What Happens When You Inverse Transform Impedance Back to Time Domain?

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Transforming impedance from the frequency domain back to the time domain does not inherently yield a purely resistive circuit, as the properties of the original circuit remain unchanged. The equivalent impedance is derived from the real part of the impedance, which represents resistance, while the imaginary components from capacitors and inductors can cancel each other out. If they cancel completely, only resistance remains; if not, the circuit will still exhibit complex impedance characteristics. The time domain analysis is frequency-dependent, with pure resistance occurring only at the resonant frequency, while other frequencies yield RC or LC combinations. Ultimately, the transformation process serves as a mathematical tool rather than altering the fundamental circuit properties.
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Homework Statement


This is a general question regarding impedance in circuit analysis.
I know that in order to find the total impedance of a circuit, you transform each element characteristic into the frequency domain, i.e.,
R\rightarrow Z_R = R,
C\rightarrow Z_C = \frac{1}{Cs},
L\rightarrow Z_L = Ls.

And then sum up the quantities properly to get an equivalent impedance.

My question is, what happens if you transform this quantity back using an inverse transform? Do you get a real resistance value?
 
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This transformation doesn't have an inverse for the imaginary part, since the effect of capacitor and inductance cancel out each other...

The resistance value is the real part of the impedance... ie Re(Z) = R,
 
what if its a circuit consisting of only a resistor and a capacitor?
 
tandoorichicken said:
My question is, what happens if you transform this quantity back using an inverse transform? Do you get a real resistance value?

well, why would you expect that in general you get just a purely resistive circuit when you started with a circuit with a resistor, a capacitor and an inductor?

The property of the circuit won't change just because you have tranformed into a more convenient domain (the freq domain in this case) for calculation purposes and then changed it back. If you begin with a 1 ohms and a 2 ohms resistor, then you can say ok, the total R is 3 ohms, so I can replace them with just one 3 ohms resistor. But that's all you have really done in finding the equivalent resistance.

same applies in the case with capacitors and inductors included. you started with R , C, L. Since C and L make up the imagary part (in freq domain) they have a chance to cancel (fully or partially). If they cancel each other fully, then you will end up having just an R in your equivalent circuit. Now if they don't cancel fully you will have either one L or one C (of different value) left pending on the phase of the equivalent resistance in freq domain.
 
In the time domain you can only calculate the impedance of the circuit for specific frequencies. At the resonant frequency, you have a pure resistance, below or over the resonant frequency, you have complex impedances, either RC or LC.
If you have a closed circuit, with sources and/or initial conditions, you can solve the resulting differential equation and have a frequency independent solution.
 

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