EXAMPLE 6.6 Exponential Fourier Series of Periodic Exponential Wave

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EXAMPLE 6.6 Exponential Fourier Series of Periodic Exponential Wave

Find the exponential Fourier series for the signal of Fig. 6.2a from Ex. 6.1.

In this case T⁰ = π, ω⁰ = 2π/T⁰ = 2, and

$$x(t) = \sum_{n = -\infty}^{\infty} D_n e^{j2nt}$$

where

$$D_n = \frac{1}{T_0} \int_{T_0} x(t) e^{-j2nt} dt = \frac{1}{\pi} \int_0^{\pi} e^{-t/2} e^{-j2nt} dt = \frac{1}{\pi} \int_0^{\pi} e^{-(1/2 + j2n)t} dt$$

= $\frac{-1}{\pi (\frac{1}{2} + j2n)} e^{-(1/2 + j2n)t} \Big|_0^{\pi} = \frac{0.504}{1 + j4n}$

$$x(t) = 0.504 \sum_{n=-\infty}^{\infty} \frac{1}{1+j4n} e^{j2nt}$$

= 0.504 $\left[ 1 + \frac{1}{1+j4} e^{j2t} + \frac{1}{1+j8} e^{j4t} + \frac{1}{1+j12} e^{j6t} + \cdots + \frac{1}{1-j4} e^{-j2t} + \frac{1}{1-j8} e^{-j4t} + \frac{1}{1-j12} e^{-j6t} + \cdots \right]$

Observe that the coefficients Dn are complex. Moreover, Dn and D−ⁿ are conjugates, as expected.

6.3-1 Exponential Fourier Spectra

In exponential spectra, we plot coefficients Dn as a function of ω. But since Dn is complex in general, we need both parts of one of two sets of plots: the real and the imaginary parts of Dn, or the magnitude and the angle of Dn. We prefer the latter because of its close connection to the amplitudes and phases of corresponding components of the trigonometric Fourier series. We therefore plot |Dn| versus ω and Dn versus ω. This requires that the coefficients Dn be expressed in polar form as |Dn|ej Dn , where |Dn| are the amplitudes and Dn are the angles of various exponential components. Equation (6.22) shows that for real x(t), the amplitude spectrum (|Dn| versus ω) is an even function of ω and the angle spectrum ( Dn versus ω) is an odd function of ω.

For the series in Ex. 6.6, for instance,

$$D_0 = 0.504$$

\n

$$D_1 = \frac{0.504}{1+j4} = 0.122e^{-j75.96^\circ} \implies |D_1| = 0.122, \angle D_1 = -75.96^\circ$$

\n

$$D_{-1} = \frac{0.504}{1-j4} = 0.122e^{j75.96^\circ} \implies |D_{-1}| = 0.122, \angle D_{-1} = 75.96^\circ$$

and

$$D_2 = \frac{0.504}{1+j8} = 0.0625e^{-j82.87^\circ} \implies |D_2| = 0.0625, \ \angle D_2 = -82.87^\circ$$ $$D_{-2} = \frac{0.504}{1-j8} = 0.0625e^{j82.87^\circ} \implies |D_{-2}| = 0.0625, \ \angle D_{-2} = 82.87^\circ$$

and so on. Note that Dn and D−ⁿ are conjugates, as expected [see Eq. (6.22)].

Figure 6.12 shows the frequency spectra (amplitude and angle) of the exponential Fourier series for the periodic signal x(t) in Fig. 6.2a.

We notice some interesting features of these spectra. First, the spectra exist for positive as well as negative values of ω (the frequency). Second, the amplitude spectrum is an even function of ω and the angle spectrum is an odd function of ω.

and

Figure 6.12 Exponential Fourier spectra for the signal in Fig. 6.2a.

At times it may appear that the phase spectrum of a real periodic signal fails to satisfy the odd symmetry: for example, when Dk = D−^k = −10. In this case, Dk = 10ej^π , and therefore, D−^k = 10e−j^π . Recall that e±j^π = −1. Here, although Dk = D−^k, their phases should be taken as π and −π.

