Trigonometric Fourier Series
While studying heat flow, Fourier discovered that a non-sinusoidal periodic function can be expressed as an infinite sum of sinusoidal functions. Recall that a periodic function is one that repeats every $ T $ seconds. In other words, a periodic function $ f(t) $ satisfies
where $ n $ is an integer and $ T $ is the period of the function. According to the Fourier theorem, any practical periodic function of frequency $ \omega_{0} $ can be expressed as an infinite sum of sine or cosine functions that are integral multiples of $ \omega_{0} $. Thus, $ f(t) $ can be expressed as
or
where $ \omega_{0}=2 \pi / T $ is called the fundamental frequency in radians per second. The sinusoid $ \sin n \omega_{0} t $ or $ \cos n \omega_{0} t $ is called the $ n $th harmonic of $ f(t) $; it is an odd harmonic if $ n $ is odd and an even harmonic if $ n $ is even. Equation $ 3 $ is called the trigonometric Fourier series of $ f(t) $.
The constants $ a_{n} $ and $ b_{n} $ are the Fourier coefficients. The coefficient $ a_{0} $ is the dc component or the average value of $ f(t) $ (Recall that sinusoids have zero average values.) The coefficients $ a_{n} $ and $ b_{n} $ (for $ n \neq 0 $ ) are the amplitudes of the sinusoids in the ac component. Thus,
A function that can be represented by a Fourier series as in Eq. (3) must meet certain requirements, because the infinite series in Eq. (3) may or may not converge. These conditions on $ f(t) $ to yield a convergent.
Fourier series are as follows:
Let us use these identities to evaluate the Fourier coefficients.
We begin by finding $ a_{0} $. We integrate both sides of Eq. (3) over one period and obtain
Invoking the identities of Eqs. (4.a) and (4.b), the two integrals involving the ac terms vanish. Hence,
or
showing that $ a_{0} $ is the average value of $ f(t) $.To evaluate $ a_{n} $, we multiply both sides of Eq. (3) by $ \cos m \omega_{0} t $ and integrate over one period:
The integral containing $ a_{0} $ is zero in view of Eq. (4.b), while the integral containing $ b_{n} $ vanishes according to Eq. (4.c). The integral containing $ a_{n} $ will be zero except when $ m=n $, in which case it is $ T / 2 $, according to Eqs. (4.e) and (4.g). Thus,
or
In a similar vein, we obtain $ b_{n} $ by multiplying both sides of Eq. (3) by $ \sin m \omega_{0} t $ and integrating over the period. The result is
Be aware that since $ f(t) $ is periodic, it may be more convenient to carry the integrations above from $ -T / 2 $ to $ T / 2 $ or generally from $ t_{0} $ to $ t_{0}+T $ instead of 0 to $ T $. The result will be the same.
An alternative form of Eq. (3) is the amplitude-phase form
we can apply the trigonometric identity
to the ac terms in Eq. (10) so that
Equating the coefficients of the series expansions in Eqs. (3) and (12) shows that
or
To avoid any confusion in determining $ \phi_{n} $, it may be better to relate the terms in complex form as
The plot of the amplitude $ A_{n} $ of the harmonics versus $ n \omega_{0} $ is called the amplitude spectrum of $ f(t) $; the plot of the phase $ \phi_{n} $ versus $ n \omega_{0} $ is the phase spectrum of $ f(t) $. Both the amplitude and phase spectra form the frequency spectrum of $ f(t) $.
