# RL High-pass Filter

$$u(t)$$Instead of $$\Delta v(t)$$, we use the European symbol for voltage difference: $$u$$. The letter ‘u’ stands for “Potentialunterschied”.

### Trigonometry method, example 2


#### Homogeneous solutions

The homogeneous solutions follows from the reduced (=homogeneous) linear differential equation where the forcing function is zero. $$L\,{i_p}^\prime(t)+R\,i_p(t)=0\label{eq:bTrigRL_homDV}$$ According the Euler, the homogeneous solutions are in the form $$i_h(t)=\mathrm{e}^{pt}\label{eq:bTrigRL_gen}$$ substituting this $$i_h(t)$$ in $$\eqref{eq:bTrigRL_homDV}$$ gives the characteristic equation with root $$p$$ \begin{align} L\,(\mathrm{e}^{pt})^\prime+R\,\mathrm{e}^{pt}&=0\nonumber\\ \Rightarrow\quad L\,p\mathrm{e}^{pt}+R\,\mathrm{e}^{pt}&=0&\div{\mathrm{e}^{pt}}\nonumber\\ \Rightarrow\quad L\,p+R&=0\nonumber\\ p&=-\frac{R}{L}\label{eq:bTrigRL_p} \end{align} The solution base $$i_{h,1}(t)$$ follows from substituting the root $$p$$ from equation $$\eqref{eq:bTrigRL_p}$$ in back in the homogeneous differential equation $$\eqref{eq:bTrigRL_gen}$$ $$i_{h1}(t)=\mathrm{e}^{pt}=\mathrm{e}^{-\frac{R}{L}t}$$ The homogeneous solution follows as a linear combination of the solution bases (only one in this case) as $$i_h(t)=c\,i_{h1}(t)=c\,\mathrm{e}^{-\frac{R}{L}t}\label{eq:bTrigRL_hSolution}$$ where the constant $$c$$ follows from the initial conditions.

#### Particular solutions

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} i_p(t)&=A\cos(\omega t+\phi)\label{eq:bTrigRL_form}\\ \Rightarrow\quad i^\prime_p(t)&=-A\,\omega\sin(\omega t+\phi)\label{eq:bTrigRL_formDer} \end{align} Substituting $$(\ref{eq:bTrigRL_form},\ref{eq:bTrigRL_formDer})$$ in the differential equation $$\eqref{eq:bTrigRL_DV}$$ \begin{align} -AL\,\omega\sin(\omega t+\phi)+AR\cos(\omega t+\phi)&=\hat{u}\cos(\omega t)\nonumber\\ \Rightarrow\quad \color{green}{R}\cos(\omega t+\phi)-\color{green}{\omega L}\,\sin(\omega t+\phi)&=\frac{\hat{u}\cos(\omega t)}{A}\label{eq:bTrigRL_part}\\ \end{align} Work towards the trigonometric identity
\begin{align} C\cos(\alpha+\beta)=C\cos\alpha\cos\beta-C\sin\alpha\sin\beta\nonumber \end{align}\nonumber
by assigning the two independent variables $$R$$ and $$\omega L$$ to two more convenient independent variables $$C\cos\alpha$$ and $$C\sin\alpha$$ \begin{align} C\cos\alpha&\triangleq R\label{eq:bTrigRL_CcosAlpha}\\ C\sin\alpha&\triangleq\omega L\label{eq:bTrigRL_CsinAlpha}\\ \end{align} to dot the ‘i’, introduce $$\beta$$ $$\beta\triangleq\omega t+\phi\label{eq:bTrigRL_beta}$$ we can rewrite $$\eqref{eq:bTrigRL_part}$$ and use the aforementioned trigonometric identity \begin{align} C\cos\alpha\cos\beta- C\sin\alpha\sin\beta&=\frac{\hat{u}\cos(\omega t)}{A}\nonumber\\ C\cos(\alpha+\beta)&=\frac{\hat{u}\cos(\omega t)}{A}\label{eq:bTrigRL_alpabetaC}\\ \end{align} Divide $$\eqref{eq:bTrigRL_CsinAlpha}$$ by $$\eqref{eq:bTrigRL_CcosAlpha}$$ to solve for $$\alpha$$, and apply the geometric identity $$\sin^2\alpha+\cos^2\alpha=1$$ to $$\eqref{eq:bTrigRL_CsinAlpha}$$ by $$\eqref{eq:bTrigRL_CcosAlpha}$$ to solve for $$C$$ \begin{align} \frac{\cancel{C}\sin\alpha}{\cancel{C}\cos\alpha}=\frac{\omega L}{R} \quad\Rightarrow\quad \alpha&=\arctan\left(\frac{\omega L}{R}\right)\label{eq:bTrigRL_alpha} \\ \left(\frac{R}{C}\right)^2+\left(\frac{\omega L}{C}\right)^2=1 \quad\Rightarrow\quad C&=\sqrt{R^2+(\omega L)^2}\label{eq:bTrigRL_C} \end{align} Substituting $$(\ref{eq:bTrigRL_beta}, \ref{eq:bTrigRL_alpha},\ref{eq:bTrigRL_C})$$ in equation $$\eqref{eq:bTrigRL_alpabetaC}$$ \begin{align} \sqrt{R^2+(\omega L)^2}\,\cos\left(\arctan\frac{\omega L}{R}+\omega t+\phi\right)&=\frac{\hat{u}\cos(\omega t)}{A}\nonumber\\ \color{brown}{A}\,\cos\left(\color{teal}{\arctan\frac{\omega L}{R}+\omega t+\phi}\right) &=\color{brown}{\frac{\hat{u}}{\sqrt{R^2+(\omega L)^2}}}\,\cos(\color{teal}{\omega t})\nonumber\\ \end{align} and combine like terms \begin{align} A&=\frac{\hat{u}}{\sqrt{R^2+(\omega L)^2}}\label{eq:bTrigRL_A}\\ \arctan\frac{\omega L}{R}+\cancel{\omega t}+\phi=\cancel{\omega t} \quad\Rightarrow\quad \phi&=-\arctan\left(\frac{\omega L}{R}\right)\label{eq:bTrigRL_Phi}\\ \end{align} The particular solution follows from substituting $$(\ref{eq:bTrigRL_A},\ref{eq:bTrigRL_Phi})$$ in $$\eqref{eq:bTrigRL_form}$$ \begin{align} i_p(t)&=A\cos(\omega t+\phi)\nonumber\\ \text{where}\quad A&=\frac{\hat{u}}{\sqrt{R^2+(\omega L)^2}}\nonumber\\ \text{and}\quad&\phi=-\arctan\left(\frac{\omega L}{R}\right)\nonumber \end{align}\label{eq:bTrigRL_pSolution}