EXAMPLE 6.7 Plotting Fourier Series Spectra with MATLAB

Using MATLAB and the results of Ex. 6.6, compute and plot the exponential Fourier spectra for the periodic signal x(t) shown in Fig. 6.2a. The result should match Fig. 6.12.

The expression for Dn is derived in Ex. 6.6.

>> clf; n = (-5:5); D_n = 0.504./(1+4j*n);

• subplot(1,2,1); stem(n,abs(D_n),‘.k’);

• xlabel(‘n’); ylabel(‘|D_n|’);

• subplot(1,2,2); stem(n,angle(D_n),‘.k’);

• xlabel(‘n’); ylabel(‘\angle D_n [rad]’);

Except that it is plotted as a function of n rather than ω, the result in Fig. 6.13 matches Fig. 6.12.

WHAT IS A NEGATIVE FREQUENCY?

The existence of the spectrum at negative frequencies is somewhat disturbing because, by definition, the frequency (number of repetitions per second) is a positive quantity. How do we interpret a negative frequency? We can use a trigonometric identity to express a sinusoid of a negative frequency −ω⁰ as

$$\cos(-\omega_0 t + \theta) = \cos(\omega_0 t - \theta)$$

This equation clearly shows that the frequency of a sinusoid cos(ω0t + θ ) is |ω0|, which is a positive quantity. The same conclusion is reached by observing that

$$e^{\pm j\omega_0 t} = \cos \omega_0 t \pm j \sin \omega_0 t$$

Thus, the frequency of exponentials e±jω0*^t* is indeed |ω0|. How do we then interpret the spectral plots for negative values of ω? A more satisfying way of looking at the situation is to say that exponential spectra are a graphical representation of coefficients Dn as a function of ω*. Existence of the spectrum at* ω = −nω⁰ is merely an indication that an exponential component e−jnω0*^t exists in the series*. We know that a sinusoid of frequency nω⁰ can be expressed in terms of a pair of exponentials ejn^ω0*^t* and e−jnω0*^t* .

We see a close connection between the exponential spectra in Fig. 6.12 and the spectra of the corresponding trigonometric Fourier series for x(t) (Figs. 6.2b, 6.2c). Equation (6.22) explains the reason for the close connection, for real x(t), between the trigonometric spectra (Cn and θn) with exponential spectra (|Dn| and Dn). The dc components D⁰ and C⁰ are identical in both spectra. Moreover, the exponential amplitude spectrum |Dn| is half the trigonometric amplitude spectrum Cn for n ≥ 1. The exponential angle spectrum Dn is identical to the trigonometric phase spectrum θ*ⁿ* for n ≥ 0. We can therefore produce the exponential spectra merely by inspection of trigonometric spectra, and vice versa. The following example demonstrates this feature.

EXAMPLE 6.8 Relating Exponential to Trigonometric Fourier Series Spectra

The trigonometric Fourier spectra of a certain periodic signal x(t) are shown in Fig. 6.14a. After inspecting these spectra, sketch the corresponding exponential Fourier spectra and verify your results analytically.

Figure 6.14 Fourier series spectra for Ex. 6.8.

The trigonometric spectral components exist at frequencies 0, 3, 6, and 9. The exponential spectral components exist at 0, 3, 6, 9, and −3, −6, −9. Consider first the amplitude spectrum. The dc component remains unchanged: that is, D⁰ = C⁰ = 16. Now |Dn| is an even function of ω and |Dn|=|D−ⁿ| = Cn/2. Thus, all the remaining spectrum |Dn| for positive n is half the trigonometric amplitude spectrum Cn, and the spectrum |Dn| for negative n is a reflection about the vertical axis of the spectrum for positive n, as shown in Fig. 6.14b.

