6.1 PERIODIC SIGNAL [REPRESENTATION](#page-12-0) BY TRIGONOMETRIC FOURIER SERIES

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6.1 PERIODIC SIGNAL REPRESENTATION BY TRIGONOMETRIC FOURIER SERIES

As seen in Sec. 1.3-3 [Eq. (1.7)], a periodic signal x(t) with period T⁰ (Fig. 6.1) has the property

$$x(t) = x(t + T_0) \qquad \text{for all } t$$

The smallest value of T⁰ that satisfies this periodicity condition is the fundamental period of x(t). As argued in Sec. 1.3-3, this equation implies that x(t) starts at −∞ and continues to ∞. Moreover, the area under a periodic signal x(t) over any interval of duration T⁰ is the same; that is, for any

Figure 6.1 A periodic signal of period T0.

real numbers a and b

$$\int_{a}^{a+T_0} x(t) dt = \int_{b}^{b+T_0} x(t) dt$$

This result follows from the fact that a periodic signal takes the same values at intervals of T0. Hence, the values over any segment of duration T⁰ are repeated in any other interval of the same duration. For convenience, the area under x(t) over any interval of duration T⁰ will be denoted by

$$\int_{T_0} x(t) \, dt$$

The frequency of a sinusoid cos 2πf0t or sin 2πf0t is f0, and the period is T⁰ = 1/f0. These sinusoids can also be expressed as cosω0t or sinω0t, where ω⁰ = 2πf⁰ is the radian frequency, although for brevity, it is often referred to as frequency (see Sec. B.2). A sinusoid of frequency nf⁰ is said to be the nth harmonic of the sinusoid of frequency f0.

Let us consider a signal x(t) made up of a sines and cosines of frequency ω⁰ and all of its harmonics (including the zeroth harmonic; i.e., dc) with arbitrary amplitudes† :

$$x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + b_n \sin n\omega_0 t \tag{6.1}$$

The frequency ω⁰ is called the fundamental frequency.

We now prove an extremely important property: x(t) in Eq. (6.1) is a periodic signal with the same period as that of the fundamental, regardless of the values of the amplitudes an and bn. Note that the period T⁰ of the fundamental satisfies

$$T_0 = \frac{1}{f_0} = \frac{2\pi}{\omega_0} \quad \text{and} \quad \omega_0 T_0 = 2\pi \tag{6.2}$$

^† In Eq. (6.1), the constant term ^a0 corresponds to the cosine term for ^{n =} 0 because cos(^{0 × ω}0)^{t =} 1. However, sin(0×ω0)t = 0. Hence, the sine term for n = 0 is nonexistent.

To prove the periodicity of x(t), all we need is to show that x(t) = x(t +T0). From Eq. (6.1),

$$x(t+T_0) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 (t+T_0) + b_n \sin n\omega_0 (t+T_0)$$

= $a_0 + \sum_{n=1}^{\infty} a_n \cos(n\omega_0 t + n\omega_0 T_0) + b_n \sin(n\omega_0 t + n\omega_0 T_0)$

From Eq. (6.2), we have nω0T⁰ = 2πn, and

$$x(t + T_0) = a_0 + \sum_{n=1}^{\infty} a_n \cos(n\omega_0 t + 2\pi n) + b_n \sin(n\omega_0 t + 2\pi n)$$

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

We could also infer this result intuitively. In one fundamental period T0, the nth harmonic executes n complete cycles. Hence, every sinusoid on the right-hand side of Eq. (6.1) executes a complete number of cycles in one fundamental period T0. Therefore, at t = T0, every sinusoid starts as if it were the origin and repeats the same drama over the next T⁰ seconds, and so on, ad infinitum. Hence, the sum of such harmonics results in a periodic signal of period T0.

This result shows that any combination of sinusoids of frequencies 0, f0, 2f0, …, kf⁰ is a periodic signal of period T⁰ = 1/f⁰ regardless of the values of amplitudes ak and bk of these sinusoids. By changing the values of ak and bk in Eq. (6.1), we can construct a variety of periodic signals, all of the same period T⁰ (T⁰ = 1/f⁰ = 2π/ω0).

The converse of this result is also true. We shall show in Sec. 6.5-4 that a periodic signal x(t) with a period T⁰ can be expressed as a sum of a sinusoid of frequency f⁰ (f⁰ = 1/T0*) and all its harmonics, as shown in Eq. (6.1)*. † The infinite series on the right-hand side of Eq. (6.1) is known as the trigonometric Fourier series of a periodic signal x(t).

COMPUTING THE COEFFICIENTS OF A FOURIER SERIES

To determine the coefficients of a Fourier series, consider an integral I defined by

$$I = \int_{T_0} \cos n\omega_0 t \cos m\omega_0 t \, dt$$

where $ ^T0 stands for integration over any contiguous interval of T⁰ seconds. By using a trigonometric identity (see Sec. B.8-6), this integral can be expressed as

$$I = \frac{1}{2} \left[ \int_{T_0} \cos(n+m)\omega_0 t \, dt + \int_{T_0} \cos(n-m)\omega_0 t \, dt \right] \tag{6.3}$$

^† Strictly speaking, this statement applies only if a periodic signal x(t) is a continuous function of t. However, Sec. 6.5-4 shows that it can be applied even for discontinuous signals, if we interpret the equality in Eq. (6.1) in the mean-square sense instead of in the ordinary sense. This means that the power of the difference between the periodic signal x(t) and its Fourier series on the right-hand side of Eq. (6.1) approaches zero as the number of terms in the series approaches infinity.

