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I am sending a sine wave through a unknown RC circuit. At the other side I have a current-to-voltage converter (transimpedance amplifier) measuring the current indirectly as voltage.

I am using IQ demodulation technique and get the Real and Imaginary part of the voltage representing the current.

I want to calculate the admittance, impedance, conductance, resistance, susceptance and reactance.

First of all the values Iv and Qv (Iv and Qv is the voltage representing current) need to be converted to current:

Ii = Iv / Rf Qi = Qv / Rf

where Rf is the transimpedance amplifier feedback resistor.

Since I am exciting with voltage and reading the current I assume I have a ratio of current to voltage which is admittance.

If the excitation voltage is Vex admittance Y is then given as:

Y = G + jB = (Ii/Vex) + (jQi/Vex)

The amplitude is given as:

|Y| = sqrt(G^2 + B^2)

By using complex number one can derive the relationship between conductance and resistance:

R = ( G / (G^2 + B^2) )

and reactance and susceptance:

X = -j * ( B / (G^2 + B^2) )

Impedance is then:

Z = R + jX

and amplitude:

|Z| = sqrt(R^2 + X^2)

Do I convert to current correctly? Do I find conductance and susceptance correctly? Did I understand correctly that the values I measure is admittance? Is the Z, R, X conversions correct?

Lets say that the unknown RC circuit was only a 1000 Ohm resistor. I would then find out that Z = R = 1000 right?

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Since I am exciting with voltage and reading the current I assume I have a ratio of current to voltage which is admittance.

More specifically, you can call this a trans-admittance, since the voltage is applied at one port of the filter and the response current is measured at a different port.

If you just talk about an "admittance" people will at least at first assume you are talking about the voltage/current relationship at a single port.

Do I convert to current correctly? Do I find conductance and susceptance correctly? Did I understand correctly that the values I measure is admittance?

I didn't follow through your calculations entirely, but you did not come up with the correct result for this case. Your result may be correct for the normal admittance to impedance conversion for a one-port network.

For converting transadmittance to transimpedance you have to consider both ports of the network.

The admittance representation is like this:

\$\begin{bmatrix}I_1 \\ I_2\end{bmatrix}=\begin{bmatrix}Y_{11} & Y_{12} \\ Y_{21} & Y_{22}\end{bmatrix}\begin{bmatrix}V_1 \\ V_2\end{bmatrix}\$

To get the Z-parameters you need to invert the Y matrix. Rather than calculate the inverse by hand I looked it up on Wikipedia, where I found the result for the Z21 term is

\$Z_{21} = \frac{-Y_{21}}{Y_{11}Y_{22}-Y_{12}Y_{21}}\$

Separating this out into real and imaginary parts is of course another algebraic effort, which I'd much rather do using software (Mathematica or whatever) than sort through by hand.

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  • \$\begingroup\$ OK, so I have V1 and I2. How do I find the rest of the parameters? This is new to me so if you could take an example of further explain I would appreciate it a lot. \$\endgroup\$ – iQt Nov 7 '13 at 9:03
  • \$\begingroup\$ Do I need to measure I1, I2, V1 and V2? \$\endgroup\$ – iQt Nov 7 '13 at 11:40
  • \$\begingroup\$ According to this link I can measure it by simply dividing i/v = Y. I am using the 3-electrode setup shown in the illustration as well. link.springer.com/chapter/10.1007%2F978-3-540-89208-3_172 \$\endgroup\$ – iQt Nov 7 '13 at 14:40
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What about converting from admittance to impedance for a one port?

You got the right answer, but you can show it without having to examine the magnitude of Y using complex conjugates:

\$Y = G + jB\$

\$Z = \dfrac{1}{Y} = \dfrac{1}{G + jB}\$

\$Z = \left(\dfrac{1}{G + jB}\right)\left(\dfrac{G - jB}{G - jB}\right)\$

\$Z = \dfrac{G - jB}{G^2 + B^2}\$

\$R = \mathrm{Re}(Z) = \dfrac{G}{G^2 + B^2}\$

\$X = \mathrm{Im}(Z) = \dfrac{-B}{G^2 + B^2}\$

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