The angle spectrum is Dn = θ*ⁿ* for positive n and is −θ*ⁿ* for negative n, as depicted in Fig. 6.14b. We shall now verify that both sets of spectra represent the same signal.

Signal x(t), whose trigonometric spectra are shown in Fig. 6.14a, has four spectral components of frequencies 0, 3, 6, and 9. The dc component is 16. The amplitude and the phase of the component of frequency 3 are 12 and −π/4, respectively. Therefore, this component can be expressed as 12cos(3t − π/4). Proceeding in this manner, we can write the Fourier series for x(t) as

$$x(t) = 16 + 12\cos\left(3t - \frac{\pi}{4}\right) + 8\cos\left(6t - \frac{\pi}{2}\right) + 4\cos\left(9t - \frac{\pi}{4}\right)$$

Consider now the exponential spectra in Fig. 6.14b. They contain components of frequencies 0 (dc), ±3, ±6, and ±9. The dc component is D⁰ = 16. The component ej3*^t* (frequency 3) has magnitude 6 and angle −π/4. Therefore, this component strength is 6e−jπ/4, and it can be expressed as (6e−jπ/4)ej3*^t* . Similarly, the component of frequency −3 is (6ejπ/4)e−j3*^t* . Proceeding in this manner, xˆ(t), the signal corresponding to the spectra in Fig. 6.14b, is

$$\hat{x}(t) = 16 + \left[6e^{-j\pi/4}e^{j3t} + 6e^{j\pi/4}e^{-j3t}\right] + \left[4e^{-j\pi/2}e^{j6t} + 4e^{j\pi/2}e^{-j6t}\right]$$

\n

$$+ \left[2e^{-j\pi/4}e^{j9t} + 2e^{j\pi/4}e^{-j9t}\right]$$

\n

$$= 16 + 6\left[e^{j(3t - \pi/4)} + e^{-j(3t - \pi/4)}\right] + 4\left[e^{j(6t - \pi/2)} + e^{-j(6t - \pi/2)}\right]$$

\n

$$+ 2\left[e^{j(9t - \pi/4)} + e^{-j(9t - \pi/4)}\right]$$

\n

$$= 16 + 12\cos\left(3t - \frac{\pi}{4}\right) + 8\cos\left(6t - \frac{\pi}{2}\right) + 4\cos\left(9t - \frac{\pi}{4}\right)$$

Clearly both sets of spectra represent the same periodic signal.

BANDWIDTH OF A SIGNAL

The difference between the highest and the lowest frequencies of the spectral components of a signal is the bandwidth of the signal. The bandwidth of the signal whose exponential spectra are shown in Fig. 6.14b is 9 (in radians). The highest and lowest frequencies are 9 and 0, respectively. Note that the component of frequency 12 has zero amplitude and is nonexistent. Moreover, the lowest frequency is 0, not −9. Recall that the frequencies (in the conventional sense) of the spectral components at ω = −3, −6, and −9 in reality are 3, 6, and 9.† The bandwidth can be more readily seen from the trigonometric spectra in Fig. 6.14a.

EXAMPLE 6.9 Fourier Series Spectra of an Impulse Train

Find the exponential Fourier series and sketch the corresponding spectra for the impulse train δ*^T0 (t) depicted in Fig. 6.15a. From this result, sketch the trigonometric spectrum and write the trigonometric Fourier series for δ^T*⁰ (t).

^† Some authors do define bandwidth as the difference between the highest and the lowest (negative) frequency in the exponential spectrum. The bandwidth according to this definition is twice that defined here. In reality, this phrasing defines not the signal bandwidth but the spectral width (width of the exponential spectrum of the signal).

