Circuit Application of The Fourier Series

We find that in practice, many circuits are driven by nonsinusoidal periodic functions. To find the steady-state response of a circuit to a nonsinusoidal periodic excitation requires the application of a Fourier series, ac phasor analysis, and the superposition principle. The procedure usually involves three steps.

Steps for Applying Fourier Series:

  • Express the excitation as a Fourier series.
  • Find the response of each term in the Fourier series.
  • Add the individual responses using the superposition principle.
The first step is to determine the Fourier series expansion of the excitation. For the periodic voltage source shown in Fig. 1(a), for example, the Fourier series is expressed as $$v(t)=V_{0}+\sum_{n=1}^{\infty} V_{n} \cos \left(n \omega_{0} t+\theta_{n}\right) \tag{1}$$ (The same could be done for a periodic current source.) Equation (1) shows that $ v(t) $ consists of two parts: the dc component $ V_{0} $ and the ac component $ \mathbf{V}_{n}=V_{n} \angle \theta_{n} $ with several harmonics. This Fourier series representation may be regarded as a set of series-connected sinusoidal sources, with each source having its own amplitude and frequency, as shown in Fig. 1(b).
Fig. 1: (a) Linear network excited by a periodic voltage source, (b) Fourier series representation (time-domain).
The second step is finding the response to each term in the Fourier series. The response to the dc component can be determined in the frequency domain by setting $ n=0 $ or $ \omega=0 $ as in Fig. 2(a), or in the time domain by replacing all inductors with short circuits and all capacitors with open circuits. The network is represented by its impedance $ \mathbf{Z}\left(n \omega_{0}\right) $ or admittance $ \mathbf{Y}\left(n \omega_{0}\right). \mathbf{Z}\left(n \omega_{0}\right) $ is the input impedance at the source when $ \omega $ is everywhere replaced by $ n \omega_{0} $, and $ \mathbf{Y}\left(n \omega_{0}\right) $ is the reciprocal of $ \mathbf{Z}\left(n \omega_{0}\right) $.
Fig. 2: Steady-state responses: (a) dc component, (b) ac component (frequency domain).
Finally, following the principle of superposition, we add all the individual responses. For the case shown in Fig. 2, $$\begin{aligned}i(t) &=i_{0}(t)+i_{1}(t)+i_{2}(t)+\cdots \\&=\mathbf{I}_{0}+\sum_{n=1}^{\infty}\left|\mathbf{I}_{n}\right| \cos \left(n \omega_{0} t+\psi_{n}\right)\end{aligned}$$ where each component $ \mathbf{I}_{n} $ with frequency $ n \omega_{0} $ has been transformed to the time domain to get $ i_{n}(t) $, and $ \psi_{n} $ is the argument of $ \mathbf{I}_{n} $.
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. $$f (t) = a_0 + \sum_{n=1}^{\infty}(a_n \cos nω_0t + b_n \sin nω_0t)$$
Fig. 3: For Example 1.
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 $$\begin{aligned}v_{o}(t)=& 0.4981 \cos \left(\pi t-51.49^{\circ}\right)+0.2051 \cos \left(3 \pi t-75.14^{\circ}\right) \\&+0.1257 \cos \left(5 \pi t-80.96^{\circ}\right)+\cdots \mathrm{V}\end{aligned}$$
Fig. 4: For Example 1: Amplitude spectrum of the output voltage.
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.
Fig. 5: For Example 2.
$$v(t)=1+\sum_{n=1}^{\infty} \frac{2(-1)^{n}}{1+n^{2}}(\cos n t-n \sin n t)$$ 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, $$i_{o}(t)=\frac{1}{4}+\sum_{n=1}^{\infty} \frac{(-1)^{n}}{2\left(1+n^{2}\right)} \cos n t \mathrm{~A}$$