Differential equation for a RC low-pass filter

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.

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 $\eqref{eq:bTrigRC_hp}$.

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

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