The unit impulse train shown in Fig. 6.15a can be expressed as

$$\sum_{n=-\infty}^{\infty} \delta(t - nT_0)$$

Following Papoulis, we shall denote this function as δ*^T*⁰ (t) for the sake of notational brevity. The exponential Fourier series is given by

$$\delta_{T_0}(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0}$$

\n(6.23)

where

$$D_n = \frac{1}{T_0} \int_{T_0} \delta_{T_0}(t) e^{-jn\omega_0 t} dt$$

Choosing the interval of integration (−T0/2,T0/2) and recognizing that over this interval δ*^T*⁰ (t) = δ(t), we get

$$D_n = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} \delta(t) e^{-jn\omega_0 t} dt$$

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

In this integral, the impulse is located at t = 0. From the sampling property of Eq. (1.11), the integral on the right-hand side is the value of e−jnω0*^t* at t = 0 (where the impulse is located). Therefore,

$$D_n = \frac{1}{T_0} \tag{6.24}$$

From this result, we see that the exponential spectrum is constant for all frequencies, as shown in Fig. 6.15b. The spectrum, being real, requires only the amplitude plot. All phases are zero.

Substituting Dn ^{= 1 T0} into Eq. (6.23) yields the desired exponential Fourier series

$$\delta_{T_0}(t) = \frac{1}{T_0} \sum_{n = -\infty}^{\infty} e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0}$$

To sketch the trigonometric spectrum, we use Eq. (6.22) to obtain

$$C_0 = D_0 = \frac{1}{T_0}$$

\n

$$C_n = 2|D_n| = \frac{2}{T_0} \qquad n = 1, 2, 3, ...$$

\n

$$\theta_n = 0$$

Figure 6.15c shows the trigonometric Fourier spectrum. From this spectrum we can express δ*^T*⁰ (t) as

$$\delta_{T_0}(t) = \frac{1}{T_0} \left[ 1 + 2(\cos \omega_0 t + \cos 2\omega_0 t + \cos 3\omega_0 t + \cdots) \right] \qquad \omega_0 = \frac{2\pi}{T_0} \tag{6.25}$$

EFFECT OF SYMMETRY IN EXPONENTIAL FOURIER SERIES

When x(t) has an even symmetry, bn = 0, and from Eq. (6.20), Dn = an/2, which is real (positive or negative). Hence, Dn can only be 0 or ±π. Moreover, we may compute Dn = an/2 by using Eq. (6.14), which requires integration over a half-period only. Similarly, when x(t) has an odd symmetry, an = 0, and Dn = −jbn/2 is imaginary (positive or negative). Hence, Dn can only be 0 or ±π/2. Moreover, we may compute Dn = −jbn/2 by using Eq. (6.15), which requires integration over a half-period only. Note, however, that in the exponential case, we are using the symmetry property indirectly by finding the trigonometric coefficients. We cannot apply it directly in finding Dn from Eq. (6.19) since the function ejn^ω0*^t* is neither even nor odd.

DR ILL 6.4 Relating Trigonometric to Exponential Fourier Series Spectra

The exponential Fourier spectra of a certain periodic signal x(t) are shown in Fig. 6.16. Determine and sketch the trigonometric Fourier spectra of x(t) by inspection of Fig. 6.16. Now write the (compact) trigonometric Fourier series for x(t).

DR ILL 6.5 Fourier Series Spectrum of a Full-Wave Rectified Sine Wave

Find the exponential Fourier series and sketch the corresponding Fourier spectrum Dn versus ω for the full-wave rectified sine wave depicted in Fig. 6.17.

ANSWER

DR ILL 6.6 Exponential Fourier Series and Spectra

Find the exponential Fourier series and sketch the corresponding Fourier spectra for the periodic signals shown in Fig. 6.7.

ANSWERS (a) ^x(t) ^{= 1 3 +} 2 π2 ”∞ n=−∞(n=0) (−1)^{n n2} ejnπ*^t* (b) ^x(t) ⁼ jA π ”∞ n=−∞(n=0) (−1)^{n n} ejnπ*^t*

6.3-2 Parseval’s Theorem

The trigonometric Fourier series of a periodic signal x(t) is given by

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

Every term on the right-hand side of this equation is a power signal. As shown in Ex. 1.2, Eq. (1.3), the power of x(t) is equal to the sum of the powers of all the sinusoidal components on the right-hand side.

$$P_x = C_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} C_n^2$$

\n(6.26)

This result is one form of Parseval’s theorem, as applied to power signals. It states that the power of a periodic signal is equal to the sum of the powers of its Fourier components.

