Solution:

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$$$$

\int_0^T \sin n\omega_0 t \cos m\omega_0 t , dt = 0 \tag{17.4c}

$$$$

\int_0^T \sin n\omega_0 t \sin m\omega_0 t , dt = 0, \qquad (m \neq n) \tag{17.4d}

$$$$

\int_0^T \cos n\omega_0 t \cos m\omega_0 t , dt = 0, \qquad (m \neq n) \tag{17.4e}

$$$$

\int_0^T \sin^2 n\omega_0 t , dt = \frac{T}{2} \tag{17.4f}

$$$$

\int_0^T \cos^2 n\omega_0 t , dt = \frac{T}{2} \tag{17.4g}

$$Let us use these identities to evaluate the Fourier coefficients. We begin by finding *a*0. We integrate both sides of Eq. (17.3) o ver one period and obtain$$

\int_0^T f(t) dt = \int_0^T \left[ a_0 + \sum_{n=1}^\infty (a_n \cos n\omega_0 t + b_n \sin n\omega_0 t) \right] dt

$$=$$

\int_0^T a_0 dt + \sum_{n=1}^\infty \left[ \int_0^T a_n \cos n\omega_0 t dt + \int_0^T b_n \sin n\omega_0 t dt \right] dt

$$(17.5) Invoking the identities of Eqs. (17.4a) and (17.4b), the tw o inte grals involving the ac terms vanish. Hence,$$

\int_0^T f(t) , dt = \int_0^T a_0 , dt = a_0 , T

$$or$$

a_0 = \frac{1}{T} \int_0^T f(t) , dt \tag{17.6}

$$showing that *a*0 is the average value of *f*(*t*). To evaluate *an*, we multiply both sides of Eq. (17.3) by cos *mω*0*t* and integrate over one period:$$

\int_0^T f(t) \cos m\omega_0 t , dt

$$\n$$

= \int_0^T \left[ a_0 + \sum_{n=1}^\infty (a_n \cos n\omega_0 t + b_n \sin n\omega_0 t) \right] \cos m\omega_0 t , dt

$$\n$$

= \int_0^T a_0 \cos m\omega_0 t , dt + \sum_{n=1}^\infty \left[ \int_0^T a_n \cos n\omega_0 t \cos m\omega_0 t , dt \right]

$$\n$$

• \int_0^T b_n \sin n\omega_0 t \cos m\omega_0 t , dt \right] dt \tag{17.7}

$$The inte gral containing *a*0 is zero in vie w of Eq. (17.4b), while the integral containing *bn* vanishes according to Eq. (17.4c). The inte gral containing *an* will be zero e xcept when *m* = *n*, in which case it is *T*∕2, according to Eqs. (17.4e) and (17.4g). Thus,$$

\int_0^T f(t) \cos m\omega_0 t , dt = a_n \frac{T}{2}, \qquad \text{for } m = n

$$or$$

a_n = \frac{2}{T} \int_0^T f(t) \cos n\omega_0 t , dt \qquad (17.8)

$$In a similar vein, we obtain *bn* by multiplying both sides of Eq. (17.3) by sin *mω*0*t* and integrating over the period. The result is$$

b_n = \frac{2}{T} \int_0^T f(t) \sin n\omega_0 t , dt \qquad (17.9)

$$Be aware that because *f*(*t*) is periodic, it may be more convenient to carry the integrations above from −*T*∕2 to *T*∕2 or generally from *t*0 to *t*0 + *T* instead of 0 to *T*. The result will be the same. An alternative form of Eq. (17.3) is the *amplitude-phase* form$$

f(t) = a_0 + \sum_{n=1}^{\infty} A_n \cos(n\omega_0 t + \phi_n)

$$(17.10) We can use Eqs. (9.11) and (9.12) to relate Eq. (17.3) to Eq. (17.10), or we can apply the trigonometric identity$$