596 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES

Because cos ω0t executes one complete cycle during any interval of duration T0, cos(n + m)ω0t executes (n + m) complete cycles during any interval of duration T0. Therefore, the first integral in Eq. (6.3), which represents the area under n+m complete cycles of a sinusoid, equals zero. The same argument shows that the second integral in Eq. (6.3) is also zero, except when n = m. Hence, I in Eq. (6.3) is zero for all n = m. When n = m, the first integral in Eq. (6.3) is still zero, but the second integral yields

$$I = \frac{1}{2} \int_{T_0} dt = \frac{T_0}{2}$$

Thus,

$$\int_{T_0} \cos n\omega_0 t \cos m\omega_0 t \, dt = \begin{cases} 0 & n \neq m \\ \frac{T_0}{2} & m = n \neq 0 \end{cases} \tag{6.4}$$

Using similar arguments, we can show that

$$\int_{T_0} \sin n\omega_0 t \sin m\omega_0 t \, dt = \begin{cases} 0 & n \neq m \\ \frac{T_0}{2} & n = m \neq 0 \end{cases} \tag{6.5}$$

and

$$\int_{T_0} \sin n\omega_0 t \cos m\omega_0 t \, dt = 0 \qquad \text{for all } n \text{ and } m \tag{6.6}$$

To determine a⁰ in Eq. (6.1), we integrate both sides of Eq. (6.1) over one period T⁰ to yield

$$\int_{T_0} x(t) dt = a_0 \int_{T_0} dt + \sum_{n=1}^{\infty} \left[ a_n \int_{T_0} \cos n\omega_0 t dt + b_n \int_{T_0} \sin n\omega_0 t dt \right]$$

Recall that T⁰ is the period of a sinusoid of frequency ω0. Therefore, functions cos nω0t and sin nω0t execute n complete cycles over any interval of T⁰ seconds so that the area under these functions over an interval T⁰ is zero, and the last two integrals on the right-hand side of the foregoing equation are zero. This yields

$$\int_{T_0} x(t) dt = a_0 \int_{T_0} dt = a_0 T_0 \quad \text{and} \quad a_0 = \frac{1}{T_0} \int_{T_0} x(t) dt$$

Next we multiply both sides of Eq. (6.1) by cosmω0t and integrate the resulting equation over an interval T0:

$$\int_{T_0} x(t) \cos m\omega_0 t \, dt = a_0 \int_{T_0} \cos m\omega_0 t \, dt + \sum_{n=1}^{\infty} \left[ a_n \int_{T_0} \cos n\omega_0 t \cos m\omega_0 t \, dt + b_n \int_{T_0} \sin n\omega_0 t \cos m\omega_0 t \, dt \right]$$

The first integral on the right-hand side is zero because it is an area under m integral number of cycles of a sinusoid. Also, the last integral on the right-hand side vanishes because of Eq. (6.6). This leaves only the middle integral, which is also zero for all n = m because of Eq. (6.4). But n takes on all values from 1 to ∞, including m. When n = m, this integral is T0/2, according to Eq. (6.4). Therefore, from the infinite number of terms on the right-hand side, only one term survives to yield anT0/2 = amT0/2 (recall that n = m). Therefore,

$$\int_{T_0} x(t) \cos m\omega_0 t \, dt = \frac{a_m T_0}{2} \qquad \text{and} \qquad a_m = \frac{2}{T_0} \int_{T_0} x(t) \cos m\omega_0 t \, dt$$

Similarly, by multiplying both sides of Eq. (6.1) by sin nω0t and then integrating over an interval T0, we obtain

$$b_m = \frac{2}{T_0} \int_{T_0} x(t) \sin m\omega_0 t \, dt$$

To sum up our discussion, which applies to real or complex x(t), we have shown that a periodic signal x(t) with period T⁰ can be expressed as a sum of a sinusoid of period T⁰ and its harmonics:

$$x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + b_n \sin n\omega_0 t \tag{6.7}$$

where ^{ω0 = 2}πf^{0 = 2π T0} and

$$a_0 = \frac{1}{T_0} \int_{T_0} x(t) dt, \quad a_n = \frac{2}{T_0} \int_{T_0} x(t) \cos n\omega_0 t dt, \quad \text{and} \quad b_n = \frac{2}{T_0} \int_{T_0} x(t) \sin n\omega_0 t dt \tag{6.8}$$

COMPACT FORM OF FOURIER SERIES

The results derived so far are general and apply whether x(t) is a real or a complex function of t. However, when x(t) is real, coefficients an and bn are real for all n, and the trigonometric Fourier series can be expressed in a compact form, using the results in Eq. (B.16):

$$x(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos(n\omega_0 t + \theta_n)$$

\n(6.9)

where Cn and θ*ⁿ* are related to an and bn, as [see Eq. (B.17)]

$$C_0 = a_0$$

, $C_n = \sqrt{a_n^2 + b_n^2}$ , and $\theta_n = \tan^{-1} \left( \frac{-b_n}{a_n} \right)$ (6.10)

These results are summarized in Table 6.1.

