Application of the Fourier Series to Filters

Filters are an important component of electronics and communications systems. Chapter 21 presented a full discussion on passive and active filters. Here, we investigate how to design filters to select the fundamental component (or any desired harmonic) of the input signal and reject other harmonics. This filtering process cannot be accomplished without the Fourier series expansion of the input signal. For the purpose of illustration, we will consider two cases, a lowpass filter and a bandpass filter.
The output of a lowpass filter depends on the input signal, the transfer function $ H(\omega) $ of the filter, and the corner or half-power frequency $ \omega_{c} $. We recall that $ \omega_{c}=1 / R C $ for an $ R C $ passive filter. As shown in Fig. 1(a), the lowpass filter passes the dc and low-frequency components, while blocking the high-frequency components. By making $ \omega_{c} $ sufficiently large $ \left(\omega_{c} \gg \omega_{0}\right. $, e.g., making $ C $ small), a large number of the harmonics can be passed.
Fig. 1: (a) Input and output spectra of a lowpass filter, (b) the lowpass filter passes only the dc component when $ω_c \ll ω_0$.
On the other hand, by making $ \omega_{c} $ sufficiently small $ \left(\omega_{c} \ll \omega_{0}\right) $, we can block out all the ac components and pass only dc, as shown typically in Fig. 1(b). Similarly, the output of a bandpass filter depends on the input signal, the transfer function of the filter $ H(\omega) $, its bandwidth $ B $, and its center frequency $\omega_{c}$. As illustrated in Fig. 2(a), the filter passes all the harmonics of the input signal within a band of frequencies $\left(\omega_{1}<\omega<\omega_{2}\right)$ in centered around $ \omega_{c} $.
Fig. 2: (a) Input and output spectra of a bandpass filter, (b) the bandpass filter passes only the fundamental component when $B \ll ω_0$.
We have assumed that $ \omega_{0}, 2 \omega_{0} $, and $ 3 \omega_{0} $ are within that band. If the filter is made highly selective $ \left(B \ll \omega_{0}\right) $ and $ \omega_{c}=\omega_{0} $, where $ \omega_{0} $ is the fundamental frequency of the input signal, the filter passes only the fundamental component $ (n=1) $ of the input and blocks out all higher harmonics. As shown in Fig. 2(b), with a square wave as input, we obtain a sine wave of the same frequency as the output.
Example 1: If the sawtooth waveform in Fig. 3(a) is applied to an ideal lowpass filter with the transfer function shown in Fig. 3(b), determine the output.
Fig. 3: For Example 1.
From the input signal in Fig. 3(a), we know that the Fourier series expansion is $$x(t)=\frac{1}{2}-\frac{1}{\pi} \sin \omega_{0} t-\frac{1}{2 \pi} \sin 2 \omega_{0} t-\frac{1}{3 \pi} \sin 3 \omega_{0} t-\cdots$$ where the period is $ T=1 \mathrm{~s} $ and the fundamental frequency is $$ \omega_{0}=2 \pi \mathrm{rad} / \mathrm{s} $$ Since the corner frequency of the filter is $$ \omega_{c}=10 \mathrm{rad} / \mathrm{s} $$ only the dc component and harmonics with $$ n \omega_{0}<10 $$ will be passed. For $ n=2 $, $$ n \omega_{0}=4 \pi=12.566 \mathrm{rad} / \mathrm{s} $$ which is higher than $ 10 \mathrm{rad} / \mathrm{s} $ meaning that second and higher harmonics will be rejected. Thus, only the dc and fundamental components will be passed. Hence the output of the filter is $$y(t)=\frac{1}{2}-\frac{1}{\pi} \sin 2 \pi t$$