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Impedance measurement from admittance

Mohammad Aftab Uzzaman

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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


1 Reply Last Post 16 feb 2023, 23:15 GMT-5
Robert Koslover Certified Consultant

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Posted: 2 years ago 16 feb 2023, 23:15 GMT-5

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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Scientific Applications & Research Associates (SARA) Inc.
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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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