Solves the differential equation for a RC low-pass filter. Gives the homogeneous and particular solutions. This supplements the article RC Low-pass filter.\(\)

## Appendix B

For old times sake, we show the traditional method to solve the differential equation for the passive filters consisting of a resistor and capacitor in series.

The output is the voltage over the capacitor \(y(t)\) as shown in the schematic below.

Assume a switch between the input and the resistor that closes at \(t=t_1\). Further assume \(y(t\leq t_1)=Y_0\).

According to Kirchhoff’s Voltage Law, for \(t\geq t_1\)

The relations between voltage and current for the

Two differential equations follow from substituting \(\eqref{eq:bIt}\) and \(\eqref{eq:bUr}\) in \(\eqref{eq:bUt}\) or its derivative

If \(u(t)\) is continuous, we can choose either differential equation, but when \(u(t)\) is non-continuous we can’t use \(\eqref{eq:bDV2}\).

Assume the non-homogeneous linear differential equation of a first order High-pass LC-filter, where \(u(t)=\hat{u}\cos(\omega t)\) is the forcing function and the current \(i(t)\) through the inductor is the response. The differential equation for this system is

The solution is a superposition of the natural response and a forced response. The so called, homogeneous solution \(y_h(t)\) and the particular solution \(y_p(t)\)

### Homogeneous solution

The homogeneous solutions follows from the reduced (=homogeneous) linear differential equation where the forcing function is zero.

According the Euler, the homogeneous solutions are in the form $$ y_h(t)=\mathrm{e}^{pt}\label{eq:bTrigRC_gen} $$

substituting this \(y_h(t)\) in \(\eqref{eq:bTrigRC_homDV}\) gives the characteristic equation with root \(p\)

The solution base \(y_{h,1}(t)\) follows from substituting the root \(p\) from equation \(\eqref{eq:bTrigRC_p}\) in back in the homogeneous differential equation \(\eqref{eq:bTrigRC_gen}\)

The homogeneous solution follows as a linear combination of the solution bases (only one in this case) as

### Particular solution

We will show how to get the particular solution using both trigonometry and complex arithmetic.

#### Using the trigonometry method

If we force a signal \(\hat{u}\cos(\omega t)\) on a linear system, the output will have the same frequency but with a different phase \(\phi\) and amplitude \(A\).

Substituting \((\eqref{eq:bTrigRC_form},\eqref{eq:bTrigRC_formDer})\) in the differential equation \(\eqref{eq:bTrigRC_DV}\)

Work towards the trigonometric identity

$$ \begin{align} \gamma\cos(\alpha+\beta)=\gamma\cos\alpha\cos\beta-\gamma\sin\alpha\sin\beta\nonumber \end{align} \nonumber $$

by assigning the two independent variables \(R\) and \(\omega L\) to two more convenient independent variables \(\gamma\cos\alpha\) and \(\gamma\sin\alpha\)

to dot the ‘i’, introduce \(\beta\)

we can rewrite \(\eqref{eq:bTrigRC_part}\) and use the aforementioned trigonometric identity

Divide \(\eqref{eq:bTrigRC_CsinAlpha}\) by \(\eqref{eq:bTrigRC_CcosAlpha}\) to solve for \(\alpha\), and apply the geometric identity \(\sin^2\alpha+\cos^2\alpha=1\) to \(\eqref{eq:bTrigRC_CsinAlpha}\) by \(\eqref{eq:bTrigRC_CcosAlpha}\) to solve for \(C\)

Substituting \(\eqref{eq:bTrigRC_beta}, \eqref{eq:bTrigRC_alpha},\eqref{eq:bTrigRC_C}\) in equation \(\eqref{eq:bTrigRC_alpabetaC}\)

and combine like terms

The particular solution follows from substituting \((\eqref{eq:bTrigRC_A}, \eqref{eq:bTrigRC_Phi})\) in \(\eqref{eq:bTrigRC_form}\)

#### Using the complex arithmetic method

Using a complex forcing function \(\underline{u}(t)\) provides a less involved method of finding the particular solution as introduced in Linear Differential Equations. Using a complex forcing function

the corresponding complex response is of the form

Substituting \(\underline{y}_p(t)\) and \({\underline{y}_p}^\prime(t)\) in the differential equation \(\eqref{eq:bTrigRC_DV}\)

The amplitude \(A\) and phase \(\phi\) of the complex response follow as

Now that \(A\) and \(\phi\) are known, the complex particular response follows as equation \(\eqref{eq:bCaRC_yp}\)

Since the forcing function was only the real part of \(\underline{u}(t)\), are only interested in the real part of the complex particular solution \(\eqref{eq:bRLSol}\) as well

### General solution

The general solution follows from substituting \(\eqref{eq:bTrigRC_hSolution}\) and \(\eqref{eq:bTrigRC_pSolution}\text{ or }\eqref{eq:bCaRC_pSolution}\) in equation \(\eqref{eq:bTrigRC_hp}\).