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  UNDER CONSTRUCTION

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Fourier Series Circuit Analysis

  • [1] Express \(v_{\t{s}}(t)\) in terms of an amplitude/phase Fourier series:
    \(\ds \qq{v}{s}(t)=a_0+\sum_{n=1}^{\infty}c_n\cos{(n\omega_0t+\phi_n)}\quad \)
  • [2] Transform circuit to phasor domain
  • [3] Obtain the transfer function at frequency \(\omega\):
    \(\ds \bn{H}(\omega)=\bns{V}{out},\quad \qq{v}{s}=1\cos{(\omega t)}\)
  • (a) Set \(\vpl \qq{v}{s}(t)=1\cos{\left(\omega t\right)}\), which corresponds to \(\sml{\bns{V}{s}(\omega)=1}\) in the phasor domain
    (b) Apply circuit analysis to obtain the expression for \(\vphantom{\int} \sml{\bns{V}{out}/\bns{V}{s}=\bns{V}{out}=\bn{H}}\)

  • [4] Compile the time-domain solution:
  • \(\ds \qq{v}{out}(t)=a_0\ \bn{H}(\omega = 0)+\sum_{n=1}^{\infty}c_n\Ree\left\{\bn{H}(\omega=n\omega_0 )e^{\lrg{}j(n\omega_0 t+\phi_n)}\right\}\)

\(\begin{align} a_0 \vphantom{\ds \int_{0}^{\int^{\int}}} & =\frac{1}{T_0}\int_{0}^{T_0}x(t)dt \\ a_n & = \frac{2}{T_0}\int_{0}^{T_0}x(t)\cos{(n\omega_0 t)} \\ b_n & = \frac{2}{T_0}\int_{0}^{T_0}x(t)\sin{(n\omega_0 t)}dt \end{align}\)
\(\vpL c_n\angle{\phi_n}=a_n-jb_n\)
\(\vpl \ds c_n=\sqrt{a_n^2+b_n^2}\quad \)
\(\ds {\phi }_n=\left\{ \begin{array}{c} -{{\mathrm{tan}}^{-1} \left(\lfrac{b_n}{a_n}\right)},\ a_n>0 \\ \pi -{{\mathrm{tan}}^{-1} \left(\lfrac{b_n}{a_n}\right)},\ a_n<0 \end{array} \right.\)



[Ex]: RC Circuit driven by Pulse Train

GIVEN:
\(R=20\ k\Omega,\ C=0.1\ m\t{F}\)


DETERMINE: \(v_{\t{out}}(t)\)
SOLUTION:

Step 1: The period of \(v_s(t)\) is \(4\ \t{s}\). Hence, \(\omega_0=2\pi/4=\pi/2\text{ rad/s}\), and by Eq.[1] we have
\begin{align} \ds a_0 & = \frac{1}{T}\int_0^T v_s(t)dt=\frac{1}{4}\int_0^1 10dt=2.5\t{ [V],} \\ a_n & = \frac{2}{4}\int_0^1 10\Cos{\frac{n\pi}{2}t}dt=\frac{10}{n\pi}\Sinf{n\pi}{2}\t{ [V],} \\ b_n & = \frac{2}{4}\int_0^1 10\Sin{\frac{n\pi}{2}t}dt=\frac{10}{n\pi}\left(1-\Cosf{n\pi}{2}\right)\t{ [V], and} \\ c_n\angle{\phi_n} & = a_n-jb_n=\frac{10}{n\pi}\left[\Sinf{n\pi}{2}-j\left(1-\Cosf{n\pi}{2}\right)\right]. \end{align}
The values of \(c_n\angle{\phi_n}\) for the first four terms are
\(\begin{align} \ds c_1\angle{\phi_1} & = \frac{10\sqrt{2}}{\pi}\adegg{-45}\t{,} \\ c_2\angle{\phi_2} & = \frac{10}{\pi}\adegg{-90}\t{,} \\ c_3\angle{\phi_3} & = \frac{10\sqrt{2}}{3\pi}\adegg{-135}\t{,} \\ c_4\angle{\phi_4} & = 0. \vphantom{\frac{10\sqrt{2}}{3\pi}} \end{align}\)
Step 2: The phasor domain circuit equivalent is
Step 3: Applying voltage division in the phasor domain, the circuit transfer function is
\(\begin{align} \ds \bn{H}(\omega) & = \bns{V}{out} \vphantom{\frac{{\int}^{\int}}{A}}\\ & = \frac{1}{1+j\omega RC} \\ & = \frac{1}{\sqrt{1+\omega^2 R^2C^2}}e^{-j\atan{\omega RC}}\Rightarrow \end{align}\)
\(\vpL \hspace{17px}\boxed{\bn{H}(\omega) = \frac{1}{\sqrt{1+4\omega^2}}e^{-j\atan{2\omega}}}\)
where we used \(RC=(20k\Omega)(0.1m\t{F}) = 2\t{ s}\).

Step 4: The time-domain output voltage is
\(\ds v_{\t{out}}(t)=2.5+\sum_{n=1}^{\infty}\Ree \left\{ c_n\frac{1}{\sqrt{1+4n^2\omega_0^2}}e^{\lrg{\ j[n\omega_0 t\ +\ \phi_n-\ \atan{2n\omega_0}]}}\right\}\Rightarrow\)
Using the values of \(c_n\angle{\phi_n}\) determined earlier for the first four terms and replacing \(\omega_0\) with its numerical value of \(\pi/2\t{ rad/s}\), the expression becomes
\(\begin{align} \ds && v_{\t{out}}(t) & = 2.5\\ && + & \frac{10\sqrt{2}}{\pi \sqrt{1+\pi^2}}\cos{\left[\frac{\pi t}{2}-\deg{45}-\atan{\pi}\right]} \\ && + & \frac{10}{\pi \sqrt{1+4\pi^2}}\cos{[\pi t-\deg{90}-\atan{2\pi}]} \\ && + & \frac{10\sqrt{2}}{\pi \sqrt{1+9\pi^2}}\cos{\left[\frac{3\pi t}{2}-\deg{135}-\atan{3\pi}\right]}\t{ ...}\Rightarrow \\ \end{align}\) \[\Bxred{v_{\t{out}}(t)=\ 2.5+1.37\Cos{\frac{\pi t}{2}-\deg{117}}+0.5\cos{(\pi t-\deg{171})}+0.16\Cos{\frac{3\pi t}{2}+\deg{141}}\text{ ... [V]}}\]
The output response \(v_{\t{out}}(t)\), computed using the series solution with \(n_{\t{max}}=1000\), is shown below






Dragon Notes,   Est. 2018     About

By OverLordGoldDragon