The compact form in Eq. (6.9) uses the cosine form. We could just as well have used the sine form, with terms sin(nω0t + θn) instead of cos(nω0t + θn). The literature overwhelmingly favors the cosine form, for no apparent reason except possibly that the cosine phasor is represented by the horizontal axis, which happens to be the reference axis in phasor representation.

Equation (6.8) shows that a⁰ (or C0) is the average value of x(t) (averaged over one period). This value can often be determined by inspection of x(t).

Because an and bn are real, Cn and θ*ⁿ* are also real. In the following discussion of trigonometric Fourier series, we shall assume real x(t), unless mentioned otherwise.

Series Form	Coefficient Computation	Conversion Formulas
Trigonometric	# 1 a0 f(t)dt = T0 T0	a0 = C0 = D0
“∞ f(t) = a0+ an cosnω0t+bn sinnω0t n=1	# 2 an f(t) cosnω0t dt = T0 T0	= Cnejθn = an−jbn 2Dn
	# 2 bn f(t)sinnω0t dt = T0	= Cne−jθn = an+jbn 2D−n
Compact trigonometric	T0 C0 = a0	C0 = D0
“∞ f(t) = C0 Cn cos(nω0t +θn) +	= 2 2 +bn Cn an	Cn = 2 Dn n ≥ 1
n=1	−bn = tan−1 θn an	θn = Dn
Exponential
f(t) = “∞ Dnejnω0t n=−∞	# 1 f(t)e−jnω0t Dn dt = T0 T0

TABLE 6.1 Fourier Series Representation of a Periodic Signal of Period T⁰ (ω⁰ = 2π/T0)

6.1-1 The Fourier Spectrum

The compact trigonometric Fourier series in Eq. (6.9) indicates that a periodic signal x(t) can be expressed as a sum of sinusoids of frequencies 0 (dc), ω0, 2ω0, …, nω0, …, whose amplitudes are C0, C1, C2, …, Cn, …, and whose phases are 0, θ1, θ2, …, θn, …, respectively. We can readily plot amplitude Cn versus n (the amplitude spectrum) and θ*ⁿ* versus n (the phase spectrum).† Because n is proportional to the frequency nω0, these plots are scaled plots of Cn versus ω and θ*ⁿ* versus ω. The two plots together are the frequency spectra of x(t). These spectra show at a glance the frequency contents of the signal x(t) with their amplitudes and phases. Knowing these spectra, we can reconstruct or synthesize the signal x(t) according to Eq. (6.9). Therefore, frequency spectra, which are an alternative way of describing a periodic signal x(t), are in every way equivalent to the plot of x(t) as a function of t. The frequency spectra of a signal constitute the frequency-domain description of x(t), in contrast to the time-domain description, where x(t) is specified as a function of time.

In computing θn, the phase of the nth harmonic from Eq. (6.10), the quadrant in which θ*ⁿ* lies should be determined from the signs of an and bn. For example, if an = −1 and bn = 1, θ*ⁿ* lies in the third quadrant, and

$$\theta_n = \tan^{-1}\left(\frac{-1}{-1}\right) = -135^\circ$$

Observe that

$$\tan^{-1}\left(\frac{-1}{-1}\right) \neq \tan^{-1}(1) = 45^{\circ}$$

^† The amplitude Cn, by definition here, is nonnegative. Some authors define amplitude An that can take positive or negative values and magnitude Cn = |An| that can only be nonnegative. Thus, what we call amplitude spectrum becomes magnitude spectrum. The distinction between amplitude and magnitude, although useful, is avoided in this book in the interest of keeping definitions of essentially similar entities to a minimum.

Although Cn, the amplitude of the nth harmonic as defined in Eq. (6.10), is positive, we shall find it convenient to allow Cn to take on negative values when bn = 0. This will become clear in later examples.

EXAMPLE 6.1 Compact Trigonometric Fourier Series of Periodic Exponential Wave

Find the compact trigonometric Fourier series for the periodic signal x(t) shown in Fig. 6.2a. Sketch the amplitude and phase spectra for x(t).

Figure 6.2 (a) A periodic signal and (b, c) its Fourier spectra.