#### General solution

Substituting equation $$(\ref{eq:bTrigRL_hSolution},\ref{eq:bTrigRL_pSolution})$$ in equation $$\eqref{eq:bTrigRL_hp}$$ gives the general solution for $$i(t)$$ \begin{align} i(t)&=c\,\mathrm{e}^{-\frac{R}{L}t}+A\cos(\omega t+\phi)\nonumber\\ \text{where}\quad A&=\frac{\hat{u}}{\sqrt{R^2+(\omega L)^2}}\nonumber\\ \text{and}\quad&\phi=-\arctan\left(\frac{\omega L}{R}\right)\nonumber \end{align} This seems like a good moment in time to start looking for a less involved method of solving these non-homogeneous differential equations. In the next section we will find such a method by using a complex forcing function.

### Complex arithmetic method

We will demonstrate this method using a RL high-pass filter, with input $$u(t)=\hat{u}\cos(\omega t)$$ and output current $$i(t)$$ through the inductor. The differential equation for this system is $$L\,{i_p}^\prime(t)+R\,i_p(t)=\hat{u}\cos(\omega t)\label{eq:bRLDV}$$ Using the complex source $$\hat{u}\,\mathrm{e}^{j\omega t}$$, the corresponding complex response is of the form \begin{align} i_p(t)&=A\,\mathrm{e}^{j(\omega t+\phi)}\\ \Rightarrow\quad {i_p}^\prime(t)&=j\omega\,A\,\mathrm{e}^{j(\omega t+\phi)}\\ \end{align} Substituting these in $$\eqref{eq:bRLDV}$$ \begin{align} L\,j\omega\,A\,\mathrm{e}^{j(\omega t+\phi)}+R\,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 L\,A\,\mathrm{e}^{j\phi} +R\,A\,\mathrm{e}^{j\phi}&=\hat{u}\nonumber\\ \Rightarrow\quad A\,\mathrm{e}^{j\phi}(j\omega L+R)&=\hat{u}\nonumber\\ \Rightarrow\quad A\,\mathrm{e}^{j\phi}&=\frac{\hat{u}}{j\omega L+R} \end{align} The amplitude $$A$$ and phase $$\phi$$ of the response are found as \begin{align} A&=\left|\frac{\hat{u}}{j\omega L+R}\right|\\ \phi&=\angle\hat{u}-\angle(j\omega L+R) =0-\mathrm{atan2}\left(\omega L,R \right) =\arctan\left(\frac{\omega L}{R}\right) \end{align} So that the complete response of the system is \begin{align} i_{ss}(t)&=A\,\mathrm{e}^{j(\omega t+\phi)},\nonumber\\[6mu] \text{where}\quad A&=\left|\frac{\hat{u}}{j\omega L+R}\right|=\hat{u}\frac{1}{\sqrt{(\omega L)^2+R^2}},\nonumber\\ \text{and}\quad\phi&=\angle\hat{u}-\angle(j\omega L+R) =0-\mathrm{atan2}\left(\omega L,R \right) =-\arctan\left(\frac{\omega L}{R}\right)\nonumber \end{align}\label{eq:bRLSol} Since the real forcing function was $$\cos$$, the real part of $$\hat{u}\cos(\omega t)+j\,\hat{u}\sin(\omega t)$$ we need to extract the real part of the solution equation $$\eqref{eq:bRLSol}$$ \begin{align} i_{ss}(t)&=A\,\cos(\omega t+\phi),\nonumber\\[6mu] \text{where}\quad A&=\frac{\hat{u}}{\sqrt{(\omega L)^2+R^2}},\nonumber\\ \text{and}\quad\phi&=-\arctan\left(\frac{\omega L}{R}\right)\nonumber \end{align} [link]
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