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As we know in frequency domain V(s)= I(s)Z(s) then in time domain it should be V(t)= I(t) Convolution Z(t) instead of V(t)= I(t)Z(t) ?

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    \$\begingroup\$ What are you asking for? It is unclear. \$\endgroup\$ – Unknown123 Feb 23 '19 at 9:12
  • \$\begingroup\$ in simple RC circuit with R time varying, if we write KVL equation then voltage across resistance will be R(t)I(t) or R(t) convolution I(t) ? As per my knowledge in frequency domain we write R(s)I(s)= V(s), so It should be R(t) convolution I(t), But at many places I saw I(t) R(t) i.e. simple multiplication instead of convolution \$\endgroup\$ – Hariom Feb 23 '19 at 9:30
  • \$\begingroup\$ What does the frequency domain have to do with this? \$\endgroup\$ – Edgar Brown Feb 23 '19 at 12:11
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The expression \$V(s) = I(s) Z(s)\$ is only valid if the impedance \$Z(s)\$ is time-invariant (i.e. it is not \$Z(s,t)\$). For a time-varying resistor, the time-domain expression \$v(t) = R(t) i(t)\$ is correct.

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Yes, \$ V(s) = I(s) Z(s) \$ is only valid for LTI (linear, time-invariant) circuits.

For an LTI circuit, all current-voltage relations are linear expressions using linear operations (multiplying by constant, addition/subtraction, time derivative, and time integral). Therefore, conversion to frequency (Laplace) domain only involves linear equations, making analysis easy because properties of linearity can be exploited.

For general circuits, you would have to apply convolution to solve, but these cases are rare if any. Usually, you “linearize” a nonlinear system by analyzing it under specific conditions/assumptions where linear approximations are valid. An example is small-signal AC analysis for transistors.

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  • \$\begingroup\$ Also, if you really need to be able to support time-variant or nonlinear system, then use state-space representation. \$\endgroup\$ – Unknown123 Feb 24 '19 at 6:00

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