In this case the period T⁰ = π and the fundamental frequency f⁰ = 1/T⁰ = 1/π Hz, and

$$\omega_0 = \frac{2\pi}{T_0} = 2 \,\text{rad/s}$$

Therefore,

$$x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos 2nt + b_n \sin 2nt$$

where

$$a_0 = \frac{1}{\pi} \int_{T_0} x(t) dt$$

In this example the obvious choice for the interval of integration is from 0 to π. Hence,

$$a_0 = \frac{1}{\pi} \int_0^{\pi} e^{-t/2} dt = 0.504$$ $$a_n = \frac{2}{\pi} \int_0^{\pi} e^{-t/2} \cos 2nt dt = 0.504 \left(\frac{2}{1 + 16n^2}\right)$$

and

$$b_n = \frac{2}{\pi} \int_0^{\pi} e^{-t/2} \sin 2nt \, dt = 0.504 \left( \frac{8n}{1 + 16n^2} \right)$$

Therefore,

$$x(t) = 0.504 \left[ 1 + \sum_{n=1}^{\infty} \frac{2}{1 + 16n^2} (\cos 2nt + 4n \sin 2nt) \right]$$

Also from Eq. (6.10),

$$C_0 = a_0 = 0.504$$

\n

$$C_n = \sqrt{a_n^2 + b_n^2} = 0.504 \sqrt{\frac{4}{(1 + 16n^2)^2} + \frac{64n^2}{(1 + 16n^2)^2}} = 0.504 \left(\frac{2}{\sqrt{1 + 16n^2}}\right)$$

\n

$$\theta_n = \tan^{-1}\left(\frac{-b_n}{a_n}\right) = \tan^{-1}(-4n) = -\tan^{-1} 4n$$

Amplitude and phases of the dc and the first seven harmonics are computed from the above equations as

n	0	1	2	3	4	5	6	7
Cn	0.504	0.244	0.125	0.084	0.063	0.0504	0.042	0.036
θn	0◦	−75.96◦	−82.87◦	−85.24◦	−86.42◦	−87.14◦	−87.61◦	−87.95◦

We can use these numerical values to express x(t) as

$$x(t) = 0.504 + 0.504 \sum_{n=1}^{\infty} \frac{2}{\sqrt{1 + 16n^2}} \cos(2nt - \tan^{-1} 4n)$$

= 0.504 + 0.244 cos (2t - 75.96°) + 0.125 cos (4t - 82.87°)

• 0.084 cos (6t - 85.24°) + 0.063 cos (8t - 86.42°) + … (6.11)

PLOTTING FOURIER SERIES SPECTRA USING MATLAB

MATLAB is well suited to compute and plot Fourier series spectra. The results in Fig. 6.3, which plot Cn and θ*ⁿ* as functions of n, match Figs. 6.2b and 6.2c, which plot Cn and θ*ⁿ* as functions of ω = nω⁰ = 2n. Plots of an and bn are similarly simple to generate.

The amplitude and phase spectra for x(t), in Figs. 6.2b and 6.2c, tell us at a glance the frequency composition of x(t), that is, the amplitudes and phases of various sinusoidal components of x(t). Knowing the frequency spectra, we can reconstruct x(t), as shown on the right-hand side of Eq. (6.11). Therefore the frequency spectra (Figs. 6.2b, 6.2c) provide an alternative description—the frequency-domain description of x(t). The time-domain description of x(t) is shown in Fig. 6.2a. A signal, therefore, has a dual identity: the time-domain identity x(t) and the frequency-domain identity (Fourier spectra). The two identities complement each other; taken together, they provide a better understanding of a signal.

An interesting aspect of Fourier series is that whenever there is a jump discontinuity in x(t), the series at the point of discontinuity converges to an average of the left-hand and right-hand limits of x(t) at the instant of discontinuity.† In the present example, for instance, x(t) is discontinuous at t = 0 with x(0+) = 1 and x(0−) = x(π ) = e−π/² = 0.208. The corresponding Fourier series converges to a value (1 + 0.208)/2 = 0.604 at t = 0. This is easily verified from Eq. (6.11) by setting t = 0.

^† This behavior of the Fourier series is dictated by its convergence in the mean, discussed later in Secs. 6.2 and 6.5.

602 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES

EXAMPLE 6.2 Compact Trigonometric Fourier Series of a Periodic Triangle Wave

Find the compact trigonometric Fourier series for the triangular periodic signal x(t) shown in Fig. 6.4a, and sketch the amplitude and phase spectra for x(t).

Figure 6.4 (a) A triangular periodic signal and (b, c) its Fourier spectra.

In this case the period T⁰ = 2. Hence,

$$\omega_0 = \frac{2\pi}{2} = \pi$$

and

$$x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\pi t + b_n \sin n\pi t$$

where

$$x(t) = \begin{cases} 2At & |t| < \frac{1}{2} \\ 2A(1-t) & \frac{1}{2} < t < \frac{3}{2} \end{cases}$$

Here it will be advantageous to choose the interval of integration from −1/2 to 3/2 rather than 0 to 2.

A glance at Fig. 6.4a shows that the average value (dc) of x(t) is zero so that a⁰ = 0. Also,

$$a_n = \frac{2}{2} \int_{-1/2}^{3/2} x(t) \cos n\pi t dt$$

=

$$\int_{-1/2}^{1/2} 2A t \cos n\pi t dt + \int_{1/2}^{3/2} 2A(1-t) \cos n\pi t dt$$

Detailed evaluation of these integrals shows that both have a value of zero. Therefore an = 0. Next,

$$b_n = \int_{-1/2}^{1/2} 2At \sin n\pi t \, dt + \int_{1/2}^{3/2} 2A(1-t) \sin n\pi t \, dt$$