We can apply the same argument to the exponential Fourier series (see Prob. 1.1-11). The power of a periodic signal x(t) can be expressed as a sum of the powers of its exponential components. In Eq. (1.4), we showed that the power of an exponential Dej^ω0*^t* is |D2|. We can use this result to express the power of a periodic signal x(t) in terms of its exponential Fourier series coefficients as

$$P_x = \sum_{n=-\infty}^{\infty} |D_n|^2 \tag{6.27}$$

For a real x(t), |D−ⁿ|=|Dn|. Therefore,

$$P_x = D_0^2 + 2\sum_{n=1}^{\infty} |D_n|^2$$

\n(6.28)

EXAMPLE 6.10 Harmonic Distortion of Clipped Sinusoid

The input signal to an audio amplifier of gain 100 is given by x(t) = 0.1 cosω0t. Hence, the output is a sinusoid 10 cos ω0t. However, the amplifier, being nonlinear at higher amplitude levels, clips all amplitudes beyond ±8 volts, as shown in Fig. 6.18a. We shall determine the harmonic distortion incurred in this operation.

Figure 6.18 (a) A clipped sinusoid cos ω0t. (b) The distortion component xd(t) of the signal in (a).

The output y(t) is the clipped signal in Fig. 6.18a. The distortion signal yd(t), shown in Fig. 6.18b, is the difference between the undistorted sinusoid 10cosω0t and the output signal y(t). The signal yd(t), whose period is T⁰ [the same as that of y(t)], can be described over the first cycle as

$$y_d(t) = \begin{cases} 10 \cos \omega_0 t - 8 & |t| \le 0.1024T_0 \\ 10 \cos \omega_0 t + 8 & \frac{T_0}{2} - 0.1024T_0 \le |t| \le \frac{T_0}{2} + 0.1024T_0 \\ 0 & \text{everywhere else} \end{cases}$$

Observe that yd(t) is an even function of t and its mean value is zero. Hence, a⁰ = C⁰ = 0, and bn = 0. Thus, Cn = an and the Fourier series for yd(t) can be expressed as

$$y_d(t) = \sum_{n=1}^{\infty} C_n \cos n\omega_0 t$$

As usual, we can compute the coefficients Cn (which is equal to an) by integrating yd(t) cos nω0t over one cycle (and then dividing by 2/T0). Because yd(t) has even symmetry, we can find an by integrating the expression over a half-cycle only using Eq. (6.14). The

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

straightforward evaluation of the appropriate integral yields†

$$C_n = \begin{cases} \frac{20}{\pi} \left[ \frac{\sin\left[0.6435(n+1)\right]}{n+1} + \frac{\sin\left[0.6435(n-1)\right]}{n-1} \right] - \frac{32}{\pi} \left[ \frac{\sin\left(0.6435n\right)}{n} \right] & n \text{ odd} \\ 0 & n \text{ even} \end{cases}$$

Computing the coefficients C1, C2, C3, … from this expression, we can write

yd(t) = 1.04 cos ω0t +0.733 cos 3ω0t +0.311 cos 5ω0t +···

COMPUTING HARMONIC DISTORTION

We can compute the amount of harmonic distortion in the output signal by computing the power of the distortion component yd(t). Because yd(t) is an even function of t and because the energy in the first half-cycle is identical to the energy in the second half-cycle, we can compute the power by averaging the energy over a quarter-cycle. Thus,

$$P_{y_d} = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} y_d^2(t) dt = \frac{1}{T_0/4} \int_0^{T_0/4} y_d^2(t) dt$$

