In a circuit with sinusoidal source and charged capacitors and inductors we have initial conditions:

$$u_{C}=u(0)+\frac{1}{C} \int_{t_{0}}^{t}i(\tau)d\tau$$

$$i_{L}=i(0)+\frac{1}{L} \int_{t_{0}}^{t}u(\tau)d\tau$$

and when we work in time domain we account for those in the equations.

If instead of phasors we use Laplace transform then we account for those initial conditions too



But when we move from time to frequency domain using phasors then I don't understand where those initial conditions appear. In the transformed equations I don't see initial conditions anywhere.

  • \$\begingroup\$ Did you want to show the frequency domain spectrum as it changes in time? or the steady state transfer function? \$\endgroup\$ – Sunnyskyguy EE75 Jan 28 at 21:43
  • \$\begingroup\$ @SunnyskyguyEE75 I don't want to show something, this is a question it popped in my mind. \$\endgroup\$ – Adam Jan 28 at 21:45

The S domain is the "complex frequency" domain, which describes both the transient and the steady-state behavior.

However "using phasors" implies a simplified, partial analysis that describes the steady-state behavior only — and only for a particular frequency. The initial conditions are only relevant to the transient behavior, and get ignored in this kind of analysis.


The reason said in another way:

Frequency domain analysis is valid only for pure sinewaves. Those signals have never started, they have been alive always without any change. They also will never die nor change. So, no initial nor final conditions are needed.

Approximately at the same time when Napoleon Bonaparte had stopped to be anyone's harm, Mr. Fourier in France published how arbitary signals can be presented as sums of pure sinewaves and how to use that idea also in practice. Fourier's idea makes frequency domain analysis cover more signals than only pure sinewaves. But that's another story.


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