Detailed evaluation of these integrals yields, in turn,

$$b_n = \frac{8A}{n^2 \pi^2} \sin\left(\frac{n\pi}{2}\right) = \begin{cases} 0 & n \text{ even} \\ \frac{8A}{n^2 \pi^2} & n = 1, 5, 9, 13, \dots \\ -\frac{8A}{n^2 \pi^2} & n = 3, 7, 11, 15, \dots \end{cases}$$

Therefore,

$$x(t) = \frac{8A}{\pi^2} \left[ \sin \pi t - \frac{1}{9} \sin 3\pi t + \frac{1}{25} \sin 5\pi t - \frac{1}{49} \sin 7\pi t + \dots \right]$$

(6.12)

To plot Fourier spectra, the series must be converted into compact trigonometric form as in Eq. (6.9). In this case this is readily done by converting sine terms into cosine terms with a suitable phase shift. For example,

$$\sin kt = \cos (kt - 90^\circ) \qquad \text{and} \qquad -\sin kt = \cos (kt + 90^\circ)$$

By using these identities, Eq. (6.12) can be expressed as

$$x(t) = \frac{8A}{\pi^2} \bigg[ \cos(\pi t - 90^\circ) + \frac{1}{9} \cos(3\pi t + 90^\circ) + \frac{1}{25} \cos(5\pi t - 90^\circ) + \frac{1}{49} \cos(7\pi t + 90^\circ) + \cdots \bigg]$$

In this series all the even harmonics are missing. The phases of the odd harmonics alternate from −90◦ to 90◦. Figure 6.4 shows amplitude and phase spectra for x(t).

EXAMPLE 6.3 Converting a Trigonometric FS to a Compact Trigonometric FS

A periodic signal x(t) is represented by a trigonometric Fourier series

$$x(t) = 2 + 3\cos 2t + 4\sin 2t + 2\sin (3t + 30^\circ) - \cos (7t + 150^\circ)$$

Express this series as a compact trigonometric Fourier series, and sketch amplitude and phase spectra for x(t).

In compact trigonometric Fourier series, the sine and cosine terms of the same frequency are combined into a single term and all terms are expressed as cosine terms with positive amplitudes. Using Eqs. (6.9) and (6.10), we have

$$3\cos 2t + 4\sin 2t = 5\cos (2t - 53.13^{\circ})$$

Also,

$$\sin(3t + 30^{\circ}) = \cos(3t + 30^{\circ} - 90^{\circ}) = \cos(3t - 60^{\circ})$$

and

$$-\cos(7t + 150^{\circ}) = \cos(7t + 150^{\circ} - 180^{\circ}) = \cos(7t - 30^{\circ})$$

Therefore,

$$x(t) = 2 + 5\cos(2t - 53.13^{\circ}) + 2\cos(3t - 60^{\circ}) + \cos(7t - 30^{\circ})$$

Figure 6.5 Fourier spectra of the signal.

In this case only four components (including dc) are present. The amplitude of dc is 2. The remaining three components are of frequencies ω = 2, 3, and 7 with amplitudes 5, 2, and 1 and phases −53.13◦, −60◦, and −30◦, respectively. The amplitude and phase spectra for this signal are shown in Figs. 6.5a and 6.5b, respectively.

EXAMPLE 6.4 Compact Trigonometric Fourier Series of a Periodic Square Wave

Find the compact trigonometric Fourier series for the square-pulse periodic signal shown in Fig. 6.6a and sketch its Fourier spectrum.

Figure 6.6 (a) A square pulse periodic signal and (b) its Fourier spectrum.

Here the period is T⁰ = 2π and ω⁰ = 2π/T⁰ = 1. Therefore,

$$x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos nt + b_n \sin nt$$

where

$$a_0 = \frac{1}{T_0} \int_{T_0} x(t) dt$$

From Fig. 6.6a, it is clear that a proper choice of region of integration is from −π to π. But since x(t) = 1 only over (−π/2, π/2), and x(t) = 0 over the remaining segment,

$$a_0 = \frac{1}{2\pi} \int_{-\pi/2}^{\pi/2} dt = \frac{1}{2}$$

We could have found a0, the average value of x(t), to be 1/2 merely by inspection of x(t) in Fig. 6.6a. Also,

$$a_n = \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \cos nt \, dt = \frac{2}{n\pi} \sin\left(\frac{n\pi}{2}\right)$$

=

$$\begin{cases} 0 & n \text{ even} \\ \frac{2}{\pi n} & n = 1, 5, 9, 13, \dots \\ -\frac{2}{\pi n} & n = 3, 7, 11, 15, \dots \end{cases}$$ $$b_n = \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \sin nt \, dt = 0$$

Therefore

$$x(t) = \frac{1}{2} + \frac{2}{\pi} \left( \cos t - \frac{1}{3} \cos 3t + \frac{1}{5} \cos 5t - \frac{1}{7} \cos 7t + \cdots \right)$$

(6.13)

Observe that bn = 0 and all the sine terms are zero. Only the cosine terms appear in the trigonometric series. The series is therefore already in the compact form except that the amplitudes of alternating harmonics are negative. Now by definition, amplitudes Cn are positive [see Eq. (6.10)]. The negative sign can be accommodated by associating a proper phase, as seen from the trigonometric identity†

−cos x = cos(x −π )

Using this fact, we can express the series in Eq. (6.13) as

$$x(t) = \frac{1}{2} + \frac{2}{\pi} \left[ \cos \omega_0 t + \frac{1}{3} \cos (3\omega_0 t - \pi) + \frac{1}{5} \cos 5\omega_0 t + \frac{1}{7} \cos (7\omega_0 t - \pi) + \frac{1}{9} \cos 9\omega_0 t + \cdots \right]$$

^† Because cos(x±π ) = −cos ^x, we could have chosen the phase ^π or ⁻π. In fact, cos(x±Nπ ) = −cos ^x for any odd integral value of N. Therefore the phase can be chosen as ±Nπ, where N is any convenient odd integer.

