Appendix B

Coert Vonk

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.

own work
Schematic RC filter

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 \(t\geq t_1\)

$$ \begin{align} u(t)&=u_R(t)+u_C(t)\nonumber\\[6mu] &=i(t)\,R+y(t)\quad\Rightarrow\nonumber\\[6mu] R\,i(t)+y(t)&=u(t)\label{eq:bUt} \end{align} $$

The relations between voltage and current for the resistor and capacitor are

$$ \begin{align} u_R(t)&=R\,i(t)&\text{resistor}\label{eq:bUr}\\[6mu] i(t)&=C\frac{\mathrm{d}y(t) }{\mathrm{d}t}&\text{capacitor}\label{eq:bIt} \end{align} $$

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

$$ \begin{align} R\,i(t)+y(t)&=u(t)\nonumber\\[6mu] R\,C\frac{\text{d}y(t)}{\text{d}t}+y(t)&=u(t)\label{eq:bDV1}\\[20mu] \frac{\text{d}i(t)}{\text{d}t}\,R+\frac{i(t)}{C}&=\frac{\text{d}u(t)}{\text{d}t}\nonumber\\[10mu] RC\frac{\text{d}i(t)}{\text{d}t}+i(t)&=C\frac{\text{d}u(t)}{\text{d}t}\label{eq:bDV2} \end{align} $$

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

$$ RC\,{y_p}^\prime(t)+\,y_p(t)=\hat{u}\cos(\omega t)\label{eq:bTrigRC_DV} $$

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 \(y_p(t)\)

$$ y(t)=y_h(t)+y_p(t)\label{eq:bTrigRC_hp} $$

Homogeneous solution

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

$$ RC\,{y_p}^\prime(t)+\,y_p(t)=0\label{eq:bTrigRC_homDV} $$

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 \(p\)

$$ \begin{align} RC\,(\mathrm{e}^{pt})^\prime+\,\mathrm{e}^{pt}&=0\nonumber\\ \Rightarrow\quad RC\,p\mathrm{e}^{pt}+\mathrm{e}^{pt}&=0&\div{\mathrm{e}^{pt}}\nonumber\\ \Rightarrow\quad RC\,p+1&=0\nonumber\\ p&=-\frac{1}{RC}\label{eq:bTrigRC_p} \end{align} $$

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 \(\eqref{eq:bTrigRC_gen}\)

$$ y_{h1}(t)=\mathrm{e}^{-\frac{t}{RC}t} $$

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

$$ \shaded{ y_h(t)=c\,y_{h1}(t)=c\,\mathrm{e}^{-\frac{t}{RC}} \label{eq:bTrigRC_hSolution} } $$
where the constant \(c\) follows from the initial conditions.

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 \(A\).

$$ \begin{align} y_p(t)&=A\cos(\omega t+\phi)\label{eq:bTrigRC_form}\\ \Rightarrow\quad y^\prime_p(t)&=-A\,\omega\sin(\omega t+\phi)\label{eq:bTrigRC_formDer} \end{align} $$

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

$$ \begin{align} -ARC\,\omega\sin(\omega t+\phi)+A\cos(\omega t+\phi)&=\hat{u}\cos(\omega t)\nonumber\\ \Rightarrow\quad \color{green}{1}\cos(\omega t+\phi)-\color{green}{\omega RC}\,\sin(\omega t+\phi)&=\frac{\hat{u}\cos(\omega t)}{A}\label{eq:bTrigRC_part}\\ \end{align} $$

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 \(\gamma\sin\alpha\)

$$ \begin{align} \gamma\cos\alpha&\triangleq 1\label{eq:bTrigRC_CcosAlpha}\\ \gamma\sin\alpha&\triangleq\omega RC\label{eq:bTrigRC_CsinAlpha}\\ \end{align} $$

to dot the ‘i’, introduce \(\beta\)

$$ \beta\triangleq\omega t+\phi\label{eq:bTrigRC_beta} $$

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

$$ \begin{align} \gamma\cos\alpha\cos\beta- \gamma\sin\alpha\sin\beta&=\frac{\hat{u}\cos(\omega t)}{A}\nonumber\\ \gamma\cos(\alpha+\beta)&=\frac{\hat{u}\cos(\omega t)}{A}\label{eq:bTrigRC_alpabetaC}\\ \end{align} $$