= $\frac{4}{T_0} \int_0^{0.1024T_0} (10 \cos \omega_0 t - 8)^2 dt = 0.865$

The power of the desired signal 10cos ω0t is (10)²/2 = 50. Hence, the total harmonic distortion is‡

$$D_{\text{tot}} = \frac{0.865}{50} \times 100 = 1.73\%$$

The power of the third harmonic components of yd(t) is (0.733)²/2 = 0.2686. The third harmonic distortion is

$$D_3 = \frac{0.2686}{50} \times 100 = 0.5372\%$$ $$C_n = a_n = \frac{8}{T_0} \int_0^{0.1024T_0} [10 \cos \omega_0 t - 8] \cos n\omega_0 t dt$$

^† In addition, yd(t) exhibits half-wave symmetry (see Prob. 6.1-6), where the second half-cycle is the negative of the first. Because of this property, all the even harmonics vanish, and the odd harmonics can be computed by integrating the appropriate expressions over the first half-cycle only (from −T0/4 to T0/4) and doubling the resulting values. Moreover, because of even symmetry, we can integrate the appropriate expressions over 0 to T0/4 (instead of from −T0/4 to T0/4) and double the resulting values. In essence, this allows us to compute Cn by integrating the expression over the quarter-cycle only and then quadrupling the resulting values. Thus,

^‡ In the literature, the harmonic distortion often refers to the rms distortion rather than the power distortion. The rms values are the square-root values of the corresponding powers. Thus, the third harmonic distortion in this sense is ^√(0.2686/50) ^{× 100 =} 7.33%. Alternately, we may also compute this value directly from the amplitudes of the third harmonic 0.733 and that of the fundamental as 10. The ratio of the rms values is (0.733/ ^√2) : (10/ √ 2) = 0.0733 and the percentage distortion is 7.33%.

6.3-3 Properties of the Fourier Series

As with the Laplace and z-transforms, the Fourier series has a variety of properties that can simplify work and help provide a more intuitive understanding of signals. Table 6.2 provides the most important properties of the Fourier series for a periodic signal x(t) and its spectrum Dn. Properties that involve two signals require that the two signals have a common fundamental frequency ω0. While not given here, the proofs of these properties are straightforward and parallel the proofs of the Fourier transform properties given in Ch. 7.

To demonstrate the utility of Fourier series properties, let us consider an example where we use a selection of properties to simplify the work of finding a piecewise polynomial signal’s spectrum.

Operation	x(t)	Dn
Scalar multiplication	kx(t)	kDn
Addition	x1(t)+x2(t)	D1,n +D2,n
	x1(t), x2(t) require same ω0
Conjugation	x∗(t)	D∗ −n
Reversal	x(−t)	D−n
Time shifting	x(t −t0)	Dne−jnω0t0
Frequency shifting	x(t)ejn0ω0t	Dn−n0
Frequency convolution	x1(t)x2(t)	D1,n ∗ D2,n
	x1(t), x2(t) require same ω0
Time differentiation	dkx(t) dtk	(jnω0)kDn

TABLE 6.2 Selected Fourier Series Properties

EXAMPLE 6.11 Using Fourier Series Properties

Use properties rather than integration to compute the exponential Fourier series coefficients Dn of the triangular signal x(t) shown in Fig. 6.4. Verify the correctness of Dn for A = 1 by synthesizing x(t) with a suitable truncation of Eq. (6.19).

From Fig. 6.4, we see that x(t) is a piecewise linear function that is T⁰ = 2 periodic. To compute Dn directly using Eq. (6.19) would therefore require tedious integration by parts. Fortunately, we can compute Dn without integration by instead using Fourier series properties. First, however, we must compute the dc component D⁰ separately from other Dn. By simple inspection of Fig. 6.4, we see that x(t) has no dc component, so D⁰ = 0.