This is the desired form of the compact trigonometric Fourier series. The amplitudes are

$$C_0 = \frac{1}{2}$$

and $C_n = \begin{cases} 0 & n \text{ even} \\ \frac{2}{\pi n} & n \text{ odd} \end{cases}$

The phases are

$$\theta_n = \begin{cases} 0 & \text{for all } n \neq 3, 7, 11, 15, \dots \\ -\pi & n = 3, 7, 11, 15, \dots \end{cases}$$

We might use these values to plot amplitude and phase spectra. However, we can simplify our task in this special case if we allow amplitude Cn to take on negative values. If this is allowed, we do not need a phase of −π to account for the sign as seen from Eq. (6.13). This means that phases of all components are zero, and we can discard the phase spectrum and manage with only the amplitude spectrum, as shown in Fig. 6.6b. Observe that there is no loss of information in doing so and that the amplitude spectrum in Fig. 6.6b has the complete information about the Fourier series in Eq. (6.13). Therefore, whenever all sine terms vanish (bn = 0*), it is convenient to allow Cn to take on negative values*. This permits the spectral information to be conveyed by a single spectrum.†

Let us investigate the behavior of the series at the points of discontinuities. For the discontinuity at t =π/2, the values of x(t) on either sides of the discontinuity are x((π/2)−)=1 and x((π/2)+) = 0. We can verify by setting t = π/2 in Eq. (6.13) that x(π/2) = 0.5, which is a value midway between the values of x(t) on either side of the discontinuity at t = π/2.

6.1-2 The Effect of Symmetry

The Fourier series for the signal x(t) in Fig. 6.2a (Ex. 6.1) consists of sine and cosine terms, but the series for the signal x(t) in Fig. 6.4a (Ex. 6.2) consists of sine terms only, and the series for the signal x(t) in Fig. 6.6a (Ex. 6.4) consists of cosine terms only. This is no accident. We can show that the Fourier series of any even periodic function x(t) consists of cosine terms only and the series for any odd periodic function x(t) consists of sine terms only. Moreover, because of symmetry (even or odd), the information of one period of x(t) is implicit in only half the period, as seen in Figs. 6.4a and 6.6a. In these cases, knowing the signal over a half-period and knowing the kind of symmetry (even or odd), we can determine the signal waveform over a complete period. For this reason, the Fourier coefficients in these cases can be computed by integrating over only half the period rather than a complete period. To prove this result, recall that

$$a_0 = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} x(t) dt, \quad a_n = \frac{2}{T_0} \int_{-T_0/2}^{T_0/2} x(t) \cos n\omega_0 t dt, \quad \text{and} \quad b_n = \frac{2}{T_0} \int_{-T_0/2}^{T_0/2} x(t) \sin n\omega_0 t dt$$

Recall also that cosnω0t is an even function and sinnω0t is an odd function of t. If x(t) is an even function of t, then x(t) cosnω0t is also an even function and x(t)sin nω0t is an odd function of t

^† Here, the distinction between amplitude An and magnitude Cn = |An^| would have been useful. But, for the reasons mentioned in the footnote on page 598, we refrain from this distinction formally.

608 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES

(see Sec. 1.5-1). Therefore, following from Eq. (1.16),

$$a_0 = \frac{2}{T_0} \int_0^{T_0/2} x(t) dt, \quad a_n = \frac{4}{T_0} \int_0^{T_0/2} x(t) \cos n\omega_0 t dt, \quad \text{and} \quad b_n = 0 \tag{6.14}$$

Similarly, if x(t) is an odd function of t, then x(t) cos nω0t is an odd function of t and x(t)sin nω0t is an even function of t. Therefore,

$$a_n = 0$$

and $b_n = \frac{4}{T_0} \int_0^{T_0/2} x(t) \sin n\omega_0 t dt$ (6.15)

Observe that because of symmetry, the integration required to compute the coefficients need be performed over only half the period.

If a periodic signal x(t) shifted by half the period remains unchanged except for a sign—that is, if

$$x\left(t - \frac{T_0}{2}\right) = -x(t)$$

then the signal is said to have a half-wave symmetry. It can be shown that for a signal with a half-wave symmetry, all the even-numbered harmonics vanish (see Prob. 6.1-6). The signal in Fig. 6.4a is an example of such a symmetry. The signal in Fig. 6.6a also has this symmetry, although it is not obvious owing to a dc component. If we subtract the dc component of 0.5 from this signal, the remaining signal has half-wave symmetry. For this reason, this signal has only odd harmonics and a dc component of 0.5.

