convert an impedence to an admittance by taking its reciprocal:
$$\dfrac{1}{Z_{L} } = \dfrac{1 \angle{0} }{ \dfrac{A}{B} \angle{ - \phi } \,\, \Omega} = \dfrac{B}{A}\angle{ \phi } \,\, S = Y_{L}$$
The admittance is simply the quotient of the complex current over voltage.

The units of admittance are Siemens, same as conductance. You can convert the impedence response by taking the reciprocal of the impedence function. But you cannot convert an impedence to a conductance, just that its reciprocal has the same units as conductance. You can them compare the magnitudes.
For example, the impedence of a capacitor can be proven to be:
$$-j \cdot \dfrac{1}{ \omega C} \,\, \Omega$$ Where $$\dfrac{1}{ \omega C } \,\, \Omega$$ is the capacitive reactance.
The reciprocal of capacitive and inductive reactance is called capacitive susceptance $$\dfrac{1}{\dfrac{1}{ \omega C } \Omega } = \omega C \,\, S$$ and inductive susceptance respectively, they have units of Siemens.

As an example, take a series RLC circuit:
$$R + j( \omega L - \dfrac{1}{\omega C} ) \,\, (\Omega) = Z(\omega)$$
The magnitude response is:
$$| Z(\omega)| \,\, \Omega = \sqrt{ {R^2 + ( \omega L - \dfrac{1}{\omega C} )^2} } \,\, (\Omega)$$
Taking the inverse of this:
$$\boxed{|Y(\omega)| S = \dfrac{1}{| Z(\omega)| (\Omega) } = \dfrac{1}{\sqrt{ {R^2 + ( \omega L - \dfrac{1}{\omega C} )^2} } \,\, (\Omega) }}$$
You can convert an impedence function of angular velocity to an admittance function of angular velocity, as stated in my earlier post. To convert it to the time domain is not possible.
The complex numbers that represent current and voltage can indeed be converted to the time domain by multiplying them with the exponential that was factored out previously:

Last edited: Jul 14, 2018
8. Jul 14, 2018

### Baluncore

Maybe it would help if we knew where the curve data came from?
Why do you need to invert the graph of the data?

9. Jul 15, 2018

### Staff: Mentor

Okay. But there is no length dimension in your impedance data.

I think the best you can do is convert impedance vs frequency to admittance vs period.

10. Jul 15, 2018

### K Murty

Next time please post the curve and more information, it is not right to leave everyone guessing as to your intentions and with zero data provided.
The units you mentioned, Siemens per meter, is this something related to transmission lines?

11. Jul 16, 2018

### Alba19

I want to model a system with a black box. my transfer function is conductivity and i just have Magnitude of impedance vs frequency in different temperature and magnitude of current vs frequency.
I would like how i get to it

12. Jul 16, 2018

### Alba19

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13. Jul 16, 2018

### CWatters

Is that...
Output impedance Vs input frequency?
Output current Vs input frequency?

If that's all you have then you can't work out a transfer function ...

Output conductance Vs input conductance.

14. Jul 16, 2018

### Baluncore

It seems the two graphs are of different signals. Is one the input, the other the output?
What is the input to the black box? What is the output?
Please provide a block diagram showing inputs, outputs and which signal is swept in frequency.

15. Jul 16, 2018

### CWatters

What is the system?

16. Jul 16, 2018

### Alba19

The output is conductivity vs time. i have some graphs of the system. 2graphs are attached above. the graphs include: Current vs time, magnitude of current vs frequency, current vs voltage, magnitude of impedance vs frequency and magnitude of current vs temperature.that's all. the graph of current vs time is dependent on temperature.

17. Jul 16, 2018

### Alba19

all graphs are related to a system. temperature, frequency, time are inputs of graphs

18. Jul 16, 2018

### f95toli

That makes no sense. There is no way to calculate the conductivity using that information. Even if you had a complete description (i.e. including phase information) of your "black box" as a 2-port system you would at the very least need a circuit model (as well as the geometric dimensions of you sample) to calculate a parameter such as conductance.

19. Jul 16, 2018

### Alba19

You mean i should have equivalent circuit of system? how come?

20. Jul 16, 2018

### Baluncore

What is the input variable?
What is the output variable?

That does not make sense. You have a frequency swept input and you have an output. The transfer function is the output divided by the input. That might have the same dimensions as conductivity.

Since you have only the magnitude of the complex variables, you have lost the phase or time information. The problem may therefore be insoluble.