# Solving RC filter by differential equation

I'm trying to solve the differential equation of a simple RC filter. What i got so far is:

$$CRV'+V=V_0\sin(\omega t)$$

Solving i get:

$$V == \frac{V_0 \sin(\omega t)}{1 + R^2C^2 \omega^2} - \frac{RCV_0 \omega \cos(\omega t)}{1 + R^2C^2 \omega^2} +c_1e^{-\frac{t}{RC}}$$

Can someone please help me understand how to interpret this with the usual concepts of reactance and impedance?

simulate this circuit – Schematic created using CircuitLab

• It is probably a lot easier to understand using complex numbers. Complex numbers make solving trig equation a lot easier. en.wikipedia.org/wiki/RC_circuit – SamFisher83 Sep 15 '15 at 5:29

For steady state sinusoidal response, the exponential term goes to zero and you are left with sine and cosine terms that can be expressed as amplitude and phase angle. The same result can be obtained by using Laplace analysis (replace C with the reactance 1/sC , followed by $s\rightarrow j\omega$) to find the steady state response.

What you ask turns out to be contradictory if taken literally: the answer would be "you can't understand that result reasoning with reactances and impedances".

The very concepts of reactance and impedance stem from a simplified analysis approach that focus on steady-state response and neglects the transient response. That's why you need s-domain (Laplace transforms) or differential equations to work out the transient response.

Said in other words: reactance and impedance are the wrong tool to understand why you get that exponential term in your circuit. That exponential term is the transient response of the circuit.

As Chu said in his answer, in that result you can see also the steady-state terms which you can relate to usual impedances.