DR ILL 6.1 Compact Trigonometric Fourier Series

Find the compact trigonometric Fourier series for periodic signals shown in Fig. 6.7. Sketch their amplitude and phase spectra. Allow Cn to take on negative values if bn = 0 so that the phase spectrum can be eliminated. [Hint: Use Eqs. (6.14) and (6.15) for appropriate symmetry conditions.]

ANSWERS

(a)

$$x(t) = \frac{1}{3} - \frac{4}{\pi^2} \left( \cos \pi t - \frac{1}{4} \cos 2\pi t + \frac{1}{9} \cos 3\pi t - \frac{1}{16} \cos 4\pi t + \cdots \right)$$

\n

$$= \frac{1}{3} + \frac{4}{\pi^2} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos n\pi t$$

\n(b) $x(t) = \frac{2A}{\pi} \left[ \sin \pi t - \frac{1}{2} \sin 2\pi t + \frac{1}{3} \sin 3\pi t - \frac{1}{4} \sin 4\pi t + \cdots \right]$
\n

$$= \frac{2A}{\pi} \left[ \cos (\pi t - 90^\circ) + \frac{1}{2} \cos (2\pi t + 90^\circ) + \frac{1}{3} \cos (3\pi t - 90^\circ) + \frac{1}{4} \cos (4\pi t + 90^\circ) + \cdots \right]$$

6.1-3 Determining the Fundamental Frequency and Period

We have seen that every periodic signal can be expressed as a sum of sinusoids of a fundamental frequency ω⁰ and its harmonics. One may ask whether a sum of sinusoids of any frequencies represents a periodic signal. If so, how does one determine the period? Consider the following three functions:

$$x_1(t) = 2 + 7\cos(\frac{1}{2}t + \theta_1) + 3\cos(\frac{2}{3}t + \theta_2) + 5\cos(\frac{7}{6}t + \theta_3)$$

\n

$$x_2(t) = 2\cos(2t + \theta_1) + 5\sin(\pi t + \theta_2)$$

\n

$$x_3(t) = 3\sin(3\sqrt{2}t + \theta) + 7\cos(6\sqrt{2}t + \phi)$$

Recall that every frequency in a periodic signal is an integer multiple of the fundamental frequency ω0. Therefore the ratio of any two frequencies is of the form m/n, where m and n are integers. This means that the ratio of any two frequencies is a rational number. When the ratio of two frequencies is a rational number, the frequencies are said to be harmonically related.

The largest number of which all the frequencies are integer multiples is the fundamental frequency. In other words, the fundamental frequency is the greatest common factor (GCF) of all the frequencies in the series. The frequencies in the spectrum of x1(t) are 1/2, 2/3, and 7/6 (we do not consider dc). The ratios of the successive frequencies are 3:4 and 4:7, respectively. Because both these numbers are rational, all the three frequencies in the spectrum are harmonically related, and the signal x1(t) is periodic. The GCF, that is, the greatest number of which 1/2, 2/3, and 7/6 are integer multiples, is 1/6.† Moreover, 3(1/6) = 1/2, 4(1/6) = 2/3, and 7(1/6) = 7/6. Therefore the fundamental frequency is 1/6, and the three frequencies in the spectrum are the third, fourth, and seventh harmonics. Observe that the fundamental frequency component is absent in this Fourier series.

^† The greatest common factor of a1/b1, a2/b2, …, am/bm is the ratio of the GCF of the numerators set (a1,a2,…,am) to the LCM (least common multiple) of the denominator set (b1,b2,…,bm). For instance, for the set (2/3, 6/7, 2), the GCF of the numerator set (2, 6, 2) is 2; the LCM of the denominator set (3, 7, 1) is 21. Therefore, 2/21 is the largest number of which 2/3, 6/7, and 2 are integer multiples.

610 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES

The signal x2(t) is not periodic because the ratio of two frequencies in the spectrum is 2/π, which is not a rational number. The signal x3(t) is periodic because the ratio of frequencies 3 √ 2 and 6^√ 2 is 1/2, a rational number. The greatest common factor of 3√2 and 6√2 is 3√2. Therefore, the fundamental frequency ω⁰ = 3 ^√2, and the period

$$T_0 = \frac{2\pi}{(3\sqrt{2})} = \frac{\sqrt{2}}{3}\pi$$

DR ILL 6.2 Determining Periodicity, Fundamental Frequency, and Harmonic Content

Determine whether the signal

$$x(t) = \cos\left(\frac{2}{3}t + 30^{\circ}\right) + \sin\left(\frac{4}{5}t + 45^{\circ}\right)$$

is periodic. If it is periodic, find the fundamental frequency and the period. What harmonics are present in x(t)?

ANSWERS

Periodic with ω⁰ =2/15 and period T⁰ =15π. Signal x(t) contains the fifth and sixth harmonics.

A HISTORICAL NOTE: BARON JEAN-BAPTISTE-JOSEPH FOURIER (1768–1830)

The Fourier series and integral comprise a most beautiful and fruitful development, which serves as an indispensable instrument in the treatment of many problems in mathematics, science, and engineering. Maxwell was so taken by the beauty of the Fourier series that he called it a great mathematical poem. In electrical engineering, it is central to the areas of communication, signal processing, and several other fields, including antennas, but its initial reception by the scientific world was not enthusiastic. In fact, Fourier could not get his results published as a paper.