Thus, the Fourier analysis is also a mathematical tool for finding the spectrum of a periodic signal. To evaluate the Fourier coefficients $ a_{0}, a_{n} $, and $ b_{n} $, we often need to apply the following integrals:
It is also useful to know the values of the cosine, sine, and exponential functions for integral multiples of $ \pi $. These are given in Table $ 1 $, where $ n $ is an integer.
$$\bbox[10px,border:1px solid grey]{f(t)=f(t+n T)} \tag{1}$$
$$\begin{aligned}f(t)=& a_{0}+a_{1} \cos \omega_{0} t+b_{1} \sin \omega_{0} t+a_{2} \cos 2 \omega_{0} t \\&+b_{2} \sin 2 \omega_{0} t+a_{3} \cos 3 \omega_{0} t+b_{3} \sin 3 \omega_{0} t+\cdots\end{aligned} \tag{2}$$
$$f(t)=\underbrace{a_{0}}_{\mathrm{dc}}+\underbrace{\sum_{n=1}^{\infty}\left(a_{n} \cos n \omega_{0} t+b_{n} \sin n \omega_{0} t\right)}_{\mathrm{ac}} \tag{3}$$
- $ f(t) $ is single-valued everywhere.
- $ f(t) $ has a finite number of finite discontinuities in any one period.
- $ f(t) $ has a finite number of maxima and minima in any one period.
- The integral $ \int_{t_{0}}^{t_{0}+T}|f(t)| d t<\infty $ for any $ t_{0} $.
$$\begin{array}{c}
\int_{0}^{T} \sin n \omega_{0} t d t=0 \quad (4.a) \\
\int_{0}^{T} \cos n \omega_{0} t d t=0 \quad (4.b) \\
\int_{0}^{T} \sin n \omega_{0} t \cos m \omega_{0} t d t=0 \quad (4.c) \\
\int_{0}^{T} \sin n \omega_{0} t \sin m \omega_{0} t d t=0, \quad(m \neq n) \quad (4.d)\\
\int_{0}^{T} \cos n \omega_{0} t \cos m \omega_{0} t d t=0, \quad(m \neq n) \quad (4.e)\\
\int_{0}^{T} \sin ^{2} n \omega_{0} t d t=\frac{T}{2} \quad (4.f)\\
\int_{0}^{T} \cos ^{2} n \omega_{0} t d t=\frac{T}{2} \quad (4.g)
\end{array}
$$
$$\begin{aligned}\int_{0}^{T} f(t) d t=& \int_{0}^{T}\left[a_{0}+\sum_{n=1}^{\infty}\left(a_{n} \cos n \omega_{0} t+b_{n} \sin n \omega_{0} t\right)\right] d t \\=\int_{0}^{T} a_{0} d t+\sum_{n=1}^{\infty} & {\left[\int_{0}^{T} a_{n} \cos n \omega_{0} t d t\right.} \\&\left.+\int_{0}^{T} b_{n} \sin n \omega_{0} t d t\right] d t\end{aligned} \tag{5}$$
$$\int_{0}^{T} f(t) d t=\int_{0}^{T} a_{0} d t=a_{0} T$$
$$a_{0}=\frac{1}{T} \int_{0}^{T} f(t) d t \tag{6}$$
$$\begin{aligned}
& \int_{0}^{T} f(t) \cos m \omega_{0} t d t \\
=& \int_{0}^{T}\left[a_{0}+\sum_{n=1}^{\infty}\left(a_{n} \cos n \omega_{0} t+b_{n} \sin n \omega_{0} t\right)\right] \cos m \omega_{0} t d t \\=& \int_{0}^{T} a_{0} \cos m \omega_{0} t d t+\sum_{n=1}^{\infty}\left[\int_{0}^{T} a_{n} \cos n \omega_{0} t \cos m \omega_{0} t d t\right.\\&\left.+\int_{0}^{T} b_{n} \sin n \omega_{0} t \cos m \omega_{0} t d t\right] d t\end{aligned} \tag{7}$$
$$\int_{0}^{T} f(t) \cos m \omega_{0} t d t=a_{n} \frac{T}{2}, \quad \text { for } m=n$$
$$a_{n}=\frac{2}{T} \int_{0}^{T} f(t) \cos n \omega_{0} t d t \tag{8}$$
