Circuit Application of The Fourier Series

Example 1: Let the function $ f(t) $ be the voltage source $ v_{s}(t) $ in the circuit of Fig. 3. Find the response $ v_{o}(t) $ of the circuit. $$f (t) = a_0 + \sum_{n=1}^{\infty}(a_n \cos nω_0t + b_n \sin nω_0t)$$ Solution: $$v_{s}(t)=\frac{1}{2}+\frac{2}{\pi} \sum_{k=1}^{\infty} \frac{1}{n} \sin n \pi t, \quad n=2 k-1$$ where $ \omega_{n}=n \omega_{0}=n \pi \mathrm{rad} / \mathrm{s} $. Using phasors, we obtain the response $ \mathbf{V}_{o} $ in the circuit of Fig. $ 3 $ by voltage division: $$\mathbf{V}_{o}=\frac{j \omega_{n} L}{R+j \omega_{n} L} \mathbf{V}_{s}=\frac{j 2 n \pi}{5+j 2 n \pi} \mathbf{V}_{s}$$ For the dc component $ \left(\omega_{n}=0\right. $ or $ \left.n=0\right) $ $$\mathbf{V}_{s}=\frac{1}{2} \quad \Longrightarrow \quad \mathbf{V}_{o}=0$$ This is expected, since the inductor is a short circuit to dc. For the $ n $th harmonic, $$\mathbf{V}_{s}=\frac{2}{n \pi} \angle-90^{\circ} \tag{1.1}$$ and the corresponding response is $$\begin{aligned}\mathbf{V}_{o} &=\frac{2 n \pi / 90^{\circ}}{\sqrt{25+4 n^{2} \pi^{2}} / \tan ^{-1} 2 n \pi / 5} \frac{2}{n \pi} \angle-90^{\circ} \\&=\frac{4 /-\tan ^{-1} 2 n \pi / 5}{\sqrt{25+4 n^{2} \pi^{2}}}\end{aligned} \tag{1.2}$$ In the time domain, $$v_{o}(t)=\sum_{k=1}^{\infty} \frac{4}{\sqrt{25+4 n^{2} \pi^{2}}} \cos \left(n \pi t-\tan ^{-1} \frac{2 n \pi}{5}\right), \quad n=2 k-1$$ The first three terms ( $ k=1,2,3 $ or $ n=1,3,5 $ ) of the odd harmonics in the summation give us $$\begin{aligned}v_{o}(t)=& 0.4981 \cos \left(\pi t-51.49^{\circ}\right)+0.2051 \cos \left(3 \pi t-75.14^{\circ}\right) \\&+0.1257 \cos \left(5 \pi t-80.96^{\circ}\right)+\cdots \mathrm{V}\end{aligned}$$ Figure $ 4 $ shows the amplitude spectrum for output voltage $ v_{o}(t) $. Notice that the two spectra are close. Why? We observe that the circuit in Fig. $ 3 $ is a high-pass filter with the corner frequency $ \omega_{c}=R / L=2.5 \mathrm{rad} / \mathrm{s} $, which is less than the fundamental frequency $ \omega_{0}=\pi \mathrm{rad} / \mathrm{s} $. The dc component is not passed and the first harmonic is slightly attenuated, but higher harmonics are passed. In fact, from Eqs. (1.1) and (1.2), $ \mathbf{V}_{o} $ is identical to $ \mathbf{V}_{s} $ for large $ n $, which is characteristic of a high-pass filter.

Example 2: Find the response $ i_{o}(t) $ in the circuit in Fig. $ 5 $ if the input voltage $ v(t) $ has the Fourier series expansion. $$v(t)=1+\sum_{n=1}^{\infty} \frac{2(-1)^{n}}{1+n^{2}}(\cos n t-n \sin n t)$$ Solution: Using Eq. (16.13), we can express the input voltage as $$\begin{aligned}v(t)=& 1+\sum_{n=1}^{\infty} \frac{2(-1)^{n}}{\sqrt{1+n^{2}}} \cos \left(n t+\tan ^{-1} n\right) \\=& 1-1.414 \cos \left(t+45^{\circ}\right)+0.8944 \cos \left(2 t+63.45^{\circ}\right) \\&-0.6345 \cos \left(3 t+71.56^{\circ}\right)-0.4851 \cos \left(4 t+78.7^{\circ}\right)+\cdots\end{aligned}$$ We notice that $ \omega_{0}=1, \omega_{n}=n \mathrm{rad} / \mathrm{s} $. The impedance at the source is $$\mathbf{Z}=4+j \omega_{n} 2 \| 4=4+\frac{j \omega_{n} 8}{4+j \omega_{n} 2}=\frac{8+j \omega_{n} 8}{2+j \omega_{n}}$$ The input current is $$\mathbf{I}=\frac{\mathbf{V}}{\mathbf{Z}}=\frac{2+j \omega_{n}}{8+j \omega_{n} 8} \mathbf{V}$$ where $ \mathbf{V} $ is the phasor form of the source voltage $ v(t) $. By current division, $$\mathbf{I}_{o}=\frac{4}{4+j \omega_{n} 2} \mathbf{I}=\frac{\mathbf{V}}{4+j \omega_{n} 4}$$ Since $ \omega_{n}=n, \mathbf{I}_{o} $ can be expressed as $$\mathbf{I}_{o}=\frac{\mathbf{V}}{4 \sqrt{1+n^{2}} / \tan ^{-1} n}$$ For the dc component $ \left(\omega_{n}=0\right. $ or $ \left.n=0\right) $ $$\mathbf{V}=1 \quad \Longrightarrow \quad \mathbf{I}_{o}=\frac{\mathbf{V}}{4}=\frac{1}{4}$$ For the $ n $th harmonic, $$\mathbf{V}=\frac{2(-1)^{n}}{\sqrt{1+n^{2}}} / \tan ^{-1} n$$ so that In the time domain, $$i_{o}(t)=\frac{1}{4}+\sum_{n=1}^{\infty} \frac{(-1)^{n}}{2\left(1+n^{2}\right)} \cos n t \mathrm{~A}$$