Fourier, a tailor’s son, was orphaned at age 8 and educated at a local military college (run by Benedictine monks), where he excelled in mathematics. The Benedictines prevailed upon the young genius to choose the priesthood as his vocation, but revolution broke out before he could take his vows. Fourier joined the people’s party. But in its early days, the French Revolution, like most such upheavals, liquidated a large segment of the intelligentsia, including prominent scientists such as Lavoisier. Observing this trend, many intellectuals decided to leave France to save themselves from a rapidly rising tide of barbarism. Fourier, despite his early enthusiasm for the Revolution, narrowly escaped the guillotine twice. It was to the everlasting credit of Napoleon that he stopped the persecution of the intelligentsia and founded new schools to replenish their ranks. The 26-year-old Fourier was appointed chair of mathematics at the newly created École Normale in 1794 [1].

Jean-Baptiste-Joseph Fourier and Napoleon

Napoleon was the first modern ruler with a scientific education, and he was one of the rare persons who are equally comfortable with soldiers and scientists. The age of Napoleon was one of the most fruitful in the history of science. Napoleon liked to sign himself as “member of Institut de France” (a fraternity of scientists), and he once expressed to Laplace his regret that “force of circumstances has led me so far from the career of a scientist” [2]. Many great figures in science and mathematics, including Fourier and Laplace, were honored and promoted by Napoleon. In 1798 he took a group of scientists, artists, and scholars—Fourier among them—on his Egyptian expedition, with the promise of an exciting and historic union of adventure and research. Fourier proved to be a capable administrator of the newly formed Institut d’Égypte, which, incidentally, was responsible for the discovery of the Rosetta Stone. The inscription on this stone in two languages and three scripts (hieroglyphic, demotic, and Greek) enabled Thomas Young and Jean-François Champollion, a protégé of Fourier, to invent a method of translating hieroglyphic writings of ancient Egypt—the only significant result of Napoleon’s Egyptian expedition.

Back in France in 1801, Fourier briefly served in his former position as professor of mathematics at the École Polytechnique in Paris. In 1802 Napoleon appointed him the prefect of Isère (with its headquarters in Grenoble), a position in which Fourier served with distinction. Fourier was named Baron of the Empire by Napoleon in 1809. Later, when Napoleon was exiled to Elba, his route was to take him through Grenoble. Fourier had the route changed to avoid meeting Napoleon, which would have displeased Fourier’s new master, King Louis XVIII. Within a year, Napoleon escaped from Elba and returned to France. At Grenoble, Fourier was brought before him in chains. Napoleon scolded Fourier for his ungrateful behavior but reappointed him the prefect of Rhône at Lyons. Within four months Napoleon was defeated at Waterloo and was exiled to St. Helena, where he died in 1821. Fourier once again was in disgrace as a Bonapartist and had to pawn his possessions to keep himself alive. But through the intercession of a former student, who was now a prefect of Paris, he was appointed director of the statistical bureau of the Seine, a position that allowed him ample time for scholarly pursuits. Later, in 1827, he was elected to the powerful position of perpetual secretary of the Paris Academy of Science, a section of the Institut de France [3].

While serving as the prefect of Grenoble, Fourier carried on his elaborate investigation of the propagation of heat in solid bodies, which led him to the Fourier series and the Fourier integral. On December 21, 1807, he announced these results in a prize paper on the theory of heat. Fourier claimed that an arbitrary function (continuous or with discontinuities) defined in a finite interval by an arbitrarily capricious graph can always be expressed as a sum of sinusoids (Fourier series). The judges, who included the great French mathematicians Laplace, Lagrange, Legendre, Monge, and LaCroix, admitted the novelty and importance of Fourier’s work but criticized it for lack of mathematical rigor and generality. Lagrange thought it incredible that a sum of sines and cosines could add up to anything but an infinitely differentiable function. Moreover, one of the properties of an infinitely differentiable function is that if we know its behavior over an arbitrarily small interval, we can determine its behavior over the entire range (the Taylor–Maclaurin series). Such a function is far from an arbitrary or a capriciously drawn graph [4]. Laplace had additional reason to criticize Fourier’s work. Laplace and his students had already approached the problem of heat conduction from a different angle, and Laplace was reluctant to accept the superiority of Fourier’s method [5]. Fourier thought the criticism unjustified but was unable to prove his claim because the tools required for operations with infinite series were not available at the time. However, posterity has proved Fourier to be closer to the truth than his critics. This is the classic conflict between pure mathematicians and physicists or engineers, as we saw earlier (Ch. 4) in the life of Oliver Heaviside. In 1829 Dirichlet proved Fourier’s claim concerning capriciously drawn functions with a few restrictions (Dirichlet conditions).

Although three of the four judges were in favor of publication, Fourier’s paper was rejected because of vehement opposition by Lagrange. Fifteen years later, after several attempts and disappointments, Fourier published the results in expanded form as a text, Théorie analytique de la chaleur, which is now a classic.