\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \tag{17.11}

$$to the ac terms in Eq. (17.10) so that$$

a_0 + \sum_{n=1}^{\infty} A_n \cos(n\omega_0 t + \phi_n) = a_0 + \sum_{n=1}^{\infty} (A_n \cos \phi_n) \cos n\omega_0 t

$$$$

-(A_n \sin \phi_n) \sin n\omega_0 t

$$(17.12) Equating the coef ficients of the series expansions in Eqs. (17.3) and (17.12) shows that$$

a_n = A_n \cos \phi_n, \qquad b_n = -A_n \sin \phi_n \tag{17.13a}

$$or$$

A_n = \sqrt{a_n^2 + b_n^2}, \qquad \phi_n = -\tan^{-1}\frac{b_n}{a_n}

$$(17.13b) To avoid any confusion in determining *n*, it may be better to relate the terms in complex form as$$

A_n / \underline{\phi_n} = a_n - jb_n \tag{17.14}

$$The convenience of this relationship will become evident in Section 17.6. The plot of the amplitude *An* of the harmonics v ersus *nω*0 is called the *amplitude spectrum* of *f*(*t*); the plot of the phase *n* v ersus *nω*0 is the *phase spectrum* of *f*(*t*). Both the amplitude and phase spectra form the *frequency spectrum* of *f*(*t*). The frequency spectrum of a signal consists of the plots of the amplitudes and phases of the harmonics versus frequency. Thus, the F ourier analysis is also a mathematical tool for finding the spectrum of a periodic signal. Section 17.6 will elaborate more on the spectrum of a signal. To evaluate the Fourier coefficients *a*0, *an*, and *bn*, we often need to apply the following integrals:$$

\int \cos at , dt = \frac{1}{a} \sin at \tag{17.15a}

$$$$

\int \sin at , dt = -\frac{1}{a} \cos at \tag{17.15b}

$$$$

\int t \cos at , dt = \frac{1}{a^2} \cos at + \frac{1}{a} t \sin at \tag{17.15c}

$$$$

\int t \sin at , dt = \frac{1}{a^2} \sin at - \frac{1}{a} , t \cos at \tag{17.15d}

It is also useful to kno w the v alues of the cosine, sine, and e xponential functions for inte gral multiples of *π*. These are given in Table 17.1, where *n* is an integer. For Example 17.1; a square wave. Determine the Fourier series of the waveform shown in Fig. 17.1. Obtain the amplitude and phase spectra. # **Solution:** The Fourier series is given by Eq. (17.3), namely,

f(t) = a_0 + \sum_{n=1}^{\infty} (a_n \cos n\omega_0 t + b_n \sin n\omega_0 t)

(17.1.1) The frequency spectrum is also known as the line spectrum in view of the discrete frequency components. # **TABLE 17.1** 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<br>2 | (−1)n∕2<br>,<br>n = even<br>0,<br>n = odd<br>{ | | sin ___ nπ<br>2 | (−1)(n−1)∕2<br>,<br>n = odd<br>0,<br>{<br>n = even | | e j2nπ | 1 | | e jnπ | (−1)n | | e jnπ∕2 | (−1)n∕2<br>,<br>n = even<br>n = odd<br>{<br>j(−1)(n−1)∕2<br>, | Our goal is to obtain the Fourier coefficients *a*0, *an*, and *bn* using Eqs. (17.6), (17.8), and (17.9). First, we describe the waveform as

f(t) = \begin{cases} 1, & 0 < t < 1 \ 0, & 1 < t < 2 \end{cases} \tag{17.1.2}

$$and *f*(*t*) = *f*(*t* + *T*). Because *T* = 2, *ω*0 = 2*π*∕*T* = *π*. Thus,$$

a_0 = \frac{1}{T} \int_0^T f(t) dt = \frac{1}{2} \left[ \int_0^1 1 dt + \int_1^2 0 dt \right] = \frac{1}{2} t \Big|_0^1 = \frac{1}{2}