$$b_{n}=\frac{2}{T} \int_{0}^{T} f(t) \sin n \omega_{0} t d t \tag{9}$$
$$f(t)=a_{0}+\sum_{n=1}^{\infty} A_{n} \cos \left(n \omega_{0} t+\phi_{n}\right) \tag{10}$$
$$\cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta \tag{11}$$
$$\begin{aligned}a_{0}+\sum_{n=1}^{\infty} A_{n} \cos \left(n \omega_{0} t+\phi_{n}\right)=& a_{0}+\sum_{n=1}^{\infty}\left(A_{n} \cos \phi_{n}\right) \cos n \omega_{0} t \\&-\left(A_{n} \sin \phi_{n}\right) \sin n \omega_{0} t\end{aligned} \tag{12}$$
$$a_{n}=A_{n} \cos \phi_{n}, \quad b_{n}=-A_{n} \sin \phi_{n} \tag{13.a}$$
$$A_{n}=\sqrt{a_{n}^{2}+b_{n}^{2}}, \quad \phi_{n}=-\tan ^{-1} \frac{b_{n}}{a_{n}} \tag{13.b}$$
$$A_{n} \angle \phi_{n}=a_{n}-j b_{n} \tag{14}$$
$$\begin{array}{l}\int \cos a t d t=\frac{1}{a} \sin a t \quad (15.a) \\
\int \sin a t d t=-\frac{1}{a} \cos a t\quad (15.b)
\end{array} $$
$$\begin{aligned}\int t \cos a t d t &=\frac{1}{a^{2}} \cos a t+\frac{1}{a} t \sin a t \quad (15.c)\\
\int t \sin a t d t &=\frac{1}{a^{2}} \sin a t-\frac{1}{a} t \cos a t \quad (15.d)
\end{aligned} $$
| Table 1: Values of cosine, sine, and exponential functions for integral multiples of π. | |
| Function | Value |
| $cos 2nπ$ | 1 |
| $sin 2nπ$ | 0 |
| $cos nπ$ | $(-1)^n$ |
| $sin nπ$ | 0 |
| $cos {nπ \over 2}$ | $\left\{\begin{array}{ll} (-1)^{n/2}, & n= even \\0, & n=odd \end{array}\right.$ |
| $sin {nπ \over 2}$ | $\left\{\begin{array}{ll} (-1)^{(n-1)/2}, & n= odd \\0, & n=even \end{array}\right.$ |
| $e^{j2n\pi}$ | 1 |
| $e^{jn\pi}$ | $(-1)^n$ |
| $e^{jn\pi /2}$ | $\left\{\begin{array}{ll} (-1)^{n/2}, & n= even \\j(-1)^{(n-1)/2}, & n=odd \end{array}\right.$ |
Example 1: Determine the Fourier series of the waveform shown in Fig. 1. Obtain the amplitude and phase spectra.
Solution:
The Fourier series is given by Eq. (3), namely,
Our goal is to obtain the Fourier coefficients $ a_{0}, a_{n} $, and $ b_{n} $ using Eqs. (6), (8), and (9). First, we describe the waveform as
and $ f(t)=f(t+T) $. Since $ T=2, \omega_{0}=2 \pi / T=\pi $. Thus,
Using Eq. (8) along with Eq. (15a),
From Eq. (9) with the aid of Eq. (15b),
Substituting the Fourier coefficients in Eqs. (1.3) to (1.5) into Eq. (1.1) gives the Fourier series as
Since $ f(t) $ contains only the dc component and the sine terms with the fundamental component and odd harmonics, it may be written as
By summing the terms one by one as demonstrated in Fig. 2, we notice how superposition of the terms can evolve into the original square. As more and more Fourier components are added, the sum gets closer and closer to the square wave. However, it is not possible in practice to sum the series in Eq. (1.6) or (1.7) to infinity. Only a partial sum $ (n=1,2,3, \ldots, N $, where $ N $ is finite) is possible.
