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Impedance measurement from admittance
Posted Feb 14, 2023, 3:25 p.m. EST Version 6.1 1 Reply
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Hi, I was trying to find impedacne by inverting admittance of a cylinder model in AC DC module frequency domain. I got the followng result
1/ec.Y11 [at f=10Hz] = 231489.93803605012-1886.854331111225i
again, I tried to get the impedance using the follwing equation,
[at f=10Hz]
1/(2x3.1416xec.freqx(imag(ec.Y22)/ec.omega)) [reactance part] = 2.8402312477671955E7
1/(2x3.1416xec.freqx(real(ec.Y22)/ec.omega)) [resistance part] = 231504.77626256025
My question is why these two equations give different result. The later one gives the correct result. 1/ec.Y11 doesn't give the correct reactance part.
I will aprreciate if someone helps me to understand the differences.
Thank you,
Aftab
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If I understand your notation correctly, this may help: Z = 1/Y . Let Z = R+iX . Let Y = G+jB .
In your case, R >> X
So the real part of 1/Z is fairly close to 1/R, because R >> X.
But the imaginary part of 1/Z is certainly NOT fairly close to 1/B (and it shouldn't be).
Is that what your question was about?
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