$$(17.1.3) Using Eq. (17.8) along with Eq. (17.15a),$$

a_n = \frac{2}{T} \int_0^T f(t) \cos n\omega_0 t , dt

= $\frac{2}{2} \left[ \int_0^1 1 \cos n\pi t \, dt + \int_1^2 0 \cos n\pi t \, dt \right]$ = $\frac{1}{n\pi} \sin n\pi t \Big|_0^1 = \frac{1}{n\pi} [\sin n\pi - \sin(0)] = 0$ (17.1.4) From Eq. (17.9) with the aid of Eq. (17.15b),

b_n = \frac{2}{T} \int_0^T f(t) \sin n\omega_0 t , dt

= $\frac{2}{2} \left[ \int_0^1 1 \sin n\pi t \, dt + \int_1^2 0 \sin n\pi t \, dt \right]$ = $-\frac{1}{n\pi} \cos n\pi t \Big|_0^1$ (17.1.5) = $-\frac{1}{n\pi} (\cos n\pi - 1)$ , $\cos n\pi = (-1)^n$ = $\frac{1}{n\pi} [1 - (-1)^n] = \begin{cases} \frac{2}{n\pi}, & n = \text{odd} \\ 0, & n = \text{even} \end{cases}$ Substituting the Fourier coefficients in Eqs. (17.1.3) to (17.1.5) into Eq. (17.1.1) gives the Fourier series as

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 + \dots

$$(17.1.6) Given that *f*(*t*) contains only the dc component and the sine terms with the fundamental component and odd harmonics, it may be written as$$

f(t) = \frac{1}{2} + \frac{2}{\pi} \sum_{k=1}^{\infty} \frac{1}{n} \sin n\pi t, \qquad n = 2k - 1

(17.1.7) By summing the terms one by one as demonstrated in Fig. 17.2, we notice how superposition of the terms can evolve into the original square. As more and more F ourier 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. (17.1.6) or (17.1.7) to infinity. Only a partial sum (*n* = 1, 2, 3, … , *N*, where *N* is finite) is possible. If we plot the partial sum (or truncated series) o ver one period for a lar ge *N* as in Sum of first three ac components # **Figure 17.2** Evolution of a square wave from its Fourier components. Summing the Fourier terms by hand calculation may be tedious. A computer is helpful to compute the terms and plot the sum like those shown in Fig. 17.2. # **Figure 17.4** For Example 17.1: (a) amplitude and (b) phase spectrum of the function shown in Fig. 17.1. Truncating the Fourier series at *N* = 11; Gibbs phenomenon. Fig. 17.3, we notice that the partial sum oscillates abo ve and below the actual value of *f*(*t*). At the neighborhood of the points of discontinuity (*x* = 0, 1, 2, …), there is o vershoot and damped oscillation. In f act, 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. 17.1. Since *an* = 0,

A_n = \sqrt{a_n^2 + b_n^2} = |b_n| = \begin{cases} \frac{2}{n\pi}, & n = \text{odd} \ 0, & n = \text{even} \end{cases}

$$(17.1.8) and$$

\phi_n = -\tan^{-1} \frac{b_n}{a_n} = \begin{cases} -90^\circ, & n = \text{odd} \ 0, & n = \text{even} \end{cases}

(17.1.9) The plots of *An* and *n* for different values of *nω*0 = *nπ* provide the amplitude and phase spectra in Fig. 17.4. Notice that the amplitudes of the harmonics decay very fast with frequency. **Figure 17.5** For Practice Prob. 17.1. Find the Fourier series of the square wave in Fig. 17.5. Plot the ampli tude and phase spectra. # **Figure 17.6** the amplitude and phase spectra. # **Solution:** The function is described as