If we plot the partial sum (or truncated series) over one period for a large $ N $ as in Fig. 3, we notice that the partial sum oscillates above and below the actual value of $ f(t) $. At the neighborhood of the points of discontinuity $ (x=0,1,2, \ldots) $, there is overshoot and damped oscillation. In fact, an overshoot of about 9 percent of the peak value is always present, regardless of the number of terms used to approximate $ f(t) $. This is called the Gibbs phenomenon.
Finally, let us obtain the amplitude and phase spectra for the signal in Fig. 16.1. Since $ a_{n}=0 $
and

The plots of $ A_{n} $ and $ \phi_{n} $ for different values of $ n \omega_{0}=n \pi $ provide the amplitude and phase spectra in Fig. 4. Notice that the amplitudes of the harmonics decay very fast with frequency.
Fig. 1: For Example 1; a square wave.
$$f(t)=a_{0}+\sum_{n=1}^{\infty}\left(a_{n} \cos n \omega_{0} t+b_{n} \sin n \omega_{0} t\right) \tag{1.1}$$
$$f(t)=\left\{\begin{array}{ll}1, & 0 < t < 1 \\0, & 1< t < 2\end{array}\right. \tag{1.2}$$
$$a_{0}=\frac{1}{T} \int_{0}^{T} f(t) d t=\frac{1}{2}\left[\int_{0}^{1} 1 d t+\int_{1}^{2} 0 d t\right]=\left.\frac{1}{2} t\right|_{0} ^{1}=\frac{1}{2} \tag{1.3}$$
$$\begin{aligned}a_{n} &=\frac{2}{T} \int_{0}^{T} f(t) \cos n \omega_{0} t d t \\&=\frac{2}{2}\left[\int_{0}^{1} 1 \cos n \pi t d t+\int_{1}^{2} 0 \cos n \pi t d t\right] \\&=\left.\frac{1}{n \pi} \sin n \pi t\right|_{0} ^{1}=\frac{1}{n \pi} \sin n \pi=0\end{aligned} \tag{1.4}$$
$$\begin{aligned}b_{n} &=\frac{2}{T} \int_{0}^{T} f(t) \sin n \omega_{0} t d t \\&=\frac{2}{2}\left[\int_{0}^{1} 1 \sin n \pi t d t+\int_{1}^{2} 0 \sin n \pi t d t\right] \\&=-\left.\frac{1}{n \pi} \cos n \pi t\right|_{0} ^{1} \\&=-\frac{1}{n \pi}(\cos n \pi-1), \quad \cos n \pi=(-1)^{n} \\&=\frac{1}{n \pi}\left[1-(-1)^{n}\right]=\left\{\begin{array}{ll}\frac{2}{n \pi}, & n=\text { odd } \\0, & n=\text { even }\end{array}\right.\end{aligned} \tag{1.5}$$
$$f(t)=\frac{1}{2}+\frac{2}{\pi} \sin \pi t+\frac{2}{3 \pi} \sin 3 \pi t+\frac{2}{5 \pi} \sin 5 \pi t+\cdots \tag{1.6}$$
$$f(t)=\frac{1}{2}+\frac{2}{\pi} \sum_{k=1}^{\infty} \frac{1}{n} \sin n \pi t, \quad n=2 k-1 \tag{1.7}$$
Fig. 2: Evolution of a square wave from its Fourier components.
Fig. 3: Truncating the Fourier series at
N = 11; Gibbs phenomenon.
$$A_{n}=\sqrt{a_{n}^{2}+b_{n}^{2}}=\left|b_{n}\right|=\left\{\begin{array}{ll}\frac{2}{n \pi}, & n=\text { odd } \\0, & n=\text { even }\end{array}\right.$$
$$\phi_{n}=-\tan ^{-1} \frac{b_{n}}{a_{n}}=\left\{\begin{array}{cl}-90^{\circ}, & n=\text { odd } \\0, & n=\text { even }\end{array}\right.$$

Fig. 4: For Example 1: (a) amplitude and (b) phase spectrum of the function
shown in Fig. 1
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