To determine the remaining Dn, we begin by noting that x(t) has a constant slope of either 2A or −2A. Thus, differentiating x(t) once yields a square wave with amplitudes ±2A. Here, differentiation reduces x(t) from a piecewise linear to a piecewise constant function,

thereby eliminating the need for integration by parts.† Differentiating x(t) twice yields a pair of shifted impulse trains, weighted by ±4A. This second differentiation eliminates the need for any integration whatsoever, since the Fourier series coefficients of an impulse train are known to be ^{1 T0} (see Ex. 6.9).

Stated mathematically, we see that

$$\frac{d^2}{dt^2}x(t) = 4A\delta_2(t + \frac{1}{2}) - 4A\delta_2(t - \frac{1}{2})$$

Transforming this expression and using the Fourier series properties of scalar multiplication, addition, frequency shifting, and time differentiation yield (for n = 0)

$$(in\pi)^2 D_n = 4A \frac{e^{jn\omega_0/2}}{2} - 4A \frac{e^{-jn\omega_0/2}}{2}$$

Substituting ω⁰ = 2π/T⁰ = π and solving for Dn yield

$$D_n = 4A \frac{e^{in\pi/2} - e^{-in\pi/2}}{-2n^2\pi^2}$$

Combining with the dc component and simplifying expressions, the final result is

$$D_n = \begin{cases} 0 & n = 0\\ \frac{-A4j\sin(n\pi/2)}{n^2\pi^2} & n \neq 0 \end{cases}$$

Overall, this is a neat way to determine the signal’s spectrum. Through the selective use of properties, we have determined Dn without any integration. With a little care, this basic approach can yield the spectrum of any piecewise polynomial periodic function. Since all periodic functions of practical interest to engineers can, to an arbitrary level of accuracy, be represented as piecewise polynomial functions, we can always find their spectra without integration except for the dc term D0.

USING A TRUNCATED FOURIER SERIES TO VERIFY SPECTRUM CORRECTNESS

While a signal’s spectrum Dn provides useful insight into signal character, it can be difficult to look at Dn and know that it is correct for a particular signal x(t). For example, is it at all obvious that Dn ^{= 4}Ajsin(nπ/2) ⁻n2π² is the spectrum for the triangle wave of Fig. 6.4? Probably not.

There is an easy way, however, to verify a signal’s spectrum: synthesize x(t) using a suitable truncation of Eq. (6.19). If the synthesized signal closely matches the original, we can be relatively certain that the spectrum Dn is correct. What is a suitable truncation? Well, it depends. All significant Dn terms need to be included, but not so many as to make the reconstruction impractical to compute. Since in the present case Dn decays at a rate 1/n2, a good approximation is possible with a relatively few number of terms; a 10-harmonic truncation should be just fine. Let us use MATLAB to synthesize x(t) using a 10-harmonic truncated Fourier series.

^† Differentiation also destroys the dc component of the signal, providing further justification as to why the dc component needs to be separately computed.

To begin, we define A, Dn, T0, ω0, and a time vector that spans two periods of the waveform.

>> A = 1; D = @(n) -A*4j*sin(n*pi/2)./(n.^2*pi^2);
>> T0 = 2; omega0 = 2*pi/T0; t = (-T0:.001:T0);

Next, we set the dc portion of the signal.

D0 = 0; x10 = D0*ones(size(t));

To add the desired 10 harmonics, we enter a loop for 1 ≤ n ≤ 10 and add in the Dn and D−ⁿ terms. Although x(t) should be real, small round-off errors cause the reconstruction to be complex. These small imaginary parts are removed using the real command.

>> for n = 1:10,
>> x10 = x10+real(D(n)*exp(1j*omega0*n*t)+D(-n)*exp(-1j*omega0*n*t));
>> end

Lastly, we plot the resulting truncated Fourier series synthesis of x(t).

plot(t,x10,‘k’); xlabel(‘t’); ylabel(‘x_{10}(t)’);

Since the synthesized waveform shown in Fig. 6.19 closely matches the original waveform in Fig. 6.4, we have high confidence that the computed Dn are correct.