RL High-pass Filter

Trigonometry method, example 2

\( \require{AMSsymbols} \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\lfzraised#1{\raise{10mu}{#1}} \def\laplace{\lfz{\mathscr{L}}} \def\fourier{\lfz{\mathcal{F}}} \def\ztransform{\lfz{\mathcal{Z}}} \def\crw{\lfz{}} \require{cancel} \newcommand\ccancel[2][black] {\color{#1}{\cancel{\color{black}{#2}}}} \newcommand\ccancelto[3][black] {\color{#1}{\cancelto{#2}{\color{black}{#3}}}} \)Assume the non-homogeneous lineair 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 $$ L\,{i_p}^\prime(t)+R\,i_p(t)=\hat{u}\cos(\omega t)\label{eq:bTrigRL_DV} $$ 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 \(i_p(t)\) $$ i(t)=i_h(t)+i_p(t)\label{eq:bTrigRL_hp} $$

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 homegeneous 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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Passionately curious and stubbornly persistent. Enjoys to inspire and consult with others to exchange the poetry of logical ideas.

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