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 \(t\geq t_1\)
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 \(y_p(t)\)
$$
y(t)=y_h(t)+y_p(t)\label{eq:bTrigRC_hp}
$$
Homogeneous solution
The homogeneous solutions follows from the reduced (=homogeneous) linear differential equation where the forcing function is zero.
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
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 \(A\).
by assigning the two independent variables \(R\) and \(\omega L\) to two more convenient independent variables \(\gamma\cos\alpha\) and \(\gamma\sin\alpha\)
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 \(C\)
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
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
The general solution follows from substituting \(\eqref{eq:bTrigRC_hSolution}\) and \(\eqref{eq:bTrigRC_pSolution}\text{ or }\eqref{eq:bCaRC_pSolution}\) in equation \(\eqref{eq:bTrigRC_hp}\).
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