f(t) = \begin{cases} t, & 0 < t < 1 \ 0, & 1 < t < 2 \end{cases}

$$Because *T* = 2, *ω*0 = 2*π*∕*T* = *π*. Then$$

a_0 = \frac{1}{T} \int_0^T f(t) , dt = \frac{1}{2} \left[ \int_0^1 t , dt + \int_1^2 0 , dt \right] = \frac{1}{2} \frac{t^2}{2} \Big|_0^1 = \frac{1}{4} \quad (17.2.1)

$$To evaluate *an* and *bn*, we need the integrals in Eq. (17.15):$$

a_n = \frac{2}{T} \int_0^T f(t) \cos n\omega_0 t , dt

= $\frac{2}{2} \left[ \int_0^1 t \cos n\pi t \, dt + \int_1^2 0 \cos n\pi t \, dt \right]$ = $\left[ \frac{1}{n^2 \pi^2} \cos n\pi t + \frac{t}{n\pi} \sin n\pi t \right]_0^1$ = $\frac{1}{n^2 \pi^2} (\cos n\pi - 1) + 0 = \frac{(-1)^n - 1}{n^2 \pi^2}$ (17.2.2) since cos *nπ* = (−1)*ⁿ* ; and

b_n = \frac{2}{T} \int_0^T f(t) \sin n\omega_0 t , dt

= $\frac{2}{2} \left[ \int_0^1 t \sin n\pi t \, dt + \int_1^2 0 \sin n\pi t \, dt \right]$ = $\left[ \frac{1}{n^2 \pi^2} \sin n\pi t - \frac{t}{n\pi} \cos n\pi t \right]_0^1$ = $0 - \frac{\cos n\pi}{n\pi} = \frac{(-1)^{n+1}}{n\pi}$ (17.2.3) Substituting the Fourier coefficients just found into Eq. (17.3) yields

f(t) = \frac{1}{4} + \sum_{n=1}^{\infty} \left[ \frac{[(-1)^n - 1]}{(n\pi)^2} \cos n\pi t + \frac{(-1)^{n+1}}{n\pi} \sin n\pi t \right]

$$To obtain the amplitude and phase spectra, we notice that, for e ven harmonics, *an* = 0, *bn* = −1∕*nπ*, so that$$

A_n / \underline{\phi_n} = a_n - jb_n = 0 + j\frac{1}{n\pi} \tag{17.2.4}

$$Hence,$$

A_n = |b_n| = \frac{1}{n\pi}, \qquad n = 2, 4, …

$$\n$$

\phi_n = 90^\circ, \qquad n = 2, 4, …

$$\n(17.2.5) <span id="page-788-0"></span>For odd harmonics, *an* = −2∕(*n*² *π*2 ),*bn* = 1∕(*nπ*) so that$$

A_n / \underline{\phi_n} = a_n - jb_n = -\frac{2}{n^2 \pi^2} - j \frac{1}{n \pi}

$$(17.2.6) That is,$$

A_n = \sqrt{a_n^2 + b_n^2} = \sqrt{\frac{4}{n^4 \pi^4} + \frac{1}{n^2 \pi^2}}

= $\frac{1}{n^2 \pi^2} \sqrt{4 + n^2 \pi^2}$ , $n = 1, 3, ...$ (17.2.7) From Eq. (17.2.6), we observe that lies in the third quadrant, so that

\phi_n = 180^\circ + \tan^{-1} \frac{n\pi}{2}, \qquad n = 1, 3, … \tag{17.2.8}

From Eqs. (17.2.5), (17.2.7), and (17.2.8), we plot *An* and *n* for different values of *nω*0 = *nπ* to obtain the amplitude spectrum and phase spectrum as shown in Fig. 17.8. **Figure 17.8** For Example 17.2: (a) amplitude spectrum, (b) phase spectrum. # Practice Problem 17.2 **Figure 17.9** For Practice Prob. 17.2. Determine the Fourier series of the sawtooth waveform in Fig. 17.9. **Answer:**

f(t) = 4.5 - \frac{9}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} \sin 2 \pi nt

$$.$$