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 \(C\)

$$ \begin{align} \frac{\cancel{\gamma}\sin\alpha}{\cancel{\gamma}\cos\alpha}=\frac{\omega RC}{1} \quad\Rightarrow\quad \alpha&=\arctan\left(\omega RC\right)\label{eq:bTrigRC_alpha} \\ \left(\frac{\omega RC}{\gamma}\right)^2+\left(\frac{1}{\gamma}\right)^2=1 \quad\Rightarrow\quad \gamma&=\sqrt{1+(\omega RC)^2}\label{eq:bTrigRC_C} \end{align} $$

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

$$ {\sqrt{1^2+(\omega RC)^2}}\, \cos{\large(}{\arctan\left(\omega RC\right)+\beta}\,{\large)} = {\frac{\hat{u}}{A}}\cos({teal}{\omega t}) $$

and combine like terms

$$ \begin{align} A &= \frac{\hat{u}}{\sqrt{1+(\omega RC)^2}} \label{eq:bTrigRC_A} \\ \arctan\left(\omega RC\right)+\cancel{\omega t}+\phi=\cancel{\omega t} \quad\Rightarrow\quad \phi &= -\arctan\left(\frac{\omega L}{R}\right) \label{eq:bTrigRC_Phi} \end{align} $$

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

$$ \shaded{\begin{align} y_p(t)&=A\cos(\omega t+\phi)\label{eq:bTrigRC_pSolution}\\ \text{where}\quad A&=\frac{\hat{u}}{\sqrt{1+(\omega RC)^2}}\nonumber\\ \text{and}\quad\phi&=-\arctan\left(\omega RC\right)\nonumber \end{align} } $$

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

$$ \underline{u}(t)=\hat{u}\,\mathrm{e}^{j\omega t} $$

the corresponding complex response is of the form

$$ \begin{align} \underline{y}_p(t)&=A\,\mathrm{e}^{j(\omega t+\phi)}\label{eq:bCaRC_yp}\\ \Rightarrow\quad {\underline{y}_p}^\prime(t)&=j\omega\,A\,\mathrm{e}^{j(\omega t+\phi)}\label{eq:bCaRC_ypDer}\\ \end{align} $$

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

$$ \begin{align} RC\,j\omega\,A\,\mathrm{e}^{j(\omega t+\phi)}+\,A\,\mathrm{e}^{j(\omega t+\phi)}&=\hat{u}\mathrm{e}^{j\omega t},&\div\mathrm{e}^{j\omega t}\nonumber\\ \Rightarrow\quad j\omega RC\,A\,\mathrm{e}^{j\phi} +A\,\mathrm{e}^{j\phi}&=\hat{u}\nonumber\\ \Rightarrow\quad A\,\mathrm{e}^{j\phi}(j\omega RC+1)&=\hat{u}\nonumber\\ \Rightarrow\quad A\,\mathrm{e}^{j\phi}&=\frac{\hat{u}}{j\omega RC+1} \end{align} $$

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

$$ \begin{align} A&=\left|\frac{\hat{u}}{j\omega RC+1}\right|\\ \phi&=\angle\hat{u}-\angle(j\omega RC+1) =0-\mathrm{atan2}\left(\omega RC,1 \right) =-\arctan\left(\omega RC\right) \end{align} $$

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

$$ \underline{y}_p(t)=A\,\mathrm{e}^{j(\omega t+\phi)}=A\cos(\omega t)+jA\sin(\omega t) \label{eq:bRLSol} $$

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

$$ \shaded{ \begin{align} y_p(t)&=\Re\left\{\underline{y}_{\,p}\right\}=A\,\cos(\omega t+\phi),\label{eq:bCaRC_pSolution}\\[6mu] \text{where}\quad A&=\frac{\hat{u}}{\sqrt{1+(\omega RC)^2}},\nonumber\\ \text{and}\quad\phi&=-\arctan\left(\omega RC\right)\nonumber \end{align} } $$

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 \(\eqref{eq:bTrigRC_hp}\).

$$ \shaded{ \begin{align} y(t)&=c\,\mathrm{e}^{-\frac{t}{RC}}+A\cos(\omega t+\phi)\nonumber\\ \text{where}\quad A&=\frac{\hat{u}}{\sqrt{1+(\omega RC)^2}}\nonumber\\ \text{and}\quad\phi&=-\arctan\left(\omega RC\right)\nonumber \end{align} }\label{eq:bTrigRC_solution} $$