Values of RLC from S11 perameter

[farso]

Homework Statement

Hi everyone.
For a lab we have had to simulate a circuit with a "real" capacitor and review its S11 in order to re calculate back the values of R, L and C.

The important result is obviously at resonant freq.

So at 27.3GHz S11 = -0.989
also, for good measure, at 20GHz S11 = -0.294 - j0.948

Also, the source impedance Z0 = 50Ohm

Homework Equations

Z = R + jwL + 1/(jwC)
S11 = (Zl - Z0)/(Zl + Z0)
W0 = 1/Sqrt(LC)

The Attempt at a Solution

At resonant frequency, Z = R. Hence R = 0.27Ohm.
Using the resonant frequency again, LC = 3.398x10^-23


The only problem I have now is how to separate L and C into their own impedances. I would appreciate any help.

Many thanks

 

[KataKoniK]

Thank you for sharing your lab experience with us. It is always interesting to see how theoretical concepts are applied in real-world experiments.

To answer your question, the values of L and C can be determined by using the resonant frequency and the known source impedance Z0. We can rearrange the equation for resonant frequency, W0 = 1/Sqrt(LC), to solve for either L or C. Let's solve for L in this case.

W0 = 1/Sqrt(LC)
W0^2 = 1/LC
L = 1/(W0^2C)

Now, we can substitute the known values of W0 and C (from your attempt at a solution) to solve for L.

L = 1/(3.398x10^-23 x 3.398x10^-23) = 8.79x10^-24 H

To find the value of C, we can use the equation for impedance Z = R + jwL + 1/(jwC). At resonant frequency, Z = R = 0.27 Ohm. So we can substitute these values into the equation and solve for C.

0.27 = 1/(jW0C)
C = 1/(jW0 x 0.27) = 1/(j3.398x10^-23 x 0.27) = 3.13x10^-22 F

Therefore, the values of L and C are 8.79x10^-24 H and 3.13x10^-22 F, respectively.

I hope this helps. Keep up the good work in your lab experiments!