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I have plotted the real and imaginary parts of a device's input impedance with a VNA.

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What would be the best physical/electrical interpretation of the real part peak along with the imaginary part zero ? I fail to give a proper explanation, even though I know something is happening.

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  • \$\begingroup\$ The imaginary part becomes all real like : Z=sqrt((R+jwL)/(G+jwC)) with R=0,G=0 then Z=sqrt(jwL/JwC) = sqrt(L/C). Its a funny thing, that antenna made of pure inductance and capacitance (all imaginary) turns out to be all real - resistor like load on resonance frequency. What you have is a RLC circuit, that when it is at resonance freq. the XC and XL have the same magnitude with opposite sign, so they cancell each other and the pure resistance remains. \$\endgroup\$
    – Marko Buršič
    Commented Mar 4, 2016 at 10:28

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The imaginary part of the impedance is zero at the resonance frequency, which means that at that frequency the impedance becomes purely resistive. So the physical phenomenon you're observing is resonance.

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  • \$\begingroup\$ Thank you, I simply lacked the basic definition of resonance. \$\endgroup\$
    – MaximGi
    Commented Mar 4, 2016 at 12:55
  • \$\begingroup\$ @MaximGi: Also have a look at this question and its answers for an example of complex impedance behavior. \$\endgroup\$
    – Matt L.
    Commented Mar 4, 2016 at 13:01
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When a circuit is in resonance, the imaginary part of the impedance is zero, like they said. But the resistance of the circuit (real part) should keep with the same value, because It's ideally independent from imaginary part.

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  • \$\begingroup\$ The real part of the impedance is generally not independent of frequency. That's only the case with very simple topologies (such as series and parallel 2nd order RLC circuits). \$\endgroup\$
    – Matt L.
    Commented Mar 4, 2016 at 12:53

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